Class 10

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4

Question 1.
Which term of the A.P. 121, 117, 113, …………. is its first negative term?
Solution:
Given A.P is 121, 117, 113, …
Here a = T₁ = 121 ;T₂ = 117; T₃ = 113
d = T₂ – T₁ = 117 – 121 = – 4
Using formula, T_n = a + (n – 1) d
T_n = 121 + (n – 1) (- 4)
= 121 – 4n + 4
= 125 – 4n.
According to question :—
T_n < 0
or 125 – 4n < 0
or 125 < 4n or 4n > 125.
or n > \(\frac{125}{4}\)
or n > 31\(\frac{1}{4}\).
But n must be integer, for first negative term.
∴ n = 32.
Hence, 32nd term be the first negative term of given A.P.

Question 2.
The sum of the third and the seventh term of an A.P. is 6 and their product ¡s 8. Find the sum of first sixteer
terms of an A.P.
Solution:
Let ‘a’ and ‘d’ be the first term and common diftèrence of given A.P.
According to 1st condition
T₃ + T₇ = 6
[a + (3 – 1)d] + [a + (7 – 1) d] = 6
∵ [T_n = a + (n – 1) d]
or a + 2d + a + 6d = 6
or 2a + 8d = 6
or a + 4d = 3 …………….(1)
According to 2nd condition
T₃ (T₇) = 8
[a + (3 – 1) d] [a + (7 – 1)d] = 8
∵ [T_n = a + (n – 1) d]
or (a + 2d) (a + 6d) = 8
or [3 – 4d + 2d] [3 – 4d + 6d] = 8
[Using (1), a = 3 – 4d]
or (3 – 2d) (3 + 2d) = 8
or 9 – 4d² = 8
or 4d² = 98
or d² = \(\frac{1}{4}\)
d = ± \(\frac{1}{2}\)

Case I:
When d = \(\frac{1}{2}\)
Putting d = \(\frac{1}{2}\) in (1), we get:
a + 4 (\(\frac{1}{2}\)) = 3
or a + 2 = 3
or a = 3 – 2 = 1
Using formula, S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
S¹⁶ = \(\frac{16}{2}\) [2 (1) + (16 – 1) \(\frac{1}{2}\)].

Case II:
Putting d = – \(\frac{1}{2}\) in (1), we get,
When d = – \(\frac{1}{2}\)
a + 4 (-\(\frac{1}{2}\)) = 3
a – 2 = 3
or a = 3 + 2 = 5
Using formula,
S_n = \(\frac{n}{2}\) [2a + (n – 1)d]
S₁₆ = \(\frac{16}{2}\) [2(5) + (16 – 1) (-\(\frac{1}{2}\))]
= 8[10 – \(\frac{15}{2}\)]
= 8 \(\left[\frac{20-15}{2}=\frac{5}{2}\right]\)
S₁₆ = 20.

Question 3.
A ladder has rungs 25 cm apart (see fig.) The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are 2 latex]\frac{1}{2}[/latex] m apart, what is the length of the wood required for the rungs?

[Hint: Number of rungs = \(\frac{250}{25}\) + 1]
Solution:
Total length of rungs = 2 \(\frac{1}{2}\) m = \(\frac{5}{2}\) m
= (\(\frac{5}{2}\) × 100) cm = 250 cm
Length of each rung = 25 cm
∴ Number of rungs = \(\frac{\text { Total length of rungs }}{\text { Length of each rung }}\) + 1
= \(\frac{250}{25}\) + 1 = 10 + 1 = 11
Length of first rung =45 cm
Here a = 45; l = 25; n = 11
Length of the wood for rungs
= S₁₁
= \(\frac{n}{2}\) [a + l]
= \(\frac{11}{2}\) [45 + 25]
= \(\frac{1}{2}\) × 70
= 11 × 35 = 385
Hence, length of the wood for rungs has 385 cm.

Question 4.
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it and find this value of x.
[Hint: S_{x – 1} = S₄₉ – S₁]
Solution:
Let ‘x’ denotes the number of any house.
Here a = T₁ = 1 ;d = 1
According to question,
S_{x – 1} = S₄₉ – S_x
= \(\frac{x-1}{2}\) [2 (1) + (x – 1 – 1) (1)]
= \(\frac{49}{2}\) [1 + 49] – \(\frac{x}{2}\) [2 (1) + (x – 1) (1)]
[Using S_n = \(\frac{n}{2}\) [2a + (n – 1) d] and S_n = \(\frac{n}{2}\) (a + l) ]
or \(\frac{x-1}{2}\) [2 + x – 2] = \(\frac{49}{2}\) (50) – \(\frac{x}{2}\) [2 + x – 1]
or \(\frac{x(x-1)}{2}=49(25)-\frac{x(x+1)}{2}\)
or \(\frac{x}{2}\) [x – 1 + x + 1] = 1225
\(\frac{x}{2}\) × 2x = 1225
or x² = 1225
or x = 35.

Question 5.
A small terrace at a football ground comprises of 15 step each of which is 50 m long and built of solid concrete. Each step has a rise of \(\frac{1}{4}\) m and a tread of \(\frac{1}{2}\) m (see fig.) Calculate the total volume of concrete required to build the terrace.
[Hint: Volume of concrete required to build of the first step \(\frac{1}{4}\) × \(\frac{1}{2}\) × 50 m³].

Solution:
Volume of concrete required to build the first step \(\frac{1}{4}\) × \(\frac{1}{2}\) × 50 m³]
= [latex]\frac{25}{4}[/latex] m³
Volume of concrete required to build the second step = [\(\frac{25}{4}\) × \(\frac{1}{2}\) × 50] m³

= \(\frac{75}{2}\) m³
Volume of concrete required to build the third step = [\(\frac{3}{4}\) × \(\frac{1}{2}\) × 50] m³ and so on upto 15 steps.

Here a = T₁ = \(\frac{25}{4}\);
T₂ = \(\frac{25}{2}\);
T₃ = \(\frac{75}{4}\); and n = 15.
d = T₂ – T₁ = \(\frac{25}{2}\) – \(\frac{25}{4}\)
= \(\frac{50-25}{4}\) = \(\frac{25}{4}\).

Total volume of concrete required to buld the terrace = S₁₅
= \(\frac{n}{2}\) [2a + (n – 1)d]
= \(\frac{15}{2}\left[2\left(\frac{25}{4}\right)+(15-1) \frac{25}{4}\right]\)
= \(\left[\frac{25}{4} \times \frac{14 \times 25}{4}\right]\)
= \(\frac{15}{2}\left[\frac{25}{2} \times \frac{175}{2}\right]\)
= \(\frac{15}{2} \times \frac{200}{2}\) = 750
Hence, total volume of concrete required to build the terrace is 750 m³.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 1.
Find the sum of the following APs:
(i) 2, 7, 12, … to 10 terms.
(ii) – 37, – 33, – 29, ………….. to 12 terms.
(iii) 0.6, 1.7, 2.8, … to 100 terms.
(iv) \(\frac{1}{5}\), \(\frac{1}{12}\), \(\frac{1}{10}\), ………… to 11 terms.
Solution:
(i) Given AP. is 2, 7, 12, …
Here a = 2, d = 7 – 2 = 5 and n = 10
S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
∴ S₁₀ = \(\frac{10}{2}\) [2 × 2 + (10 – 1)5]
= 5 [4 + 45] = 245.

(ii) Given A.P. is – 37, – 33, – 29…
Here a = -37, d = – 33 + 37 = 4 and n = 12
S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
∴ S₁₂ = [2(-37) + (12 – 1)4]
= 6 [- 74 + 44] = – 180.

(iii) Given A.P. is 0.6, 1.7. 2.8.
Here a = 0.6, d = 1.7 – 0.6 = 1.1 and n = 100
S_n = \(\frac{n}{2}\) [2a+(n – 1) d]
∴ S₁₀₀ = \(\frac{100}{2}\) [2(0.6) + (100 – 1) 1.1]
= 50 [1.2 + 108.9] = 5505.

(iv) Given AP. is \(\frac{1}{5}\), \(\frac{1}{12}\), \(\frac{1}{10}\), ………… to 11 terms.
Here a = \(\frac{1}{5}\), d = \(\frac{1}{5}\) and n = 11
S_n = \(\frac{n}{2}\) [2a +(n – 1)d]
∴ S₁₁ = \(\frac{11}{2}\left[2\left(\frac{1}{15}\right)+(11-1) \frac{1}{60}\right]\)

= \(\frac{11}{2}\left[\frac{2}{15}+\frac{10}{60}\right]=\frac{11}{2}\left[\frac{2}{15}+\frac{1}{6}\right]\)

= \(\frac{11}{2}\left[\frac{4+5}{30}=\frac{9}{30}\right]=\frac{33}{20}\).

Question 2.
Find the sums given below:
(i) 7+ 10\(\frac{1}{2}\) + 14 + …………… + 84
(ii) 34 + 32 + 30 + ………. + 10
(iii) – 5 + (- 8) + (- 11) +………+ (- 230)
Solution:
(i) Given A.P. is
7 + 10\(\frac{1}{2}\) + 14 + ………….. + 84
Here a = 7, d = 10\(\frac{1}{2}\) – 7 = \(\frac{21}{2}\) – 7
= \(\frac{21-14}{2}=\frac{7}{2}\)
and l = T_n = 84
or a + (n – 1) d = 84
or 7 + (n – 1) \(\frac{7}{2}\) = 84
or (n – 1) \(\frac{7}{2}\) = 84 – 7 = 77
or n – 1 = 77 × \(\frac{2}{7}\) =22
n = 22 + 1 = 23
∵ S_n = \(\frac{n}{2}\) [a + l]
Now, S₂₃ = \(\frac{23}{2}\) [7 + 84]
= \(\frac{23}{2}\) × 91 = \(\frac{2093}{2}\).

(ii) Given A.P. is 34 + 32 + 30 + …………… + 10
Here a = 34, d = 32 – 34 = -2
l = T_n = 10
a + (n – 1) d = 10
or 34 + (n – 1) (- 2) = 10
or – 2(n – 1) = 10 – 34 = -24
or n – 1 = 12
n = 12 + 1 = 13
[∵ S_n = \(\frac{n}{2}\) [a + l]]
Now, S₁₃ = \(\frac{13}{2}\) [34 + 10]
= \(\frac{23}{2}\) × 44
= 13 × 22 = 286.

(iii) Give A.P. is – 5 + (- 8) + (- 11) + ……………. + (- 230)
Here a = – 5, d = – 8 + 5 = – 3
and l = T_n = – 230
a + (n – 1) d = – 230
or – 5 + (n – 1) (- 3) = – 230
or – 3(n – 1) = – 230 + 5 = – 225
or n – 1 = \(\frac{225}{3}\) = 75
or n = 75 + 1 = 76
Now, S₇₆ = \(\frac{76}{2}\) [- 5 + (- 23o)]
= 38 (- 235) = – 8930.

Question 3.
In an AP:
(i) given a = 5, d = 3, a_n = 50 find n and S_n.
(ii) given a = 7. a₁₃ = 35, find d and S₁₃.
(iii) given a₁₂ = 37, d = 3. find a and S₁₂.
(iv) given a₃ = 15, S₁₀ = 125, find d and a₁₀.
(y) given d = 5, S₉ = 75, find a and a₉.
(vi) given a = 2, d = 8, S_n = 90, find n and a_n.
(vii) given a = 8, a_n = 62, S_n = 210, find n and d.
(viii)given a_n = 4, d = 2, S_n = -14, find n and a.
(ix) given a = 3, n = 8, S = 192, find d.
(x) given l = 28, S = 144, and there are total 9 terms. Find a.
Solution:
(i) Given a = 5, d = 3, a_n = 50
a_n = 50
a + (n – 1) d = 50
or 5 + (n – 1) 3 = 50
or 3 (n – 1) = 50 – 5 = 45
or n – 1 = \(\frac{45}{3}\) = 15
or n = 15 + 1 = 16
Now, S_n = latex]\frac{n}{2}[/latex] [a + l]
= latex]\frac{16}{2}[/latex] [5 + 50] = 8 × 55 = 440.

(ii) Given a = 7, a₁₃ = 35
a₁₃ = 35
a + (n – 1) d = 35
or 7 + (13 – 1) d = 35
or 12 d = 35 – 7 = 28
or d = \(\frac{7}{3}\).
∵ [S_n = latex]\frac{n}{2}[/latex] [a + l]
Now, S₁₃ = \(\frac{13}{2}\) [7 + 35]
= \(\frac{13}{2}\) × 42 = 273

(iii) Given a₁₂ = 37, d = 3
∵ a₁₂ = 37
a + (n – 1) d = 37
or a + (12 – 1) 3 = 37
a + 33 = 37
a = 37 – 33 = 4
∵ [S_n = latex]\frac{n}{2}[/latex] [a + l]
Now, S₁₂ = \(\frac{13}{2}\) [4 + 37]
= 6 × 41 = 246.

(iv)Given a₃ = 15, S₁₀ = 125
∵ a₃ = 15
a + (n – 1) d = 15
a + (3 – 1) d = 15
or a + 2d = 15 …………….(1)
∵ S₁₀ = 125
∵ [S_n = latex]\frac{n}{2}[/latex] [2a + (n – 1)d]
\(\frac{10}{2}\) [2a + (10 – 1) d] = 125
or 5[2a + 9d] = 125
Note this
or 2a + 9d = \(\frac{125}{5}\) = 25
or 2a + 9d = 25 …………(2)
From (1), a = 15 – 24 …………..(3)
Substitute this value of (a) in (2i, we c1
2(15 – 2d) + 9d = 25
or 30 – 4d + 9d = 25
5d = 25 – 30
or d = \(\frac{-5}{5}\) = -1
Substitute this value of d in (3), we get
a = 15 – 2(- 1)
a = 15 + 2 = 17
Now, a₁₀ = 17 + (10 – 1)(- 1)
∵ T_n = a + (n – 1) d = 17 – 9 = 8.

(v) Given d = 5, S₉ = 75
∵ S₉ = 75
∵ [S_n = \(\frac{n}{2}\) [2a + (n – 1)d]
\(\frac{9}{2}\) [2a + 40] = 75
or [2a + 40] = \(\frac{50}{3}\)
or 2a = \(\frac{50}{3}\) – 40
or a = \(-\frac{70}{3} \times \frac{1}{2}\)
or a = – \(\frac{35}{3}\)
Now, a₉ = a + (n – 1) d
= – \(\frac{35}{3}\) + (9 – 1)⁰ × 5
= – \(\frac{35}{3}\) + 4o = \(\frac{-35+120}{3}\)
a₉ = \(\frac{85}{3}\).

(vi) Given a = 2, d = 8, S_n = 90
∵ S_n = 90
\(\frac{n}{2}\) [2a + (n – 1)d] = 9o
or \(\frac{n}{2}\) [2 × 2 + (n – 1) 8] = 90
or n [2 + 4n – 4] = 90
or n (4n – 2) = 90
or 4n² – 2n – 90 = 0
or 2n² – 10n + 9n – 45 = 0
S = – 2
P = – 45 × 2 = – 90
or 2n [n – 5] + 9(n – 5) = 0
or (2n + 9) (n – 5) = 0
Either 2n + 9 = 0 or n – 5 = 0
Either n = – \(\frac{9}{2}\) or n = 5
∵ n cannot be negative so reject n = – \(\frac{9}{2}\)
∴ n = 5
Now, a_n = a₅ = a + (n – 1) d
= 2 + (5 – 1) 8 = 2 + 32 = 34.

(vii) Given a = 8, a_n = 62, S_n = 210
∵ S_n = 210
\(\frac{n}{2}\) [a + a_n] = 210
or \(\frac{n}{2}\) [8 + 62] = 210
or \(\frac{n}{2}\) × 70 = 210
or n = \(\frac{210}{35}\) = 6
Now, a_n = 62
[∵ T_n = a + (n – 1) d]
or 5d = 62 – 8 = 54
or d = \(\frac{54}{5}\).

(viii) Given a_n = 4, d = 2, S_n = – 14
∵ a_n = 4
a + (n – 1) d = 4
or a + (n – 1)2 = 4
or a + 2n – 2 = 4
or a = 6 – 2n ……………(1)
and S_n = – 14
\(\frac{n}{2}\) [a + a_n] = – 14
or \(\frac{n}{2}\) [6 – 2n +4] = – 14 (Using (1))
or \(\frac{n}{2}\) [10 – 2n] = – 14
or 5n – n² + 14 = 0
or n² – 5n – 14 = 0
S = – 5
P = 1 × – 14 = – 14
or n² – 7n + 2n – 14 = 0
or n(n – 7) + 2 (n – 7) = 0
or (n – 7) (n + 2) = 0
either n – 7 = 0
or n + 2 = 0
n = 7 or n = -2
∵ n cannot be negative so reject n = – 2
∴ n = 7
Substitute this value of n in (1), we get
a = 6 – 2 × 7
a = 6 – 14 = – 8.

(ix) Given a = 3, n = 8, S = 192
∵ S = 192
S₈ = 192 [∵ n = 8]
∵ S_n = \(\frac{n}{2}\) [2a + (n – 1)d]
or \(\frac{8}{2}\) [2 × 3 + (8 – 1) d] = 192
or 4 [6 + 7d] = 192
6 + 7d = \(\frac{192}{4}\) = 48
or 7d = 48 – 6 = 42
or d = \(\frac{42}{6}\) = 6.

(x) Given l = 28, S = 144 and there are total 9 terms
∴ n = 9; l = a₉ = 28; S₉ = 144
∵ a₉ = 28
or a + (9 – 1) d = 28
∵ [a_n = T_n = a + (n – 1) d ]
or a + 8d = 28 …………….(1)
and S₉ = 144
∵ [S_n = \(\frac{n}{2}\) [a + a_n]]
\(\frac{9}{2}\) [a + 28] = 144
or a + 28 = \(\frac{144 \times 2}{9}\) = 32
a = 32 – 28 = 4.

Question 4.
How many terms of the A.P : 9, 17, 5… must be taken to give a sum of 636?
Solution:
Given A.P. is 9, 17, 25, …………
Here a = 9, d = 17 – 9 = 8
But S_n = 636
\(\frac{n}{2}\) [2a + (n – 1) d] = 636
or \(\frac{n}{2}\) [2(9) + (n – 1) 8] = 636
or \(\frac{n}{2}\) [18 + 8n – 8] = 636
or n [4n + 5] = 636
or 4n² + 5n – 636 = 0
a = 4, b = 5, c = – 636
D = (5) – 4 × 4 × (- 636)
= 25 + 10176 = 10201
∴ n = \(\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}\)

= \(\frac{-5 \pm \sqrt{10201}}{2 \times 4}=\frac{-5 \pm 101}{8}\)

= \(\frac{-106}{8} \text { or } \frac{96}{8}\)

= \(-\frac{53}{4}\) or 12
∴ n cannot be negative so reject n = \(-\frac{53}{4}\)
∴ n = 12
Hence, sum of 12 terms of given A.P. has sum 636.

Question 5.
The first term ofan AP is 5, the last term is 45 and the sum Is 400. Find the number of terms and the common difference.
Solution:
Given that a = T₁ = 5; l = a_n = 45
and S_n = 400
∴ T_n = 45
a + (n – 1) d = 45
or 5 + (n – 1) d = 45
or (n – 1) d = 45 – 5 = 40
or (n – 1) d = 40 ……………….(1)
and S_n = 400
\(\frac{n}{2}\) [a + a_n] = 400
or \(\frac{n}{2}\) [5 + 45] = 400
or 25 n = 400
or n = \(\frac{400}{25}\) = 16
Substitute this value of n in (1), we get
(16 – 1) d = 40
or 15d = 40
d = \(\frac{40}{15}\) = \(\frac{8}{3}\)
Hence, n = 16 and d = \(\frac{8}{3}\)

Question 6.
The first and last terms of an AP are 17 and 350 respectively. 1f the common difference is 9, how many terms are there
and what is their sum?
Solution:
Given that a = T₁ = 17;
l = a_n = 350 and d = 9
∵ l = a_n = 350
a + (n – 1) d = 350
17 + (n – 1) 9 = 350
or 9 (n – 1) = 350 – 17 = 333
or n – 1 = \(\frac{333}{9}\) = 37
n = 37 + 1 = 38
Now, S₃₈ = \(\frac{n}{2}\) [a + l]
= \(\frac{38}{2}\) (17 + 350]
= 19 × 367 = 6973.
Hence, sum of 38 terms of given A.P. arc 6973.

Question 7.
Find the sum of first 22 terms of an AP. in which d = 7 and 22nd term is 149.
Solution:
Given that d = 7; T₂₂ = 149 and n = 22
∵ T₂₂ = 149
a + (n – 1) d = 149
or a + (22 – 1) 7 = 149
or a + 147 = 149
or a = 149 – 147 = 2
Now, S₂₂ = [a + T₂₂]
= \(\frac{22}{2}\) [2+ 149] = 11 × 151 = 1661
Hence, sum of first 22 temis of given A.P. is 1661.

Question 8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
Let ‘a’ and ‘d’ be fïrst term and common difference
Given that T₂ = 14; T₃ = 18 and n = 51
∵ T₂ = 14
a + (n – 1) d = 14
a + (2 – 1)d = 14
or a + d = 14
a = 14 – d …………….(1)
and T₃ = 18 (Given)
a + (n – 1) d = 18
a + (3 – 1) d = 18
or a + 2d = 18
or 14 – d + 2d = 18
or d = 18 – 14 = 4
or d = 4
Substitute this value of d in (1), we get
a = 14 – 4 = 10
Now, S₅₁ = \(\frac{n}{2}\) [2a + (n – 1) d]
= \(\frac{51}{2}\) [2 × 10 + (51 – 1) 4]
= \(\frac{51}{2}\) [2o + 2oo]
= \(\frac{51}{2}\) × 220 = 51 × 110 = 5610
Hence, sum of first 51 terms of given A.P. is 5610.

Question 9.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
According to 1st condition
S₇ = 49
\(\frac{n}{2}\) [2a + (n – 1) d] = 49
or \(\frac{7}{2}\) [2a + (7 – 1) d] = 49
or \(\frac{7}{2}\) [2a + 6d] = 49
or a + 3d = 7
or a = 7 – 3d
According to 2nd condition
S₁₇ = 289
\(\frac{n}{2}\) [2a+(n – 1)d]=289
\(\frac{17}{2}\) [2a + (17 – 1) d] = 289
a + 8d = \(\frac{289}{17}\) = 17
Substitute the value of a from (1), we get
7 – 3d + 8d = 17
5d = 17 – 7 = 10
d = \(\frac{10}{5}\) = 2
Substitute this value of d in (1), we get
a = 7 – 3 × 2
a = 7 – 6 = 1
Now, S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
= \(\frac{n}{2}\) [2 × 1 + (n – 1) × 2] = n [I + n – 1] = n × n = n²
Hence, sum of first n terms of given A.P. is n².

Question 10.
Show that a₁, a₂, ……., a_n, … form an AP where is defined as below.
(i) a_n = 3 + 4n
(ii) a_n = 9 – 5n
Also find the sum of the first 15 terms In each case.
Solution:
(i) Given that a_n = 3 + 4n ………..( 1)
Putting the different values of n in(1), we get
a₁ = 3 + 4(1) = 7;
a₂ = 3 + 4 (2) = 11
a₃ = 3 + 4 (3) = 15, …………
Now, a₂ – a₁ = 11 – 7
and a₃ – a₂ = 15 – 11 = 4
a₂ – a₁ = 11 – 7 = 4
and a₃ – a₂ = 4 = d(say)
∵ given sequence form an A.P.
Here a = 7, d = 4 and n = 15
∴ S₁₅ = \(\frac{n}{2}\) [2a + (n – 1)d]
= \(\frac{15}{2}\) [2(7) + (15 – 1) 4]
= \(\frac{15}{2}\) [14 + 56] = \(\frac{15}{2}\) × 70
= 15 × 35 = 525.

(ii) Given that a_n = 9 – 5n ……………….(1)
Putting the different values of n is (1), we get
a₁ = 9 – 5(1) = 4;
a₂ = 9 – 5(2) = -1;
a₃ = 9 – 5(3) = -6.
Now, a₂ – a₁ = – 1 – 4 = – 5
and a₃ – a₂ = – 6 + 1 = – 5
∵ a – a₁ = a₃ – a₂ = – 5 = d (say)
∴ given sequence form an A.P.
Here a = 4, d = – 5 and n = 15
∴ S₁₅ = \(\frac{n}{2}\) [2a + (n – 1) d]
= \(\frac{15}{2}\) [2(4) + (15 – 1) (-5)]
= \(\frac{15}{2}\) [8 – 70]
= \(\frac{15}{2}\) (-62) = – 465

Question 11.
If the sum of the first n terms of an AP is 4n – n², what is the first term (that is S₁) ? What is the sum of two terms? What is the second term ? Similarly, find the 3rd, the 10th and the nth terms.
Solution:
Given that, sum of n terms of an A.P. are
S_n = 4n – n²
Putting n = 1 in (1), we get
S₁ = 4(1) – (1)² = 4 – 1
S₁ = 3
∴ a = T₁ = S₁ = 3
Putting n = 2, in (1), we get
S₂ = 4(2) – (2)² = 8 – 4
S₂ = 4
or T₁ + T₂ = 4
or 3 + T₂ = 4
or T₂ = 4 – 3 = 1
Putting n = 3 in (1), we get
S₃ =4(3) – (3)² = 12 – 9
S₃ = 3
or S₂ + T₃ = 3
or 4+ T₃ = 3
or T₃ = 3 – 4 = – 1
Now, d = T₂ – T₁
d = 1 – 3 = -2
∴ T₁₀ = a + (n – 1) d
= 3 + (10 – 1) (- 2)
T₁₀ = 3 – 18 = – 15
and T_n = a + (n – 1) d
= 3 + (n – 1) (- 2)
= 3 – 2n + 2
T_n = 5 – 2n.

