Bhagya

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 8 Comparing Quantities Ex 8.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

1. Convert the following fractions into percents

(i). \(\frac {1}{8}\)
Solution:
\(\frac {1}{8}\) = \(\frac {1}{8}\) × 100
= \(\frac {25}{2}\)
= 12.5
Thus, \(\frac {1}{8}\) = 12.5%

(ii). \(\frac {49}{50}\)
Solution:
\(\frac {49}{50}\) = \(\frac {49}{50}\) × 100
= 98
Thus, \(\frac {49}{50}\) = 98%

(iii). \(\frac {5}{4}\)
Solution:
\(\frac {5}{4}\) = \(\frac {5}{4}\) × 100
= 125
Thus, \(\frac {5}{4}\) = 125%

(iv). 1\(\frac {3}{8}\)
Solution:
1\(\frac {3}{8}\) = \(\frac {11}{8}\) × 100
= \(\frac {275}{2}\)
= 137\(\frac {1}{2}\)
Thus, 1\(\frac {3}{8}\) = 137\(\frac {1}{2}\)%

2. Convert the following percents into fractions in simplest form

(i). 25%
Solution:
25% = \(\frac {25}{100}\)
= \(\frac {1}{4}\)

(ii). 150%
Solution:
150% = \(\frac {150}{100}\)
= \(\frac {3}{2}\)

(iii). 7\(\frac {1}{2}\)
Solution:
7\(\frac {1}{2}\) = \(\frac {15}{2}\) × \(\frac {1}{100}\)
= \(\frac {3}{40}\)

3. (i) Anita secured 324 marks out of 400 marks. Find the percentage of marks secured by Anita.
(ii) Out of 32 students, 8 are absent from the class. What is the percentage of students who are absent ?
(iii) There are 120 voters, 90 out of them voted. What percent did not vote ?
Solution:
(i) Anita scored 324 marks out of 400 marks.
∴ Percentage of marks secured by Anita = \(\left(\frac{324}{400} \times 100\right) \%\) = 50%

(ii) Out of 32 students 8 students are absent
∴ Percentage of students who are absent = \(\frac {8}{32}\) × 100% = 25%

(iii) Total voters = 120
Voters who voted = 90
The voters who did not vote = 120 – 90
= 30
Percentage voters who did not vote = \(\frac {30}{120}\) × 100%
= 25%

4. Estimate the part of figure which is shaded and hence find the percentage of the part which is shaded.

Solution:
(i) Shaded part = \(\frac {2}{4}\) = \(\frac {1}{2}\)
Percentage of shaded part = \(\left(\frac{1}{2} \times 100\right) \%\)
= 50%

(ii) Shaded part = \(\frac {2}{6}\) = \(\frac {1}{3}\)
Percentage of shaded part = \(\left(\frac{1}{3} \times 100\right) \%\)
= 33\(\frac {1}{3}\)%

(iii) Shaded part = \(\frac {5}{8}\)
Percentage of shaded part = \(\left(\frac{5}{8} \times 100\right) \%\)
= \(\frac {125}{2}\)%
= 62.5%

5. Convert the following percentages into ratios in simplest form :

(i). 14%
Solution:
14% = 14 × \(\frac {1}{100}\)
= \(\frac {7}{50}\)
= 7 : 50

(ii). 1\(\frac {3}{4}\)%
Solution:
1\(\frac {3}{4}\)% = \(\frac {7}{4}\) × \(\frac {1}{100}\)
= \(\frac {7}{400}\)
= 7 : 400

(iii). 33\(\frac {1}{3}\)%
Solution:
33\(\frac {1}{3}\)% = \(\frac {100}{3}\) × \(\frac {1}{100}\)
= \(\frac {1}{3}\)
= 1 : 3

6. Express the following ratios as percentages :

(i). 5 : 4
Solution:
5:4 = \(\frac {5}{4}\) × 100
= 125%

(ii). 1 : 1
Solution:
1 : 1 = \(\frac {1}{1}\) × 100
= 100%

(iii). 2 : 3
Solution:
2 : 3 = \(\frac {2}{3}\) × 100
= \(\frac {200}{3}\)%
= 66\(\frac {2}{3}\)%

(iv). 9 : 16
Solution:
9 : 16 = \(\frac {9}{16}\) × 100
= \(\frac {225}{4}\)%
= 56\(\frac {1}{4}\)%

