Fractions In Disguise Class 8 Maths Ganita Prakash Part 2 Chapter 1 NCERT Solutions

Fractions In Disguise Class 8 Maths Ganita Prakash Part 2 Chapter 1 NCERT Solutions Looking for the Fractions In Disguise Class 8 Maths Ganita Prakash Part 2 Chapter 1 NCERT Solutions? You are at the right place. This chapter helps students understand equivalent fractions, simplifying fractions, and solving NCERT questions with easy explanations.

Figure it Out (Page 3)

1. Express the following fractions as percentages.

Solution:
(i) $\frac{3}{5}$ = $\frac{3}{5}$ × 100% = 3 × 20% = 60%.

(ii) $\frac{7}{4}$ = $\frac{7}{4}$ × 100% = 7 × 25% = 175%.

(iii) $\frac{9}{20}$ = $\frac{9}{20}$ × 100% = 9 × 5% = 45%.

(iv) $\frac{72}{150}$ = $\frac{72}{150}$ × 100% = $\frac{72}{15}$ × 10% = $\frac{24}{5}$ × 10% = 24 × 2% = 48%.

(v) $\frac{1}{3}$ = $\frac{1}{3}$ × 100% = $\frac{100}{3}$ % = 33 $\frac{1}{3}$ %.

(vi) $\frac{5}{11}$ = $\frac{5}{11}$ × 100% = $\frac{500}{11}$ % = 45 $\frac{5}{11}$ %.

2. Nandini has 25 marbles, of which 15 are white. What percentage of her marbles are white?

Solution:
Total marbles = 25
White marbles = 15
Percentage of white marbles = $\frac{15}{25}$ × 100% = 15 × 4% = 60%.
Therefore,
(iv) 60% is the correct answer.

3. In a school, 15 of the 80 students come to school by walking. What percentage of the students come by walking?
Solution:
Total students = 80
Students coming to school by walking = 15
Percentage of students coming to school by walking = $\frac{15}{80}$ × 100% = $\frac{15}{4}$ × 5% = $\frac{75}{4}$ % = 18.75%.

4. A group of friends is participating in a long-distance run. The positions of each of them after 15 minutes are shown in the following picture. Match (among the given options) what percentage of the race each of them has approximately completed.

Solution:
A is clearly well before halfway → 38%
B is approximately at the halfway mark → 55%
C is around three-fourths of the distance → 72%
D is very near the finish → 93%

5. Pairs of quantities are shown below. Identify and write appropriate symbols ‘>’, ‘<’, ‘=’ in the blanks. Try to do it without calculations.

Solution:
(i) 50% > 5%
Clearly, 50% is greater than 5%.

(ii) $\frac{5}{10}$ ___ 50%
$\frac{5}{10}$ × 100% = 50%
∴ $\frac{5}{10}$ = 50%

(iii) $\frac{3}{11}$ ___ 61%
$\frac{3}{11}$ × 100% = $\frac{300}{11}$ % = 27.27%
∴ $\frac{3}{11}$ < 61%

(iv) 30% ____ $\frac{1}{3}$
$\frac{1}{3}$ × 100% = $\frac{100}{3}$ % = 33.3%
∴ 30% < $\frac{1}{3}$

Figure it Out (Page 12 – 14)

Estimate first before making any computations to solve the following questions. Try different methods including mental computations.
1. Find the missing numbers. The first problem has been worked out.

Solution:

2. Find the value of the following and also draw their bar models.

Solution:
(i) 25% of 160
= $\frac{25}{100}$ × 160 = $\frac{1}{4}$ × 160 = $\frac{160}{4}$ = 40.

(ii) 16% of 250
= $\frac{16}{100}$ × 250 = $\frac{16}{10}$ × 25 = $\frac{16}{2}$ × 5 = 8 × 5 = 40.

(iii) 62% of 360
= $\frac{62}{100}$ × 360 = $\frac{62}{10}$ × 36 = $\frac{62}{5}$ × 18 = $\frac{1116}{5}$ = 223.2%.

(iv) 140% of 40
= $\frac{140}{100}$ × 40 = $\frac{14}{10}$ × 40 = 14 × 4 = 56.

(v) 1% of 1 hour
= $\frac{1}{100}$ × 60 min = $\frac{1}{100}$ × 3600 sec = 36 sec.

(vi) 7% of 10 kg
= $\frac{7}{100}$ × 10000 g = 7 × 100 g = 700 g.

3. Surya made 60 ml of deep orange paint. How much red paint did he use if red paint made up $\frac{3}{4}$ of the deep orange paint?
Solution:
Quantity of orange paint = 60 ml
Quantity of red paint = $\frac{3}{4}$ × 60 ml = 3 × 15 ml = 45 ml.