Question 12.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
Positive integers divisible by 6 are 6, 12, 18, 24, 30, 36 42, …
Here a = T₁ = 6, T₂ = 12, T₃ = 18, T₄ = 24
T₂ – T₁ = 12 – 6 = 6
T₃ – T₂ = 18 – 12 = 6
T₄ – T₃ = 24 – 18 = 6
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = 6 = d (say)
Using formula, S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
S₄₀ = \(\frac{40}{2}\) [2(6) + (40 – 1)6].
= 20 [12 + 234]
= 20 (246) = 4920
Hence, sum of first 40 positive integers divisible by 6 is 4920.

Question 13.
Find the sum of first 15 multiples of 8.
Solution:
Multiples of 8 are 8, 16, 24, 32, 40, 48, …………..
Here a = T₁ = 8; T₂ = 16; T₃ = 24 ; T₄ = 32
T₂ – T₁ = 16 – 8= 8
T₃ – T₂ = 24 – 16=8
T₂ – T₁ = T₃ – T₂ = 8 = d(say)
Ùsing formula. S_n = [2a + (n – 1) d}
S₁₅ = \(\frac{15}{2}\) [2(8) + (15 – 1) 8]
= \(\frac{15}{2}\) [16 + 112]
= \(\frac{15}{2}\) × 128 = 960
Hence, sum of first 15 multiples of 8 is 960.

Question 14.
Find the sum of the odd numbers between 0 and 50.
Solution:
Odd numbers between 0 and 50 are 1, 3, 5, 7, 9, …………, 49
Here a = T₁ = 1; T₂ = 3; T₃ = 5 ; T₄ = 7
and l = T_n = 49
T₂ – T₁ = 3 – 1 = 2
T₃ – T₂ = 5 – 3 = 2
∵ T₂ – T₁ = T₃ – T₂ = 2 = d (say)
Also, l = T_n = 49
a + (n – 1) d = 49
1 + (n – 1) 2 = 49
or 2 (n – 1) = 49 – 1 = 48
n – 1 = \(\frac{48}{2}\) = 24
n = 24 + 1 = 25.
Using formula, S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
S₂₅ = \(\frac{25}{2}\) [2(1) + (25 – 1)2]
= \(\frac{25}{2}\) [2 + 48]
= \(\frac{25}{2}\) × 50 = 625
Hence, sum of the odd numbers between 0 and 50 are 625.

Question 15.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for
the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day, How much money the contractor has to pay as penalty, if he hasdelayed the work by 30 days?
Solution:
Penalty (cost) for delay of one, two, third day are ₹ 200, ₹ 250, ₹ 300
Now, penalty increase with next day with a difference of ₹ 50.
∴ Required A.P. are ₹ 200, ₹ 250, ₹ 300, ₹ 350, …
Here a = T₁ = 200; d = ₹ 50 and n = 30
Amount of penalty gives after 30 days
= S₃₀
= \(\frac{n}{2}\) [2a + (n – 1) d]
= \(\frac{30}{2}\) [2(2oo) + (30 – 1) 50)
= 15 [400 + 1450]
= 15 (1850) = 27750
Hence, ₹ 27350 pay as penalty by the contractor if he has delayed the work 30 days.

Question 16.
A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If
each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.
Solution:
Let amount of prize given to 1st student = ₹ x
Amount of prize given to 2nd student = ₹ (x – 20)
Amount of prize given to 3rd student = ₹ [x – 20 – 20] = ₹ (x – 40)
and so on.
∴ Required sequence are ₹ x, ₹ (x – 20), ₹ (x – 40), … which form on A.P. with
a = ₹ x, d = – 20 and n = 7
Using formula, S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
S₇ = \(\frac{7}{2}\) [2(x) + (7 – 1) (- 20)]
S₇ = \(\frac{7}{2}\) [2x – 120] = 7 (x – 60)
According to question,
7 (x – 60) = 700
x – 60 = 7 = 100
x – 60 = 7 = 100
x = 100 + 60
x = 160
Hence, 7 prizes are ₹ 160, ₹ 140, ₹ 120, ₹ 100, ₹ 80, ₹ 60, ₹ 40.

Question 17.
In a school, student thought of planting trees in and around the school to reduce air pollution. It was decided that
number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g. – a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Solution:
Number of trees planted by three sections of class I = 3 × 1 = 3
Number of trees planted by three sections of class II = 3 × 2 = 6
Number of trees planted by three sections of class III = 3 × 3 = 9
…………………………………………………………………..
…………………………………………………………………..
……………………………………………………………………
Number of trees planted by three sections of class XII = 3 × 12 = 36
:. Required A.P. are 3, 6, 9, …………., 36
Here a = T₁ = 3; T₂ = 6; T₃ = 9
and l = T_n = 36; n = 12
d = T₂ – T₁ = 6 – 3 = 3
Total number of trees planted by students
= S₁₂
= \(\frac{n}{2}\) [a + l]
= \(\frac{12}{2}\) [3+ 36]
= 6 × 39 = 234
Hence, 234 trees will be planted by students to reduce air pollution.

Question 18.
A spiral is made up of successive semicircics, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, …. as shown in Fig. What is the total length of such a spiral made up of thirteen consecutive semicircies? (Take π = \(\frac{22}{7}\))
[Hint: Length of successive semicircies is ‘l₁’ ‘l₂’ ‘l₃‘ ‘l₄‘ … wIth centres at A, B, A, B, …, respectively.]

Solution:
Let l₁ = length of first semi circle = πr₁ = π(0.5) = \(\frac{\pi}{2}\)
l₂ = length of second semi circlem = πr₂= π(1) = π
l₃ = length of third semi circle = πr₃ = π(1.5) = \(\frac{3 \pi}{2}\)
and l₄ = length of fourth senil circle = πr₄ = π(2) = 2π and so on
∵ length of each successive semicircle form an A.P.
Here
a = T₁ = \(\frac{\pi}{2}\); T₂ = π;
T₃ = \(\frac{3 \pi}{2}\); T₄ = 2π……… and n = 13
d = T₂ – T₁ = π – \(\frac{\pi}{2}\)
= \(\frac{2 \pi-\pi}{2}=\frac{\pi}{2}\)
Length of whole spiral = S₁₃

Hence, total length of a spiral made up of thirteen consecutive semi circies is 143 cm

Question 19.
200 logs are stacked in the following manner : 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see Fig). In how many rows are the 200 logs placed and how many logs are in the top row?

Solution:
Number of logs in the bottom (1st row) = 20
Number of logs in the 2nd row = 19
Number of log in the 3rd row = 18 and so on.
∴ Number of logs in the each steps form an
Here a = T₁ = 20; T₂ = 19; T₃ = 18…
d = T₂ – T₁ = 19 – 20 = – 1
Let S, denotes the number logs.
Using formula, S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
= \(\frac{n}{2}\) [2 (20) + (n – 1) (-1)]
∴ S_n = \(\frac{n}{2}\) [40 – n + 1]
= \(\frac{n}{2}\) [41 – n]
According to question,
\(\frac{n}{2}\) [41 – n] = 200
or 41 – n² = 400
or – n² + 41n – 400 = 0
or n² – 41n + 400 = 0
S = – 41, P = 400
or n² – 16n – 25n + 400 = 0
or n (n – 16) – 25 (n – 16)=0
or (n – 16) (n – 25) = 0
Either n – 16 = 0 or n – 25 = 0
Either n = 16 or n = 25
∴ n = 16, 25.
Case I:
When n = 25
T₂₅ = a + (n – 1) d
= 20 + (25 – 1)(- 1)
= 20 – 24 = – 4;
which is impossible
∴ n = 25 rejected.

Case II. When n = 16
T₁₆ = a + (n – 1) d
= 20 + (16 – 1)(- 1)
= 20 – 15 = 5
Hence, there are 16 row and 5 logs are in the top row.

Question 20.
In a potato race a bucket ¡s placed at the starting point which Ls 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Fig.)

Each competitor starts from the bucket, picks up the nearest potato, runs back with It, drops It in the bucket, runs back to pick up the next potato, runs to the bucket to drop it In, and the continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
(Hint : To pick up the first potato and second potato, the distance run is [2 × 5 + 2 × (5 + 3)] m)
Solution:
Distance covered to pick up the Ist potato = 2(5) m = 10 m
Distance between successive potato = 3 m
distance covered to pick up the 2nd potato = 2(5 + 3) m = 16 m
Distance covered to pick up the 3rd potato = 2 (5 + 3 + 3) m = 22 m
and this process go on. h is clear that this situation becomes an A.P. as 10 m, 16 m, 22 m, 28 m, …
Here a = T₁ = 10; T₂ = 16; T₃ = 22, …
d = T₂ – T₁ = 16 – 10 = 6 and n = 10
∴ Total distance the competitor has to run = S₁₀
= \(\frac{n}{2}\) [2a + (n – 1) d]
= \(\frac{10}{2}\) [2(10) + (10 – 1) 6]
= 5 [20 + 54]
= 5 × 74 = 370
Hence, 370 m is the total distance run by a competitor.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

Question 1
Fill in the blanks in the following table, given that a is the first term, d the common difference and a the nth term of the
AP:

Solution:
(i) Here a = 7, d = 3, n = 8
∵ a_n = a + (n – 1)d
∴ a₈ = 7 + (8 – 1)3
= 7 + 21 = 28.

(ii) Here a = – 18, n = 10, a_n = 0
∵ a_n = a + (n – 1)d
∴ a₁₀ = – 18 + (10 – 1)d
or 0 = – 18 + 9d .
or 9d = 18
d = \(\frac{18}{2}\) = 2.

(iii) Here d = – 3, n = 18, a_n = – 5
∵ a_n = a + (n – 1)d
∴ a₁₈ = a + (18 – 1)(-3)
or -5 = a – 51
or a = – 5 + 51 = 46.

(iv) Here a = – 18.9, d = 2.5 a_n = 3.6
∵ a_n = a + (n – 1)d
∴ 3.6 = – 18.9 + (n – 1) 2.5
or 3.6 + 18.9 = (n – 1) 2.5
or (n – 1) 2.5 = 22.5
or n – 1 = \(\frac{22.5}{2.5}\)
or n = 9 + 1 = 10.

(v) Here a = 3.5, d = 0, n = 105
∵ a_n = a + (n – 1) d
∴ a_n = 3.5 + (105 – 1) 0
a_n = 3.5 + 0 = 3.5.

Question 2.
Choose the correct choice in the following and justify:
(i) 30th term of the AP: 10, 7, 4, …………….. is
(A) 97 (B) 77 (C) – 77 (D) – 87

(ii) 11th term of the AP: – 3, – \(\frac{1}{2}\), 2, ………. is
(A) 28 (B) 22 (C) – 38 (D) – 48\(\frac{1}{2}\)

Solution:
(i) Given A.P. is 10, 7, 4 ……………
T₁ = 10, T₂ = 7, T₃ = 4
T₂ – T₁ = 7 – 10 = – 3
T₃ – T₂ = 4 – 7 = – 3
∵ T₂ – T₁ = T₃ – T₂ = – 3 = d(say)
∵ T_n = a + (n – 1) d
Now, T₃₀ = 10 + (30 – 1)(-3)
= 10 – 87 = – 77
∴ Correct choice is (C).

(ii) Given A.P. is – 3, –\(\frac{1}{2}\), 2, ……….
T₁ = – 3 T₂ = –\(\frac{1}{2}\), T₃ = 2, …………..
T₂ – T₁ = –\(\frac{1}{2}\) + 3 = \(\frac{-1+6}{2}=\frac{5}{2}\)
T₃ – T₂ = 2 + \(\frac{1}{2}\) = \(\frac{4+1}{2}=\frac{5}{2}\)
∵ T₂ – T₁ = T₃ – T₂ = \(\frac{5}{2}\) = d(say)
∵ T_n = a + (n – 1) d
Now, T₁₁ = -3 + (11 – 1) \(\frac{5}{2}\)
= -3 + 10 × \(\frac{5}{2}\) = – 3 + 25 = 22
∴ Correct choice is (B).

Question 3.
In the following APs, find the missing terms in the boxes:
(i) 2, , 26
(ii), 13, , 3
(iii) 5,, , 9
(iv) – 4, , , , , 6
(v) , 38, , , , – 22
Solution:
Let a be the first term and ‘d’ be the common difference of given A.P.
(i) Here T₁ = a = 2
and T₃ = a + 2d = 26
or 2 + 2d = 26
or 2d = 26 – 2 = 24
or d = 12
∴ Missing term = T₂ = a + d = 2 + 12 = 14.

(ii) Here, T₂ = a + d = 13 ……………(1)
and T₄ = a + 3d = 3 …………….(3)
Now, (2) – (1) gives

Substitute this value of d in (1), we get
a – 5 = 13
a = 13 + 5 = 18.
∴ T₁ = a = 18
T₃ = a + 2d = 18 + 2(-5)
= 18 – 10 = 8.

(iii)Here T₁ = a = 5
and T₄ = a + 3d = 9
or a + 3d = \(\frac{19}{2}\)
or 5 + 3d = \(\frac{19}{2}\)
or 3d = \(\frac{19}{2}\) – 5
or 3d = \(\frac{19-10}{2}=\frac{9}{2}\)
or d = \(\frac{9}{2} \times \frac{1}{3}=\frac{3}{2}\)
T₂ = a + d = 5 + \(\frac{3}{2}\)
= \(\frac{10+3}{2}=\frac{13}{2}\)
T₃ = a + 2d = 5 + 2(\(\frac{3}{2}\)) = 5 + 3 = 8.

(iv) Here T₁ = a = —
T₆ = a + 5d = 6
or -4 + 5d = 6
or 5d = 6 + 4
or 5d = 10
or d = \(\frac{10}{2}\) = 2
Now, T₂ = a + d = -4 + 2 = -2
T₃ = a + 2d = – 4 + 2(2)
= – 4 + 4 = 0
T₄ = a + 3d = – 4 + 3(2)
= – 4 + 6 = 2
T₅ = a + 4d = – 4 + 4(2)
= – 4 + 8 = 4

(v) Here T₂ = a + d = 38 ………….(1)
and T₆ = a + 5d = -22 ……………(2)
Now, (2) – (1) gives

Substitute this value of d in (1), we get
a + (-15) = 38
a = 38 + 15 = 53
∴ T₁ = a = 53
T₃ = a + 2d = 53 + 2(-15) = 53 – 30 = 23.
T₄ = a + 3d = 53 + 3(-15) = 53 – 45 = 8
T₅ = a + 4d = 53 + 4(-15) = 53 – 60 = – 7.

Question 4
Which term of the A.P. 3, 8, 13, 18, …………… is 78?
Solution:
Given A.P. is 3, 8, 13, 18, ………….
T₁ = 3, T₂ = 8, T₃ = 13, T₄ = 18
T₂ – T₁ =8 – 3=5
T₃ – T₂= 13 – 8=5
T₂ – T₁ = T₃ – T,= 5 = d (say)
Using, T_n = a + (n – I) d
or 78 = 3 + (n – 1) 5
or 5(n – 1) = 78 – 3 = 75
or n – 1 = 15
or n = 15 + 1 = 16
Hence, 16th term of given AP. is 78.

Question 5.
Find the number of terms in each of the following APs:
(i) 7, 13, 19,…, 205
(ii) 18, 15\(\frac{1}{2}\), 13, ………….., – 47
Solution:
(i) Given A.P. is 7, 13, 19, …………..
T₁ = 7, T₂ = 13, T₃ = 19
T₂ – T₁ = 13 – 7 = 6
T₃ – T₂ = 19 – 13 = 6
T₂ – T₁ = T₃ – T₂ = 6 = d(say)
Using formula, T_n = a + (n – 1) d
205 = 7 + (n – 1) 6
or (n – 1) 6 = 205 – 7 = 198
or (n – 1) = \(\frac{198}{6}\)
or n – 1 = 33
n = 33 + 1 = 34
Hence, 34th term of an AP. is 205.

(ii) Given A P. is 18, 15\(\frac{1}{2}\), 13, …………..
T₁ = 18, T₂ = 15\(\frac{1}{2}\) = \(\frac{31}{2}\), T₃ = 13
T₂ – T₁ = \(\frac{31}{2}\) – 18 = \(\frac{31-36}{2}=-\frac{5}{2}\)
T₃ – T₂ = 13 – \(\frac{31}{2}\) = \(\frac{26-31}{2}=-\frac{5}{2}\)
∵ T₂ – T₁ = T₃ – T₂ = \(\frac{-5}{2}\) = d (say)
Using formula. T_n = a + (n – 1) d
– 47 = 18 + (n – 1) \(\frac{-5}{2}\)
or (n – 1) (\(\frac{-5}{2}\)) = – 47 – 18
or (n – 1) (\(\frac{-5}{2}\)) = – 65
or n – 1 = – 65 × – \(\frac{2}{5}\)
or n – 1 = 26
or n = 26 + 1 = 27
Hence, 27th term of an A.P. is – 47.

Question 6.
Is – 150 a term of 11, 8, 5, 2….? why?
Solution:
Given sequence is 11, 8, 5, 2, ………..
T₁ = 11, T₂ = 8, T₃ = 5, T₄ = 2
T₂ – T₁ = 8 – 11 = – 3
T₃ – T₂ = 5 – 8 = – 3
T₄ – T₃ = 2 – 5 = – 3
T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = – 3 = d (say).
Let – 150 be any term of given A.P.
then T_n = – 150
a+(n – 1)d = – 150
or 11 +(n – 1)(- 3) = – 150
or (n – 1)( – 3) = – 150 – 11 = – 161
or n – 1 = \(\frac{161}{3}\)
or n = \(\frac{161}{3}\) + 1 = \(\frac{161+3}{3}\)
n = \(\frac{164}{3}\) = 54\(\frac{2}{3}\),
which is not a natural number.
Hence, – 150 cannot be a term of given A.P.

Question 7.
Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
Solution:
Let ‘a’ and 4d’ be the first term and common difference of given A.P.
Given that T₁₁ = 38
a +(11 – 1) d = 38
[∵ T_n = a + (n – 1) d]
a + 10 d = 38
and T₁₆ = 73
a + (16 – 1) d = 73
[∵ T_n = a + (n – 1) d]
a + 15 d = 73
Now, (2) – (1) gives

Substitute this value of d in (1), we get
a + 10 (7) = 38
or a + 70 = 38
or a = 38 – 70 = – 32
Now, T₃₁ = a + (31 – 1) d
= – 32 + 30 (7) = – 32 + 210 = 178.

Question 8.
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 291h term.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
Given that, T₃ = 12
a + (3 – 1) d = 12
∵ T_n = a + (n – 1) d
or a + 2d = 12 ………………(1)
and Last term = T₅₀ = 106
a + (50 – 1) d = 106
∵ T_n = a + (n – 1) d
a + 49 d = 106 ……………(2)
Now, (2) – (1) gives

Substitute this value of d in (1), we get
a + 2(2) = 12
or a + 4 = 12
or a + 12 – 4 = 8
Now, T₂₉ = a + (29 – 1) d
= 8 + 28 (2) = 8 + 56 = 64.

Question 9.
If the 3rd and 9th tenus of an A.P. are 4 and – 8 respectively, which term of this A.P. is zero.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given AP.
Given that: T₃ = 4
a + (3 – 1) d = 4
∵ T_n = a + (n – 1) d
a + 2d = 4 …………..(1)
and T₉ = – 8
a + (9 – 1) d = 8
and T₉ = – 8
a + (9 – 1)d = 8
∵ T_n = a + (n – 1) d
or a + 8d = – 8
Now, (2) – (1) gives

Substitute this value of d in (1), we get
a + 2(- 2) = 4
or a – 4 = 4
or a = 4 + 4 = 8
Now, T_n = 0 (Given)
a + (n – 1) d = 0
or 8 + (n – 1)(- 2)=0
or -2 (n – 1) = – 8
or n – 1 = 4
or n = 4 + 1 = 5
Hence, 5th term of an AP. is zero.

Question 10.
The 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
Now, T₁₇ = a + (17 – 1) d = a + 16 d
and T₁₀ = a + (10 – 1) d = a + 9 d
According to question
T₁₇ – T₁₀ = 7
(a + 16 d) – (a + 9 d) = 7
or a + 16 d – a – 9 d = 7
7 d = 7
or d = \(\frac{7}{7}\) = 1
Hence, common difference is 1.

Question 11.
Which term of the A.P. 3, 15, 27, 39, …………. will be 132 more than its 54th term?
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
Given A.P. is 3, 15, 27, 39, …
T₁ = 3, T₂ = 15, T₃ = 27, T₄ = 39
T₂ – T₁ = 15 – 3 = 12
T₃ – T₂ = 27 – 15 = 12
:. d=T₂ – T₁ = T₃ – T₂ =12
Now, T₅₄ = a + (54 – 1) d
= 3 + 53 (12) = 3 + 636 = 639
According to question
T = T₅₄ + 132
a + (n – 1)d = 639 + 132
3 + (n – 1)(12) = 771
(n – 1) 12 = 771 – 3 = 768
or n – 1 = \(\frac{768}{12}\) = 64
or n = 64 + 1 = 65
Hence, 65th term of A.P. is 132 more than its 54th term.

Question 12.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of first AP.
Also, ‘A’ and ‘d’ be the first term and common difference of second A.P.
According to question
[T₁₀₀ of second A.P.] – [T₁₀₀ of first A.P.] = 100
or[A +(100 – 1)d] – [a +(100 – 1)d] = 100
or A + 99d – a – 99d = 100
or A – a = 100
Now, [T₁₀₀₀ of second A.P.] – [T₁₀₀₀ of first A.P.]
= [A + (1000 – 1) d) – (a + (1000 – 1) d]
= A + 999 d – a – 999 d
= A – a = 100 [Using (I)].

Question 13.
How many three-digits numbers are divisible by 7?
Solution:
Three digits numbers which divisible by 7 are 105, 112, 119 , 994
Here a = T₁ = 105, T₂ = 112, T₃ = 119 and T_n = 994
T₂ – T₁ = 112 – 105=7
T₃ – T₂ = 119 – 112=7
∴ d = T₂ – T₁ = T₃ – T₂ = 7
Given that, T_n = 994
a + (n – 1) d = 994
or 105 + (n – 1) 7 = 994
or (n – 1) 7 = 994 – 105
or (n – 1) 7 = 889
or n – 1 = \(\frac{889}{7}\) = 127
or n = 127 + 1 = 128.
Hence, 128 terms of three digit number are divisible by 7.

Question 14.
How many multiples of 411e between 10 and 250?
Solution:
Multiples of 4 lie between 10 and 250 are 12, 16, 20, 24, … 248
Here a = T₁ = 12, T₂ = 16, T₃ = 20 and T_n = 248
T₂ – T₁ = 16 – 12 = 4
T₃ – T₂ = 20 – 16 = 4
∴ d = T₂ – T₁ = T₃ – T₂ = 4
Given that, T_n = 248
a + (n – 1) d = 248
or 12 + (n – 1)4 = 248
or 4(n – 1) = 248 – 12 = 236
or n – 1 = \(\frac{236}{4}\) = 59
or n = 59 + 1 = 60
Hence, there are 60 terms which are multiples of 4 lies between 10 and 250.

Question 15.
For what value of n, are the n terms of two A.P.s 63, 65, 67, …………. and 3, 10, 17, …………….. equal?
Solution:
Given A.P. is 63, 65, 67, ……………..
Here a = T₁ = 63, T₂ = 65, T₃ = 67
T₂ – T1 = 65 – 63 = 2
T₃ – T₂ = 67 – 65 = 2
∴ d = T₂ – T₁ = T₃ – T₂ = 2
and second A.P. is 3, 10, 17, …
Here a = T₁ = 3, T₂ = 10, T₃ = 17
T₂ – T₁ = 10 – 3 = 7
T₃ – T₂ = 17 – 10 = 7
According to question.
[nth term of first A.P.] = [nth term of second A.P.]
63 + (n – 1)2 = 3 + (n – 1) 7
or 63 + 2n – 2 = 3 + 7n – 7
or 61 + 2n = 7n – 4
or 2n – 7n = – 4 – 61
– 5n = – 65
n = \(\frac{65}{5}\) = 13.

Question 16.
Determine the AP. whose third term is 16 and 7 term exceeds the by 12.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
Given that T₃ = 16
a + (3 – 1) d = 16
a + 2d = 16
According to question
T₇ – T₅ = 12
[a + (7 – 1) d] – [a + (5 -1) d] = 12
a + 6 d – a – 44 = 12
2d = 12
d = \(\frac{12}{2}\) = 6
Substitute this value of d in (1), we get
a + 2(6) = 16
a = 16 – 12 = 4 .
Hence, given A.P. are 4, 10, 16, 22, 28, ………….