7. Chalk contains calcium, carbon and sand in the ratio 12 : 3 : 10. Find the percentage of carbon in the chalk.
Solution:
Calcium : Carbon : sand
= 12 : 3 : 10
Total of ratios = 12 + 3 + 10
= 25
Percentage of carbon is chalk
= \(\frac {3}{25}\) × 100
= 12%

8. Convert each part of the following ratios into percentage :

(i). 3 : 1
Solution:
Total of ratios = 3 + 1 = 4
Percentage of first part = \(\frac {3}{4}\) × 100
= 75%
Percentage of second part = \(\frac {1}{4}\) × 100
= 25%

(ii). 1 : 4
Solution:
Total of ratios = 1 + 4 = 5
Percentage of first part = \(\frac {1}{5}\) × 100
= 20%
Percentage of second part = \(\frac {4}{5}\) × 100
= 80%

(iii). 4 : 5 : 6.
Solution:
Total of ratios = 4 + 5 + 6 = 15
Percentage of first part = \(\frac {4}{15}\) × 100
= \(\frac {8}{3}\)%
= 26\(\frac {2}{3}\)%
Percentage of second part = \(\frac {5}{15}\) × 100
= \(\frac {100}{3}\)%
= 33\(\frac {1}{3}\)%
Percentage of third part = \(\frac {6}{15}\) × 100
= 40%

9. Convert the following percentages to decimals :

(i). 28%
Solution:
28% = \(\frac {28}{100}\)
0.28

(ii). 3%
Solution:
3% = \(\frac {3}{100}\)
= 0.03

(iii). 37\(\frac {1}{2}\)%
Solution:
37\(\frac {1}{2}\)%
= \(\frac {75}{2}\) × \(\frac {1}{100}\)=
= \(\frac {3.75}{100}\)
= 0.375

10. Convert the following decimals to percentage :

(i). 0.65
Solution:
0.65 = (0.65 × 100)%
= \(\left(\frac{65}{100} \times 100\right) \%\)
= 65%

(ii). 0.9
Solution:
0.9 = (0.9 × 100)%
= \(\left(\frac{9}{10} \times 100\right) \%\)
= 90%

(iii). 2.1
Solution:
2.1 = (2.1 × 100)%
= \(\left(\frac{21}{10} \times 100\right) \%\)
= 210%

11. (i) If 65% of students in a class have a bicycle, then what percent of the students do not have a bicycle ?
(ii) We have a basket full of apples, oranges and mangoes. If 50% are apples, 30% are oranges, then what percent are mangoes ?
Solution:
(i) Percentage of student having a bicycle = 65%
Percentage of students do not have a bicycle = (100 – 65)%
= 35%

(ii) Percentage of apples = 50%
Percentage of oranges = 30%
Percentage Of mangoes = (100 – (50% + 30%)
= (100 – 80)%
= 20%

12. The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.
Solution:
Original population = 25000
Decreased population = 24,500
Decrease in population = (25000 – 24500)
= 500
Percentage decrease = \(\frac{\text { Decrease in population }}{\text { Original population }} \times 100 \%\)
= \(\frac {500}{25000}\) × 100%
= 2%

13. Arun bought a plot for ₹ 3,50,000. The next year, the price went upto ₹ 3,70,000. What was the percentage of price increase ?
Solution:
Original price of plot = ₹ 3,50,000
The increased price of plot = ₹ 3,70,000
Increase in price = ₹ 3,70,000 – ₹ 3,50,000
= ₹ 20,000.
Percentage of price increase
= \(\left(\frac{\text { Increase in price }}{\text { Original price }} \times 100\right)\)
= \(\frac {20,000}{350000}\) × 100%
= \(\frac {40}{7}\)%
= 5\(\frac {5}{7}\)

14. Find :

(i). 15% of 250
Solution:
15% of 250 = \(\frac {15}{100}\) × 250
= \(\frac {375}{10}\)
= 37.5

(ii). 25% of 120 litres
Solution:
25% of 120 litres = \(\frac {25}{100}\) × 20 litres
= 30 liters.

