PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 6 Triangles Ex 6.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2

Question 1.
In fig. (i) and (ü), DE U BC. Find EC in (i) and AD in (ii).

PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2 1

Solution:
(i) In ∆ABC, DE || BC ……………(given)
∴ \(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
[By using Basic Proportionality Theorem]
\(\frac{1.5}{3}=\frac{1}{\mathrm{EC}}\)
EC = \(\frac{3}{1.5}\)
EC = \(\frac{3 \times 10}{15}\) = 2
∴ EC = 2 cm.

(ii) In ∆ABC,
DE || BC ……………(given)
∴ \(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
[By using Basic Proportionality Theorem]
\(\frac{\mathrm{AD}}{7.2}=\frac{1.8}{5.4}\)
AD = \(\frac{1.8 \times 7.2}{5.4}\)
= \(\frac{1.8}{10} \times \frac{72}{10} \times \frac{10}{54}=\frac{24}{10}\)
AD = 2.4
∴ AD = 2.4 cm.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2

Question 2.
E and F are points on the sides PQ and PR respectively of a APQR. For each of the following cases, state whether EF || QR :
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE =4 cm, QE = 4.5 cm, PF =8 cm and RF = 9cm.
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.
Solution:
In ∆PQR, E and F are two points on side PQ and PR respectively.

PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2 2

(i) PE = 3.9 cm, EQ = 3 cm
PF = 3.6 cm, FR = 2.4 cm
\(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{3.9}{3}=\frac{39}{30}=\frac{13}{10}=1.3\)

\(\frac{P F}{F R}=\frac{3.6}{2.4}=\frac{36}{24}=\frac{3}{2}=1.5\) \(\frac{\mathrm{PE}}{\mathrm{EQ}} \neq \frac{\mathrm{PF}}{\mathrm{FR}}\)

∴ EF is not parallel to QR.

(ii) PE = 4 cm, QE = 4.5 cm,
PF = 8 cm, RF = 9 cm.
\(\frac{\mathrm{PE}}{\mathrm{QE}}=\frac{4}{4.5}=\frac{40}{45}=\frac{8}{9}\) ………….(1)
\(\frac{P F}{R F}=\frac{8}{9}\) ……………..(2)
From (1) and (2),
\(\frac{\mathrm{PE}}{\mathrm{QE}}=\frac{\mathrm{PF}}{\mathrm{RF}}\)
∴ By converse of Basic Proportionality theorem EF || QR.

(iii) PQ = 1.28 cm, PR = 2.56 cm
PE = 0.18 cm, PF = 0.36 cm.
EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm
ER = PR – PF = 2.56 – 0.36 = 2.20 cm
Here \(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{0.18}{1.10}=\frac{18}{110}=\frac{9}{55}\) …………..(1)

and \(\frac{\mathrm{PF}}{\mathrm{FR}}=\frac{0.36}{2.20}=\frac{36}{220}=\frac{9}{55}\) …………….(2)

From (1) and (2), \(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}\)
∴ By converse of Basic Proportionality Theorem EF || QR.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2

Question 3.
In fig., LM || CB; and LN || CD. Prove that \(\frac{\mathbf{A M}}{\mathbf{A B}}=\frac{\mathbf{A N}}{\mathbf{A D}}\).

PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2 3

Solution:
In ∆ABC,
LM || BC (given)
∴ \(\frac{\mathrm{AM}}{\mathrm{MB}}=\frac{\mathrm{AL}}{\mathrm{LC}}\) ………..(1)
(By Basic Proportionality Theorem)
Again, in ∆ACD
LN || CD (given)
∴ \(\frac{A N}{N D}=\frac{A L}{L C}\) …………..(2)
(By Basic Proportionality Theorem)
From (1) and (2),
\(\frac{\mathrm{AM}}{\mathrm{MB}}=\frac{\mathrm{AN}}{\mathrm{ND}}\)

or \(\frac{\mathrm{MB}}{\mathrm{AM}}=\frac{\mathrm{ND}}{\mathrm{AN}}\)

or \(\frac{\mathrm{MB}}{\mathrm{AM}}+1=\frac{\mathrm{ND}}{\mathrm{AN}}+1\)
or \(\frac{\mathrm{MB}+\mathrm{AM}}{\mathrm{AM}}=\frac{\mathrm{ND}+\mathrm{AN}}{\mathrm{AN}}\)

or \(\frac{\mathrm{AB}}{\mathrm{AM}}=\frac{\mathrm{AD}}{\mathrm{AN}}\)

or \(\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}\)
Hence, \(\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}\) is the required result.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2

Question 4.
In Fig. 6.19, DE || AC and DF || AE. Prove that \(\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}\).

PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2 4

Solution:
In ∆ABC, DE || AC(given)

PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2 4

∴ \(\frac{B D}{D A}=\frac{B E}{E C}\) …………….(1)
[By Basic Proportionality Theorem]
In ∆ABE, DF || AE
\(\frac{\mathrm{BD}}{\mathrm{DA}}=\frac{\mathrm{BF}}{\mathrm{FE}}\) …………….(2)
[By Basic Proportionality Theorem]
From (1) and (2),
\(\frac{\mathrm{BE}}{\mathrm{EC}}=\frac{\mathrm{BF}}{\mathrm{FE}}\)
Hence proved.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2

Question 5.
In fig. DE || OQ and DF || OR. Show that EF || QR.

PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2 5

Solution:
Given:
In ∆PQR, DE || OQ DF || OR.
To prove: EF || QR.
Proof: In ∆PQO, ED || QO (given)

PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2 6

∴ \(\frac{P D}{D O}=\frac{P E}{E Q}\)

[By Basic Proportionality Theorem]
Again in ∆POR,
DF || OR (given)
∴ \(\frac{P D}{D O}=\frac{P F}{F R}\) ………….(2)
[By Basic Proportionality Theorem]
From (1) and (2),
\(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}\)
In ∆PQR, by using converse of Basic proportionaIity Theorem.
EF || QR,
Hence proved.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2

Question 6.
In flg., A, B and C points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show thatBC || QR.

PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2 7

Solution:
Given : ∆PQR, A, B and C are points on OP, OQ and OR respectively such that AB || PQ, AC || PR.

PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2 8

To prove: BC || QR.
Proof: In ∆OPQ, AB || PQ (given)
∴ \(\frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OB}}{\mathrm{BQ}}\) …………….(1)
[BY using Basic Proportionality Theorem]
Again in ∆OPR.
AC || PR (given)
∴ \(\frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OC}}{\mathrm{CR}}\) ……………….(2)
[BY using Basic Proportionality Theorem]
From (1) and (2),
\(\frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OC}}{\mathrm{CR}}\)
∴ By converse of Basic Proportionality Theorem.
In ∆OQR, BC || QR. Hence proved.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2

Question 7.
Using Basic Proportionality theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved ¡t in class IX).
Solution:
Given: In ∆ABC, D is mid point of AB, i.e. AD = DB.
A line parallel to BC intersects AC at E as shown in figure. i.e., DE || BC.

PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2 9

To prove: E is mid point of AC.
Proof: D is mid point of AB.
i.e.. AD = DB (given)
Or \(\frac{\mathrm{AD}}{\mathrm{BD}}\) = 1 ……………..(1)
Again in ∆ABC DE || BC (given)
∴ \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
[By Basic Proportionality Theorem]
∴ 1 = \(\frac{\mathrm{AE}}{\mathrm{EC}}\) [From (1)]
∴ AE = EC
∴ E is mid point of AC. Hence proved.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2

Question 8.
Using converse of Basic Proportionality theorem prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done ¡tin Class IX).
Solution:
Given ∆ABC, D and E are mid points of AB and AC respectively such that AD = BD and AE = EC, D and Eare joined

PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2 10

To Prove, DE || BC
Proof. D is mid point of AB (Given)
i.e., AD = BD
Or \(\frac{\mathrm{AD}}{\mathrm{BD}}\) = 1 ………………(1)
E is mid point of AC (Given)
∴ AE = EC
Or \(\frac{\mathrm{AE}}{\mathrm{EC}}\) = 1 ………………(2)
From (1) and (2),
By using converse of basic proportionality Theorem
DE || BC Hence Proved.

Question 9.
ABCD is a trapeiiumin with AB || DC and its diagonals Intersect each other at the point O. Show that \(\frac{A O}{B O}=\frac{C O}{D O}\).
Solution:
Given. ABCD is trapezium AB || DC, diagonals AC and BD intersect each other at O.

PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2 11

To Prove. \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)
Construction. Through O draw FO || DC || AB
Proof. In ∆DAB, FO || AB (construction)
∴ \(\frac{\mathrm{DF}}{\mathrm{FA}}=\frac{\mathrm{DO}}{\mathrm{BO}}\) ……………..(1)
[By using Basic Proportionality Theorem]
Again in ∆DCA,
FO || DC (construction)
\(\frac{\mathrm{DF}}{\mathrm{FA}}=\frac{\mathrm{CO}}{\mathrm{AO}}\)
[By using Basic Proportionality Theorem]
From (1) and (2),
\(\frac{\mathrm{DO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{AO}} \quad \frac{\mathrm{AO}}{\mathrm{BO}} \quad \frac{\mathrm{CO}}{\mathrm{DO}}\)
Hence Proved.