Question 17.
Find the 20th term from the last term of the AP: 3, 8, 13, ………., 253.
Solution:
Given A.P. is 3, 8, 13, …………., 253
Here, a = T₁ = 3, T₂ = 8, T₃ =13 and T_n = 253
T₂ – T₁ = 8 – 3 = 5
T₃ – T2 = 13 – 8 = 5
∴ d = T₂ – T₁ = T₃ – T₁ = 5
Now, T_n = 253
3 + (n – 1)5 = 253
∵ T_n = a + (n – 1) d
(n – 1) 5 = 250
n-1 = \(\frac{250}{5}\) = 50
n = 50 + 1 = 51
20th term from the end of AP = (Total number of terms) – 20 + 1
= 51 – 20 + 1 = 32nd term
∴ 20th term from the end of AP
= 32nd term from the starting
= 3 + (32 – 1)5
∵ Tn = a + (n – 1)d
= 3 + 31 × 5
= 3 + 155 = 158.

Question 18.
The sum of the 4th and 8th term of an AP is 24 and the sum of the 6’ and 10th terms is 44. FInd the first three terms of the A.P.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
According to 1st condition
T₄ + T₈ = 24
a + (4 – 1) d + a + (8 – 1) d = 24
∵ T_n = a + (n – 1) d
or 2a + 3d + 7d = 24
2a + 10d = 24
a + 5d = 12 …………(1)
According to 2nd condition
T₆ + T₁₀ = 44
a + (6 – 1) d + a +(10 – 1) d = 44
∵ T_n = a + (n – 1) d
2a + 5d + 9d = 44
2a + 14d = 44
a + 7d = 22
Now (2) – (1) gives

Substitute this value of d in (I). we get
a + 5(5) = 12
a + 25 = 12
a = 12 – 25 = -13
T₁ = a = -13
T₂ = a + d = 13 + 5 = -8
T₂ = a + 2d = – 13 + 2(5) = – 13 + 10 = -3
Hence, given A.P. is -13, -8, -3, ……………

Question 19.
Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?
Solution:
Subba Rao’s starting salary = ₹ 5000
Annual increment = ₹ 200
Let ‘n’ denotes number of years.
∴ first term = a = ₹ 5000
Common diflerence = d = ₹ 200
and T_n = ₹ 7000
5000 + (n – 1) 200 = 7000
∵ T_n = a + (n – 1) d
(n – 1) 200 = 7000 – 5000
or (n – 1) 200 = 2000
or n – 1 = \(\frac{2000}{200}\) = 10
or n = 10 + 1 = 11
Now, in case of year the sequence are 1995. 1996, 1997, 1998, ……………
Here a = 1995, d = 1 and n = 11
Let T_n denotes the required year.
∴ T_n = 1995 + (11 – 1) 1
= 1995 + 10 = 2005
Hence, in 2005, Subba Rao’s salary becomes 7000.

Question 20.
Ramkali saved ₹ 5 in the first week of a year and then increased her weekly saving by ₹ 1.75. If in the nth week, her weekly saving becomes ₹ 20.75, find n.
Solution:
Amount saved in first week = ₹ 5
Increment in saving every week = ₹ 1.75
It is clear that, it form an A.P. whose terms are
T₁ = 5, d = 1.75
∴ T₂ = 5 + 1.75 = 6.75
T₃ = 6.75 + 1.75 = 8.50
Also. T_n = 20. 75 (Given)
5 + (n – 1) 1.75 = 20.75
∵ T_n = a + (n – 1) d
or (n – 1) 1.75 = 20.75 – 5
or (n – 1) 1.75 = 15.75
or (n – 1) = \(\frac{1575}{100} \times \frac{100}{175}\)
or n – 1 = 9
or n = 9 + 1 = 10
Hence, in 10th week, Ramkali’s saving becomes ₹ 20.75.

Punjab State Board PSEB 10th Class English Book Solutions English Reading Comprehension Unseen Passages Exercise Questions and Answers, Notes.

PSEB 10th Class English Reading Comprehension Unseen Passages

नोट : परीक्षा-पत्र में Unseen Reading Comprehension के लिए दिया जाने वाला पैरा प्रायः व्याकरण की निर्धारित पुस्तक में से ही लिया जाता है। इसलिए विद्यार्थियों को ऐसे पैरे बहुत ध्यान से तैयार कर लेने चाहिएं।

(A) Passages From Grammar Book Note :

In all the passages, questions have been changed according to new pattern of Testing.

Passage – 1

One evening a boy of three was out for a walk with his father. There was also an elderly man with the father. Chatting, they walked on and went beyond the village. The green crops delighted the eyes. The elders were walking along the edge of a field. Not hearing the footsteps of the boy, the father looked back. The boy was sitting on the ground and seemed to be planting something. The father became curious. “What are you doing?” said he. “Look, Father, I shall grow guns all over the field,” was the innocent reply of the boy. His eyes shone with the strong faith that guns would grow in the field. Both the elders were struck with wonder at the little boy’s words. The boy was Bhagat Singh who later fought like a hero for India’s freedom and sacrificed his life.

Word-meanings : 1. elderly – बुजुर्ग 2. chatting – बातचीत करते हुए ; 3. curious – उत्सुक; 4. sacrificed – बलिदान दे दिया

Choose the correct option to answer each question :

Question 1.
Where were the elders walking?
(a) Along the bank of a canal.
(b) Along the edge of a well.
(c) Along the bank of the river.
(d) Along the edge of a field.
Answer:
(d) Along the edge of a field.

Question 2.
What was the boy doing when his father looked back?
(a) He was sitting on the ground.
(b) He was planting something.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 3.
What faith did the boy have?
(a) He believed that bombs would grow in the field.
(b) He believed that guns would grow in the field.
(c) He believed that green crops would grow in the field.
(d) He believed that brave men would grow in the field.
Answer:
(b) He believed that guns would grow in the field.

Question 4.
Father looked back because …………
(a) he did not hear the footsteps of his son.
(b) his son was calling him.
(c) his son was left behind.
(d) his son asked him to do so.
Answer:
(a) he did not hear the footsteps of his son.

Question 5.
Bhagat Singh sacrificed his life for …..
(a) his family
(b) truth
(c) India’s freedom
(d) his ideals.
Answer:
(c) India’s freedom

Passage 2

Dr. C.V. Raman was, a genius? who won the Nobel Prize for Physics, using simple equipment barely worth $300. He was the first Asian scientist to win the Nobel Prize. He was a man of boundless curiosity and had a lively sense of humour. His spirit of inquiry and devotion to science laid the foundation for scientific research in India. And he won honour as a scientist, and affection as a teacher and a man. Raman was studious. He kept in touch with the latest developments in science in the world around him. He had personal contacts with many scientists. He used to read new books and research papers from different centres. “The equipment which brought me the Nobel Prize did not cost more than three hundred rupees. A table drawer can hold all my research equipment,” he used to say with pride. It was his conviction that if the research worker is not inspired from within, no amount of money can bring him or her success in research.

Word meanings : 1. genius – प्रतिभाशाली ; 2. equipment – साज-सामान, उपकरण; 3. inquiry – जांच, उत्सुकता ; 4. studious – अध्ययनशील ; 5. conviction – दृढ़-विश्वास।

Choose the correct option to answer each question :

Question 1.
What kind of a man was Dr. C.V. Raman?
(a) He was a man of unlimited curiosity.
(b) He had a lively sense of humour.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 2.
What is he famous for ?
(a) For being the first American scientist to win the Nobel Prize.
(b) For being the first Asian scientist to win the Nobel Prize.
(c) For being the first African scientist to win the Nobel Prize.
(d) For being the first Japanese scientist to win the Nobel Prize.
Answer:
(b) For being the first Asian scientist to win the Nobel Prize.

Question 3.
He believed that if the research work is not ……… any amount of money cannot make the research successful.
(a) applauded by people
(b) done deeply
(c) inspired from within
(d) done thoughtfully.
Answer:
(c) inspired from within

Question 4.
The equipment that won him the Nobel Prize
(a) did not cost more than five hundred rupees.
(b) did not cost more than three hundred rupees.
(c) did not cost more than four hundred rupees.
(d) did not cost more than six hundred rupees.
Answer:
(b) did not cost more than three hundred rupees.

Question 5.
Which word in the passage means ‘scholarly’?
(a) genius
(b) studious
(c) conviction
(d) barely.
Answer:
(b) studious

Passage 3

Early rising leads to health and happiness. The man who rises late can have little rest in the course of the day. Anyone who lies in bed late is compelled to work till late hours in the evening. He has to go without the evening exercise which is so necessary for his health. In spite of all his efforts, his work will not produce as good results as that of an early riser. The reason for this is that he cannot take advantage of the refreshing morning hours. Some people say that the quiet hours of midnight are the best time for working. Several great thinkers say that they can write best only when they burn the midnight. Yet it is true to say that few men have a clear brain at midnight when the body needs rest and sleep. Those who work at that time soon ruin their health. Bad health must, in the long run, have a bad effect on the quality of their work.

Word-meanings : 1. in the course of — के दौरान; 2. compelled — बाध्या होना ; 3. burn the midnight oil – देर रात तक काम करना

Choose the correct option to answer each question :

Question 1.
What leads to health and happiness ?
(a) Late rising.
(b) Early rising.
(c) Brisk walk.
(d) Long walk.
Answer:
(b) Early rising.

Question 2.
What is a late riser compelled to do ?
(a) To work for a short while.
(b) To work early in the morning.
(c) To stop working at all.
(d) To work till late hours in the evening.
Answer:
(d) To work till late hours in the evening.

Question 3.
Why can’t we have a clear brain at midnight?
(a) Because we are fully fresh at that time.
(b) Because we want to go for walk at that time.
(c) Because at that time our body needs rest and sleep.
(d) Because at that time we are fully disturbed.
Answer:
(c) Because at that time our body needs rest and sleep.

Question 4.
The man who rises late cannot take advantage of ………
(a) the refreshing sunny hours.
(b) the refreshing midnight hours.
(c) the refreshing evening hours.
(d) the refreshing morning hours.
Answer:
(d) the refreshing morning hours.

Question 5.
The word ‘essential’ means – …………
(a) compelled
(b) necessary
(c) advantage
(d) ruin
Answer:
(b) necessary

Passage 4.
Each one of us should have a hobby. Sometimes boys and girls are encouraged in schools to take up hobbies. They work at certain things in schools such as collecting stamps, or carpentry, but this so-called hobby is a thing for schools only. They do not pay any attention to it at their homes. Sometimes this is because of poverty, sometimes because of lack of interest. But a hobby is not really a hobby unless we are so interested in it that we want to carry it on whenever we have spare time. According to the dictionary meaning, a hobby is a favourite subject or occupation that is not one’s main business. It is something in which we are more interested than in anything else.

Word-meanings : 1. encouraged – प्रोत्साहित; 2. poverty – निर्धनता

Choose the correct option to answer each question :

Question 1.
What is the dictionary meaning of ‘hobby’ ?
(a) A skill that is one’s source of earning.
(b) Favourite subject or occupation that is not one’s main business.
(c) Both (a) and (b)
(d) Neither (a) nor (b).
Answer:
(b) Favourite subject or occupation that is not one’s main business.

Question 2.
When does a hobby become a thing for schools only?
(a) It is when no attention is paid to it at home.
(b) It is when full attention is paid to it at home.
(c) It is when no attention is paid to it at school.
(d) It is when full attention is paid to it at school.
Answer:
(a) It is when no attention is paid to it at home.

Question 3.
Name the hobbies mentioned in the passage.
(a) Stamp collecting.
(b) Carpentry.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 4.
We should carry on our hobby whenever we have ……..
(a) no time
(b) spare time
(c) golden time
(d) busy time
Answer:
(b) spare time

Question 5.
Sometimes students cannot pay attention to their hobbies due to ……….
(a) over work and studies
(b) lack of spare time
(c) poverty or lack of interest
(d) none of these three.
Answer:
(c) poverty or lack of interest

Passage – 5

Ashoka, the most trusted son of Bindusara and the grandson of Chandragupta Maurya, was a brave soldier. He was the most famous of the Mauryan kings and was one of the greatest rulers of India. Ashoka extended the Mauryan Empire to the whole of India, reaching out even into Central Asia. Ashoka succeeded in conquering Kalinga after a bloody war in which 100,000 men were killed, 150,000 injured and thousands were captured and retained as slaves. The sight of the slaughter involved in his conquest’ deeply affected his mind. He renounced war and sought peace in Buddha’s preachings of love and non-violence. So he gave up hunting.

Word-meanings: 1. extended – फैलाया; 2. in conquering- जीतने में; 3. retained – बंदी बना लिए गए ; 4. slaughter- हत्याकाण्ड; 5. conquest- जीत ; 6. renounced- त्याग दिया|

Choose the correct option to answer each question :

Question 1.
What was the name of Ashoka’s grandfather ?
(a) Bindusara Maurya.
(b) Rishabh Maurya.
(c) Chandragupta Maurya.
(d) Bhisham Maurya.
Answer:
(c) Chandragupta Maurya.

Question 2.
What happened in the war of Kalinga ?
(a) 100,000 men were killed.
(b) 150,000 men were injured.
(c) Thousands were captured and retained as slaves.
(d) All of these three.
Answer:
(d) All of these three.

Question 3.
Write the changes in Ashoka’s life after the war.
(a) He gave up war.
(b) He sought peace in love and non-violence.
(c) Both (a) and (b).
(d) Neither (a) nor (b)
Answer:
(c) Both (a) and (b).

Question 4.
Ashoka was one of the …………..
(a) greatest saints of India
(b) greatest rulers of India
(c) greatest preachers of India
(d) greatest yogis of India
Answer:
(b) greatest rulers of India

Question 5.
For Ashoka the war of Kalinga was a ……………
(a) bloody massacre
(b) great war
(c) great victory for him
(d) turning point of his life.
Answer:
(d) turning point of his life.

Passage 6

“Brothers and sisters, the long night is at last drawing to a close. Miseries and sorrows are disappearing. Ours is a sacred country. She is gradually waking up, thanks to the fresh breeze all around. Are you prepared for all sacrifices for the sake of your mother-land ? If you are, then you can rid the land of poverty and ignorance. You should develop a strong physique? You should shape your mind through study and meditation. Only then will victory be yours. I loved my motherland dearly before I went to America and England. After my return, every particle of the dust of this land seems sacred to me.” Do you know who carried this message to the whole continent of Asia ? It was Swami Vivekananda.

Word meanings : 1. miseries – दुःख ; 2. gradually-धीरे-धीरे ; 3. physique – डील-डौल ; 4. meditation – चिन्तन।

Choose the correct option to answer each question :

Question 1.
How can we rid the land of poverty and ignorance?
(a) By fighting for our rights.
(b) By doing our duties.
(c) By making sacrifices for our family.
(d) By making sacrifices for the sake of our motherland.
Answer:
(d) By making sacrifices for the sake of our motherland.

Question 2.
How can you shape your mind ?
(a) Through study.
(b) Through meditation.
(c) Both (a) and (b)
(d) Neither (a) nor (b)
Answer:
(c) Both (a) and (b)

Question 3.
Where did Swami Vivekananda carry the message ?
(a) To the whole continent of Antarctica.
(b) To the whole continent of Asia.
(c) To the whole continent of Africa.
(d) To the whole continent of Europe.
Answer:
(b) To the whole continent of Asia.

Question 4.
India is gradually. …….
(a) waking up
(b) sleeping
(c) lagging behind
(d) preparing for the war.
Answer:
(a) waking up

Question 5.
Who speaks the above lines ?
(a) Mahavira
(b) Gautama Buddha
(c) Swami Vivekananda
(d) Guru Nanak Dev.
Answer:
(c) Swami Vivekananda

Passage 7.

A Nation’s Strength :
Not gold, but only men can make
A nation great and strong
Men who, for truth and honour’s sake,
Stand fast and suffer long.

Brave men who work while others sleep,
Who darewhile others fly,
They build a nation’s pillars deep,
And lift them to the sky. – R.W. Emerson

Word-meanings : 1. stand fast – डटे रहना ; 2. dare – सामना करते हैं ; 3. fly – भाग जाते हैं ; 4. pillars – स्तम्भ।

Choose the correct option to answer each question :

Question 1.
What makes a nation great and strong ?
(a) Its youth.
(b) Its soldiers.
(c) Its people.
(d) Its government.
Answer:
(c) Its people.

Question 2.
What do brave men do while others sleep?
(a) They work hard.
(b) They suffer long.
(c) They also sleep.
(d) They run away.
Answer:
(a) They work hard.

Question 3.
How can they lift a nation to the sky ?
(a) Through their sacrifices.
(b) Through their faith.
(c) Through their duty.
(d) Through their hard work.
Answer:
(d) Through their hard work.

Question 4.
Complete the line :
Who dare while ……………..!
(a) others fly
(b) others laugh
(c) others die
(d) others challenge.
Answer:
(a) others fly

Question 5.
Who are the real pillars of a nation ?
(a) Common people.
(b) Rich people.
(c) Brave people.
(d) Poor people.
Answer:
(c) Brave people.

Question 6.
The word ‘dare’ means –
(a) endure
(b) challenge
(c) raise
(d) reality
Answer:
(b) challenge

Passage 8

My Books :

I love my books
They are the homes
of queens and fairies,
Knights? and gnomes.

Each time I read I make a call
On some quaint person large or small,
Who welcomes me with a hearty hand
And leads me through his wonderland.
Each book is like

A city street
Along whose winding
Way I meet
New friends and old who laugh and sing
And take me off adventuring.

Word-meanings : 1. knights – योद्धा ; 2. gnomes – बौने ; 3. quaint – अजीब ; 4. winding – बल-खाती।

Choose the correct option to answer each question :

Question 1.
The poet loves his books because through them he …………
(a) meets new friends
(b) sees new lands
(c) both (a) and (b)
(d) neither (a) nor (b)
Answer:
(c) both (a) and (b)

Question 2.
How does the quaint person welcome the poet ?
(a) With no smile.
(b) With a sad heart.
(c) With a hearty hand.
(d) Any of these three.
Answer:
(c) With a hearty hand.

Question 3.
The poet is led through a …….. each time he reads a book.
(a) homeland
(b) wonderland
(c) woodland
(d) farmland.
Answer:
(b) wonderland

Question 4.
In the books, the poet meets new and old friends in the ……
(a) city streets
(b) wonderland
(c) villages
(d) farmhouses.
Answer:
(a) city streets

Question 5.
The antonym of the word ‘winding’ is ……
(a) straight
(b) round
(c) curved
(d) feeble
Answer:
(a) straight

Passage 9

Stone Walls Do Not A Prison Make:
Stone walls do not a prison make
Nor iron bars a cage :
Mind’s innocence and quiet take
That for a hermitage?

If I have freedom in my love,
And in my soul am free,
Angels alone that soar above
Enjoy such liberty. -R. Lovelace

Word-meanings : 1. hermitage – आश्रम; 2. angel – देवदूत

choose the correct option to answer each question :

Question 1.
What does not make a prison ?
(a) Strong walls.
(b) Stone walls.
(c) Weak walls.
(d) Brick Walls.
Answer:
(b) Stone walls.

Question 2.
According to the poet, what does not make a cage ?
(a) Iron bars.
(b) Wooden bars.
(c) Golden bars.
(d) Silver bars.
Answer:
(a) Iron bars.

Question 3.
Which quality of the mind makes a hermitage ?
(a) Innocence of the mind.
(b) Quiet acceptance.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 4.
‘Such liberty’ is enjoyed only by ……….
(a) birds
(b) angels
(c) devils
(d) animals.
Answer:
(b) angels

Question 5.
The word ‘liberty’ means – ……….
(a) spirit
(b) freedom
(c) jail
(d) rise
Answer:
(b) freedom

Passage 10

A child’s Evening Prayer :
Erel on my bed my limbs I lay,
God grant me grace my prayers to
say:
O God ! preserve? my mother dear
In strength and health for many a year;

And, O ! preserve my father too,
And may I pay him reverence due;
And may I my best thoughts employ
To be my parents’ hope and joy;
And O ! preserve my brothers both

From evil doings and from sloth,
And may we always love each other,
Our friends, our father, and our mother :
And still, O Lord, to me impart
An innocent and grateful heart,

That after my great sleep I may
Awake to Thy eternal day! – Samuel Taylor Coleridge

Worst-meanings : 1. ere – से फूर्व ; 2. preserve – सुरक्षित रखना ; 3. reverence – सम्मान ; 4. sloth — सुस्ती; 5. Thy — your.

Choose the correct option to answer each question :

Question 1.
When does the child pray ?
(a) Before going to bed.
(b) After going to bed.
(c) Before getting up.
(d) After getting up.
Answer:
(a) Before going to bed.

Question 2.
The child prays to God that his mother ……………. .
(a) may remain strong
(b) may remain healthy
(c) live long
(d) all of these three.
Answer:
(d) all of these three.

Question 3.
What does he ask for himself ?
(a) An innocent heart.
(b) A grateful heart.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 4.
The child pays ………… to his father.
(a) reverence
(b) strength
(c) wealth
(d) attention.
Answer:
(a) reverence

Question 5.
The child wants to be the hope and joy of ………
(a) God
(b) his parents
(c) his siblings
(d) his teachers.
Answer:
(b) his parents

Question 6.
Give the rhyme-scheme of the last four lines.
(a) abab
(b) abba
(c) aabb
(d) baba.
Answer:
(c) aabb

Passage 11.

The Rainbow :
Boats sail on the rivers,
And ships sail on the seas;
But clouds that sail across the sky
Are prettier? than these.
There are bridges on the rivers,
As pretty as you please;
But the bow that bridges heaven,
And overtops the trees,
And builds a road from earth to sky,
Is prettier far than these. – Christina Rossetti

Word-meanings : 1. sail – तैरना; 2. prettier — जयादा सुन्दर; 3. bow – कमान, इन्दूधनुष ; 4. overtops – के ऊपर से गुजाती है

Choose the correct option to answer each question :

Question 1.
Where does the rainbow build a road?
(a) From the sky to the sea.
(b) From the earth to the sky.
(c) From the land to the sea.
(d) None of these three.
Answer:
(b) From the earth to the sky.

Question 2.
What are prettier than boats and ships ?
(a) Aeroplanes.
(b) Raindrops.
(c) Clouds.
(d) Trains.
Answer:
(c) Clouds.

Question 3.
According to the poet, the …… is the prettiest.
(a) sky
(b) rainbow
(c) rain
(d) cloud.
Answer:
(b) rainbow

Question 4.
The rainbow bridges ………..
(a) hell
(b) aboder
(c) heaven
(d) sky.
Answer:
(c) heaven

Question 5.
The word ‘far’ means – ……….
(a) a little.
(b) much more.
(c) many a.
(d) the few.
Answer:
(b) much more.

Passage 12 

The Noble Nature :
It is not growing like a tree
In bulk?, doth make man better be;
Or standing long an oak, three hundred year,
To fall a log at last, dry, bald and sere?:

A lily of a day
Is fairer far in May,
Although it falls and dies that night;
It was the plant and flower of light.
In small proportions we just beauties see:

And in short measures life may perfect be. -Ben Jonson

Word-meanings : 1. bulk — आकार में बड़ा ; 2. bald — गंजा; 3. sere – बेकार के; 4. proportions — मात्रा

Choose the correct option to answer each question :

Passage 1.
What does not make a man a better being ?
(a) Being wealthy
(b) Becoming famous.
(c) Growing in size.
(d) Being kind-hearted.
Answer:
(c) Growing in size.

Passage 2.
Which thing in nature can live up to three hundred years ?
(a) A lily flower.
(b) An oak tree.
(c) A sunflower.
(d) A willow tree.
Answer:
(b) An oak tree.

Passage 3.
How long does a lily live ?
(a) For a day.
(b) For a week.
(c) For a month.
(d) For a year.
Answer:
(a) For a day.

Passage 4.
The message of this poem is that a …………. but ………… life is far better than a worthless life of long years.
(a) long, virtuous.
(b) long, vicious.
(c) short, virtuous.
(d) short, vicious.
Answer:
(c) short, virtuous.

Passage 5.
In short measures also, life may be …….
(a) pious
(b) perfect
(c) impious
(d) imperfect.
Answer:
(b) perfect

Passage 13.

The great advantage of early rising is the good start it gives us in our day’s work. The early riser has done a large amount of hard work before other men have got out of bed. In the early morning, the mind is fresh and there are few sounds or other distractions, so that the work done at that time is generally well-done. In many cases, the early riser also finds time to take some exercise in fresh morning air, and this exercise supplies him with a fund of energy that will last until the evening . By beginning so early, he knows that he has plenty of time to do thoroughly all the work he can be expected to do, and is not tempted to hurry over any part of it. All his work being finished in good time, he has a long interval of rest in the evening before the timely hour when he goes to bed. He gets to sleep several hours before midnight, at the time when sleep is most refreshing and after a sound night’s rest, rises early next morning in good health and spirits for the labour of the next day.

Word-meanings : 1. advantage – लाभ ; 2. distractions – भटकाव ; 3. thoroughly – अच्छी तरह।

Choose the correct option to answer each question :

Question 1.
……….. gives our day’s work a good start.
(a) Late rising.
(b) Going to bed late.
(c) Early rising.
(d) Going to bed early.
Answer:
(c) Early rising.

Question 2.
Why is the work done in the early hours generally well-done ?
(a) Because the mind is fresh
(b) Because the mind is free from noises.
(c) Because the mind is free from distractions.
(d) All of these three.
Answer:
(d) All of these three.

Question 3.
When is the mind fresh ?
(a) Late at night.
(b) In the evening.
(c) In the early morning.
(d) At midnight.
Answer:
(c) In the early morning.