4. Pairs of quantities are shown below. Identify and write appropriate symbols ‘>’, ‘<’, ‘=’ in the boxes. Visualising or estimating can help. Compute only if necessary or for verification.
(i) 50% of 510 ____ 50% of 515
(ii) 37% of 148 ____ 73% of 148
(iii) 29% of 43 ____ 92% of 110
(iv) 30% of 40 ____ 40% of 50
(v) 45% of 200 ____ 10% of 490
(vi) 30% of 80 ____ 24% of 64
Solution:
(i) 50% of 510 ____ 50% of 515.
Since 510 < 515
∴ 50% of 510 < 50% of 515

(ii) 37% of 148 ____ 73% of 148
Since 37% < 73%
∴ 37% of 148 < 73% of 148

(iii) 29% of 43 ____ 92% of 110
29% of 43 = $\frac{29}{100}$ × 43 = $\frac{1247}{100}$ = 12.47
92% of 110 = $\frac{92}{100}$ × 110 = $\frac{10120}{100}$ = 101.2
Since 12.47 < 101.2
∴ 29% of 43 < 92% of 110

(iv) 30% of 40 ____ 40% of 50
30% of 40 = $\frac{30}{100}$ × 40 = 3 × 4 = 12
40% of 50 = $\frac{40}{100}$ × 50 = 4 × 5 = 20
Since 12 < 20
∴ 30% of 40 < 40% of 50

(v) 45% of 200 ____ 10% of 490
45% of 200 = $\frac{45}{100}$ × 200 = 45 × 2 = 90
10% of 490 = $\frac{10}{100}$ × 490 = 49
Since 90 > 49
∴ 45% of 200 > 10% of 490

(vi) 30% of 80 ____ 24% of 64
30% of 80 = $\frac{30}{100}$ × 80 = 3 × 8 = 24
24% of 64 = $\frac{24}{100}$ × 64 = $\frac{1536}{100}$ = 15.36
Since 24 > 15.36
∴ 30% of 80 > 24% of 64

5. Fill in the blanks appropriately:
(i) 30% of k is 70, 60% of k is ____, 90% of k is ____, 120% of k is ____
(ii) 100% of m is 215, 10% of m is ____, 1% of m is ____, 6% of m is ____
(iii) 90% of n is 270, 9% of n is ____, 18% of n is ____, 100% of n is ____
(iv) Make 2 more such questions and challenge your peers.
Solution:
(i) 30% of k is 70
(30% of k is 70) × 2 = 60% of k = 140
(30% of k is 70) × 3 = 90% of k = 210
(30% of k is 70) × 4 = 120% of k = 280

(ii) 100% of m is 215
$\frac{100}{100}$ × m = 215
m = 215
10% of m = $\frac{10}{100}$ × 215 = 21.5
1% of m = $\frac{1}{100}$ × 215 = 2.15
6% of m = $\frac{6}{100}$ × 215 = $\frac{1290}{100}$ = 12.9

(iii) 90% of n is 270
$\frac{90}{100}$ × n = 270
n = 270 × $\frac{10}{9}$ = 30 × 10 = 300
9% of n = $\frac{9}{100}$ × 300 = 9 × 3 = 27
18% of n = $\frac{18}{100}$ × 300 = 18 × 3 = 54
100% of n = $\frac{100}{100}$ × 300 = 300

(iv) 25% of x is 50. 50% of x is ___, 75% of x is ___, 200% of x is ___.
80% of y is 160. 10% of y is ___, 5% of y is ___, 100% of y is ___.

6. Fill in the blanks:
(i) 3 is ____ % of 300.
(ii) _____ is 40% of 4.
(iii) 40 is 80% of _____.
Solution:
(i) 3 is ____ % of 300
Let the percentage be a.
a% of 300 = 3
$\frac{a}{100}$ × 300 = 3
a × 3 = 3
a = 1
∴ 3 is 1 % of 300.

(ii) _____ is 40% of 4
Let the number be b.
b = 40% of 4
b = $\frac{40}{100}$ × 4
b = $\frac{4}{10}$ × 4 = 1.6
∴ 1.6 is 40% of 4.

(iii) 40 is 80% of _____.
Let the number be y.
80% of y = 40
$\frac{80}{100}$ × y = 40
y = 40 × $\frac{10}{8}$ = 5 × 10 = 50.
∴ 40 is 80% of 50.

7. Is 10% of a day longer than 1% of a week? Create such questions and challenge your peers.
Solution:
1 day = 24 hours → 10% of a day = $\frac{10}{100}$ × 24 = 2.4 hours
1 week = 7 days = 168 hours → 1% of a week = $\frac{1}{100}$ × 168 = 1.68 hours
Since 2.4 > 1.68,
10% of a day is longer than 1% of a week.