(iii). 4% of 12.5
Solution:
4% of 12.5 = \(\frac {4}{100}\) × \(\frac {125}{10}\)
= \(\frac {5}{10}\)
= 0.5

(iv). 12% of ₹ 250.
Solution:
12% of ₹ 250 = ₹\(\frac {12}{100}\) × 250
= ₹300

15. Multiple Choice Questions :

Question (i).
The ratio 2 : 3 expressed as percentage is
(a) 40%
(b) 60%
(c) 66\(\frac {2}{3}\)%
(d) 33\(\frac {1}{3}\)%
Answer:
(c) 66\(\frac {2}{3}\)%

Question (ii).
If 30% of x is 72, then x is equal to
(a) 120
(b) 240
(c) 360
(d) 480
Answer:
(b) 240

Question (iii).
0.025 when expressed as a percent is
(a) 250%
(b) 25%
(c) 4%
(d) 2.5%
Answer:
(d) 2.5%

Question (iv).
In a class, 45% of students are girls. If there are 22 boys in the class, then the total number of students in the class is
(a) 30
(b) 36
(c) 40
(d) 44
Answer:
(c) 40

Question (v).
What percent of \(\frac {1}{7}\) is \(\frac {2}{35}\) ?
(a) 20%
(b) 25%
(c) 30%
(d) 40%
Answer:
(d) 40%

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 2 Polynomials Ex 2.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3

Question 1.
Find the remainder when x³ + 3x² + 3x + 1 is divided by
Answer:
The remainder theorem states that when polynomial p (x) of degree greater than or equal to 1 is divided by linear polynomial x – a, the remainder is p (a).
Here. p(x) = x³ + 3x² + 3x + 1

(i) x + 1
Answer:
Divisor g (x) = x + 1.
Comparing x + 1 with zero, we get x = – 1.
Then, remainder
= p(- 1)
= (- 1)³ + 3(- 1)² + 3(- 1) + 1
= – 1 + 3 – 3 + 1
= 0

(ii) x – \(\frac{1}{2}\)
Answer:
Divisor g (x) = x – \(\frac{1}{2}\)
x – \(\frac{1}{2}\) = 0 gives x = \(\frac{1}{2}\)
Then, remainder
= p\(\left(\frac{1}{2}\right)\)
= \(\left(\frac{1}{2}\right)^{3}+3\left(\frac{1}{2}\right)^{2}+3\left(\frac{1}{2}\right)+1\)
= \(\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1\)
= \(\frac{27}{8}\)

(iii) x
Answer:
Divisor g (x) = x.
x = 0 gives x = 0.
Then, remainder = p (0)
= (0)³ + 3(0)² + 3(0) + 1
= 0 + 0 + 0 + 1
= 1

(iv) x + π
Answer:
Divisor g (x) = x + π.
x + π = 0 gives x = – π.
Then, remainder
= p(- π)
= (- π)³ + 3(- π)² + 3(- π) + 1
= – π³ + 3π² – 3π + 1

(v) 5 + 2x
Answer:
Divisor g(x) = 5 + 2x.
5 + 2x = 0 gives x = – \(\frac{5}{2}\)
Then, remainder
= P\(\left(-\frac{5}{2}\right)\)
= \(\left(-\frac{5}{2}\right)^{3}+3\left(-\frac{5}{2}\right)^{2}+3\left(-\frac{5}{2}\right)+1\)
= \(-\frac{125}{8}+\frac{75}{4}-\frac{15}{2}+1\)
= \(\frac{-125+150-60+8}{8}\)
= \(-\frac{27}{8}\)

Question 2.
Find the remainder when x³ – ax² + 6x – a is divided by x – a.
Answer:
Here, p (x) = x³ – ax² + 6x – a and divisor
g (x) = x – a.
x – a = 0 gives x = a.
Then, remainder = p (a)
= (a)³ – a(a)² + 6(a) – a
= a³ – a³ + 6a – a
= 5a

Question 3.
Check whether 7 + 3x is a factor of 3x³ + 7x.
Answer:
Here, p (x) = 3x³ + 7x and divisor g (x) = 7 + 3x.
7 + 3x = 0 gives x = –\(\frac{7}{3}\).
Then, remainder = p\(\left(-\frac{7}{3}\right)\)
= \(3\left(-\frac{7}{3}\right)^{3}+7\left(-\frac{7}{3}\right)\)
= \(-\frac{343}{9}-\frac{49}{3}\)
= \(\frac{-343-147}{9}\)
= – \(\frac{490}{9}\) ≠ 0
Since the remainder is not zero when
p (x) = 3x³ + 7x is divided by 7 + 3x, it is clear that 7 + 3x is not a factor of 3x³ + 7x.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 2 Polynomials Ex 2.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2