Question 4.
We can find ample time of rest in the evening by …
(a) taking exercise
(b) finishing our work at night
(c) rising late
(d) finishing our work in good time.
Answer:
(d) finishing our work in good time.

Question 5.
The word ‘tempted’ means –
(a) suitable
(b) benefit
(c) usually
(d) attracted
Answer:
(d) attracted

(B) Some Other Passages

Passage 1

Of all the trees, of southern Asia, the banyan is unique, not only for the manner of its growth, but for the area of shade it provides from the burning sun. Its close relationship with man has evolved? over the years to make the banyan a popular meeting place, a focal point of worship and a source of practical materials for commerce. Known as the strangler fig because of its unusual manner of growth, the banyan is an epiphyte6 or air plant, that has its birth in the branches of a host tree and lives on airborne moisture and nutrients. Banyan seeds are deposited by birds, bats or monkeys in the rich soil collected in the crevices? of host-tree branches. As the banyan grows, it sends aerial roots down the trunk of the supporting tree. In time, the roots that reach the ground choke the host tree by preventing its trunk from enlarging. The two best-known species of banyans are : the Indian (Ficus benghalensis), one of the world’s largest tropicalo trees; and the Chinese (Ficus retusa), a smaller species with fewer aerial roots.

Word-meanings : 1. provides – देता है, टिंटा 7 ; 2. evolved – धीरे-धीरे विकसित हुआ ; 3. focal – केन्द्रीय ; 4. strangler – जो किसी चीज़ की वृद्धि को दबाता या रोकता है ; 5. fig – चौड़े पत्तों वाला पेड़; 6. epiphyte – एक पौधा जो दूसरे पौधे या कभी-कभी दूसरी चीज़ पर बढ़ता है; 7. crevices – दरारें; 8. aerial – आकाश में लटका हुआ ; 9. tropical – उष्ण कटिबन्धी।

Choose the correct option to answer each question :

Question 1.
Why is the banyan called the strangler fig?
(a) Because it chokes the host tree.
(b) Because it chokes other trees.
(c) Because it chokes the drains.
(d) Any of these three.
Answer:
(a) Because it chokes the host tree.

Question 2.
In what ways is the banyan tree unique ?
(a) It is unique in the manner of its growth.
(b) It is unique in terms of the area of shade it provides.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 3.
The banyan tree is a popular meeting place because ……….
(a) it provides cool shade.
(b) it provides sweet fruit.
(c) it provides a focal point of worship.
(d) none of these three.
Answer:
(c) it provides a focal point of worship.

Question 4.
The Indian banyan is one of the world’s tropical trees.
(a) oldest
(b) smallest
(c) youngest
(d) largest
Answer:
(d) largest

Question 5.
The antonym of the word ‘enlarging’ is –
(a) enhancing
(b) following
(c) contracting
(d) forgiving.
Answer:
(c) contracting

Passage 2

Sir Alexander Fleming was a very humble? and modest? man. He found penicillin by chance. He was very hard-working. He tried all his life to fight for killing germs in the human body. Before the invention of penicillin, carbolic acid and sulpha drugs were used as disinfectants. These things killed the germs of many diseases, but they also harmed the cells of human body. Penicillin saved the lives of millions of human beings.

Word-meanings : 1. humble – नम्र, विनीत ; 2. modest – विनम् संकोची; 3. by chance – संयोग से ; 4. cells – कोशिकाएं।

Choose the correct option to answer each question :

Question 1.
What was the nature of Sir Alexander Fleming ?
(a) A very humble man.
(b) A very hardworking man.
(c) A very modest man.
(d) All of these three.
Answer:
(d) All of these three.

Question 2.
What did he find ?
(a) Penicillin.
(b) Radium.
(c) Carbolic acid.
(d) Sulpha drugs.
Answer:
(a) Penicillin.

Question 3.
Penicillin saved the lives of ……………….. human beings.
(a) hundreds
(b) thousands
(c) millions
(d) any of these three.
Answer:
(c) millions

Question 4.
……… harmed the cells of human body.
(a) Carbolic acid
(b) Sulpha drugs
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 5.
The word ‘invention’ means –
(a) medicines
(b) Sulpha drugs
(c) noble
(d) discovery
Answer:
(d) discovery

Passage 3

The elephant is the only animal with a trunk?. It uses its trunk in many ways. It pulls leaves of trees with its trunk and then puts them into its mouth. It can use its trunk to get water. The trunk can hold a lot of water to drink. An elephant needs to drink more than three pints of water every day. In Africa, men have hunted elephants for their tusks. The ivory from tusks is made into many beautiful things.
heavy things over long distances.

Word-meanings : 1. trunk- हाथी की सूंड ; 2. pint- तरल पदार्थ नापने की इकाई ; 3. tusk- हाथी का लम्बा दांत जो मुंह से बाहर निकला होता है ; 4. ivory- हाथीदांत।

Choose the correct option to answer each question :

Question 1.
An elephant uses its trunk to …..
(a) pull leaves
(b) put food into its mouth
(c) get water
(d) all of these three.
Answer:
(d) all of these three.

Question 2.
How much water does an elephant need to drink in a day?
(a) More than one pint of water.
(b) More than three pints of water.
(c) Less than two pints of water.
(d) Less than four pints of water.
Answer:
(b) More than three pints of water.

Question 3.
Which is the only animal who has a trunk ?
(a) The lion.
(b) The horse.
(c) The leopard.
(d) The elephant
Answer:
(d) The elephant

Question 4.
The ivory from ………… is used to make many beautiful things.
(a) trunk
(b) tusks
(c) tail
(d) trumpet.
Answer:
(b) tusks

Question 5.
Which word in the passage means ‘teeth?
(a) tusks
(b) trunk
(d) ivory.
Answer:
(b) tusks

Passage 4.

Guru Gobind Singh, the last of the ten Sikh Gurus, was a great son of India. He led India to sublime heights of glory. He brought about moral resurgences and created order out of disorderly social conditions. He was a multi-faceted personality. He was a poet, a soldier, a statesman, a leader and a true socialist, all in one. He put life into the dying social order by giving it a wholly new outlook.

Word-meanings : 1. sublime – उत्कृष्ट, भव्य; 2. glory – यश, गौरव: ; 3. resurgence – एक बार फिर से अधिक ताक़तवर तथा अधिक प्रसिद्ध होना।

Choose the correct option to answer each question :

Question 1.
Who was the last Sikh Guru ?
(a) Guru Nanak Dev.
(b) Guru Gobind Singh.
(c) Guru Arjun Dev.
(d) Guru Ram Das.
Answer:
(b) Guru Gobind Singh.

Question 2.
What did he do to improve the social order ?
(a) He brought about financial prosperity.
(b) He brought about moral resurgence.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(b) He brought about moral resurgence.

Question 3.
How did he put life into the dying social order ?
(a) By working for the poor and weak people.
(b) By leading it to heights of glory.
(c) By giving it a wholly new outlook.
(d) None of these three.
Answer:
(c) By giving it a wholly new outlook.

Question 4.
Guru Gobind Singh was ……..
(a) a great soldier
(b) a great poet
(c) a true socialist
(d) any of these three.
Answer:
(d) any of these three.

Question 5.
The antonym of the word ‘disorderly’ is ……..
(a) ordered
(b) disorder
(c) orderly
(d) disordered.
Answer:
(c) orderly

Passage 5

True Friendship:

Everyone that flatters thee
Is no friend in misery.
Words are easy like the wind
Faithful friends are hard to find.

Every man will be thy friend,
Whilst thou hast wherewithal to spend.
But if store of crowns” be scant,

No man will supply thy want’.
He that is thy friend indeed,
He will help thee in thy need.
If thou sorrow, he will weep,
If thou wake, he cannot sleep.
Thus of every grief in heart,
He with thee doth bear a part.

These are certain signs to know,
Faithful friends from flattering foe. -William Shakespeare

Word-meanings : 1. flatter – चापलूसी करना, खुशामद करना; 2. thee, thou – तुम्हें, तुम; 3. easy like the wind – कुछ खर्च नहीं होता; 4. wherewithal – सम्पत्ति; 5. crowns – पुराने समय में ब्रिटेन में प्रचलित सिक्के; 6. scant – कम, थोड़ा; 7. want – ज़रूरत, आवश्यकता; 8. doth bear – लेता है; 9. foe – शत्रु।

Choose the correct option to answer each question :

Question 1.
What is this poem about ?
(a) About the love of money.
(b) About true friendship.
(c) About great leaders.
(d) About foolish persons.
Answer:
(b) About true friendship.

Question 2.
What does the poet say about flatterers ?
(a) He calls them friends.
(b) He calls them enemies.
(c) He calls them teachers.
(d) He calls them poets.
Answer:
(b) He calls them enemies.

Question 3.
When do people become our friends ?
(a) When we have the money to spend.
(b) When we have the time to spare.
(c) When we have the power to rule.
(d) When we have the wisdom to judge.
Answer:
(a) When we have the money to spend.

Question 4.
How will a true friend feel when we are in grief?
(a) He will feel a sense of relief.
(b) He will feel bored and disturbed.
(c) He will feel very indifferent.
(d) He will feel very unhappy.
Answer:
(d) He will feel very unhappy.

Question 5.
What distinction does the poet want us to understand ?
(a) The distinction between poverty and riches.
(b) The distinction between high and low.
(c) The distinction between friends and flatterers.
(d) The distinction between purity and impurity.
Answer:
(c) The distinction between friends and flatterers.

Passage 6

The Centipede

The little creature
with a hundred feet
was on its journey
to where

only it knew
My civilised foot
dressed in polished leather
came down upon it
ever so gently

there was only a soft sound
signifyings
the end
of a creature of God
my Maker.

I looked to see
if my sole was soiled?
and walked away.

word-meanings : 1. centipede – कनखजूरा ; 2. creature – जीव ; 3. civilised – सभ्य ; 4. polished leather – पॉलिश किए चमड़े के जूते ; 5. signifying – व्यक्त करते हुए ; 6. sole – तलवा ; 7. soiled – गन्दा होना।

Choose the correct option to answer each question :

Question 1.
What little creature does the poet talk of ?
(a) The scorpion.
(b) The wasp.
(c) The ringworm.
(d) The centipede.
Answer:
(d) The centipede.

Question 2.
Why does the poet call his foot civilised ?
(a) Because he was wearing polished shoes.
(b) Because his foot had crushed the little creature.
(c) Because his polished shoes had come down gently upon the little creature.
(d) None of the above.
Answer:
(c) Because his polished shoes had come down gently upon the little creature.

Question 3.
What kind of poem would you say this is ?
(a) Romantic.
(b) Satirical.
(c) Humorous.
(d) Religious.
Answer:
(b) Satirical.

Question 4.
What happened when the poet’s foot came down upon the little creature ?
(a) There was a soft sound.
(b) There was a loud sound.
(c) There was a sharp cry.
(d) There was no sound at all.
Answer:
(a) There was a soft sound.

Question 5.
The poet walked away. What does this show ?
(a) He was full of repentance on what he had done.
(b) He was full of pity for the little creature.
(c) He was indifferent to what had happened.
(d) He was full of praise for God.
Answer:
(c) He was indifferent to what had happened.

Passage 7.

Who Loved The Best :
‘I Love you, mother,’ said little John;
Then, forgetting his work his cap went on,
And he was off to the garden swing?,
Leaving his mother, the wood to bring.
‘I love you, mother,’ said little Nell,
‘I love you better than tongue can tell.’
Then she teased and pouted half the day,
Till mother rejoiced when she went to play.
‘I love you, mother,’ said little Fan,

“To-day I’ll help you all I can’.
To the cradle then she did softly creep
And rocked the baby till it fell asleep.
Then stepping softly she took the broom
And swept floor and dusted the room.
Busy and happy all day was she,
Helpful and cheerful as child can be.
‘I love you, mother,’ again they said,
Three little children going to bed.
How do you think the mother guessed”,
Which of them really loved the best?

Word-meanings : 1. little – नन्हा; 2. swing— झूला; 3. pouted – मुंह फुला लिया; 4. rejoiced – बहुत खुश हो गई; 5. cradle – पालना; 6. cheerful- प्रसन्न guessed–अंदाज़ा लगाया

Choose the correct option to answer each question :

Question 1.
Who said, “I love you, mother’?
(a) John.
(b) Nell.
(c) Fan.
(d) All the three.
Answer:
(d) All the three.

Question 2.
Who had to bring the wood ?
(a) Mother.
(b) John
(c) Nell.
(d) Fan.
Answer:
(a) Mother.

Question 3.
Who was lying in the cradle ?
(a) John.
(b) Nell.
(c) Fan.
(d) None of these three.
Answer:
(d) None of these three.

Question 4.
Who was busy and happy all day ?
(a) Little John.
(b) Little Nell.
(c) Little Fan.
(d) The mother.
Answer:
(c) Little Fan.

Question 5.
Who do you think loved mother the best ?
(a) John.
(b) Nell.
(c) Fan
(d) None of these three.
Answer:
(c) Fan

Passage 8

The storm came on before its time,
She wandered? up and down;
And many a hill did Lucy climb;
But never reached the town.

The wretched parents all that night,
Went shouting far and wide;
But there was neither sound nor sight,
To serve them for a guide.

Word-meanings : 1. wandered — इधर-इधर भटकती रही; 2. wretched – दुःखी, शोक-संतप्त; 3. far and wide – दूर-दूर तक।

Choose the correct answer for each question :

Question 1.
When did the storm come ?
(a) Before its time.
(b) After its time.
(c) At midnight.
(d) Early in the morning.
Answer:
(a) Before its time.

Question 2.
Why did Lucy not reach the town?
(a) Because she didn’t want to travel in the storm.
(b) Because she didn’t want to leave her native place.
(c) Because she drowned in the river.
(d) Because she lost her way in the storm.
Answer:
(d) Because she lost her way in the storm.

Question 3.
What did the parents do?
(a) They went to the town.
(b) They searched for Lucy all night.
(c) They waited for her return from the town.
(d) They went to the police station for help.
Answer:
(b) They searched for Lucy all night.

Question 4.
She wandered ………..
(a) up and down
(b) here and there
(c) far and wide
(d) near and far.
Answer:
(a) up and down

Question 5.
But there was …..
(a) neither heat nor light
(b) neither sound nor sight
(c) either heat or light
(d) either sound or sight.
Answer:
(b) neither sound nor sight

Punjab State Board PSEB 10th Class Science Book Solutions Chapter 16 Management of Natural Resources Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Science Chapter 16 Management of Natural Resources

PSEB 10th Class Science Guide Management of Natural Resources Textbook Questions and Answers

Question 1.
What changes would you suggest in your home in order to be environment friendly?
Answer:
We should use the formula of three R’s to save the environment i.e. Reduce, Recycle and Reuse.

• Reduce: Use less, save water and electricity by stopping the wastage. Do not waste food.
• Recycle: Segregate the wastes so that material which can be recycled should be dumped in place for recycling.
• Reuse: Reuse certain articles e.g. reverse the used envelope and reuse it.

Question 2.
Can you suggest some changes in your school which could make it environment friendly?
Answer:
We should use the formula of three R’s to save the environment i.e. Reduce, Recycle and Reuse.

• Reduce: Use less, save water and electricity by stopping the wastage. Do not waste food.
• Recycle: Segregate the wastes so that material which can be recycled should be dumped in place for recycling.
• Reuse: Reuse certain articles e.g. reverse the used envelope and reuse it.

Question 3.
We have read in this chapter that there are four main stakeholders when it comes to forests and wild-life. Which among these should have the authority to decide the management of forest produce? Why do you think so?
Answer:
By active and willing participation of local people, the sal forests of Arabari under went a remarkable recovery as 25 per cent of final harvest was provided to people. They were allowed fuel wood and fodder collection at very nominal rates. This project helped in saving the forests. Hence to have sustainable development there should be equal participation of local community.

Question 4.
How can you as an individual contribute or make a difference to the management of
(a) forest and wild-life
Answer:

• Selective use and reuse of resources.
• Discourage the killing of wild animals.
• Educate the people about the resources provided by forests.

(b) water resources and
Answer:
Water harvesting at one’s home level so as to restore water at subsoil level.

(c) Coal and Petroleum?
Answer:
Alternative sources of energy such as use of solar energy instead of coal and petroleum.

Question 5.
What can you, as an individual, do to reduce your consumption of the various natural resources?
Answer:
(a) forest and wild-life

• Selective use and reuse of resources.
• Discourage the killing of wild animals.
• Educate the people about the resources provided by forests.

(b) water resources
Water harvesting at one’s home level so as to restore water at subsoil level.

(c) Coal and Petroleum
Alternative sources of energy such as use of solar energy instead of coal and petroleum.

Question 6.
List five things you have done over the last one week to :
(i) conserve our natural resources
Answer:
Conserve our natural resources

• Saved electricity by switching off electricity of unnecessary light and fans.
• Used disposable paper cups and plates during my journey in the train.
• Reused the envelopes by turning inside out.
• Planted five plants in my school.
• Educated the people in my locality regarding conservation of natural resources for sustainable development.

(ii) increase the pressure on our natural resources.
Answer:
Increase the pressure on natural resources

• Wasted food.
• Used scooter to consume petrol.
• Added to air pollution and sound pollution.
• Used plastic bags.
• Used D.D.T. in my home.

Question 7.
On the basis of the issues raised in this chapter, what changes would you incorporate in your life-style in a move towards a sustainable use of our resources?
Answer:
forest and wild-life

• Selective use and reuse of resources.
• Discourage the killing of wild animals.
• Educate the people about the resources provided by forests.

(b) water resources
Water harvesting at one’s home level so as to restore water at subsoil level.

(c) Coal and Petroleum
Alternative sources of energy such as use of solar energy instead of coal and petroleum.

Conserve our natural resources

• Saved electricity by switching off electricity of unnecessary light and fans.
• Used disposable paper cups and plates during my journey in the train.
• Reused the envelopes by turning inside out.
• Planted five plants in my school.
• Educated the people in my locality regarding conservation of natural resources for sustainable development.

Increase the pressure on natural resources

• Wasted food.
• Used scooter to consume petrol.
• Added to air pollution and sound pollution.
• Used plastic bags.
• Used D.D.T. in my home.

Science Guide for Class 10 PSEB Management of Natural Resources InText Questions and Answers

Question 1.
What changes can you make in your habits to become more environment friendly?
Answer:
Changes in habits to become environment friendly

• Check your careless habit of wasting. Save water by repairing leaking taps.
• Switch off unnecessary lights and fans.
• Do not waste food.
• Say ‘No’ to plastic bags, disposable plastic cups and other forms of plastic crockery.
• Reuse the things again and again. Instead of throwing away envelopes, you can reverse it and use it again.

Question 2.
What would be the advantages of exploring resources with short term aims?
Answer:
The advantages of exploiting resources with short term aims are as follows :

• We will be able to fulfil the requirement of mass population.
• We will be able to get industrial growth and hence economic development is linked to environmental conservation.

Question 3.
How would these advantages differ from the advantages of using a long term perspective in managing our resources?
Answer:
The demand for all resources is increasing due to increasing human population. The resources are limited.

• It should ensure selective careful use so as to maintain.
• There should be equal distribution of resources to all i.e. rich and poor.
• There should be safe disposal of wastes.

Question 4.
Why do you think there should be equitable distribution of resources? What forces would be working against an equitable distribution of our resources?
Answer:
Management of resources should ensure equitable distribution of resources to all and not to just a handful of rich and powerful people, benefit from the development of these resources.

Question 5.
Why should we conserve forest and wildlife?
Answer:
Forests and wild life are resources of great value.

• Forests help in protection of land.
• Forests help in retaining sub-soil water.
• Forests check floods.
• Forest and wild life maintain ecosystem.
• Wildlife is helpful in exploiting various scientific researches.
• They should be maintained for our economic and social growth and to meet our and of future generation’s material aspiration.

Question 6.
Suggest some approaches towards conservation of forests.
Answer:
Conservation of forests

• Participation of local people indeed lead to efficient management of forest as exemplified by strategy developed in Arabari Forest range of Midnapur district.
• Replantation of trees.
• Selective use of forests.
• Scientific research, monitoring and education plays vital role in conservation.
• Overgrazing should be checked.
• Protection of wildlife and banning the hunting of wild animals.
• Use of modern technique of forestry.
• Make suitable outlet channels to carry out flood water.

Question 7.
Find out about the traditional system of water harvesting/management of your region.
Answer:
Traditional Methods in water harvesting/management

• Diversion of water flowing in streams into man-made channels to cater to the demand of villagers.
• Collection of water in ponds.
• Construction of dams.

Question 8.
Compare the above system with the probable system in hilly mountainous areas or plains or plateau regions.
Answer:
Water harvesting systems are highly local specific and benefits are also localized. The present-day systems include mega projects such as dams.

Question 9.
Find out the source of water in your region/locality. Is water from this source available to all people living in that area?
Answer:
Sources of water are different in different areas. However, these sources are available to all people.

Punjab State Board PSEB 10th Class Science Important Questions Chapter 16 Management of Natural Resources Important Questions and Answers.

PSEB 10th Class Science Important Questions Chapter 16 Management of Natural Resources

Long Answer Type Questions

Question 1.
Explain the three R’s to serve the environment.
Answer:
The three R’s to save the environment : Reduce, Recycle and Reuse. What do they refer to?

• Reduce: This means that use less. Save electricity by switching off unnecessary lights and fans. Save water by repairing leaky taps. Do not waste food.
• Recycle: This means that collection of plastic, paper, glass and metal items and recycle this material to make required things instead of synthesising or extracting fresh plastic, paper, glass or metal. In order of recycle, first need to segregate waste so that the material that can be recycled is not dumped along with other waste.
• Reuse: This is actually even better than recycling because the process of recycling uses some energy. In the ‘reuse’ strategy, simply use things again and again. Instead of throwing away used envelopes, you can reverse it and use it again. The plastic bottles in which you buy various food-items like jam or pickle can be used for storing things in the kitchen.

Question 2.
Write a short note on wildlife.
Answer:
Various plants and animals found in forest are collectively called wildlife. The diverse groups of animals live in forests. Similarly, there are many varieties of plants and trees too, are there in forests. In ancient times, there were comparatively more forests where animals could live with more ease. The number of lions, panthers, rhinos, elephants etc. was exceptionally high. But from the time when human beings started shedding down the forests, the number of animals started decreasing. The number of wild animals is continuously decreasing very rapidly.

Wildlife now needs more protection because of their usefulness and more dependence on it. Wildlife sanctuaries and National Parks have been established for animals where they are provided better safety. Poaching has been strictly banned. Only by protecting wild animals we can help them in breeding and thus can save them from extinction.

Question 3.
What is the importance of forest resources?
Answer:
A. Productive functions

• Forests provide wood. Indian forests yield valuable timber wood like teak, sal, deodar, shisham etc.
• Forests provide paper. The conifers and bamboos are used as raw material to prepare paper.
• Medicinal plants of forests provide medicine of great importance.
• Forests provide number of products like resins, gums, rubber, food and insecticides.
• Forests provide shelter to variety of animals.
• Forests also provide the cork e.g. oak (Quercus).

B. Protective functions
Role of forests in the protection of lands. Forests play a vital role in the life and economy of all tribes living in the forests.

• Forests prevent erosion of soil by wind and water.
• Large trees provide shade which prevents the soil from becoming dry and friable during the summer.
• They check the velocity of rain drops or wind striking the ground and reduce dislodging of the soil particles.
• The root system of plants firmly binds the soil.

Question 4.
What are the methods for safety of forests?
Answer:

• Banning deforestation.
• Only such trees should be cut those have dried up or have been affected by some diseases. New plants/trees should be grown in place of removed trees.
• Van Mahotsav should be celebrated. In this week, thousands of new plants/trees should be planted.
• Newly planted trees should be taken care of.
• Trees should be counted every year and the target set on Van Mahotsav should be attained.
• New plans of Van Mahotsav should be followed.
• Steps should be taken to protect forest from the danger of fire.
• To protect trees from diseases, chemical and medicines must be used.

Question 5.
Discuss various intermediaries of environmental pollution.
Answer:
Human beings and environment have a strong and unbreakable natural bond. Human beings only either protect or pollute the environment and the consequences of it affect them. Clean and healthy environment is very necessary for human society. Unplanned acts of human beings harms the environment. Smoke from trucks and buses, dirty water of sewerages in rivers and rubbish on roads, pollutes the environment of cities.

Constant increase in population is one of the main reason of environment degradation. Need of houses, clothing and food increases due to increasing population. Natural resources are affected adversely in order to meet these demands like deforestation, uncontrollable use of underground water, industrialisation etc. All these are the major reasons of pollution.

When natural sources do not help properly to clean the environment again, it causes pollution. Industrial mishappenings and illegal setting of factories also help in causing pollution. Over use of harmful chemical is one of its factors. Over exploitation of natural resources also pollutes the environment. Industrial revolution causes water and air pollution. Acid rain is caused due to sulphur oxides which are expelled out from the engines of the vehicles. Ozone layer gets depleted due to excessive use of aerosols.

Waste materials are produced mainly due to various activities which cause environmental degradation. These waste products are very harmful and their effects spread very rapidly.

It is a universal issue now to get rid of it. Recycling and reuse of material only can protect our environment from degradation.
Man is the major factor of environmental pollution and he himself is being affected very adversely because of his own blunders.