Question 10.
The diagonals of a quadrilateral ABCD Intersect each other at the point O such that \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\).Show that ABCD is a
trapezium.
Solution:
Given: Quadrilateral ABCD, Diagonal AC and BD intersects each other at O
such that = \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)

PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2 12

To Prove. Quadrilateral ABCD is trapezium.
Construction. Through ‘O’ draw line EO || AB which meets AD at E.
Proof. In ∆DAB,
EO || AB [Const.]
∴ \(\frac{\mathrm{DE}}{\mathrm{EA}}=\frac{\mathrm{DO}}{\mathrm{OB}}\) ………………(1)
[By using Basic Proportionality Theoremj
But = \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\) (Given)

or \(\frac{\mathrm{AO}}{\mathrm{CO}}=\frac{\mathrm{BO}}{\mathrm{DO}}\)

or \(\frac{\mathrm{CO}}{\mathrm{AO}}=\frac{\mathrm{DO}}{\mathrm{BO}}\)

⇒ \(\frac{\mathrm{DO}}{\mathrm{OB}}=\frac{\mathrm{CO}}{\mathrm{AO}}\) …………….(2)
From (1) and (2),
\(\frac{\mathrm{DE}}{\mathrm{EA}}=\frac{\mathrm{CO}}{\mathrm{AO}}\)
∴ By using converse of basic
proportionlity Theorem,
EO || DC also EO || AB [Const]
⇒ AB || DC
∴ Quadrilateral ABCD is a trapezium with AB || CD.

PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.1

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 6 Triangles Ex 6.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.1

Question 1.
Fill in the blanks using the correct word given in brackets:
(i) All circles are ………………. (congruent, similar).
Solution:
All circles are similar.

(ii) All squares are ………………. (similar, congruent).
Solution:
MI squares are similar.

(iii) All ………………. triangles are similar. (isosceles, equilateral).
Solution:
All equilateral triangles are similar.

(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are __________ and
Solution:
equal

(b) their corresponding sides are ………………. (equal, proportional).
Solution:
proportional.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.1

Question 2.
Give two different examples of pair
(i) similar figures
(ii) non-similar figures.
Solution:
(i) 1. Pair of equilateral triangle are similar figures.
2. Pair of squares are similar figures.

(ii) 1. A triangle and quadrilateral form a pair of non-similar figures.
2. A square and rhombus form pair of non – similar figures.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.1

Question 3.
State whether the following quadrilaterals are similar or not :-

PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.1 1

Solution:
The two quadrilaterals in the figure are not similar because their corresponding angles are not equal.

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

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, Tn = a + (n – 1) d
Tn = 121 + (n – 1) (- 4)
= 121 – 4n + 4
= 125 – 4n.
According to question :—
Tn < 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.

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

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
∵ [Tn = 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
∵ [Tn = 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 = ± \(\frac{1}{2}\)

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

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, Sn = \(\frac{n}{2}\) [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,
Sn = \(\frac{n}{2}\) [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.

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

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?

PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 1

[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.

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

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: Sx – 1 = S49 – S1]
Solution:
Let ‘x’ denotes the number of any house.
Here a = T1 = 1 ;d = 1
According to question,
Sx – 1 = S49 – Sx
= \(\frac{x-1}{2}\) [2 (1) + (x – 1 – 1) (1)]
= \(\frac{49}{2}\) [1 + 49] – \(\frac{x}{2}\) [2 (1) + (x – 1) (1)]
[Using Sn = \(\frac{n}{2}\) [2a + (n – 1) d] and Sn = \(\frac{n}{2}\) (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.

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

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].

PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 2

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.

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

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
Sn = \(\frac{n}{2}\) [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
Sn = \(\frac{n}{2}\) [2a + (n – 1) d]
∴ S12 = [2(-37) + (12 – 1)4]
= 6 [- 74 + 44] = – 180.

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

(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
Sn = \(\frac{n}{2}\) [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
Sn = \(\frac{n}{2}\) [2a +(n – 1)d]
∴ S11 = \(\frac{11}{2}\left[2\left(\frac{1}{15}\right)+(11-1) \frac{1}{60}\right]\)

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

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

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

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 = Tn = 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
∵ Sn = \(\frac{n}{2}\) [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 = Tn = 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
[∵ Sn = \(\frac{n}{2}\) [a + l]]
Now, S13 = \(\frac{13}{2}\) [34 + 10]
= \(\frac{23}{2}\) × 44
= 13 × 22 = 286.