Question 1.
Find the value of the polynomial 5x – 4x² + 3 at (i) x = 0, (ii) x = – 1 and (iii) x = 2.
Answer:
Here, p (x) 5x – 4x² + 3
(i) The value of polynomial p (x) at x = 0 is given by
p(0) = 5(0) – 4(0)² + 3
= 3

(ii) The value of polynomial p (x) at x = – 1 is given by
p(- 1) = 5(- 1) – 4 (- 1)² + 3
= – 5 – 4 + 3
= – 6

(iii) The value of polynomial p (x) at x = 2 is given by
p(2) = 5(2) – 4(2)² + 3
= 10 – 16 + 3
= – 3

Question 2.
Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p (y) = y² – y + 1
Answer:
p(y) = y² – y + 1
∴ p(0) = (0)² – (0) + 1 = 1
∴ P(1) = (1)² – (1) + 1 = 1 – 1 + 1 = 1
∴ p(2) = (2)² – (2) + 1 = 4 – 2 + 1 = 3

(ii) p (t) = 2 + t + 2t² – t³
Answer:
p(t) = 2 + t + 2t² – t³
∴ p(0) = 2 + 0 + 2(0)² – (0)³ = 2
∴ p (1) = 2 + (1) + 2 (1)² – (1)³
= 2 + 1 + 2 – 1
= 4
∴ p(2) = 2 + (2) + 2(2)² – (2)³
= 2 + 2 + 8 – 8 = 4

(iii) p(x) = x³
Answer:
p(x) = x³
∴ p (0) = (0)³ = 0
∴ p ( 1) = (1)³ = 1
∴ p (2) = (2)³ = 8

(iv) p(x) = (x – 1) (x + 1)
Answer:
p(x) = (x – 1) (x + 1)
∴ p(0) = (0 – 1) (0 + 1) = (- 1) × 1 = – 1
∴ p(1) = (1 – 1) (1 + 1) = 0 × 2 = 0
∴ p(2) = (2 – 1)(2 + 1) = 1 × 3 = 3

Question 3.
Verify whether the following are zeros of the polynomial, indicated against them:
(i) p(x) = 3x + 1, x = – \(\frac{1}{3}\)
Answer:
Here, p (x) = 3x + 1
Then, p\(\left(-\frac{1}{3}\right)\) = 3\(\left(-\frac{1}{3}\right)\) + 1 = – 1 + 1 = 0
Hence, – \(\frac{1}{3}\) is a zero of polynomial
p(x) = 3x + 1

(ii) p(x) = 5x – π, x = \(\frac{4}{5}\)
Answer:
Here, p(x) = 5x – π,
Then, p\(\left(\frac{4}{5}\right)\) = 5\(\left(\frac{4}{5}\right)\) – π = 4 – π ≠ 0
Hence, \(\frac{4}{5}\) is not a zero of polynomial
p(x) = 5x – π

(iii) p(x) = x² – 1, x = 1, – 1
Answer:
Here, p(x) = x² – 1
Then, p(1) = (1)² – 1 = 1 – 1 = 0 and
p(- 1) = (- 1)² – 1 = 1 – 1 = 0.
Hence, 1 and – 1 both are zeroes of polynomial p(x) = x² – 1.

(iv) p(x) = (x + 1) (x – 2), x = – 1, 2
Answer:
Here, p (x) = (x + 1) (x – 2)
Then, p(- 1) = (- 1 + 1) (- 1 – 2) = 0 × (-3)= 0
and p (2) = (2 + 1) (2 – 2) = 3 × O = O.
Hence, – 1 and 2 both are zeros of polynomial p(x) = (x + 1) (x – 2).

(v) p(x) = x², x = 0
Answer:
Here, p(x) = x²
Then, p(0) = (0)² = 0
Hence, 0 is a zero of polynomial p (x) = x².

(vi) p(x) = lx + m, x = –\(\frac{n}{l}\)
Answer:
Here, p (x) = lx + m
Then, p \(\left(-\frac{m}{l}\right)\) = l\(\left(-\frac{m}{l}\right)\) + m = – m + m = 0
Hence, \(-\frac{m}{l}\) is a zero of polynomial
p(x) = lx + m.