Question 6.
Comment on ‘Chipko Movement’.
Answer:
‘Chipko Movement’ was the result of policy to separate people living in mountaineous region from forests. This movement started by incident that took place in ‘Raini’ village in Garhwal located in Himalayan range in 1970. The problem was started between local people and the wood contractors. The contract of shedding down the trees near the village was given to wood contractors. On the specific day, labour of the contractor came to cut down the trees. The local male population was not present at that time but women of that village reached there instantly and hugged the trees and didn’t allow them to cut the trees. Since they stuck themselves to trees and the labour of contractor was forced to stop the work.

Question 7.
Why should coal and petroleum be used carefully?
Answer:
Coal and petroleum are the results of dead and decayed fossils of plants and animals in which hydrogen, nitrogen and sulphur are present along with carbon. On burning these carbon dioxide, water, nitrogen oxides and oxides of sulphur are produced.

On burning them in the inadequate supply of air, carbon monoxide is produced in place of carbon dioxide. All these gaseous products are poisonous in nature. Carbon dioxide is a greenhouse gas. Carbon is present in large amount in coal and petroleum. If complete carbon present in them is not burnt, then availability of oxygen will become abundant on earth and it will change to carbon dioxide. It will lead to global warming of our earth as well. So, coal and petroleum should be used carefully.

Question 8.
How is Ganga river polluted? Comment on its cleanliness plan.
Answer:
Ganga covers the distance of 2500 km from Gangotri situated in Himalaya mountain to Bay of Bengal. It flows from different states with more than one hundred cities due to which it gets polluted.

Following are the types of pollution which are responsible for Ganga’s pollution :

• Industrial waste.
• Unprocessed Excreta.
• Burning the dead bodies near its banks and immersing the remains of dead bodies.
• Taking bath in Ganga because of certain superstitions.

Cleanliness plan of Ganga. Remains of dead bodies are immersed in the water of Ganga because of traditional customs that is why coliform bacteria is present in it and it goes on increasing in downstream water. Cleanliness plan of Ganga was started in 1985. Who the budget in first step was 462 crores and in second step it was 416 crores. Under this plan 873 million litres water was to be treated daily. To have a control on pollution of Ganga, immediate need is to increase the functioning of this plan.

Question 9.
What are the methods of conserving water resources?
Answer:
Conservation of water resources
For conservation and management of fresh water following methods can be employed :

• Reducing agricultural water wastage by increasing efficiency of irrigation system.
• Afforestation and protection of watersheds to improve economy.
• Formation of artificial clouds, so that a change can be brought in the atmosphere.
• Changes in the vegetation of a particular area, so that even in the dry season maximum amount of water can be stored. In addition to this, dams should be constructed to stop the drainage of rain water.
• Control the distribution of water by constructing big dams.
• Desalinization of sea water and saline ground water.
• Proper utilization of underground water.
• Attempt to extract minerals from water.
• Melt the glaciers and snow for using as a fresh water resource.
• Regular dredging and desiltation of water bodies.

Question 10.
Discuss various ways of water harvesting.
Answer:
Water harvesting can be undertaken through a variety of ways such as follows :

• Capturing run off from roof tops.
• Capturing run off form local catchments.
• Capturing floodwaters from local streams.
• Conserving water through watershed management.

Question 11.
Why use of coal and petroleum should be restricted?
Answer:
Necessity of Judicious Use of Coal and Petroleum. The fossil fuels, coal and petroleum get exhausted and their combustion pollutes our environment, so a judicious use of these resources is necessary.

• Since coal and petroleum have been formed from biomass, in addition to carbon, these contain hydrogen, nitrogen and sulphur.
• When these are burnt, the products formed are carbon dioxide, water, oxides of nitrogen and oxides of sulphur. When combustion takes place in insufficient air (oxygen), then carbon monoxide is formed instead of carbon dioxide.
• Of these products, the oxides of sulphur and nitrogen and carbon monoxide are poisonous at high concentrations and carbon dioxide is a greenhouse gas which leads to global warming.
• Another way of looking at coal and petroleum is that they are huge reservoirs of carbon and if all of this carbon is converted to carbon dioxide, then the amount of carbon dioxide in the atmosphere is going to increase leading to intense global warmings. Thus, we need to use these resources judiciously.

Question 12.
Explain any four measure for conservation of wild life.
Answer:
Some of the steps involved in the conservation of wild life are :

• The wild life should be protected in natural habitats (in-situ conservation) as well as in places under human control such as zoological and botanical gardens, genetic re-source centres, culture collection (Ex-situ conservation).
• The threatened species should be given preference over others in the conservation programme. Among these, the endangered species should get priority over the vulner¬able species, and the latter over the rare species.
• The habitats of wild relatives of useful plants and animals should be preserved in protected areas.
• The critical habitats of wild animals should be kept intact.

Short Answer Type Questions

Question 1.
Define Natural resources with example.
Answer:
Natural Resources. Resources which exist naturally in nature and are useful to human beings are called natural resources e.g. air, water, soil, coal, petroleum etc. are natural resources.

Question 2.
Explain the principles of first ‘R’ in 3R’s.
Answer:
First ‘R’ stands for Reduce. It means that we should use the resources in limited exten. For example we can switch off the fans and bulbs when not in use and can save electricity. In the same way by using water in less amount and by getting leaking pipes repaired, we can save water from wastage.

Question 3.
Explain the principle of second ‘R’ in 3R’s.
Answer:
Second ‘R’ stands for ‘recycle. It means that we should reuse plastics, paper, glass and should make these in making useful in our daily life. We should not discard them by throwing in dust bin instead should keep them separately for using again.

Question 4.
What is the importance of third ‘R’ in R’s?
Or
How is ‘reuse’ helpful in the conservation of environment?
Answer:
Third ‘R’ represents ‘ Reuse’. This is considered better than recycling. In recycling some of the energy definitely goes waste. In the process of reusing, same thing can be used again and again. For example the containers and bottles of eatables can be used for storage of other things.

Question 5.
Give in detail the journey of Ganga and discuss how is it getting polluted?
Answer:
Ganga runs its course of 2500 km from Gangotri in Himalayas to Bay of Bengal. It crosses through Uttar Pradesh, Bihar, Bengal and 100 more cities which are responsible for its pollution. Its main reason is that garbage waste and excreta from these cities are made to flow in it. Apart from this , other human activities like bathing, washing clothes, passing the remains of dead people are also the causes of its pollution and toxic and chemical waste from industries have increased its pollution. These industries contribute chemical effluents to the Ganga’s pollution load and toxicity kills fish in large section of the rivers.

Question 6.
How can we control the pollution of river Ganga?
Answer:
Pollution of Ganga can be controlled by the following steps :

• By not allowing disposal of industrial waste material in the river.
• By not disposing toxic substances and the dead remains of animals.
• By not expelling the domestic waste in river water.
• Avoid washing clothes in the river.
• By not pouring ash and remains of dead bodies.

Question 7.
What is recyling? What can we do for it?
Answer:
Old newspapers, books, notebooks, old products made of metals etc. can be easily transformed into other useful products, which is called recycling. Re-cycled product can be used again and again.

We can do the following works for recycling of materials :

• Buying those things which can be recycled.
• By using such things which are produced by recycling process.
• By selling things for recycling before they get completely destroyed.

Question 8.
Write the contribution of Bishnoi community in Rajasthan to protect environment.
Answer:
Bishnoi community of Rajasthan made applaudable efforts to protect forests and wildlife which has now become a part of their traditional culture. In 1731, in Khejreli village near Jodhpur, 363 people sacrificed their lives to protect ‘Khejri’ trees. This is why the Government of India has instituted an Amrita Devi Bishnoi National Award’ for wild life conservation in memory of Amrita Devi.

Question 9.
“The human intervention has been very much a part of the forest landscape’— what should be the nature and limit of intervention to control over it?
Answer:
We should certainly make some arrangements to control over the nature and limitation of this intervention. Forest resources should be used in such a way that environment and development both may get advantage from these. Local people should get more of its advantage where environment is conserved. It is a decentralisation system that can help both monetary development and environmental conservation. Environment cannot be accepted just as a collection of plants and animals. It is a complex and complicated system. We have many natural resources for the use of it. We must use all these resources very carefully for our financial and social development.

Question 10.
What did Chipko Movement teach to the government and people? Explain.
Answer:
Chipko Movement spread very rapid in various communities. Many social organisations and media also played a vital role in it. It compelled the government to rethink about the decisions needed to be taken in use of forest resources. People by their experience learnt that deforestation not only affects the availability of trees but also quality of soil and water level too gets spoiled. Participation of local people in management of forests must be encouraged.

Question 11.
Explain with an example the role of people in forest management.
Answer:
In 1972, the forest department of West Bengal made some changes in their policy because of getting failure in reviving the degraded Sal forests. Those sal were destroyed in south-west regions. They started their movement at Madinipura in the region of Arabari. A far sighted forest officer of the department A.K Banerjee. encouraged villagers to join the movement and with their help they protected badly degraded sal forests of about 1272 hectares area.

Question 12.
What are the three main problems in constructing of big dams?
Answer:

• Social problems: Large number of people are forced to migrate and they get affected badly. They have to leave their houses and business. Farmers are forced to leave their agricultural lands.
• Economic Problems: A huge amount of public money is used without generation of proportionate benefits.
• Environmental Problems: Deforestation is taking place on large scale. Bio-diversity is adversely affecteu and environmental problems are arising enormously.

Question 13.
What is the importance of water shed management?
Answer:
More importance is given to scientific soil and water conservation to increase the biomass production in water shed management. Its main importance is to develop primary sources of land and water such that secondary resources of plants and animals may not cause ecological imbalance.

Question 14.
What are the advantages of underground water?
Answer:

• It does not evaporate to mix in atmosphere.
• Vegetation Plants animals are not present in it.
• It improves the water level.
• It provides dampness to plants.
• It does not get polluted due to the presence of living beings.

Question 15.
How can underground water level be recharged in largely level terrain?
Answer:
In largely level terrain water harvesting structures are made. These are mainly crescentshaped earthen embankments. These may be low, straight concrete check dams built across seasonal flooded gullies. Rain water gets stored in Monsoon rains behind the structures. Most of the water dries up in six months after monsoons. It recharges the ground water beneath.

Question 16.
What are fossil fuels and how are they made? Give two examples of these.
Answer:
The remains of animals and plants got compressed under the crust of earth in the absence of oxygen million of years ago. Thus they could not get oxygen. Under the crust neither oxidation of these was possible nor destruction but because of the internal pressure water and volatile substances squeezed out. These substances are termed fossil fuels. Actually fossil fuels are the molecules of carbonic compounds. These are the results of degradation of biomass millions of years ago. The examples of fossil fuel are coal, petroleum and natural gas.

Question 17.
Water is very essential for survival of life’. Justify this statement.
Answer:
Water is essential due to following reasons :

• It takes part in all chemical reactions taking place in our body.
• It maintains the temperature of body.
• It is important for body to carry nutrients to all parts of the body.
• It helps in excretion of organic waste from the body.
• It is important for transportation of substances.
• Irrigation, industries and electricity are totally dependent on water.

Question 18.
Write some major means of water conservation.
Answer:
We can conserve water by :

• Using water for irrigation.
• Flood control, hydrological survey and construction of dams.
• Recharging of underground water and avoiding its wastage.
• Transformation of high water levels into low water levels.
• Safety of soil to avoid soil erosion.

Question 19.
Mention the names of air pollutants.
Answer:
The main air pollutants are :

• Carbon monoxide
• Carbon dioxide
• Oxides of sulphur and nitrogen
• Compounds of flourides
• Hydrocarbons

Question 20.
Write some methods by which pollution can be controlled.
Answer:

• By getting knowledge of the causes and control on pollution.
• By installing the control devices in factories to check the air pollution.
• By recycling the waste products.
• Disposal of excreta and expelled waste material in correct way.
• By using Gobar gas plants.

Question 21.
What is the role of increasing population in destruction of natural resources?
Answer:
Population is increasing constantly due to which human needs are increasing day-by-day. To meet these requirements natural resources are being continuously exhausted. Natural resources are available in large amount but even then these are limited because population has risen at its verge. Our resources are very limited to meet the needs of increasing population. If the monster of increasing population grows at this rate then it will be very difficult to maintain balance between the natural resources and ecology.

Question 22.
Give five points to control pollution.
Answer:
Pollution can be controlled in following ways :

• Gobar gas plants should be installed.
• Bio-degradable substances should be disposed in pits.
• Waste material should be recycled.
• Excreta and disposal matter should be disposed of in a correct way.
• CNG should be used in automobiles.

Question 23.
What are the harmful effects of deforestation?
Answer:
If rate of deforestation is more than the rate of afforestation, then the number of trees will reduce with the passage of time. Trees release water vapours in large amount during transpiration and clouds are formed with the help of these. Rainfall is always less in the area where trees are less in number. Due to this the fertile lands convert into deserts. Deforestation slowly causes removal of upper fertile surface of soil that gets carried away through rain water to rivers ultimately.

Question 24.
How can coal and petroleum be saved for long time?
Answer:
Use of coal and petroleum depend on the quality of machines. In vehicles used for transportation consume excess fuel. Scientists have been researching for the complete combustion of fuel. Efforts are being made to get such fuels having more efficiency, less pollution and capacity for storage of enhanced period.

Question 25.
What are the problems faced in the construction of big dams?
Answer:

• Large numbers of forest, vegetation and bio diversity are lost.
• Migration of people causes unstability and mental tension.
• Migrated persons face number of social and economic problems.
• Benefits are less than the money spent in constructing big dams.

Question 26.
Write difference between National Parks and Animal sanctuaries.
Answer:

National Parks	Animal Sanctuaries
1. These provide shelter to special wild animals like cheetah, rhino, lion, etc.	1. Animals are protected in natural environment.
2. These are spread over 100 to 50C kilometres of area.	2. These are spread over 500 to 1000 sq. kilometres of area.
3. Strong walls are erected around these.	3. Tall and high fenced temporary walls are erected around these.

Question 27.
What are the harms caused by burning of fossil fuels?
Answer:

• Burning of fossil fuel produces harmful gases like CO₂, CO, SO₂, NO₂, etc. are produced.
• Many respiratory problems are caused due to air pollution.
• Chances of skin problems increase.
• Burning sensation is felt in the eyes.

Question 28.
How does rate of development affect the environment? Explain.
Answer:
The rate of development is always beneficial but at the same time it affects the environment. Deforestation is resulted due to urbanisation. Land suitable for agriculture is used for constructing factories. Smoke expelled through chimneys of factories pollutes the air and leaves a harmful effect on health of all living beings. Due to urbanisation cities expand resulting in deleting of villages due to which agricultural land is extincting. Transportation and telecommunication are increased due to which roads and railway need more land. Deforestation is needed for these.

Question 29.
What are uses of water shed management?
Answer:

• To face the situation of drought.
• To get rid of floods.
• Long life of dams and water reservoirs.
• For conservation of water.
• To make water available for irrigation for the whole year.

Question 30.
What are Kulhs? How were these managed?
Answer:
Kulhs are the local irrigation canals evolved by the people of Himachal Pradesh about 400 years ago. The water of natural waterfalls and flowing in streams w as diverted into man-made channels which took this water to numerous villages down the hillside. The management of water flowing in these kulhs was based on a common agreement among the villages. During planting season, watjr-was first of all supplied to the village farthest away from the source of Kulh. The water flow in Kulhs was managed by two-three persons who were paid by the villagers. Apart from irrigation these Kulh’s also percolated into the soil that helped the springs to feed at various points. Government took over these kulhs later but after some time irrigation department lost control and these became non-functional.

Question 31.
What was the losses of inequality in distribution of water?
Answer:
Canal systems leading from dams transfer water from one place to the other distant places. Indra Gandhi canal has helped a big part of Rajasthan to get green now but due to unequal distribution of water and poor management many people could not get benefits from this system. Water is not equally distributed due to which people living near the water source are able to grow crops like sugarcane and wheat which need water in large amount whereas people living away cannot get sufficient amount of water. They cannot grow crops of their choice.

Question 32.
Find out the traditional methods of water storage in your nearby locations.
Answer:
The water harvesting has been a traditional system in India. In Rajasthan by Khad, tanks and Nadis, in Maharashtra by bandharas and Tals, in Madhya Pradesh and Uttar Pradesh by Bundhis the water is harvested. In Bihar Aahar and Pynes, in Himachal Pradesh Kulhs, in Jammu region Ponds and in Tamil Nadu by Eris, in Kerala by surangams and in Karnataka by Kattas rain water is harvested.

Question 33.
Explain a traditional Rain water Harvesting system.
Answer:
Water harvesting is a traditional system in India. In Rajasthan by Khadin, in, Maharashtra by bandharas and Tals, in Madhya Pradesh and Uttar Pradesh by Bundhis, in Bihar by Aahar and Pynes, in Himachal Pradesh by Kulhs and in Jammu regions of Jammu by ponds the rain water in harvested and is still in use by people living there. Rain water is stored in dams and is used for different persons. This water becomes the reason of increasing water level in wells.

Traditional water harvesting system-khadin system

Very Short Answer Type Questions

Question 1.
What is Pollution?
Answer:
Substances which are found in natural state or when they get mixed with dust or toxic substances is called as pollution.

Question 2.
Write five Natural resources.
Answer:

1. Forest
2. Wildlife
3. Water
4. Coal
5. Petroleum.

Question 3.
What are 3R’s?
Answer:

1. Reduce
2. Recycle
3. Reuse.

Question 4.
Write things which can be recycled?
Answer:
Plastic, glass, paper and things made of metals.

Question 5.
Which energy do we get on earth?
Answer:
Solar energy that we get from the sun.

Question 6.
Write full form of CFC.
Answer:
Chloro-Fluoro Carbons.

Question 7.
Name the living beings responsible for biodiversity.
Answer:
Bacteria, fungi, ferns, flowers, plants, nematodes, insects, birds, reptiles.

Question 8.
When was Ganga cleanliness plan enacted?
Answer:
In 1985.

Question 9.
What is coliform?
Answer:
A coliform is a group of bacteria which is found in human intestines due to water pollution.

Question 10.
Write the origin of river Ganga and its course of run.
Answer:
Ganga runs its course of over 2500 km from Gangotri in Himalyas to Ganga Sagar in the Bay of Bengal.

Question 11.
What do we get from sea water?
Answer:
We get iodine from sea water.

Question 12.
How do get solar energy?
Answer:
We get solar energy from sunlight which comes directly from sun and we intake them through plants.

Question 13.
Name three toxic gases.
Answer:
Nitrogen oxide (NO), Sulphur dioxide (SO₂), Carbon monoxide (CO).

Question 14.
Name three trees which are found in forests.
Answer:
Pine, Eucalyptus, Timber.

Question 15.
How does the growth of grass stop in Alpine forests?
Answer:
Grass grows very tall and it falls down on ground which haults the growth of grass.

Question 16.
When and where Chipko Andolan was started?
Answer:
Chipko Andolan was started in village Reni of Garhwal in 1970.

Question 17.
Name three major resources of minerals.
Answer:
Coper, Iron and Manganese.

Question 18.
Write three renewable sources of energy.
Answer:
Wood, Water and Solar energy.

Question 19.
Write two non-renewable sources of energy.
Answer:
Coal and Petroleum.

Question 20.
Name two traditional sources of energy.
Answer:
Fossil fuel and flowing water.

Question 21.
Name two major constituent gases of atmosphere.
Answer:

1. Nitrogen 78%,
2. Oxygen 21%.

Question 22.
What is universal indicator?
Answer:
Litmus paper is called as universal indicator.

Question 23.
What is measured by litmus paper?
Answer:
pH is measured easily with the help of litmus paper.

Question 24.
Why is reuse better than recycle?
Answer:
In recycling some amount of energy is wasted.

Question 25.
Who is the main stakeholders of forests when we think about conservation of forests?
Answer:
The main stakeholders are people who live in and around the forests and are dependent on forest for different needs.

Question 26.
What is the other name of timber?
Answer:
The other name of timber is carpet-wood.

Question 27.
For what forests are considered as source of industries?
Answer:
For raw materials.

Question 28.
Mention two industrial requirements those lead to deforestation.
Answer:
Construction of roads and dams.

Question 29.
On what rains of our country are dependent?
Answer:
Monsoon. The duration of rainy season is limited for a few months.

Question 30.
What is the ancient system of water harvesting in India?
Answer:
Dams, ponds and rivers.

Question 31.
What are Kulhs?
Answer:
400 years ago a traditional harvesting system was developed in Himachal Pardesh canal water. These were called as Kulhs.

Question 32.
If these natural resources are used at the present rate, for how long will they exist?
Answer:
Petroleum will be available for about next 40 years and coal resource for next 200 years.

Question 33.
On what pressure is always maintained in watershed arrangement?
Answer:
Soil and water management.

Question 34.
Name two categories in which waste substance can be divided. Which among these two is comparatively more harmful?
Answer:

1. Biodegradable substances.
2. Non-biodegradable substances. Non-biodegradable substances are comparatively more harmful.

Question 35.
What is the main reason behind the establishment of National Parks and sanctuaries?
Answer:
To protect wildlife.

Question 36.
Write two harmful effects caused by industries on air setting up heavy industries on local climate.
Answer:

1. Air pollution.
2. Increase in temperature of atmosphere.

Question 37.
Write the main cause of global warming.
Answer:
High amount of CO₂ a green house gas, in the atmosphere due to burning of fossil fuel is resulting in global warming.

Question 38.
Name two renewable natural resources.
Answer:

1. Forests and
2. Water.

Question 39.
Name any one endangered plant species.
Answer:
Podophythum.

Question 40.
Name the person who started Appiko movement in South.
Answer:
Pandu Kumar Hegde started appiko movement in Karnataka in 1983.

Multiple Choice Questions

Question 1.
What is the desired minimum total coliforin count level in the Ganga?
(A) 500 MPN/100 ml
(B) 600 MPN/100 ml
(C) 1000 MPN/100 ml
(D) 1200 MPN/100 ml.
Answer:
(A) 500 MPN/100 ml

Question 2.
Chipko Andolan was started by:
(A) Amrita Devi Bishnoi
(B) H.N. Bahuguna
(C) Sunder Lai Bahuguna
(D) A.K. Bannerji.
Answer:
(C) Sunder Lai Bahuguna

Question 3.
Which of the following is/are example(s) of people partici pation in the management of forests?
(A) Chipko Andolan
(B) Arabara forest range of Midnapur
(C) Appiko movement
(D) Khejri tree movement.
Answer:
(B) Arabara forest range of Midnapur

Question 4.
Which of the following represents the regulative function of forest?
(A) Storage and release of gases
(B) production of wood
(C) production of essential oils
(D) conservation of soil and water.
Answer:
(D) conservation of soil and water.

Question 5.
Extensive planting of trees to increase forest cover is called :
(A) afforestation
(B) agroforestry
(C) deforestation
(D) social forestry.
Answer:
(A) afforestation

Question 6.
Enviromnent is formed of:
(A) only biotic component
(B) Only abiotic
(C) Both (A) and (B)
(D) None of above.
Answer:
(C) Both (A) and (B)

Question 7.
Environment day falls on:
(A) 28th Feb
(B) 23rd March
(C) 5th June
(D) 14th Nov.
Answer:
(C) 5th June

Question 8.
Joint Forest management is an example of:
(A) Political equality
(B) Participatory approach
(C) Food stabilization
(D) Economic equality.
Answer:
(B) Participatory approach

Fill in the Blanks :

Question 1.
Chipko Andolan was started by _________ in 1970.
Answer:
Sundar Lai Bahuguna.

Question 2.
Ganga Action Plan was launched in _________
Answer:
1985.

Question 3.
_________ (coal and petroleum) are important sources of energy.
Answer:
Fossil Fuels.

Question 4.
Three R’s to save the environment are Reduce, _________ and _________
Answer:
Recycle, Reuse.

Question 5.
_________ is the ultimate source of energy.
Answer:
Sun.

Question 6.
Tehu Dam is built on _________
Answer:
River Ganga.

Question 7.
Forests provide _________ and plants.
Answer:
Wood, medicinal.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1

Question 1.
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why ?
(i) The taxi fare after each km when the fare is 15 for the first km and 8 for each additional km.

(ii) The amount of air present in a cylinder when a vacuum pump removes of the air remaining in the cylinder at a time.

(iii) The cost of digging a well after every metre of digging, when it costs 150 for the first metre and rises by 50 for each msubsequent metre.

(iv) The amount of money in the account every year when 10000 is deposited at compound interest at 8% per annum.

Solution:
(i) Let T_n denotes the taxi fare in n^th km.
According to question,
T₁ = 15 km;
T₂ = 15 + 8 = 23;
T₃ = 23 + 8 = 31
Now, T₃ – T₂ = 31 – 23 = 8
T₂ – T₁ = 23 – 15 = 8
Here, T₃ – T₂ = T₂ – T₁ = 8
∴ given situation form an AP.

(ii) Let T_n denotes amount of air present in a cylinder.
According to question,
T₁ = x;
T₂ = x – \(\frac{1}{4}\)x
= \(\frac{4-1}{4}\)x = \(\frac{3}{4}\)x
T₃ = \(\frac{3}{4} x-\frac{1}{4}\left[\frac{3}{4} x\right]=\frac{3}{4} x-\frac{3}{16} x\)

= \(\left(\frac{12-3}{16}\right) x=\frac{9}{16}\)x and so on
Now, T₃ – T₂ = \(\frac{9}{16}\)x – \(\frac{3}{4}\)x
= \(\left(\frac{9-12}{16}\right) x=-\frac{3}{16}\)x

T₂ – T₁ = \(\frac{3}{4}\)x – x
= \(\left(\frac{3-4}{4}\right) x=-\frac{1}{4}\)x
Here, T₃ – T₂ ≠ T₂ – T₁
∴ given situation donot form an AP.