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

(iii) Give A.P. is – 5 + (- 8) + (- 11) + ……………. + (- 230)
Here a = – 5, d = – 8 + 5 = – 3
and l = Tn = – 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, an = 50 find n and Sn.
(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, Sn = 90, find n and an.
(vii) given a = 8, an = 62, Sn = 210, find n and d.
(viii)given an = 4, d = 2, Sn = -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, an = 50
an = 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, Sn = latex]\frac{n}{2}[/latex] [a + l]
= latex]\frac{16}{2}[/latex] [5 + 50] = 8 × 55 = 440.

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

(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 = \(\frac{7}{3}\).
∵ [Sn = 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
∵ [Sn = latex]\frac{n}{2}[/latex] [a + l]
Now, S12 = \(\frac{13}{2}\) [4 + 37]
= 6 × 41 = 246.

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

(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
∵ [Sn = 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)
∵ Tn = a + (n – 1) d = 17 – 9 = 8.

(v) Given d = 5, S9 = 75
∵ S9 = 75
∵ [Sn = \(\frac{n}{2}\) [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 = \(\frac{85}{3}\).

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

(vi) Given a = 2, d = 8, Sn = 90
∵ Sn = 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, an = a5 = a + (n – 1) d
= 2 + (5 – 1) 8 = 2 + 32 = 34.

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

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

(viii) Given an = 4, d = 2, Sn = – 14
∵ an = 4
a + (n – 1) d = 4
or a + (n – 1)2 = 4
or a + 2n – 2 = 4
or a = 6 – 2n ……………(1)
and Sn = – 14
\(\frac{n}{2}\) [a + an] = – 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]
∵ Sn = \(\frac{n}{2}\) [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.

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

(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
∵ [an = Tn = a + (n – 1) d ]
or a + 8d = 28 …………….(1)
and S9 = 144
∵ [Sn = \(\frac{n}{2}\) [a + an]]
\(\frac{9}{2}\) [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 Sn = 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.

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

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 = an = 45
and Sn = 400
∴ Tn = 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 Sn = 400
\(\frac{n}{2}\) [a + an] = 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 = an = 350 and d = 9
∵ l = an = 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.

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

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.

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

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, Sn = \(\frac{n}{2}\) [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, ……., an, … form an AP where is defined as below.
(i) an = 3 + 4n
(ii) an = 9 – 5n
Also find the sum of the first 15 terms In each case.
Solution:
(i) Given that an = 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.

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

(ii) Given that an = 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
Sn = 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 Tn = a + (n – 1) d
= 3 + (n – 1) (- 2)
= 3 – 2n + 2
Tn = 5 – 2n.

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

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, Sn = \(\frac{n}{2}\) [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. Sn = [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.

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

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 = Tn = 49
T2 – T1 = 3 – 1 = 2
T3 – T2 = 5 – 3 = 2
∵ T2 – T1 = T3 – T2 = 2 = d (say)
Also, l = Tn = 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, Sn = \(\frac{n}{2}\) [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.

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

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, Sn = \(\frac{n}{2}\) [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 = Tn = 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.

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

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.]

PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 1

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

PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 2

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?

PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 3

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, Sn = \(\frac{n}{2}\) [2a + (n – 1) d]
= \(\frac{n}{2}\) [2 (20) + (n – 1) (-1)]
∴ Sn = \(\frac{n}{2}\) [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.

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

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.)

PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 4

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.

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

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:

PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 1

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

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

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

(iii) Here d = – 3, n = 18, an = – 5
∵ an = 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 an = 3.6
∵ an = 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
∵ an = a + (n – 1) d
∴ an = 3.5 + (105 – 1) 0
an = 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)
∵ Tn = a + (n – 1) d
Now, T30 = 10 + (30 – 1)(-3)
= 10 – 87 = – 77
∴ Correct choice is (C).

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

(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)
∵ Tn = 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, PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 2, 26
(ii)PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 2, 13, PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 2, 3
(iii) 5,PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 2, PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 2, 9
(iv) – 4, PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 2, PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 2, PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 2, PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 2, 6
(v) PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 2, 38, PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 2, PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 2, PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 2, – 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.

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

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

PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 3

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

PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 4.

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.

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

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, Tn = 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, Tn = 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.

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

(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. Tn = 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 Tn = – 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.

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

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
[∵ Tn = a + (n – 1) d]
a + 10 d = 38
and T16 = 73
a + (16 – 1) d = 73
[∵ Tn = a + (n – 1) d]
a + 15 d = 73
Now, (2) – (1) gives

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

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
∵ Tn = a + (n – 1) d
or a + 2d = 12 ………………(1)
and Last term = T50 = 106
a + (50 – 1) d = 106
∵ Tn = a + (n – 1) d
a + 49 d = 106 ……………(2)
Now, (2) – (1) gives

PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 6

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.