(vii) p(x) = 3x² – 1, x = – \(\frac{1}{\sqrt{3}}\), \(\frac{2}{\sqrt{3}}\)
Answer:
Here, p(x) = 3x² – 1
Then, p\(\left(-\frac{1}{\sqrt{3}}\right)\) = 3\(\left(-\frac{1}{\sqrt{3}}\right)^{2}\) – 1
= 3\(\left(\frac{1}{3}\right)\) – 1 = 1 – 1 = 0
and p\(\left(\frac{2}{\sqrt{3}}\right)\) = 3\(\left(\frac{2}{\sqrt{3}}\right)^{2}\) – 1
= 3\(\left(\frac{4}{3}\right)\) – 1 = 4 – 1 = 3 ≠ 0
Hence, –\(\frac{1}{\sqrt{3}}\) is a zero of polynomial p (x) = 3x² – 1, but \(\frac{2}{\sqrt{3}}\) is not a zero of polynomial p(x) = 3x² – 1.

(viii) p (x) = 2x + 1, x = \(\frac{1}{2}\)
Answer:
Here, p(x) = 2x + 1
Then, p\(\left(\frac{1}{2}\right)\) = 2\(\left(\frac{1}{2}\right)\) + 1 = 1 + 1 = 2 ≠ 0
Hence, \(\frac{1}{2}\) is not a zero of polynomial
p(x) = 2x + 1.

Question 4.
Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5
Answer:
To find the zero of polynomial p (x) = x + 5,
we solve the equation p (x) = 0.
∴ x + 5 = 0
∴ x = – 5
Thus, – 5 is the zero of polynomial
p(x) = x + 5.

(ii) p(x) = x – 5
Answer:
To find the zero of polynomial p(x) = x – 5,
we solve the equation p (x) = 0.
∴ x – 5 = 0
∴ x = 5
Thus, 5 is the zero of polynomial
p(x) = x – 5.

(iii) p (x) = 2x + 5
Answer:
To find the zero of polynomial p(x) = 2x + 5,
we solve the equation p (x) = 0.
∴ 2x + 5 = 0
∴ 2x = – 5
Thus, –\(\frac{5}{2}\) is the zero of polynomial
p(x) = 2x + 5.

(iv) p (x) = 3x – 2
Answer:
To find the zero of polynomial p (x) = 3x – 2,
we solve the equation p (x) = 0.
∴ 3x – 2 = 0
∴ 3x = 2
Thus, \(\frac{2}{3}\) is the zero of polynomial
p(x) = 3x – 2.

(v) p(x) = 3x
Answer:
To find the zero of polynomial p (x) = 3x.
we solve the equation p (x) = 0.
∴ 3x = 0
∴ x = 0
Thus, 0 is the zero of polynomial P(x) = 3x.

(vi) p(x) = ax, a ≠ 0
Answer:
To find the zero of polynomial p (x) = ax,
a ≠ 0, we solve the equation p (x) = 0.
∴ ax = 0
∴ x = 0 (∵ a ≠ 0)
Thus, 0 is the zero of polynomial
p(x) = ax, a ≠ 0.

(vii) p(x) = cx + d, c ≠ 0, C, d are real numbers.
Answer:
To find the zero of polynomial
p(x) = cx + d, c ≠ 0, c, d are real numbers, we solve the equation p (x) = 0.
∴ cx + d = 0
∴ cx = – d
∴ x = – \(\frac{d}{c}\)
Thus, – \(\frac{d}{c}\) is the zero of polynomial p (x) = cx + d, c ≠ 0, c, d are real numbers.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 1 Knowing Our Numbers MCQ Questions with Answers.

PSEB 6th Class Maths Chapter 1 Knowing Our Numbers MCQ Questions

Multiple Choice Questions

Question 1.
The number of digits are:
(a) 9
(b) 10
(c) 8
(d) Infinite.
Answer:
(b) 10

Question 2.
The greatest 4 digit number using 1, 5, 2, 9 once is:
(a) 9215
(b) 9512
(c) 5912
(d) 9521.
Answer:
(b) 9512

Question 3.
The smallest 4 digit number using 2, 0, 3, 7 once is:
(a) 0237
(b) 2037
(c) 7320
(d) 7023.
Answer:
(b) 2037

Question 4.
Which of the following are in ascending order?
(a) 217, 271, 127, 721
(b) 217, 127, 721, 271
(c) 127, 217, 271, 721
(d) 721, 271, 217, 127.
Answer:
(c) 127, 217, 271, 721

Question 5.
The face value of digit 4 in 23468 is:
(a) 4
(b) 400
(c) 40
(d) 468.
Answer:
(a) 4