(iii) Let T_n denotes cost of digging a well for the nth metre,
According to question,
T₁ = ₹ 150; T₂ = (150 + 50) = ₹ 200;
T₃ = ₹ (200 + 5o) = 250 and so on
Now, T₃ – T₂ = ₹ (250 – 200) = 50
T₂ – T₁ = ₹ (200 – 150) = 50
Here, T₃ – T₂ = T₂– T₁ = 50
∴ given situation form an A.P.

(iv) Let T_n denotes amount of money in the nth year.
According to question
T₁ = ₹ 10,000
T₂ = ₹ 10,000 + ₹ \(\left[\frac{10,000 \times 8 \times 1}{100}\right]\)
= ₹ 10,000 + ₹ 800 = ₹ 10,800
T₃ = ₹ 10,800 + ₹ \(\left[\frac{10,800 \times 8 \times 1}{100}\right]\)
= ₹ 10,800 + ₹ 864
= ₹ 11,640 and so on.
Now, T₃ – T₂ = ₹ (11,640 – 10,800) = ₹ 840
T₂ – T₁ = ₹ (10,800 – 10,000) = ₹ 800
Here, T₃ – T₂ ≠ T₂ – T₁
∴ given situation do not form an A.P.

Question 2.
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(1) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = -3
(iv) a = -1, d = \(\frac{1}{2}\)
(w) a = -1.25, d = -0.25
Solution:
(i) Given that first term = a = 10
and common difference = d = 10
∴ T₁ = a = 10;
T₂ = a + d = 10 + 10 = 20;
T₃ = a + 2d
= 10 + 2 × 10 = 10 + 20 = 30;
T₄ = a + 3d = 10 + 3 × 10
= 10 + 30 = 40
Hence, first four terms of an A.P. are 10, 20, 30, 40………….

(ii) Given that first term = a = -2
and common iifference = d = 0
∴ T₁ = a = -2;
T₂ = a + d = -2 + 0 = -2
T₃ = a + 2d = -2 + 2 × 0 = -2
T₄ = a + 3d = -2 + 3 × 0 = -2
Hence, first four terms of an A.P. are -2, -2, -2, -2,…………….

(iii) Given that first term = a = 4
and common difference = d = -3
∴ T₁ = a = 4;
T₂= a + d = 4 – 3 = 1
T₃ = a + 2d = 4 + 2(-3) = 4 – 6 = -2
T₄ = a + 3d = 4 + 3(-3) = 4 – 9 = -5
Hence, first four terms of an A.P. are 4, 1, -2, -5, ……….

(iv) Given that first term = a = -1
and common difference = d = \(\frac{1}{2}\)
∴ T₁ = a = -1; T₂ = a + d
= -1 + \(\frac{1}{2}\) = \(-\frac{1}{2}\)
T₃ = a + 2d = -1 + 2(\(\frac{1}{2}\))
= -1 + 1 = 0
T₄ = a + 3d = -1 + 3(\(\frac{1}{2}\))
= \(\frac{-2+3}{2}=\frac{1}{2}\)
Hence, first four terms of an AP are -1, –\(\frac{1}{2}\), 0, \(\frac{1}{2}\), …………..

(v) Given that first term = a = – 1.25
and common difference = d = – 0.25
∴ T₁ = a = – 1.25;
T₂ = a + d = – 1.25 – 0.25 = -1.50
T₃ = a + 2d = – 1.25 + 2(- 0.25)
= – 1.25 – 0.50 = – 1.75
T₄ = a + 3d = – 1.25 + 3(- 0.25)
= – 1.25 – 0.75 = – 2
Hence, first four terms of an A.P. are – 1.25, – 1.50, – 1.75, – 2, ……………..

Question 3.
For the following APs, wilte the first term and the common difference:
(i) 3, 1, -1, -3, …………
(ii) 5, -1, 3, 7, ………….
(iii) \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}\), …………..
(iv) 0.6, 1.7, 2.8, 3.9, ………..
Solution:
(i) Given A.P., is 3, 1, -1, -3, ………
Here T₁ = 3, T₂ = 1,
T₃ = -1, T₄ = -3
First term = T₁ = 3
Now, T₂ – T₁ = 1 – 3 = – 2
T₃ – T₂ = – 1 – 1 = -2
T₄ – T₃ = -3 + 1 = -2
∴ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = – 2
Hence, common difference = – 2 and first term = 3.

(ii) Given A.P. is – 5, – 1, 3, 7, ………….
Here T₁ = – 5, T₂ = – 1,
T₃ = 3, T₄ = 7
First term T₁ = -5
Now, T₂ – T₁ = -1 + 5 = 4
T₃– T₂ = 3 + 1 = 4
T₄ – T₃ = 7 – 3 = 4
∴ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = 4
Hence, common difference = 4 and first term = – 5.

(iii) Given AP. is:
\(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots\)
Here T₁ = \(\frac{1}{3}\), T₂ = \(\frac{5}{3}\),
T₃ = \(\frac{9}{3}\), T₄ = \(\frac{13}{3}\)
First term = T₁ = \(\frac{1}{3}\)
Now, T₂ – T₁ = \(\frac{5}{3}-\frac{1}{3}=\frac{5-1}{3}=\frac{4}{3}\)
T₃ – T₂ = \(\frac{9}{3}-\frac{5}{3}=\frac{9-5}{3}=\frac{4}{3}\)
T₄ – T₃ = \(\frac{13}{3}-\frac{9}{3}=\frac{13-9}{3}=\frac{4}{3}\)
∴ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = \(\frac{4}{3}\)

Hence, common difference = \(\frac{4}{3}\) and first term = \(\frac{1}{3}\).

(iv) Given A.P. is 0.6, 1.7, 2.8, 3.9,…
Here, T₁ = 0.6, T₂ = 1.7, T₃ = 2.8, T₄ = 3.9
First term = T₁ = 0.6
Now, T₂ – T₁ = 1.7 – 0.6 = 1.1
T₃ – T₂ = 2.8 – 1.7 = 1.1
T₄ – T₃ = 3.9 – 2.8 = 1.1
Hence, common difference = 1.1 and first term = 0.6.

Question 4.
WhIch of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16
(ii) 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), ………
(iii) – 1.2, – 3.2, – 5.2, – 7.2, ………….
(iv) – 10, – 6, – 2, 2, ………….
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, ……….
(vi) 0.2, 0.22, 0.222, 0.2222, ………….
(vii) 0, -4, -8, -12, …………..
(viii) \(-\frac{1}{2}\), \(-\frac{1}{2}\), \(-\frac{1}{2}\), \(-\frac{1}{2}\), ………..
(ix) 1, 3, 9, 27 …………….
(x) a, 2a, 3a, 4a, ………………
(xi) a, a², a³, a⁴, ……………….
(xii) √2, √8, √18, √32, …………
(xiii) √3, √6, √9, √12, ……………..
(xiv) 1², 3², 5², 7², ………..
(xv) 1², 5², 7², 73, ………….

Solution:
(i) Given terms are 2, 4, 8, 16 ………………
Here T₁ = 2, T₂ = 4, T₃ = 8, T₄ = 16
T₂ – T₁ = 4 – 2 = 2
T₃ – T₂ = 8 – 4 = 4
∵ T₂ – T₁ ≠ T₃ – T₂
Hence, given terms do not form an A.P.

(ii) Given terms are 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), ………
Here T₁ = 2, T₂ = 4, T₃ = 3, T₄ = 16
T₂ – T₁ = \(\frac{4}{4}\) – 2 = \(\frac{5-4}{2}\) = \(\frac{1}{2}\)
T₃ – T₂ = 3 – \(\frac{5}{2}\) = \(\frac{6-5}{2}=\frac{1}{2}\)
T₄ – T₃ = \(\frac{7}{2}-3=\frac{7-6}{2}=\frac{1}{2}\)
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = \(\frac{1}{2}\)
∴ Common difference = d = \(\frac{1}{2}\)
Now, T₅ = a + 4d = 2 + 4\(\frac{1}{2}\) = 4

T₆ = a + 5d = 2 + 5(\(\frac{1}{2}\)) = \(\frac{4+5}{2}=\frac{9}{2}\)

T₇ = a + 6d = 2 + 6(\(\frac{1}{2}\)) = 2 + 3 = 5.

(iii) Given terms are – 1.2, – 3.2, – 5.2, – 7.2, …………
Here T₁ = – 1.2, T₂ = – 3.2,
T₃ = – 5.2, T₄ = – 7.2
T₂ – T₁ = – 3.2 + 1.2 = – 2
T₃ – T₂ = – 5.2 + 3.2 = – 2
T ₄ – T₃ = – 7.2 + 5.2 = – 2
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = – 2
∴ Common difference = d = – 2
Now, T₅ = a + 4d = – 1.2 + 4(-2) = – 1.2 – 8 = – 9.2
T₆ = a + 5d = – 1.2 + 5(-2) = – 1.2 – 10 = – 11.2
T₇ = a + 6d = – 1.2 + 6(-2) = -1.2 – 12 = – 13.2

(iv) Given terms are – 10, – 6, – 2, 2, ………..
Here T₁ = – 10,T₂ = – 6
T₃ = – 2, T₄=2 .
T₂ – T₁ = – 6 + 10 = 4
T₃ – T₂ = – 2 + 6 =4
T₄ – T₃ = 2 + 2 = 4
∵ T₂ – T₁=T₃ – T₂ = T₄ – T₃ = 4 .
∴ Common difference = d = 4
Now, T₅ = a + 4d = – 10 + 4(4) = – 10 + 16 = 6
T₆ = a + 5d = – 10 + 5(4) = – 10 + 20 = 10
T₇ = a + 6d = – 10 + 6(4) = – 10 + 24 = 14.

(v) Given terms are 3, 3 + √2, 3 + 2√2, 3 + 3√2, …………
Here T₁ = 3, T₂ = 3 + √2,
T₃ = 3 + 2√2, T₄= 3 + 3√2
T₂ – T₁ = 3 + √2 – 3 = √2
T₃ – T2 = 3 + 2√2 – (3 + √2)
= 3 + 2√2 – 3 – √2 = √2
T₄ – T₃ = 3 + 3√2 – (3 + 2√2)
= 3 + 3√2 – 3 – 2√2 = √2
∵ T₂ -T₁ = T₃ – T₂ = T₄ – T₃ = √2
∴ Common difference = d = √2
Now, T₅ = a + 4d = 3 + 4(√2) = 3 + 4√2
T₆ = a + 5d = 3 + 5√2
T₇ = a + 6d = 3 + 6√2

(vi) Given terms are 0.2, 0.22, 0.222, 0.2222, …………..
Here Here T₁ = 0.2, T₂ = 0.22,
T₃ = 0.222, T₄ = 0.2222.
T₂ – T₁ = 0.22 – 0.2 = 0.02
T₃ – T₂ = 0.222 – 0.22 = 0.002
∵ T₂ – T₁ ≠ T₃ – T₂
∴ given terms do not form an A.P.

(vii) Given terms are 0, -4, -8, -12
Here T₁ = 0, T₂ = -4,
T₃ = -8, T₄ = -12
T₂ – T₁ = – 4 – 0 = -4
T₃ – T₂= – 8 + 4 = -4
T₄ – T₃= – 12 + 8 = -4.
T₂ – T₁ = T₃ – T₂ = T₄ – T₃
∴ Common difference = d = -4
Now, T₅= a + 4d = 0 + 4(-4) = -16
T₆ = a + 5d = 0 + 5(-4) = -20
T₇ = a + 6d = 0 + 6(-4) = -24.

(viii) Given terms are \(-\frac{1}{2}\), \(-\frac{1}{2}\), \(-\frac{1}{2}\), \(-\frac{1}{2}\), ……….
Here T₁ = \(-\frac{1}{2}\), T₂ = –\(\frac{1}{2}\)
T₃ = \(-\frac{1}{2}\), T₄ = \(-\frac{1}{2}\)
T₂ – T₁ = \(-\frac{1}{2}\) + \(\frac{1}{2}\) = 0
T₃ – T₂ = \(-\frac{1}{2}\) + \(\frac{1}{2}\) = 0
∵ T₂ – T₁ = T₃ – T₂ = 0
∴ Common difference = d = 0
Now, T₅ = T₆ = T₇ = –\(\frac{1}{2}\)
[∵ a = –\(\frac{1}{2}\), d = 0]

(ix) Given terms are 1, 3, 9, 27
T₁ = 1, T₂ = 3, T₃ = 9, T₄ = 27
T₂ – T₁ = 3 1 = 2
T₃ – T₂ = 9 – 3 = 6.
∵ T₂ – T₁ ≠ T₃ – T₂
∴ Given terms do not form an A.P.

(x) Given terms are a, 2a, 3a, 4a, …
T₁ = a, T₂ = 2a, T₃ = 3a, T4 = 4a
T₂ – T₁ = 2a – a = a
T₃ – T₂ = 3a – 2a = a
T₄ – T₃ = 4a – 3a = a
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = a
∴ Common difference = d = a
Now T₅ = a + 4d = a + 4(a) = a + 4a = 5a
T₆ = a + 5d = a + 5a = 6a
T₇ = a + 6d = a + 6a = 7a

(xi) Given terms are a, a², a³, a⁴, …………
T₁ = a, T₂ = a², T3 = a³, T₄ = a⁴
T₂ – T₁ = a² – a
T₃ – T₂ = a³ – a²
∵ T₂ – T₁ ≠ T₃ – T₂
∴ Given terms do not form an A.P.

(xii) Given terms are √2, √8, √18, √32, …………
T₁ = √2, T₂ = √8, T₃ = √18, T₄ = √32
or T₁ = √2, T₂ = 2√2 T₃ = 3√2, T₄ = 4√2
T₂ – T₁ = 2√2 – √2 = √2
T₃ – T = 3√2 – 2√2 = √2
T₄ – T₃ = 4√2 – 3√2 = √2
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃= √2
∴ Common difference = d = √2
Now, T₅ = a + 4d = √2 + 4√2 = 5√2
T₆ = a + 5d = √2 + 5√2 = 6√2
T₇ = a + 6d = √2 + 6√2 = 7√2

(xiii) Given terms are √3, √6, √9, √12, ……………..
T₁ = √3, T₂= √6, T₃= √9, T₄= √12
or T₁ = √3, T₂ = √6, T₃ = 3, T₄ = 2√3
T₄ – T₁ = √6 – √3
T₃ – T₂ = 3 – √6
∵ T₂ – T₁ ≠ T₃ – T₂
∴Given terms do not form an A.P.

(xiv) Given terms are 1², 3², 5², 7², ………..
T₁ = 1², T₂ = 3², T₃ = 5², T₄ = 7²
or T₁ = 1, T₂ = 9, T₃ = 25, T₄ = 49
T₄ – T₁ = 9 – 1 = 8
T₃ – T₂ = 25 – 9 = 16
∵ T₂ – T₁ ≠ T₃ – T₂
∴ Given terms do not form an A.P.

(xv) Given terms are 12, 52, 72, 73
T₁ = 12, T₂ = 52, T₃ = 72, T₄ = 73
or T₁ = 1, T₂ = 25, T₃ = 49, T₄ = 73
T₂ – T₁ = 25 – 1 = 24
T₃ – T₂ =49 – 24= 24
T₄ – T₃ = 73 – 49 = 24
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = 24
∴ Common difference = d = 24
T₅ = a + 4d = 1 + 4(24) = 1 + 96 = 97
T₆ = a + 5d = 1 + 5(24) = 1 + 120 = 121
T₇ = a + 6d = 1 +6(24) = 1 + 144 = 145

Punjab State Board PSEB 10th Class Hindi Book Solutions Hindi Grammar prapatra poorti प्रपत्र पूर्ति Exercise Questions and Answers, Notes.

PSEB 10th Class Hindi Grammar प्रपत्र पूर्ति

प्रपत्र पाठ्यक्रम-बैंक, डाकघर तथा रेलवे से संबंधित प्रपत्र पूर्ति
‘प्रपत्र’ अंग्रेज़ी शब्द फार्म का अनुवाद है। आम बोलचाल की भाषा में प्रपत्र को फार्म भी कहते हैं। प्रपत्र शब्द पत्र के आगे ‘प्र’ उपसर्ग लगाने से बना है। बालक के जन्म से लेकर मृत्यु तक के प्रपत्र भरकर बालक का जन्म प्रमाण-पत्र तथा मृतक का मृत्यु प्रमाण-पत्र प्राप्त करना होता है। इसके अतिरिक्त स्कूल, कॉलेज में प्रवेश के लिए, परीक्षा देने के लिए, नौकरी के लिए, प्रतियोगी परीक्षाओं के लिए, राशन कार्ड, आधार कार्ड, बिजली-पानीटेलीफोन-मोबाइल, गैस कनेक्शन लेने के लिए, पासपोर्ट बनवाने, बैंक/डाकखाने में पैसे जमा कराने या निकालने के लिए, रेल-हवाई यात्रा के समय आरक्षण कराने के लिए, विवाह के पंजीकरण, वाहन पंजीकरण, व्यवसाय पंजीकरण आदि अनेक कार्यों के लिए भी प्रपत्र भरने होते हैं। कोई भी प्रपत्र भरते समय निम्नलिखित तथ्यों का ध्यान रखना चाहिए-

(क) प्रपत्र के साथ दिए गए निर्देशों को अच्छी प्रकार से पढ़ कर ही प्रपत्र भरना चाहिए।
(ख) प्रपत्र में दिए गए निर्देश के अनुसार ही प्रपत्र को पैन/बॉलपैन/पेंसिल से भरना चाहिए।
(ग) प्रपत्र में जिन स्तंभों (कॉलमों) को अंग्रेज़ी के बड़े (कैपिटल) अक्षरों में भरना हो, उन्हें बड़े अक्षरों में भरना चाहिए।
(घ) प्रपत्र पर अपना छायाचित्र (फोटो) उचित स्थान पर चिपकाएं या पिन से नत्थी कीजिए। यदि फ़ोटो को सत्यापित (अटैस्ट) कराना है अथवा अपने हस्ताक्षर करने हैं, तो वह भी कीजिए।
(ङ) प्रपत्र के साथ निर्देशानुसार स्वहस्ताक्षरित अथवा राजपत्रित अधिकारी द्वारा सत्यापित (अटैस्टिड) दस्तावेज़ लगाएँ।
(च) प्रपत्र का कोई भी स्तम्भ (कॉलम) रिक्त न छोड़ें। यदि वह स्तम्भ आप पर लागू नहीं होता तो वहां लागू नहीं लिख दीजिए।
(छ) प्रपत्र में दिए गए स्थान पर अपने हस्ताक्षर कीजिए तथा स्थान और दिनांक लिखिए।
(ज) प्रपत्र साफ़, स्पष्ट, सुंदर तथा पढ़ा जा सके-ऐसे अक्षरों में भरा जाना चाहिए।
(झ) प्रपत्र भरकर व्यक्तिगत रूप से जमा कराते समय उसकी जमा करने वाले कार्यालय से रसीद ले लें। डाक से रजिस्ट्री अथवा स्पीड-पोस्ट से प्रपत्र भरकर भेज सकते हैं। यहाँ कुछ प्रपत्रों को भरने के उदाहरण दिए जा रहे हैं।

1. उदाहरण-किसी खेल में चयन के लिए दिए गए प्रपत्र को भरना-

2. उदाहरण-नौकरी के लिए दिए गए प्रपत्र को भरना
(क) व्यक्तिगत तथा पारिवारिक विवरण

(ग) व्यावसायिक अनुभव (यदि कोई हो)

दसवीं के पाठ्यक्रम में बैंक, डाकघर तथा रेलवे से संबंधित प्रपत्रों की पूर्ति निर्धारित है। इसलिए यहाँ इन्हीं तीनों विभागों से संबंधित प्रपत्रों पर विस्तार से चर्चा की जा रही है-

(क) बैंक से संबंधित प्रपत्र
आधुनिक युग में बैंकों का हमारे दैनिक जीवन में बहुत महत्त्व है। इससे हमारे लेन-देन तथा व्यवसाय में बहुत सहायता मिलती है। बैंक में खाता खुलवाने के लिए कुछ प्रमाण-पत्रों की आवश्यकता होती है, जो निम्नलिखित हैं-

1. आवासीय पते का प्रमाण-पत्र; जैसे-राशन कार्ड, आधार कार्ड, ड्राइविंग लाइसेंस, पासपोर्ट, वरिष्ठ नागरिक प्रमाण-पत्र आदि।
2. जिस बैंक में खाता खुलवाना हो वहाँ के किसी खाताधारी से गवाही।
3. फ़ोटो।
4. पैनकार्ड की प्रतिलिपि।
5. जन-धन योजना में शून्य राशि से खाता खुल सकता है।
6. विद्यार्थियों को अपने विद्यालय के पहचान-पत्र तथा प्राचार्य के हस्ताक्षरित खाता खोलने के फार्म के आधार पर बैंक में शून्य राशि से खाता खोलने की सुविधा है।
7. बैंक में खाता खोलने वाले को बैंक में अपने नमूने के हस्ताक्षर भी देने पड़ते हैं, जिससे कोई अन्य व्यक्ति उसके खाते से रुपए नहीं निकाल सके।
8. बैंक में खाता खोलने के बाद मिलने वाले ए०टी०एम० कार्ड से खाता धारक किसी भी ए०टी०एम० बूथ से कभी भी पैसे निकलवा सकता है तथा खरीददारी भी कर सकता है।

बैंक में खातों के प्रकार-बैंक में खातों के अनेक प्रकार होते हैं; जैसे-बचत खाता, चालू खाता, आवर्ती खाता, सावधि जमा खाता आदि। विद्यार्थियों को बचत खाता खुलवाना चाहिए, जिसमें उनकी छात्रवृत्ति, अपनी बचत आदि जमा हो सकती है। बचत खाता, आवर्ती खाता तथा सावधि खाता पर बैंक जमा राशि पर नियमानुसार ब्याज भी देते हैं, जो खाताधारक के खाते में जमा होता रहता है। बैंक के लेन-देन में मुख्य रूप से निम्नलिखित प्रपत्र प्रयोग में आते हैं-

1. बैंक में खाता खोलने का प्रपत्र
2. रुपए नकद जमा करने का प्रपत्र
3. राशि चैक द्वारा जमा करने का प्रपत्र
4. रुपए निकलवाने का प्रपत्र/चैक
5. ड्राफ्ट बनवाने का प्रपत्र।

यहां प्रमुख प्रपत्रों की रूपरेखा तथा उनके भरने के उदाहरण दिए जा रहे हैं-
1. बैंक में खाता खोलने का प्रपत्र

बैंक में खाता खोलने के प्रपत्र के प्रत्येक कॉलम को ध्यान से पढ़कर साफ़-साफ़ शब्दों में भरिए। खाता एक व्यक्ति अकेले या दो-तीन व्यक्ति मिलकर भी खोल सकते हैं तथा इस संबंध में परिचालन का विकल्प चुन सकते हैं। किसी खाता धारक से परिचय भी दिया जाता है तथा नामांकन का पूरा विवरण भी भरिए। नामिती नाबालिग हो तो उसके संरक्षक का नाम भी लिखना होता है। खाता खोलने हेतु आवेदन फार्म का एक उदाहरण यहां दिया जा रहा है-
1. खाता खोलने का प्रपत्र

प्रश्न 1.
विद्यार्थियों को बैंक में खाता खुलवाने के लिए किन-किन दस्तावेज़ों की आवश्यकता होती है?
उत्तर:
आधुनिक युग में बैंकों का हमारे दैनिक जीवन में बहुत महत्त्व है। इससे हमारे लेन-देन तथा व्यवसाय में बहुत सहायता मिलती है। बैंक में खाता खुलवाने के लिए कुछ प्रमाण-पत्रों की आवश्यकता होती है, जो अग्रलिखित हैं-

1. आवासीय पते का प्रमाण-पत्र ; जैसे-राशन कार्ड, आधार कार्ड, ड्राइविंग लाइसेंस, पासपोर्ट वरिष्ठ नागरिक प्रमाण-पत्र आदि।
2. जिस बैंक में खाता खुलवाना हो वहाँ के किसी खाताधारी से गवाही।
3. फ़ोटो।
4. पैनकार्ड की प्रतिलिपि।
5. जन-धन योजना में शून्यराशि से खाता खुल सकता है।
6. विद्यार्थियों को अपने विद्यालय के पहचान-पत्र तथा प्राचार्य के हस्ताक्षरित खाता खोलने के फार्म के आधार पर बैंक में शून्य राशि से खाता खोलने की सुविधा है।
7. बैंक में खाता खोलने वाले को बैंक में अपने नमूने के हस्ताक्षर भी देने पड़ते हैं, जिससे कोई अन्य व्यक्ति उसके ख़ाते से रुपए नहीं निकाल सके।

2. बैंक में पैसे जमा कराने का प्रपत्र
बैंक में नकद राशि जमा करने के लिए जो प्रपत्र भरना होता है, उसके दो हिस्से होते हैं। एक बैंक की प्रति तथा दूसरी ग्राहक की प्रति होती है। इन दोनों हिस्सों को भरना होता है, जिसमें दिनांक, खाता संख्या, खाताधारी का नाम, जमा राशि तथा जमा राशि का विवरण कि किस-किस राशि के कितने नोट हैं लिखना होता है। दिए गए स्थान पर मोबाइल नंबर लिखकर जमाकर्ता के हस्ताक्षर होते हैं। यहाँ एक उदाहरण दिया जा रहा है जिसमें विजय सिंह और सुरजीत कौर के खाते में एक हज़ार रुपए नकद जमा करने के लिए प्रपत्र भरा गया है-