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

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
∵ Tn = a + (n – 1) d
a + 2d = 4 …………..(1)
and T9 = – 8
a + (9 – 1) d = 8
and T9 = – 8
a + (9 – 1)d = 8
∵ Tn = a + (n – 1) d
or a + 8d = – 8
Now, (2) – (1) gives

PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 7

Substitute this value of d in (1), we get
a + 2(- 2) = 4
or a – 4 = 4
or a = 4 + 4 = 8
Now, Tn = 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.

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

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)].

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

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 Tn = 994
T2 – T1 = 112 – 105=7
T3 – T2 = 119 – 112=7
∴ d = T2 – T1 = T3 – T2 = 7
Given that, Tn = 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 Tn = 248
T2 – T1 = 16 – 12 = 4
T3 – T2 = 20 – 16 = 4
∴ d = T2 – T1 = T3 – T2 = 4
Given that, Tn = 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.

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

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, ………….

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

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 Tn = 253
T2 – T1 = 8 – 3 = 5
T3 – T2 = 13 – 8 = 5
∴ d = T2 – T1 = T3 – T1 = 5
Now, Tn = 253
3 + (n – 1)5 = 253
∵ Tn = 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
∵ Tn = 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
∵ Tn = a + (n – 1) d
2a + 5d + 9d = 44
2a + 14d = 44
a + 7d = 22
Now (2) – (1) gives

PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 8

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, ……………

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

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 Tn = ₹ 7000
5000 + (n – 1) 200 = 7000
∵ Tn = 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 Tn denotes the required year.
∴ Tn = 1995 + (11 – 1) 1
= 1995 + 10 = 2005
Hence, in 2005, Subba Rao’s salary becomes 7000.

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

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. Tn = 20. 75 (Given)
5 + (n – 1) 1.75 = 20.75
∵ Tn = 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.

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

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.

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

Solution:
(i) Let Tn denotes the taxi fare in nth 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 Tn 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.

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

(iii) Let Tn 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 Tn 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.

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

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,…………….

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

(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, \(\frac{1}{2}\), …………..

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

(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.

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

(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 = \(\frac{1}{3}\).

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

(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, ………….

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.1
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.

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

(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

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

(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.

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

(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.

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

(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.

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

(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

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

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.

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

(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 \(\frac{3+\sqrt{3}}{2}\).

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

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.

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

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.

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

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.

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

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, \(\frac{1}{2}\).

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

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

PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 1

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 \(\frac{-1-\sqrt{33}}{4}\).

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

(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.

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

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.

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

(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.

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

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.

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

(ii) Given equation is

PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 2

-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.

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

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.

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

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.

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

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

PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 3

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.

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

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.

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

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

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

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.

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

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.

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

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,

PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 5

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.

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

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.

PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2

Question 1.
Find the roots of the following quadratic equations by factorisation:
(i) x2 – 3x – 10 = 0
(ii) 2x2 + x – 6 = 0
(iii) √2x2 + 7x + 5√2 = 0
(iv) 2 x2 – x + \(\frac{1}{8}\) = 0
(v) 100x2 – 20x + 1 = 0
Solution:
(i) Given quadratic
x2 – 3x – 10 = 0
Or x2 – 5x + 2x – 10 = 0
S = -3, p = -10
Or x (x – 5) + 2 (x – 5) = 0
Or (x – 5) (x + 2) = 0
Either x – 5 = 0 Or x + 2 = 0
x = 5 Or x = -2
Hence, 5 and -2 are roots of given Quadratic Equation.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

(ii) Given quadratic equation
2x2 + x – 6 = 0 =1
0r 2x2 + 4x – 3x – 6 = 0
S = 1 P = -6 × 2 = -12
Or 2x (x + 2) -3 (x + 2) = 0
Or (x + 2) (2x – 3) = 0
Either x + 2 = 0 Or 2x – 3 = 0
x = -2 Or x = –\(\frac{3}{2}\)
Hence, – 2 and \(\frac{3}{2}\) are roots of given quadratic equation.

(iii) Given Quadratic Equation,
√2x2 + 7x + 5√2 = 0
Or √2x2 + 2x + 5x + 5√2 = 0
S = 7, P = √2 × 5√2 = 10
Or √2x (x + √2) + 5 (x + √2) = 0
Or (x + √2) (√2x + 5) = 0
Either x + √2 = 0 Or √2x + 5 = 0
x = -√2 Or x = –\(\frac{-5}{\sqrt{2}}\)
Hence, -√2 and \(\frac{-5}{\sqrt{2}}\) are roots of given quadratic equation.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

(iv) Given quadratic equation
2x2 – x + \(\frac{1}{8}\) = 0
Or \(\frac{16 x^{2}-8 x+1}{8}\) = 0
Or 16x2 – 8x + 1 = 0
S = -8, P = 16 × 1 = 16
Or 16x2 – 8x + 1 = 0
Or 16x2 – 4x – 4x + 1 = 0
Or 4x(4x – 1) -1(4x – 1) = 0
Or (4x – 1) (4x – 1) = 0
Either 4x – 1 = 0
Or 4x – 1 = 0
x = \(\frac{1}{4}\) Or x = \(\frac{1}{4}\)
Hence, \(\frac{1}{4}\) and \(\frac{1}{4}\) are roots of given quadratic equation.