Question 6.
The place value of digit 2 in 4123 is:
(a) 23
(b) 2
(c) 20
(d) 200.
Answer:
(c) 20

Question 7.
The difference between place value and face value of 5 in 76542 is:
(a) 537
(b) 45
(c) 0
(d) 495
Answer:
(d) 495

Question 8.
5 × 10000 + 3 × 100 + 2 × 10 + 2 = …………..
(a) 5322
(b) 53022
(c) 50322
(d) 53202.
Answer:
(c) 50322

Question 9.
Four lakh two thousand three hundred fifty-one = …………..
(a) 42351
(b) 402351
(c) 420351
(d) 4002351.
Answer:
(b) 402351

Question 10.
How many four-digit numbers are there?
(a) 9999
(b) 9900
(c) 9000
(d) 9990.
Answer:
(c) 9000

Question 11.
Seventeen million twenty-four thousand fifty-four = …………….
(a) 172454
(b) 170024054
(c) 170240054
(d) 17024054.
Answer:
(d) 17024054.

Question 12.
1 Crore = …………….. million.
(a) 1
(b) 10
(c) 100
(d) 1000.
Answer:
(b) 10

Question 13.
Rounded off 7213 to nearest thousands.
(a) 7200
(b) 7000
(c) 7210
(d) 7213.
Answer:
(b) 7000

Question 14.
Rounded off 45553 to nearest hundreds.
(a) 45500
(b) 45550
(c) 45600
(d) 45650.
Answer:
(c) 45600

Question 15.
Solve : (9 – 4) × 6 = …………….. .
(a) 30
(b) 54
(c) 78
(d) 64.
Answer:
(a) 30

Question 16.
Which of the following number does not have symbol in Roman numerals?
(a) 0
(b) 1
(c) 10
(d) 1000.
Answer:
(a) 0

Question 17.
How many symbols are used in Roman Numerals?
(a) 5
(b) 8
(c) 9
(d) 7.
Answer:
(d) 7

Question 18.
Which of the following are meaningless?
(a) LXIX
(b) XC
(c) IL
(d) LI.
Answer:
(c) IL

Question 19.
CLXVI = ………..
(a) 164
(b) 144
(c) 176
(d) 166.
Answer:
(d) 166

Question 20.
XCIX + XLVI = …………….
(a) CVL
(b) CLV
(c) CXLV
(d) CXLIV.
Answer:
(c) CXLV

Question 21.
Using the digits 4, 5, 7 and 0 without repetition which of the following is the smallest four-digit number?
(a) 0457
(b) 4057
(c) 4507
(d) 4075.
Answer:
(b) 4057

Question 22.
Using the digits 2, 8, 7 and 4 without repetition which of the following is the greatest four-digit number?
(a) 2874
(b) 8742
(c) 8472
(d) 8274.
Answer:
(b) 8742

Question 23.
Which is the smallest four digits number made from the digits 3, 8, 7 by using one-digit twice?
(a) 3378
(b) 3783
(c) 3873
(d) 3837.
Answer:
(a) 3378

Question 24.
Make the greatest four-digit number from the digits 9, 0, 5 by using one-digit twice.
(a) 9005
(b) 9905
(c) 9950
(d) 9050.
Answer:
(c) 9950

Question 25.
Take two digits, 2 and 3, from diem make smallest four digit number, using both the digits equal number of time.
(a) 3232
(b) 2323
(c) 3223
(d) 2233.
Answer:
(d) 2233.

Question 26.
Take two digits, 2 and 3 from them make greatest four-digit number, using both the digits equal number of time.
(a) 3232
(b) 3322
(c) 3223
(d) 2323.
Answer:
(b) 3322

Question 27.
The greatest number from 4536, 4892, 4370, 4452 is:
(a) 4536
(b) 4892
(c) 4370
(d) 4452.
Answer:
(b) 4892

Question 28.
Out of 15623, 15073, 15189, 15800 the smallest number is:
(a) 15623
(b) 15073
(c) 15189
(d) 15800.
Answer:
(b) 15073

Question 29.
The ascending order of the numbers 847, 9754, 8320, 571 is:
(a) 847, 9754, 8320, 571
(b) 9754, 8320, 847, 571
(c) 571, 847, 8320, 9754
(d) 571, 8320, 847, 9754.
Answer:
(c) 571, 847, 8320, 9754

Fill in the blanks:

Question 1.
1 lakh = ten thousands.
Answer:
Ten

Question 2.
1 million = ……………… hundred thousand.
Answer:
Ten

Question 3.
1 crore = ……………….. million.
Answer:
Ten

Question 4.
1 crore = …………… ten lakh.
Answer:
Ten

Question 5.
1 million = ……………. lakh.
Answer:
Ten

Write True/False:

Question 1.
The number of digits are 10. (True/False)
Answer:
True

Question 2.
The greatest four-digit number is 1000. (True/False)
Answer:
False

Question 3.
The place value of digit 5 in 3564 is 50. (True/False)
Answer:
False

Question 4.
0 does not have symbol in Roman numbers. (True/False)
Answer:
True

Question 5.
IL is meaningless. (True/False)
Answer:
True

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 8 Comparing Quantities Ex 8.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.1

1. Find the ratio of :

Question (i).
₹ 5 to 50 paise
Solution:
To find ratio of ₹ 5 to 50 paise.
Firstly convert both the quantities into same units.
₹ 1 = 100 paise
₹ 5 = 5 × 100 paise = 500 paise.
So, ratio of 500 paise to 50 paise
= \(\frac{500}{50}=\frac{10}{1}\)
= 10 : 1
Hence, required ratio is 10 : 1

Question (ii).
15 kg to 210 g
Solution:
To find ratio of 15 kg to 210 g
Firstly, convert both the quantities into same unit
1 kg = 1000 g
15 kg = 15 × 1000 g = 15000 g.
So, ratio of 15000 to 210 g
= \(\frac{15000}{210}=\frac{500}{7}\)
= 500 : 7
Hence, required ratio is 500 : 7

Question (iii).
4 m to 400 cm
Solution:
To find ratio of 4 m to 400 cm
Firstly convert both the quantities into same unit.
1 m = 100 cm
4 m = 4 × 100 cm = 400 cm
So, ratio of 400 cm to 400 cm
= \(\frac{400}{400}=\frac{1}{1}\)
= 1 : 1
Hence, required ratio is 1 : 1

Question (iv).
30 days to 36 hours
Solution:
To find ratio of 30 days to 36 hours
1 day = 24 hours
30 days = 30 × 24 hours
= 720 hours
So, ratio of 720 hours to 36 hours
= \(\frac{720}{36}=\frac{20}{1}\)
= 20 : 1
Hence required ratio is 20 : 1

2. Are the ratios 1: 2 and 2 :3 equivalent ?
Solution:
To check this we need to know whether
1 : 2 and 2 : 3 are equal
First convert the ratios into fractions
1 : 2 is written as \(\frac {1}{2}\)
2 : 3 is written as \(\frac {2}{3}\)
Now to change these fraction into like fraction.
Make denominator of both the fraction same.
\(\frac{1}{2}=\frac{1}{2} \times \frac{3}{3}=\frac{3}{6}\) and \(\frac{2}{3}=\frac{2}{3} \times \frac{2}{2}=\frac{4}{6}\)
4 > 3
\(\frac{4}{6}>\frac{3}{6}\)
Hence 1 : 2 and 2 : 1 are not equivalent

3. If the cost of 6 toys is ₹ 240, find the cost of 21 toys.
Solution:
The more the number of toys one purchases, the more is the amount to be paid.
Therefore, there is a direct proportion between the number of toys and the amount to be paid.
Let x be number of toys to be purchased
∴ 6 : 240 : : 21 : x
\(\frac{6}{240}=\frac{21}{x}\)
x = \(\frac{21 \times 240}{6}\) = ₹ 840
Thus cost of 21 toys is ₹ 840.

4. The car that I own can go 150 km with 25 litres of petrol. How far can it go with 30 litres of petrol ?
Solution:
More km ↔ More petrol
So, the consumption of petrol and the distance travelled by a car is a case of direct proportion.
Let the distance travelled be x km.
∴ 150 : 25 : : x : 30
\(\frac{150}{25}=\frac{x}{30}\)
x = \(\frac{150 \times 30}{25}\)
x = 180
Thus, it can go 180 km is 30 litres of petrol.

5. In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students ?
Solution:
The more the number of students, the more is the number of computer needed.
More students ↔ More will be the computed needed.
There is a direct proportion between the number of students and the number of computers needed. Let computer needed will be x.
6 : 3 : : 24 : x
\(\frac{6}{3}=\frac{24}{x}\)
6 × x = 24 × 3
x = \(\frac{24 \times 3}{6}\) = 12
Thus, 12 computers will be needed.