3. बैंक में चैक द्वारा राशि जमा कराने का प्रपत्र बैंक में चैक जमा कराने का प्रपत्र भी दो भागों में होता है। एक प्रति बैंक की तथा दूसरी ग्राहक की होती है। इसमें दिनांक, खाते का प्रकार तथा संख्या, खातेदार का नाम, चैक संख्या, चैक जारी करने वाले बैंक तथा शाखा का नाम, दिनांक तथा राशि भरनी होती है। यहाँ एक उदाहरण दिया जा रहा है जिसमें मनमोहन सिंह अपने खाते में दस हज़ार रुपए का चैक जमा कर रहे हैं-

4. बैंक से रुपए निकलवाने का प्रपत्र/चैक बैंक से रुपए निकलवाने के लिए बैंक के भगतान प्रपत्र तथा बैंक द्वारा दिए गए चैक का प्रयोग किया जाता है। चैक द्वारा किसी संस्था अथवा व्यक्ति को भी भुगतान कर सकते हैं। बैंक से भुगतान प्रपत्र पर रुपए निकलवाने के लिए पासबुक साथ लगानी पड़ती है। इसमें खातेदार अपने खाते की संख्या, निकाले जाने वाली राशि, दिनांक तथा अपने हस्ताक्षर कर के राशि निकलवा सकता है। प्रीतम कौर अपने खाते से दस हज़ार रुपए भुगतान प्रपत्र पर निकाल रही है। इसका उदाहरण यहाँ दिया जा रहा है-

चैक द्वारा किसी को भुगतान करने अथवा स्वयं रुपए निकालने के लिए चैक में दिनांक, खाता संख्या, किसे भुगतान करना है, राशि तथा खाताधारी के हस्ताक्षर करना आवश्यक होता है। यहाँ एक चैक राजनाथ सिंह द्वारा भारतीय जीवन बीमा निगम को आठ हजार नौ सौ बीस रुपए का जारी किया गया है उसका प्रारूप यहाँ दिया जा रहा है। चैक को रेखांकित कर देना चाहिए इस चैक को कोई अन्य नहीं भुना सकता।

5. बैंक ड्राफ्ट बनवाने का प्रपत्र
बैंक ड्राफ्ट एक मांग-प्रपत्र होता है, जिसे कोई भी व्यक्ति बैंक को निर्धारित शुल्क देकर बनवा सकता है। यह बैंक की एक शाखा द्वारा अपनी उस शाखा के नाम जारी किया जाता है जहाँ ग्राहक ने अपनी राशि भेजनी है। बैंक ड्राफ्ट पाने वाला व्यक्ति उस शाखा से ड्राफ्ट की राशि नकद अथवा अपने बैंक खाते में जमा करवा कर प्राप्त कर सकता है। बैंक ड्राफ्ट के प्रपत्र में दिनांक, आवेदन का नाम पता, किसके पक्ष में, कहाँ भेजना है, ड्राफ्ट की राशि, शुल्क आवेदक के हस्ताक्षर, मोबाइल नंबर भरने होते हैं। यहाँ राजेन्द्र सिंह द्वारा श्री रघुराज भालेराव पाठक को नागपुर पाँच हज़ार रुपए का ड्राफ्ट बनवाने का प्रपत्र उदाहरण के रूप में भरकर दिया जा रहा है। ड्राफ्ट के लिए राजेन्द्र सिंह ने दिए पाँच हजार तीस रुपए हैं परन्तु उसे ड्राफ्ट पांच हज़ार रुपए का ही मिलेगा। तीस रुपए ड्राफ्ट बनवाने का शुल्क है। इसके भी दो भाग होते हैं। एक ग्राहक के लिए तथा दूसरा बैंक के लिए-

बोर्ड परीक्षाओं में पूछे गए प्रपत्र-पूर्ति सम्बन्धी प्रश्नोत्तर

प्रश्न 1.
मान लो आपका नाम निर्मला देवी है। आपको सेविंग्स बैंक खाता नं0 79684 में से दस हज़ार रुपये निकलवाने हैं। अत: आप नीचे दिए गए प्रपत्र की रूपरेखा को अपनी उत्तर-पुस्तिका पर उतार कर भरें-

उत्तर:

प्रश्न 2.
मान लो आपका नाम तमन्ना है। आपका मुख्य डाकघर, अमृतसर में बचत खाता नं0 764329 है। आपके इस खाते में ₹4,500/- हैं। आपको इस खाते में ₹ 1,500/- जमा करवाने हैं। इसलिए नीचे दिए गए प्रपत्र के प्रारुप को अपनी उत्तर-पुस्तिका पर उतार कर भरें-

उत्तर:

प्रश्न 3.
मान लीजिए आपका नाम विनोद कुमार है। आप पटियाला में रहते हैं। आपको अपने बचत खाता नं० 17321986 में र दो हज़ार जमा करवाने हैं। अत: आप नीचे दिए गए प्रपत्र की रूपरेखा को अपनी उत्तरपुस्तिका पर उतार कर भरें-

उत्तर:

(ख) डाकघर से संबंधित प्रपत्र

डाकघर में भी बैंक की तरह अनेक प्रकार से लेन-देन होता है तथा यहाँ भी रुपए जमा करने, निकालने, मनीआर्डर से रुपए भेजने आदि से संबंधित अनेक कार्य होते हैं, जिनके लिए अनेक प्रपत्रों का प्रयोग करना पड़ता है। डाकखाने में प्रयोग में आने वाले मुख्य प्रपत्र अग्रलिखित हैं-

1. खाता खोलने का प्रपत्र
2. रुपए जमा कराने का प्रपत्र
3. रुपए निकलवाने का प्रपत्र
4. मनीआर्डर से रुपए भेजने का प्रपत्र

1. डाकघर में खाता खोलने का प्रपत्र
डाकघर में खाता खोलने के लिए दिए गए प्रपत्र को ध्यानपूर्वक पढ़कर सभी कॉलम साफ़-साफ़ शब्दों में भरने चाहिए। खाता एक व्यक्ति अथवा संयुक्त रूप से दो-तीन व्यक्ति भी खोल सकते हैं। परिचायक का नाम, पता तथा खातेदारों के हस्ताक्षर करने होते हैं। खाते में नामांकन भी किया जा सकता है। उदाहरण के लिए एक भरा हुआ प्रपंत्र यहाँ दिया जा रहा है-
डाकघर बचत-बैंक
खाता खोलने के लिए आवेदन पत्र

4. मैंहम किसी भी समय सुसंगत नियम में विनिर्दिष्ट सीमा के भीतर अपने सभी एकल या संयुक्त बचत बैंक/सा० स० ज० खातों में अतिशेष बनाए रखने का और डाकघर बचत बैंक से मांग किए जाने पर ऐसे सभी खातों की विशिष्टियां भी देने का वचन देता हूँ/देते हैं।

5. मैं/हम केंद्रीय सरकार द्वारा बनाए गए ऐसे नियमों का पालन करने के लिए सहमत हूँ/हैं जो समय-समय पर खाते को लागू हों।

6. मैं हम सरकारी बचत बैंक अधिनियम, 1873 (1873 का 5) की धारा 4 के अधीन नीचे व्यक्ति (व्यक्तियों) को अपनी मृत्यु हो जाने की दशा में खाते में जमा रकम के लिए एकमात्र प्राप्तकर्ता (प्राप्तकर्ताओं) के रूप में नाम निर्दिष्ट करता हूँ करते हैं।

2. डाकघर में पैसे जमा कराने का प्रपत्र
डाकघर में धनराशि जमा करवाने के प्रपत्र में खाताधारी को डाकघर का पता, दिनांक, खातेधारी का नाम, खाता संख्या, जमा कराने की राशि शब्दों और अंकों में, जमा के बाद कुल राशि लिख कर अपने हस्ताक्षर करने होते हैं। इसके साथ डाकघर बचत बैंक की पास बुक भी लगानी होती है, जिसमें बचत बैंक सहायक जमा की गई राशि लिखकर, डाकघर की मोहर लगाकर तथा अपने हस्ताक्षर कर जमाकर्ता को वापिस कर देता है। यहां जालंधर के डाकघर में सतीश कुमार शर्मा के खाता संख्या 1901 में एक हज़ार रुपए जमा करने के प्रपत्र को भरकर उदाहरण के रूप में दिया जा रहा है।

3. डाकघर बचत बैंक से पैसे निकलवाने का प्रपत्र
डाकघर बचत बैंक से पैसे निकलवाने के लिए प्रपत्र में जमाकर्ता को डाकघर का नाम, दिनांक, खाता संख्या, निकाली गई राशि, शेष राशि, यदि आवश्यकता हो तो संदेशवाहक का नाम, हस्ताक्षर देने होते हैं। अदायगी आदेश में राशि मिलने के हस्ताक्षर खातेदार अथवा संदेशवाहक द्वारा किए जाते हैं। यहां एक उदाहरण के रूप में भरा हुआ प्रपत्र दिया जा रहा है-

प्रश्न 3.
मान लीजिए आपका नाम नरेश कुमार है। आपको लोकेश कुमार को दिनांक 10.04.2016 को ₹10,000/- का स्वहस्ताक्षरित रेखांकित किया हुआ चेक लिखकर देना है। इस अनुसार निम्नलिखित चेक के प्रपत्र को भरें-

उत्तर:

4. मनीऑर्डर से रुपये भेजने का प्रपत्र
मनीऑर्डर से रुपए भेजने का काम डाकघर में वर्षों से हो रहा है। डाकघर से मनीऑर्डर प्रपत्र लेकर भरना होता है। इसके 6 भाग होते हैं। पहला भाग जिसे रुपये भेजने हैं रुपयों की संख्या शब्दों और अंकों में प्राप्तकर्ता का नाम, पता, पिनकोड, दिनांक लिखकर भेजने वाला अपने हस्ताक्षरकर्ता है। चौथा भाग में भेजने वाले का नाम पता तथा पिन कोड लिखना होता है। छठे भाग में भेजने वाला संदेश लिख सकता है। दूसरा, तीसरा और पांचवां भाग डाक विभाग ने भरना होता है। डाविभाग भेजने वाली राशि पर निर्धारित शुल्क भी लेता है। यहाँ मनीऑर्डर प्रपत्र को भर कर प्रस्तुत किया जा रहा है-

(संदेश के लिए स्थान)
जालन्धर शहर प्रिय, सुजान
02-12-2000 तुम्हारे पत्र के अनुसार 800 सौ रुपए भेज रहे हैं । मिलने पर सूचना देना और रुपयों की ज़रूरत हो तो लिखना। सब को सत श्री अकाल कहना।
तुम्हारा मित्र,
सुखदेव सिंह

5. इलेक्ट्रानिक मनीआर्डर (ई०एम०ओ०) आधुनिक युग में इंटरनेट की प्रगति के साथ-साथ अपनी धनराशि किसी दूसरे को भेजने का इलेक्ट्रानिक मनीऑर्डर जल्दी से जल्दी और सरलता से भेजने का डाक विभाग द्वारा चलाया गया मनीऑर्डर का नया रूप है। इससे आम मनीआर्डर भेजने की अपेक्षा कम समय लगता है। जिस दिन ई०एम०ओ० किया जाता है, वह उसी दिन प्राप्तकर्ता को मिल जाता है। इस पर भी डाक विभाग निर्धारित शुल्क लेता है। इसके प्रपत्र की रूपरेखा यहाँ प्रस्तुत की जा रही है-

6. मोबाइल मनीट्रांसफर सर्विस (इंसटेंट मनीऑर्डर) मोबाइल मनीट्रांसफर सर्विस धन भेजने का सबसे तेज़, सुविधाजनक, विश्वसनीय और सरल माध्यम है। इसे इंसटेट मनीऑर्डर अथवा तत्काल मनीआर्डर भी कहते हैं। इस माध्यम से कुछ ही मिनटों में धनराशि प्राप्तकर्ता को मिल जाती है। डाकघर से प्राप्त प्रपत्र को भर कर अधिकारी को देने पर प्रेषक को कम्प्यूटर द्वारा सोलह अंकों का सीलबंद गुप्त नम्बर दिया जाता है, जिसे प्रेषक प्राप्तकर्ता को फोन, एस०एम०एस० अथवा ई-मेल द्वारा बता देता है, जो इंसटेंट मनीऑर्डर के अपने नगर के केंद्र पर जाकर वहाँ के अधिकारी को बताता है तथा अपनी पहचान का फोटो पहचानपत्र दिखाकर प्रेषक द्वारा उसे भेजी गई राशि प्राप्त कर सकता है। इस सेवा में उन्नीस हज़ार रुपए तक की राशि नकद प्राप्त की जा सकती है जबकि बीस हज़ार रुपए अथवा उससे अधिक राशि का भुगतान चैक द्वारा किया जाता है। इस सेवा को प्राप्त करने का प्रपत्र अग्रलिखित है-

प्रश्न 1.
इन्सटेंट मनीऑडर्र (आई०एम०ओ) किसे कहते हैं?
उत्तर:
मोबाइल मनीट्रांसफर सर्विस धन भेजने का सबसे तेज़, सुविधाजनक, विश्वसनीय और सरल माध्यम है। इसे इंसटेट मनीऑर्डर अथवा तत्काल मनीआर्डर भी कहते हैं। इस माध्यम से कुछ ही मिनटों में धनराशि प्राप्तकर्ता को मिल जाती है। डाकघर से प्राप्त प्रपत्र को भरकर अधिकारी को देने पर प्रेषक को कम्प्यूटर द्वारा सोलह अंकों का सीलबंद गुप्त नम्बर दिया जाता है, जिसे प्रेषक प्राप्तकर्ता को फोन, एस०एम०एस० अथवा ई०-मेल द्वारा बता देता है, जो इंसटेंट मनीऑर्डर के अपने नगर के केंद्र पर जाकर वहाँ के अधिकारी को बताता है तथा अपनी पहचान का फोटो पहचानपत्र दिखाकर प्रेषक द्वारा उसे भेजी गई राशि प्राप्त कर सकता है। इस सेवा में उन्नीस हजार रुपए तक की राशि नकद प्राप्त की जा सकती है जबकि बीस हज़ार रुपए अथवा उससे अधिक राशि का भुगतान चैक द्वारा किया जाता है।

7. अंतर्राष्ट्रीय धन अंतरण
डाकघर में विदेशों से धन मंगवाने अथवा भेजने के लिए मनीऑर्डर सेवाओं के अन्तर्गत अंतर्राष्ट्रीय धन अंतरण सेवा, मनीग्राम, इलेक्ट्रानिक क्लियरेंस सेवाएं (ई०सी०एस०) आदि भी समुचित मूल्य पर उपलब्ध हैं।

8. इंडियन पोस्टल ऑर्डर
इंडियन पोस्टल ऑर्डर डाकघर से निश्चित राशि के प्राप्त किये जा सकते हैं। इनके द्वारा एक व्यक्ति किसी दूसरे व्यक्ति अथवा संस्था को धनराशि भेज सकता है। ये भारत के सभी डाकघरों से प्राप्त किए जा सकते हैं। विभिन्न प्रतियोगी परीक्षाओं, नौकरियों के आवेदन-पत्रों आदि का शुल्क इंडियन पोस्टल ऑर्डर द्वारा मांगा जाता है। इन्हें रेखांकित भी कर सकते हैं। इसकी खरीद तिथि से वैधता दो वर्षों की होती है। इसे प्रयोग नहीं करने पर निश्चित अवधि में वापस करने पर खरीदी हुई राशि वापस भी ली जा सकती है।

इसके दो भाग फॉइल और काउंटर फॉइल होते हैं जिसके नाम इंडियन पोस्टल ऑर्डर भेजा जाता है उसे फॉइल कहते हैं और जो हिस्सा भेजने वाले के पास रहता है उसे काउंटर फॉइल कहते हैं। पोस्टल आर्डर में राशि प्राप्तकर्ता का नाम, पता, शहर, प्रेषक का नाम पता, दिनांक तथा जिस डाकघर से प्राप्तकर्ता को धनराशि प्राप्त करनी है उसका नाम लिखा जाता है। यदि इसे रेखाकिंत करना है तो वह भी किया जाता है। यहाँ इंडियन पोस्टल ऑर्डर का एक उदाहरण दिया जा रहा है, जिसमें गुरनाम सिंह, फिरोजपुर, पंजाब लोक सेवा आयोग, लुधियाना को एक रुपए का पोस्टल ऑर्डर भेज रहा है-

9. रजिस्ट्री-पत्र की पावती
डाकघर से किसी को रजिस्ट्री-पत्र भेजते समय उसके साथ रजिस्ट्री-पत्र की पावती भी लगानी होती है। यह पावती रजिस्ट्री-पत्र प्राप्तकर्ता हस्ताक्षर कर प्रेषक के पास भेज देता है। इसमें भेजने वाले को अपना तथा प्राप्तकर्ता का पूरा पता लिखना होता है। इसके आगे-पीछे दो भाग होते हैं। इसका प्रारूप निम्नलिखित है-
रजिस्ट्री-पावती प्रपत्र

(ग) रेलवे आरक्षण प्रपत्र
रेलवे में आरक्षण करवाने अथवा आरक्षण रद्द करवाने के लिए प्रपत्र भरना होता है। दोनों ही स्थितियों में एक प्रकार का ही प्रपत्र प्रयोग में आता है। भरने से पहले उस पर लिखना होता है कि आरक्षण लेने के लिए प्रपत्र भर रहे हैं । अथवा आरक्षण रद्द करवाने के लिए। इस प्रपत्र में तीन भाग होते हैं । प्रथम भाग में आरक्षण लेने अथवा रद्द की जाने वाली रेलगाड़ी का नाम, क्रमांक, यात्रा करने की तिथि, श्रेणी, कहां से कहाँ तक, यात्रियों के नाम, आयु, लिंग, आवेदक का नाम, पता, मोबाइल नम्बर तथा हस्ताक्षर होते हैं। दूसरा भाग में वापसी यात्रा का विवरण तथा तीसरा भाग रेलवे के कर्मचारी भरते हैं। इसका एक उदाहरण यहाँ दिया जा रहा है-

रेलवे आरक्षण कराने के लिए प्रपत्र का नमूना
उत्तर:

आगे की यात्रा/वापसी यात्रा का विवरण

टिप्पणी : 1. अधिकतम अनुपेय यात्रियों की संख्या प्रति मांग पत्र 6 व्यक्ति।
2. एक बार एक व्यक्ति से केवल एक ही मांग पत्र स्वीकार किया जाएगा।
3. कृपया खिड़की छोड़ने से पहले अपने टिकट और शेष राशि की जांच कर लें।
4. ठीक ढंग से न भरे हुए तथा अपठनीय फार्म स्वीकार नहीं किए जाएंगे।
5. विशेष मांग पर उपलब्धता के आधार पर विचार किया जाएगा।

अब रेलवे आरक्षण/आरक्षण रद्द करने की सुविधा इंटरनेट/मोबाइल फोन द्वारा भी उपलब्ध है, जिसकी पूरी जानकारी आई०आर०सी०टी०सी० की बेबसाइट पर उपलब्ध है।

1. मान लीजिए आपका नाम तमन्ना शर्मा है।आपका गुड बैंक, शाखा गंगानगर में एक बचत खाता है। आपका खाता नम्बर-45632789 है। आपका मोबाइल नं0 1009321432 है। आपको दिनांक 22.10.2015 को अपने इस खाते में 12,000 रुपये नकद जमा करवाने हैं। आपके पास जमा करवाने के लिए एक हज़ार के 8 नोट, पांच सौ के नोट तथा पचास के 20 नोट हैं। इस अनुसार निम्नलिखित प्रपत्र भरें:
उत्तर:
गुड बैंक, गंगानगर
बैंक में रुपये जमा करवाने के लिए प्रपत्र ( बैंक प्रति)

गुड बैंक, गंगानगर
बैंक में रुपये जमा करवाने के लिए प्रपत्र (ग्राहक प्रति)

2. मान लीजिए आपका नाम रत्नलीन कौर है। आपका मोबाइल नम्बर 1890876535 है। आपका हिमालय बैंक, शाखा सोलन में एक बचत खाता नम्बर 96237180 है। आपको दिनांक 11.08.2015 को अपने इस खाते में से 7,000 रुपये निकालवाने हैं। इस अनुसार निम्नलिखित प्रपत्र भरें
उत्तर:

3. मान लीजिए आपका नाम प्रकाश सिंह है। आपका भारतीय डाक के सेक्टर 15 चंडीगढ़ में 344565 नम्बर एक बचत खाता है। आपके इस खाते में अब तक 45,000 रुपये जमा हैं। आपको दिनांक 02.01.2016 को अपने इस खाते में से 4,000 रुपये निकलवाने हैं। इस अनुसार निम्नलिखित प्रपत्र भरें:
उत्तर:

4. मान लीजिए आपका नाम तुषार कपूर है। आप मकान नं० 987, पुरानी दिल्ली रोड, गुड़गाँव (हरियाणा)-122001 में रहते हैं। आपका बेटा मकान नं० 456 सिविल लाइन्स बरेली (उत्तर प्रदेश) 243001 में रहता है। आप अपने बेटे को भारतीय डाक के माध्यम से दिनांक 05.11.2015 को मनीआर्डर के द्वारा 5,000/- रुपये भेजना चाहते हैं। इस प्रपत्र में उसे संदेश भी लिखें ‘अपनी सेहत का ध्यान रखना’। इस अनुसार निम्नलिखित प्रपत्र भरें।

5. मान लीजिए आपका नाम शिखर कुमार है। आपको लोकेश कुमार को दिनांक 6.12.15 को 10,000/- रु० का स्वहस्ताक्षरित रेखांकित किया हुआ चेक लिखकर देना है। इस अनुसार निम्नलिखित चैक के प्रपत्र को भरें:

बोर्ड परीक्षा में पूछे गए प्रश्न

निम्नलिखित में से किसी एक प्रपत्र को अपनी उत्तर-पुस्तिका पर उतारकर भरें
1. (क) मान लीजिए आपका नाम सुलोचना कुमारी है। आपका हिमालय बैंक, शाखा-शिमला में एक बचत खाता है, जिसका नम्बर, 18954326781 है। आपको अपने इस खाते में से 2500 रुपये निकलवाने हैं। इस अनुसार निम्नलिखित प्रपत्र भरें-

उत्तर:

(ख) मान लीजिए आपका नाम रूप सिंह है। आपका भारतीय डाक के सेक्टर 15, करनाल में एक बचत खाता है, जिसका नम्बर 128947654431 है। आपको अपने इस खाते में से 4000/- रुपये निकलवाने हैं। इस अनुसार निम्नलिखित प्रपत्र भरें:

उत्तर:

2. (क) मान लीजिए आपका नाम पीयूष सिन्हा है। आपका कल्याण बैंक, शाखा-भुवनेश्वर में एक बचत खाता है, जिसका नम्बर 1150000234567 है। आपको अपने इस खाते में 18000/- रुपये जमा करवाने हैं। इस अनुसार निम्नलिखित प्रपत्र भरें:

उत्तर:

(ख) मान लीजिए आपका नाम दीदार सिंह है। आपका भारतीय डाक के सेक्टर-25, देहरादून में एक बचत खाता है, जिसका नम्बर 18634956744 है। आपको अपने इस खाते में से 10000/- रुपये निकलवाने हैं। इस अनुसार निम्नलिखित प्रपत्र भरें:

उत्तर:

3. (क) मान लीजिए आपका नाम निर्मला कौर है। आपका भारतीय डाक के सेक्टर-42, जम्मू में एक बचत खाता है, जिसका नम्बर 62745367823 है। आपको अपने इस खाते से 8200/- रुपये निकलवाने हैं। इस अनुसार निम्नलिखित प्रपत्र भरें-

उत्तर:

(ख) मान लीजिए आपका नाम दिनेश कुमार है। आपको मुनीश कुमार को 12000/- रु० का स्वहस्ताक्षरित चेक लिखकर देना है। इस अनुसार निम्नलिखित चेक के प्रपत्र को भरें:

उत्तर:

4. (क) निम्नलिखित में से किसी एक प्रपत्र को अपनी उत्तर-पुस्तिका पर उतारकर भरें-
(i) मान लीजिए आपका नाम पूनम कुमारी है। आपका भविष्य बैंक, शाखा-शिमला में एक बचत खाता है, जिसका नम्बर 373748488 है। आपको अपने इस खाते में से ₹ 5500/- निकलवाने हैं। इस अनुसार
निम्नलिखित प्रपत्र की रूपरेखा अपनी उत्तर पुस्तिका पर उतार कर भरें-

उत्तर:

(ख) मान लीजिए आपका नाम सुप्रिया देवी है। आपका भारतीय डाक के सेक्टर-15 नोएडा में एक बचत खाता है, जिसका नम्बर 834747993 है। आपको अपने इस खाते में से दिनांक 23.05.2018 को ₹4500/- निकलवाने हैं। इस अनुसार अग्रलिखित प्रपत्र की रूपरेखा अपनी उत्तर-पुस्तिका पर उतार कर भरें:

उत्तर:

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4

Question 1.
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i)2x² – 3x + 5 = 0
(ii) 3x² – 4√3x + 4 = o
(iii) 2x² – 6x + 3 = 0
Solution:
(i) Given quadratic equation is, 2x² – 3x + 5 = 0
Compare it with ax² + bx + c = 0
a = 2, b = -3, c = 5
D = b² – 4ac
= (-3)² 4 × 2 × 5
= 9 – 40 = -31 < 0
Hence, given quadratic equation has no real roots.