(v) Given quadratic equation,
100x2 – 20x + 1 = 0
Or 100x2 – 10x – 10x + 1 = 0
S = -20, P = 100 × 1 = 100
Or 10x(10x – 1) – 1 (10x – 1) = 0
Or (10x – 1)(10x – 1) = 0
Either 10x – 1 = 0 Or 10x – 1 = 0
x = \(\frac{1}{10}\) Or x = \(\frac{1}{10}\)
Hence, \(\frac{1}{10}\) and \(\frac{1}{10}\) are roots of given quadratic equation.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

Question 2.
Solve the problems given in Example I. Statements of these problems are given below:
(i) John and Jivanti together have 45 marbles. Both of them lost S marbles each, and the product of the number of marbles they now have is 124. We would lfke to find out how many marbles they had to start with.

(ii) A cottage Industry produces a certain number of toys in a day. The cost of production of each toy (In rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was 750. We would like to find out the number of toys produced on that day.

Solution:
(i) Let the number of marbles John had be x.
Then the number of marbles Jivanti had = 45 – x
The number of marbles Íeft withJohn, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5 = 40 – x
Therefore, their product = (x – 5) (40 – x)
= 40x – x2 – 200 + 5x
= -x2 + 45x – 200
According to question,
-x2 + 45x – 200 = 124
Or -x2 + 45x – 324 = 0
Or x2 – 45x + 324 =0
Or x2 – 36x – 9x + 324 = 0
S = -45, P = 324
Or x(x – 36) – 9(x – 36) = 0
Or (x – 36)(x – 9) = 0
Either x – 36 = 0, Or x – 9 = 0
x = 36 Or x = 9
∴ x = 36, 9
Hence, number of marbles they had to start with were 36 and 9 or 9 and 36.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

(ii) Let the number of toys produced on that day be x.
Therefore, the cost of production (in rupees) of each toy that day = 55 – x
So, the total cost of production (in rupees) that day = x (55 – x)
According to question.
x(55 – x) = 750
Or 55x – x2 = 750
Or -x2 + 55x – 750 = 0
Or x2 – 55x – 750 = 0
Or x2 – 30x – 25x + 750=0
S = -33, P = 750
Or x(x – 30) – 25(x – 30) = 0
Or (x – 30)(x – 25) = 0
Either x – 30 = 0 Or x – 25 = 0
x = 30 Or x = 25
∴ x = 30, 25
Hence, number of toys produced on that day were 30 and 25 or 25 and 30.

Question 3.
Find two numbers whose sum is 27 and product is 182.
Solution:
Let one number = x
2nd number = 27 – x
Their product = x (27 – x) = 27x – x2
According to question,
27x – x2 = 182
Or – x2 + 27x – 182 = 0
Or x2 – 27x + 182 = 0
S = -27, P = 182
Or x2 – 13x – 14x + 182 = 0
Or x(x – 13) – 14(x – 13) = 0
Or (x – 13) (x – 14) = 0
Either x – 13 = 0 Or x – 14 = 0
x = 13 Or x = 14
x = 13, 14
Hence, two numbers are 13 and 14 Or 14 and 13.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

Question 4.
Find two consecutive positive integers, sum of whose squares is 365.
Solution:
Let one positive integer = x
2nd positive integer = x + 1
According to question,
(x)2 + (x + 1)2 = 365
Or x2 + x2 + 1 + 2x = 365
Or 2x2 + 2x + 365 = 0
Or 2x2 + 2x – 364 = 0
Or x2 + x – 182 = 0
Or x2 + 14x – 13x – 182 = 0
S = 1, P = -182
Or x(x + 14) – 13(x + 14) = 0
(x + 14)(x— 13) = O
Either x + 14 = 0
Or x = -14
Or
x – 13 = 0
x = 13
∵ We have positive integers.
So, we reject x = – 14.
∴ x = 13
∴ One positive integer = 13
and 2nd positive integer = 13 + 1 = 14
Hence, required consecutive positive integers are 13 and 14.

