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%