(ii) Given quadratic equation is, 3x² – 4√3x + 4 = 0
Compare it with ax² + bx + c = 0
a = 3, b = -4√3, c = 4
D = b² – 4ac
= (-4√3)² – 4 × 3 × 4
= 48 – 48 = 0
given equation has real and equal roots.
Now, x = \(\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}\)
= \(\frac{-(-4 \sqrt{3}) \pm \sqrt{0}}{2 \times 3}\) = \(\frac{2}{\sqrt{3}}\)
Hence, roots of given quadratic equation are \(\frac{2}{\sqrt{3}}\) and \(\frac{2}{\sqrt{3}}\).

(iii) Given quadratic equation is :
2x² – 6x + 3 = 0
Compare it with ax² + bx + c = 0
∴ a = 2, b = -6, c = 3
D = b² – 4ac
= (-6)² 4 × 2 × 3
= 36 – 24 = 12 > 0
∴ given equation has real and distinct roots.
Now, x = \(\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}\)

= \(\frac{-(-6) \pm \sqrt{12}}{2 \times 2}\)

= \(\frac{6 \pm 2 \sqrt{3}}{4}\)

= \(\frac{3 \pm \sqrt{3}}{2}\)

= \(\frac{3+\sqrt{3}}{2}\) and \(\frac{3-\sqrt{3}}{2}\)
Hence, roots of given quadratic equation are \(\frac{3+\sqrt{3}}{2}\) and \(\frac{3+\sqrt{3}}{2}\).

Question 2.
Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x² + kx + 3 = 0
(ii) kx(x – 2) + 6 = 0
Solution:
(i) Given quadratic equation is : 2x² + kx + 3 = 0
Compare it with ax² + bx + c = 0
∴ a = 2, b = k, c = 3
∵ roots of the given quadratic equation are equal.
∴ D = 0
b² – 4ac = 0
Or(k)² – 4 × 2 × 3 = 0
Or k² – 24 = 0
Or k² = 24
Or k = ±√24
Or k = ±2√6.

(ii) Given quadratic equation is:
kx (x – 2) + 6 = 0
Or k – 2kx + 6 = 0
Compare it with ax² + bx + c = 0
∴ a = k, b = -2k, c = 6
∵ roots of the given quadratic equation are equal
∴ b² – 4ac = 0
Or(-2k)² – 4 × k × 6 = 0
Or 4k² – 24k = 0
Or 4k[k – 6]= 0
Either 4k = 0 Or k- 6 = 0
k = 0 Or k = 6
∴ k = 0, 6.

Question 3.
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find
its length and breadth.
Solution:
Let breadth of rectangular grove = x m
and length of rectangular grove = 2x m
Area of rectangular grove = length × breadth
= [x × 2x] m² = 2 × 2 m²
According to question
2x² = 800
x² = \(\frac{800}{2}\) = 400
x = ± √400
x = ± 20.
∵ length of rectangle cannot be negative.
So, we reject x = -20
∴ x = 20
∴ breadth of rectangular grove = 20 m
and length of rectangular grove = (2 × 20) m = 40 m.

Question 4.
Is the following situation possible?If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Solution:
Let age of one friend = x years
and age of 2nd friend = (20 – x) years
Four years ago,
Age of 1st friend = (x – 4) years
Age of 2nd friend = (20 – x – 4) years = (16 – x) years
∴ Their product = (x – 4) (16 – x)
= 16x – x2 – 64 + 4x
= – x² + 20x – 64
According to Question
– x² + 20x – 64 = 48
Or – x² + 20x – 64 – 48 = 0
Or – x² + 20x – 112 = 0
Or x² – 20x + 112 = 0 …………….(1)
Compare it with ax² + bx + c = 0
∴ a = 1, b = -20, c = 112
D = b² – 4ac
= (-20)² – 4× 1 × 112
= 400 – 448 = -48 < 0
∴ roots are not real
then no real value of x satisfies the quadratic equation (1).
Hence, given situation is not possible.

Question 5.
Is it possible to design a rectangular park of perimeter 80 m and area 400 m2 ? If so, find its length and breadth.
Solution:
Let length of rectangular park = x m
Breadth of rectangular park = y m
∴ Perimeter of rectangular park = 2 (x + y) m
and area of rectangular park = xy m²
According to 1st condition
2 (x + y) = 80
x + y = \(\frac{80}{2}\) = 40
y = 40 – x …………(1)
According to 2nd condition,
xy = 400
x (40 – x) = 400 [using (1)]
Or 40x – x² = 400
Or 40x – x² – 400 = 0
Or x² – 40x + 400 = 0
Compare it with ax² +bx + c = 0
a = 1, b = -40, c = 400
D = b² – 4ac
= (-40)² – 4 × 1 × 400
= 1600 – 1600 = 0
Now, x = \(\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}\)

= \(\frac{-(-40) \pm \sqrt{0}}{2 \times 1}\)

= \(\frac{40}{2}\) = 20
When x = 20 then from (1)
y = 40 – 20 = 20
∴ Length and breadth of rectangular park are equal of measure 20 m.
Hence, given rectangular park exist and it is a square.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3

Question 1.
Find the roots of the following quadratic equations if they exist, by the method of completing the square:
(i) 2x² + 7x + 3
(ii) 2x² + x – 4 = 0
(ili) 4x² + 4√3x + 3 = 0
(iv) 2x² + x + 4 = 0
Solution:
(i) Given quadratic equation is
2x² – 7x + 3 = 0
Or 2x² – 7x = -3
Or x² – \(\frac{7}{2}\)x = –\(\frac{3}{2}\)
Or x² – \(\frac{7}{2}\)x + (\(\frac{-7}{4}\))² = \(\frac{-3}{2}+\left(\frac{-7}{4}\right)^{2}\)

Or \(\left(x-\frac{7}{4}\right)^{2}=\frac{-3}{2}+\frac{49}{16}\)

Or \(\left(x-\frac{7}{4}\right)^{2}=\frac{-24+49}{16}\)

Or \(\left(x-\frac{7}{4}\right)^{2}=\frac{25}{16}\)

Or x – \(\frac{7}{4}\) = \(\pm \sqrt{\frac{25}{16}}=\pm \frac{5}{4}\)

Case I:
When x – \(\frac{7}{4}\) = \(\frac{5}{4}\)
Or x = \(\frac{5}{4}+\frac{7}{4}=\frac{5+7}{4}\)
Or x = \(\frac{12}{4}\) = 3

Case II:
When x – \(\frac{7}{4}\) = \(\frac{-5}{4}\)
Or x = \(\frac{-5}{4}+\frac{7}{4}=\frac{-5+7}{4}\)
Or x = \(\frac{2}{4}=\frac{1}{2}\)
Hence, roots of given quadratic equation is 3, \(\frac{1}{2}\).

(ii) Given Quadratic Equation is
2x² + x – 4 = 0
Or 2x² + x = 4
Or x² + \(\frac{1}{2}\)x = \(\frac{4}{2}\)

Case I:
When x + \(\frac{1}{4}\) = \(\frac{-\sqrt{33}}{4}\)
Or x = \(-\frac{\sqrt{33}}{4}-\frac{1}{4}\)
Or x = \(\frac{-\sqrt{33}-1}{4}\)
Hence, roots of given quadratic equation are \(\frac{-1+\sqrt{33}}{4}\) and \(\frac{-1-\sqrt{33}}{4}\).

(iii) Given quadratic equation is
4x² + 4√3x + 3 = 0
Or 4x² + 4√3x = -3
Or x² + \(\frac{4 \sqrt{3}}{4}\)x = \(\frac{-3}{4}\)
Or x² + √3x = \(\frac{-3}{4}\)
Or x² + √3x + \(\left(\frac{\sqrt{3}}{2}\right)^{2}=\frac{-3}{4}+\left(\frac{\sqrt{3}}{2}\right)^{2}\)
Or \(\left(x+\frac{\sqrt{3}}{2}\right)^{2}=\frac{-3}{4}+\frac{3}{4}\)
or (x + \(\frac{\sqrt{3}}{2}\))² = 0
(x + \(\frac{\sqrt{3}}{2}\)) (x + \(\frac{\sqrt{3}}{2}\)) = 0
Either x + \(\frac{\sqrt{3}}{2}\) = 0
x = –\(\frac{\sqrt{3}}{2}\)
Or x + \(\frac{\sqrt{3}}{2}\) = 0
Or x = –\(\frac{\sqrt{3}}{2}\)
Hence, roots of given quadratic equation are –\(\frac{\sqrt{3}}{2}\) and –\(\frac{\sqrt{3}}{2}\).

(iv) Given quadratic equation is
2x² + x + 4 = 0
2x² + x = -4
x² + \(\frac{1}{2}\)x = \(-\frac{4}{2}\)
Or x² + \(\frac{1}{2}\)x + (\(\frac{1}{4}\))² = -2 + (\(\frac{1}{4}\))²

Or \(\left(x+\frac{1}{4}\right)^{2}=-2+\frac{1}{16}\)
Or \(\left(x+\frac{1}{4}\right)^{2}=\frac{-32+1}{16}\)
Or \(\left(x+\frac{1}{4}\right)^{2}=\frac{-31}{16}<0\)

∴ square of any number cannot be negative. So, (x + \(\frac{1}{4}\))² cannot be negative for any real x.
∴ There is no real x whith satisfied the given quadratic equation.
Hence, given quadratic equation has no real roots.

Question 2.
Find the roots of the quadratic equations given in Q. 1 by applying the quadratic formula. Which of the above two
methods do you prefer, and why?
Solution:
(i) Given quadratic equation is
2x² – 7x + 3 = 0
Compare it with ax² + bx + c = 0
a = 2, b = -7, c = 3
Now, b² – 4ac = (-7)² 4 x 2 x 3
= 49 – 24
= 25 > 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-(-7) \pm \sqrt{25}}{2 \times 2}=\frac{7 \pm 5}{4}\)
= \(\frac{7+5}{4} \text { and } \frac{7-5}{4}\)
= \(\frac{12}{4} \text { and } \frac{2}{4}\)
= 3 and \(\frac{1}{2}\)
Hence, 3 and \(\frac{1}{2}\) are the roots of given quadratic equation.

(ii) Given quadratic equation is
2x² + x – 4 = 0
Compare it with ax² + bx + c = 0
∴ a = 2, b = 1, c = -4
Now,
b² – 4ac = (1)² – 4 × 2 × 4
= 1 + 32 = 33 > 0
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-1 \pm \sqrt{33}}{2 \times 2}=\frac{-1 \pm \sqrt{33}}{4}\)
= \(\frac{-1+\sqrt{33}}{4}\) and \(\frac{-1-\sqrt{33}}{4}\)
Hence, \(\frac{-1+\sqrt{33}}{4}\) and \(\frac{-1-\sqrt{33}}{4}\) are the roots of given quadratic equation.

(iii) Given quadratic equation is
4x² + 4√3x + 3 = 0
Compare it with ax² + bx + c = 0
a = 4, b = 4√3, c = 3
b² – 4ac = (4√3)² – 4 × 4 × (3)
= 48 – 48 = 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

= \(\frac{-4 \sqrt{3} \pm \sqrt{0}}{2 \times 4}\)

= \(-\frac{4 \sqrt{3}}{8}\), \(-\frac{4 \sqrt{3}}{8}\)

= –\(\frac{\sqrt{3}}{2}\), –\(\frac{\sqrt{3}}{2}\)

Hence, –\(\frac{\sqrt{3}}{2}\), –\(\frac{\sqrt{3}}{2}\) are the roots of given quadratic equation.

(iv) Given quadratic equation is 2x² + x + 4 = 0
Compare it with ax² + bx + c = 0
∴ a = 2, b = 1, c = 4
Now, b² – 4ac = (1)² – 4 × 2 × 4
= 1 – 32 = -31 < 0
But
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
Since the square of a real number cannot be negative, therefore x will not have any real value.
Hence, there are no real roots for the given quadratic equation.

From above two questions, we used two methods to find the roots of the quadratic equations. It is very clear from above discussion that quadratic formula method is very convenient as compared to method of completing the square.

Question 3.
Find the roots of the following equations:

(i) x – \(\frac{1}{x}\) = 3, x ≠ 0
(ii) \(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}\), x ≠ -4, 7
Solution:
(i) Given Equation is
x – \(\frac{1}{x}\) = 3
Or \(\frac{x^{2}-1}{x}\) = 3
Or x² – 1 = 3x
Or x² – 3x – 1 = 0
Compare it with ax² + bx + c = 0
∴ a = 1, b = -3, c = -1
Now, b² – 4ac = (-3)² – 4 . 1 . (-1)
= 9 + 4 = 13 > 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-(-3) \pm \sqrt{13}}{2 \times 1}\)
= \(\frac{3 \pm \sqrt{13}}{2 \times 1}\)
= \(\frac{3+\sqrt{13}}{2}\) and \(\frac{3-\sqrt{13}}{2}\)
Hence, \(\frac{3+\sqrt{13}}{2}\) and \(\frac{3-\sqrt{13}}{2}\) are the roots of given quadratic equation.

(ii) Given equation is

-11 × 30 = 11 (x2 – 3x – 28)
Or -30 = x² – 3x – 28
Or x² – 3x – 28 + 30 = 0
Or x² – 3x + 2 = 0
Compare it with ax² + bx + c = 0
a = 1, b = – 3, c = 2
Now, b² – 4ac = (-3)² – 4 × 1 ×2
= 9 – 8 = 1 > 0
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-(-3) \pm \sqrt{1}}{2 \times 1}=\frac{3 \pm 1}{2}\)
= \(\frac{3+1}{2}\) and \(\frac{3+1}{2}\)
= \(\frac{4}{2}\) and\(\frac{2}{2}\) and 1
Hence, 2 and 1 are the roots of given quadratic equation.

Question 4.
The sum of the reciprocals of Rehman’s age (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\). F1nd his present age.
Solution:
Let Rehman’s present age = x years
3 years ago Rehman’s age (x – 3) years
5 years from now Rehman’s age =(x + 5) years
According to question,
\(\frac{1}{x-3}+\frac{1}{x+5}=\frac{1}{3}\)

Or \(\frac{x+5+x-3}{(x-3)(x+5)}=\frac{1}{3}\)

Or \(\frac{2 x+2}{x^{2}+5 x-3 x-15}=\frac{1}{3}\)

Or \(\frac{2 x+2}{x^{2}+2 x-15}=\frac{1}{3}\)
Or 6x + 6 = x² + 2 -15
Or x² + 2x – 15 – 6x – 6 = 0
Or x² – 4x – 21 = 0, which is quadratic in x.
So compare it with ax² + bx + c =0
a = 1, b = -4, c = -21
Now, b² – 4ac = (- 4)² 4 × 1 × (-21)
= 16 + 84 = 100 > 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
x = \(\frac{-(-4) \pm \sqrt{100}}{2 \times 1}\)
= \(\frac{4 \pm 10}{2}\)
= \(\frac{4+10}{2}\) and \(\frac{4-10}{2}\)
\(\frac{14}{2}\) and \(\frac{-6}{2}\)
= 7 and -3
∵ age cannot be negative,
so, we reject x = – 3
∴ x = 7
Hence, Rehman’s present age = 7 years.

Question 5.
In a class test, the sum of Shefall’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less ¡n English, the product of their marks would have been 210. Find her marks In the two subjects.
Solution:
Let Shefali get marks in Mathematics = x
Shefali’s marks in English = 30 – x
According to 1st condition,
Shefali’s marks in Mathematics = x + 2
and Shefali’s marks in English = 30 – x – 3 = 27 – x
∴ Their product = (x + 2) (27 – x)
= 27x – x² + 54 – 2x
= x² + 25x + 54
According to 2nd condition,
-x²+ 25x+ 54 = 210
Or -x² + 25x + 54 – 210 = 0
Or -x² + 25x – 156 = 0
Or x² – 25x+ 156 = o
Compare it with ax² + bx + c = O
a = 1, b = -25, c = 156
Now, b² – 4ac = (-25)² – 4 × 1 × 156
= 625 – 624 = 1 > 0
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-(-25) \pm \sqrt{1}}{2 \times 1}\)
= \(\frac{25 \pm 1}{2}\)
= \(\frac{25+1}{2}\) and \(\frac{25-1}{2}\)
= \(\frac{26}{2}\) and \(\frac{24}{2}\)
= 13 and 12.

Case I:
When x = 13
then Shefaiis marks in Maths = 13
Shefali’s marks in English = 30 – 13 = 17.

Case II:
When x = 12
then Shefalis marks in Maths = 12
Shefali’s marks in English = 30 – 12 =18.
Hence, Shefalis marks in two subjects are 13 and 17 Or 12 and 18.

Question 6.
The diagonal of a rectangular field is 60 metres more than the shorter side. if the longer side is 30 metres more than the shorter side, find the sides of the field.
Solution:
Let shorter side of rectangular field = AD = x m

Longer side of rectangular field = AB = (x + 30) m
and diagonal of rectangular field = DB = (x + 60) m
In rectangle. the angle between the length and breadth is right angle.
∴ ∠DAB = 90°
Now, in right angled triangle DAB, using Pythagoras Theorem,
(DB)² = (AD)² + (AB)²
(x + 60)² = (x)² + (x + 30)²
Or x² + 3600 + 120x = x² + x² + 900 + 60x
Or x² + 3600 + 120x – 2x² – 900 – 60x = 0
Or -x² + 60x + 2700 = 0
Or x² – 60x – 2700 = 0
Compare it with ax² + bx + e = O
∴ a = 1, b = -60, c = -2700
and b² – 4ac = (-60)² – 4. 1 . (-2700)
= 3600 + 10800 = 14400 > 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-(-60) \pm \sqrt{14400}}{2 \times 1}\)

= \(\frac{60 \pm 120}{2}\)

= \(\frac{60+120}{2}\) and \(\frac{60-120}{2}\)
= \(\frac{180}{2}\) and \(\frac{-60}{2}\)
= 90 and – 30
∴ length of any side cannot be negative
So, we reject x = -30
∴ x = 90
Hence, shorter side of rectangular field = 90 m
Longer side of rectangular field = (90 + 30) m = 120 m.

Question 7.
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find
the two numbers.
Solution:
Let larger number = x .
Smaller number = y
According to 1st condition,
x² – y² = 180 ……………(1)
According to 2nd condition,
y² = 8x
From (1) and (2), we get
x² – 8x = 180
Or x² – 8x – 180 = 0
Compare it with ax² + bx + c = 0
∴ a = -1, b = -8, c = -180
and b² – 4ac = (-8)² – 4 × 1 × (-180)
= 64 + 720 = 784 > 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-(-8) \pm \sqrt{784}}{2 \times 1}\)
= \(\frac{8 \pm 28}{2}\)
= \(\frac{8+28}{2}\) and \(\frac{8-28}{2}\)
= \(\frac{36}{2}\) and \(\frac{-20}{2}\)
= 18 and -10
When x = – 10 then from (2),
y² = 8 (- 10) = – 80, which is impossible.
So, we reject x = – 10
When x = 18 then from (2).
y² = 8(18) = 144
Or y = ±√144
Or y = ± 12
Hence, required numbers are 18 and 12 Or 18 and -12.

Question 8.
A train travels 360 km ¡t a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:
Let constant speed of the train = x km/hour
Distance covered by the train = 360 km
Time taken by the train = \(=\frac{\text { distance }}{\text { speed }}\)
(∵ speed = \(\frac{\text { Distance }}{\text { Time }}\))
= \(\frac{360}{x}\)
Increased speed of the train = (x + 5) km/hour
∴ Time taken by the train with increased speed = \(\frac{360}{x+5}\) hour
According to question

Or 1800 = x² + 5x
Or x² + 5x – 1800 = 0
Compare it with, ax² + bx + c = 0
a = 1, b = 5, c = – 1800
and b² – 4ac = (5)² 4 × 1 × (- 1800)
= 25 + 7200 = 7225 > 0

x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-5 \pm \sqrt{7225}}{2 \times 1}\)
= \(\frac{-5 \pm 85}{2}\)
= \(\frac{-5+85}{2}\) and \(\frac{-5-85}{2}\)
= \(\frac{80}{2}\) and \(\frac{-90}{2}\)
= 40 and – 45
∵ speed of any train cannot be negative.
So, we reject x = – 45
x = 40
Hence, speed of train = 40 km/hour.

Question 9.
Two water taps together can fill a tank in 9\(\frac{3}{8}\) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Solution:
Let time taken by larger tap to fill the tank = x hours.
Time taken by smaller tap to fill the tank = (x + 10) hours
In case of one hour:
Larger tap can fill the tank = \(\frac{1}{x}\)
Smaller tap can fill the tank = \(\frac{1}{x+10}\)
∴ Larger and smaller tap fill the tank = \(\frac{1}{x}\) + \(\frac{1}{x+10}\) ………….(1)
But, two taps together can fill the tank = 9\(\frac{3}{8}\)hour = \(\frac{75}{8}\) hour
Now, two taps together can fill the tank in one hour = \(\frac{8}{75}\) ……………..(2)
From (1) and (2), we get
\(\frac{1}{x}+\frac{1}{x+10}=\frac{8}{75} \)

Or \(\frac{x+10+x}{x(x+10)}=\frac{8}{75}\)

Or \(\frac{2 x+10}{x^{2}+10 x}=\frac{8}{75}\)

Or 75(2x + 10) = 8(x² + 10x)
Or 150x + 750 = 8x² + 80x
Or 8x² + 80x – 150x – 750 = 0
Or 8x² – 70x – 750 = 0
Or 4x² – 35x – 375 = 0
Compare it with ax² + bx + c = 0
∴ a = 4, b = -35, c = -375
and b² – 4ac = (35)² -4 × 4 × (-375)
= 1225 + 6000 = 7225 > 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

= \(\frac{-(-35) \pm \sqrt{7225}}{2 \times 4}\)

= \(\frac{35 \pm 85}{8}\)

= \(\frac{35+85}{8}\) and \(\frac{35-85}{8}\)

= \(\frac{120}{8}\) and \(\frac{-50}{8}\)

= 15 and \(\frac{-25}{4}\)

∵ time cannot be negative.
So,we reject x = \(\frac{-25}{4}\)
∴ x = 15
Hence, larger water tap fills the tank = 15 hours
and smaller water tap fills the tank = (15 + 10) hours = 25 hours.

Question 10.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop to intermediate stations), if the average speed of the express train is 11 km/hr more than that of the passenger train, find the average speed of the two trains.
Solution:
Let average speed of passenger train = x km/hour
Average speed of express train = (x+ 11) km/hour
Distance between Mysore and Bangalore = 132 km
Time taken by passenger train = \(\frac{132}{x}\) hour
[∵ Speed = \(=\frac{\text { Distance }}{\text { Time }}\) ]
Time taken by express train‚ = \(\frac{132}{x+11}\) hour
According to question,

Or 1452 = x² + 11x
Or x² + 11x – 1452 = 0
Compare it with ax² + bx + c = 0
∴ a = 1, b = 11, c = -1452
and b² – 4ac = (11)² – 4 × 1 × (- 1452)
= 121 + 5808 = 5929 > 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-11 \pm \sqrt{5929}}{2 \times 1}\)

= \(\frac{-11 \pm 77}{2}\)

= \(\frac{-11+77}{2}\) and \(\frac{-11-77}{2}\)

= \(\frac{66}{2}\) and \(\frac{-88}{2}\) = 33 and -44

∵ speed of any train cannot be negative
∴ x = 33
Hence, speed of passenger train = 33 km/hour
and speed of express train = (33 + 11) km/hour = 44 km/hour.

Question 11.
Sum of the areas of two squares is 468 m². If the difference of their perimeters
is 24 m, find the sides of the two squares.
Solution:
In case of larger square
Let length of each side of square = x m
Area of square = x² m²
Perimeter of square = 4x m

In case of smaller square:

Let lenth of each side of square = y m
Area of square = y² m²
Perimeter of square = 4y m
According to 1st condition,
x² + y² = 468 …………….(1)
According to 2nd condition,
4x – 4y = 24
Or 4(x – y) = 24
Or x – y = 6
x = 6 + y
From (1) and (2), we get
(6 + y)² + y² = 468
Or 36 + y² + 12y + y² = 468
Or 2y² + 12y + 36 – 468 = 0
Or 2y² + 12y – 432 = 0
Or y² + 6y – 216 = 0
Compare it with ay² + by + c = 0
∴ a = 1, b = 6, c = -216
and b² – 4ac = (6)² – 4 × 1 × (- 216) = 36 + 864 = 900 > 0
∴ y = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

= \(\frac{-6 \pm \sqrt{900}}{2 \times 1}\)

= \(\frac{-6 \pm 30}{2}\)

= \(\frac{-6+30}{2}\) and \(\frac{-6-30}{2}\)

= \(\frac{24}{2}\) and \(\frac{-36}{2}\) = 12 and -18

∵ length of square cannot be negative
So, we reject y = – 18
∴ y = 12
From (2), x = 6 + 12 = 18
Hence, sides of two squares are 12 m and 18 m.