Question 5.
The altitude of a right triangle is 7 cm less than its base. 1f the hypotenuse is 13 cm, find the other two sides.
Solution:
Let base of right triangle = x cm
Altitude of right triangle = (x – 7) cm
and hypotenuse of right triangle = 13 cm (Given)
According to Pythagoras Theorem,
(Base)2 + (Altitude)2 = (Hypotenuse)2
(x)2 ÷ (x – 7)2 = (13)2
Or x2 + x2 + 49 – 14x = 169
Or 2x2 – 14x + 49 – 169 = 0
Or 2x2 – 14x – 120 = 0
Or 2[x2 – 7x – 60] = 0
Or x2 – 7x – 60 = 0
Or x2 – 12x + 5x – 60 = 0
S = – 7 P = – 60
Or x(x – 12) + 5(x – 12) = 0
Or (x – 12) (x + 5) = 0
Either x – 12 = 0 Or x + 5 = 0
x = 12 Or x= – 5
∵ Length of any triangle cannot be negative.
So, we reject x = – 5
∴ x = 12
Hence, base of right triangle = 12 cm
Altitude of right triangle = (12 – 7) cm = 5 cm.

Question 6.
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was 90, find the number of articles produced and the cost of each article.
Solution:
Let, number of pottery articles produced by industry in one day = x
Cost of production of each article = ₹ (2x + 3)
∴ Total cost of production in panicular day = ₹ [x(2x + 3)] = ₹ (2x2 + 3x)
According to question,
2x2 + 3x = 90
2x2 + 3x – 90 = 0
S = 3, P = 2 × -90 = -180
Or 2x2 – 12x + 15x – 90 = 0
Or 2x (x – 6) + 15 (x – 6) = 0
Or (x – 6) (2x + 15) = 0
Either x – 6 = 0 Or 2x + 15 = 0
x = 6 Or x = \(\frac{-15}{2}\)
∵ number of articles cannot be negative
So, we reject x = 2
∴ x = 6
Hence, number of articles produced on certain day = 6
and cost of production of each article = ₹ [2 × 6 + 3] = ₹ 15.

PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.1

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.1

Question 1.
Check whether the following are quadratic equations:
(i) (x + 1)2 = 2(x – 3)
(ii) x2 – 2x = (-2) (3 – x)
(iii) (x – 2) (x + 1) = (x – 1) (x + 3)
(iv) (x – 3)(2x + 1) = x (x + 5)
(v) (2x – 1) (x – 3) = (x + 5) (x – 1)
(vi) x2 + 3x + 1 = (x – 2)
(vii) (x + 2)3 = 2x(x2 – 1)
(viii) x3 – 4x2 – x + 1 = (x – 2)3

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.1

Solution:
(i) Given that
(x + 1)2 = 2(x – 3)
Or x2 + 1 + 2x = 2x – 6
Or x2 + 1 + 2x – 2x + 6 = 0
Or x2 + 7 = 0
Or x2 + 0x + 7 = 0
which is in the formof ax2 + bx + c = 0;
∴ It is a quadratic equation.

(ii) Given that
x2 – 2x = (-2) (3 – x)
Or x2 – 2x = -6 + 2x
Or x2 – 2x + 6 – 2x = 0
Or x2 – 4x + 6 = 0
which is the form of ax2 + bx + c = 0; a ≠ 0
∴ It is the quadratic equation.

(iii) Given that ,
(x – 2) (x + 1) = (x – 1) (x + 3)
Or x2 + x – 2x – 2 = x2 + 3x – x – 3
Or x2 – x – 2 = x2 + 2x – 3
Or x2 – x – 2 – x2 -2x + 3 = 0
Or -3x + 1 = 0 which have no term of x2.
So it is not a quadratic equation.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.1

(iv) Given that
(x – 3)(2x + 1) = x(x + 5)
Or 2x2 + x – 6x – 3 = x2 + 5x
Or 2x2 – 5x – 3 – x2 – 5x = 0
Or x2 – 10x – 3 = 0
which is a form of ax2 + bx + c = 0; a ≠ 0
∴ It is a quadratic equation.

(v) Given that ,
(2x – 1) (x – 3) = (x + 5) (x – 1)
0r2x2 – 6x – x + 3 = x2 – x + 5x – 5
Or 2x2 – 7x + 3 = x2 + 4x – 5
Or 2x2 – 7x + 3 – x2 – 4x + 5 = 0
Or x2 – 11x + 8 = 0
which is a form of ax2 + bx + c = 0; a ≠ 0
∴ It is a quadratic equation.

(vi) Given that
x2+3x+1 = (x – 2)2
Or x2 + 3x + 1 = x2 + 4 – 4x
Or x2 + 3x + 1 – x2 – 4 + 4x = 0
Or 7x – 3 = 0
which have no term of x2.
So it is not a quadratic equation.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.1

(vii) Given that
(x + 2)3 = 2x(x2 – 1)
Or x3 + (2)3 + 3 (x)2 2 + 3(x)(2)2 = 2x3 – 2x
Or x3 + 8 + 6x2 + 12x = 2x3 – 2x
Or x3 + 8 + 6x2 + 12x – 2x3 + 2x = 0
Or -x3 + 6x2 + 14x + 8 = 0
Here the highest degree of x is 3. which is a cubic equation.
∴ It is not a quadratic equation.

(viii) Given that
x3 – 4x2 – x+ 1= (x – 2)3
Or x3 – 4x2 – x + 1 = x3 – (2)3 + 3(x)2 (-2) + 3 (x) (-2)2
Or x3 – 4x2 – x + 1 = x3 – 8 – 6x2 + 12x
Or x3 – 4x2 – x + 1 – x3 + 8 + 6x2 – 12x = 0
Or 2x2 – 13x + 9 = 0
which is in the form of ax2 + bx +c = 0; a ≠ 0
∴ It is a quadratic equation.

Question 2.
Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.1

Solution:
(i) Let Breadth of rectangular plot = x m
Length of rectangular plot= (2x + 1) m
∴ Area of rectangular plot = [x (2x + 1)] m2 = (2x2 + x) m2
According to question,
2x2 + x = 528
S = 1
P = -528 × 2 = -1056
0r 2x2 + x – 528 = 0
Or 2x2 – 32x + 33x – 528 = 0
Or 2x(x – 16) + 33(x – 16) = 0
Or (x – 16) (2x + 33) = 0
Either x – 16 = 0 Or 2x + 33 = 0
x = 16 Or x = 2
∵ breadth of any rectangle cannot be negative, so we reject x = \(\frac{-33}{2}\), x = 16
Hence, breadth of rectangular plot = 16 m
Length of rectangular plot = (2 ×16 + 1)m = 33m
and given problem in the form of Quadratic Equation are 2x2 + x – 528 = 0.

(ii) Let two consecutive positive integers are x and x + 1.
Product of Integers = x (x + 1) = x2 + x
According to question,
Or x2 + x – 306 = 0
S = 1, P = – 306
Or x2 + 18x – 17x – 306 = 0
Or x(x + 18) -17 (x + 18) = 0
Or (x + 18) (x – 17) = 0
Either x + 18 = 0 Or x – 17 = 0
x = -18 Or x = 17
∵ We are to study about the positive integers, so we reject x = – 18.
x = 17
Hence, two consecutive positive integers are 17, 17 + 1 = 18
and given problem in the form of Quadratic Equation is x2 + x – 306 = 0.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.1

(iii) Let present age of Rohan = x years
Rohan’s mother’s age = (x + 26) years
After 3 years, Rohan’s age = (x + 3) years
Rohan’s mother’s age = (x + 26 + 3) years = (x + 29) years
∴ Their product = (x + 3) (x + 29)
= x2 + 29x + 3x + 87
= x2 + 32x + 87
According to question,
x2 + 32x + 87 = 360
Or x2 + 32x + 87 – 360 = 0
Or x2 + 32x – 273 = 0
Or x2 + 39x – 7x – 273 = 0
S = 32, P = – 273
Or x(x + 39) – 7(x + 39) = 0
Or (x + 39) (x – 7) =
Either x + 39 = Or x – 7 = 0
x = -39 Or x = 7
∵ age of any person cannot be negative so, we reject x = -39
∴ x = 7
Hence, Rohans present age = 7 years
and given problem in the form of Quadratic Equation is x2 + 32x – 273 = 0.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.1

(iv) Let u km/hour be the speed of train.
Distance covered by train = 480 km
Time taken by train = \(\frac{480}{u}\) hour
[ Using, Speed = \(\frac{\text { Distance }}{\text { Time }}\)
or Time = \(=\frac{\text { Distance }}{\text { Speed }}\) ]

If speed of train be decreased 8km/hr.
∴ New speed of train = (u – 8) km/hr.
and time taken by train = \(\frac{480}{u-8}\) hour
According to question.

PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 1

or 3840 = 3 (u2 – 8u)
or u2 – 8u = 1280
or u2 – 8u – 1280=0
or u2 – 40u + 32u – 1280 = 0
S = -8, P = – 1280
or u(u – 40) + 32 (u – 40) = 0
or (u – 40)(u + 32) = 0
Either u – 40 = 0
or u + 32 = 0
u = 40 or u = -32
But, speed cannot be negative so we reject
u = – 32
∴ u = 40.
Hence speed of train is 40 km/hr Ans.