8. Mariam’s farm has a peculiar bull. One day she gave the bull 2 units of fodder and the bull ate 1 unit. The next day, she gave the bull 3 units of fodder and the bull ate 2 units. The day after, she gave the bull 4 units and the bull ate 3 units. This continued, and on the 99th day she gave the bull 100 units and the bull ate 99 units. Represent these quantities as percentages. This task can be distributed among the class. What do you observe?
Solution:
Day 1 = $\frac{1}{2}$ × 100% = 50%
Day 2 = $\frac{2}{3}$ × 100% = 66.67%
Day 3 = $\frac{3}{4}$ × 100% = 75%
Day 4 = $\frac{4}{5}$ × 100% = 80%
…………….
…………….
Day 99 = $\frac{99}{100}$ × 100% = 99%
The fraction follows the pattern $\frac{n}{n + 1}$ which approaches 1 (or 100%) as n increases.
The percentage of fodder eaten is increasing every day and is getting closer and closer to 100%, but never actually reaches 100%.

9. Workers in a coffee plantation take 18 days to pick coffee berries in 20% of the plantation. How many days will they take to complete the picking work for the entire plantation, assuming the rate of work stays the same? Why is this assumption necessary?
Solution:
Days taken to complete 20% of the plantation = 18.
Ddays taken to complete 100% (5 × 20%) of the plantation = 5 × 18 = 90.
Therefore, workers will take 90 days to complete the entire plantation.

10. The badminton coach has planned the training sessions such that the ratio of warm up : play : cool down is 10% : 80% : 10%. If he wants to conduct a training of 90 minutes. How long should each activity be done?

Solution:
Warm up time = 10% of 90 min = $\frac{10}{100}$ × 90 = 9 min.
Play time = 80% of 90 min = $\frac{80}{100}$ × 90 = 8 × 9 = 72 min.
Down time = 10% of 90 min = $\frac{10}{100}$ × 90 = 9 min.

11. An estimated 90% of the world’s population lives in the Northern Hemisphere. Find the (approximate) number of people living in the Northern Hemisphere based on this year’s worldwide population.
Solution:
World population = 8.3 billion
90% of 8.3 billion = $\frac{90}{100}$ × 8.3 billion = 7.47
Therefore, there are 7.47 billion people living in the Northern Hemisphere.

12. A recipe for the dish, halwa, for 4 people has the following ingredients in the given proportions — Rava: 40%, Sugar: 40%, and Ghee: 20%.
(i) If you want to make halwa for 8 people, what is the proportion of each of the above ingredients?
(ii) If the total weight of the ingredients is 2 kg, how much rava, sugar and ghee are present?
Solution:
(i) The number of people has increased from 4 to 8 (i.e., doubled), but proportions remain the same in a recipe.
(ii) Total weight = 2 kg
Rava = 40% of 2 kg = $\frac{40}{100}$ × 2000 g = 800 g = 0.8 kg
Sugar = 40% of 2 kg = $\frac{40}{100}$ × 2000 g = 800 g = 0.8 kg
Ghee = 20% of 2 kg = $\frac{20}{100}$ × 2000 g = 400 g = 0.4 kg

Figure it Out (Page 19 – 20)

1. If a shopkeeper buys a geometry box for ₹75 and sells it for ₹110, what is his profit margin with respect to the cost?
Solution:
Profit = ₹110 – ₹75 = ₹35.
Profit percentage = $\frac{35}{75}$ × 100% = $\frac{7}{15}$ × 100% = 46.77%

2. I am a carpenter, and I make chairs. The cost of materials for a chair is ₹ 475, and I want to have a profit margin of 50%. At what price should I sell a chair?
Solution:
Cost of material = ₹475
Profit = 50% of ₹475 = $\frac{50}{100}$ × 475 = $\frac{1}{2}$ × 475 = ₹237.50
Selling price = ₹475 + ₹237.50 = ₹ 712.50

3. The total sales of a company (also called revenue) was ₹2.5 crore last year. They had a healthy profit margin of 25%. What was the total expenditure (costs) of the company last year?
Solution:
Let the total expenditure be x.
Total revenue = ₹2.5 crore
Profit percentage = 25%
Profit = 25% of x = 0.25x
Expenditure + Profit = Revenue
x + 0.25x = 2.5 crore
1.25x = 2.5 crore
x = $\frac{2.5}{1.25}$ crore
x = $\frac{250}{125}$ crore
x = 2 crore.
Therefore, 2 crore was the total expenditure of the company.

4. A clothing shop offers a 25% discount on all shirts. If the original price of a shirt is ₹300, how much will Anwar have to pay to buy this shirt?
Solution:
Discount = 25%
Marked price = ₹300
Discount = 25% of 300 = $\frac{25}{100}$ × 300 = 25 × 3 = ₹75
Final price = ₹300 – ₹75 = ₹225
Therefore, Anwar will have to pay ₹225.

5. The petrol price in 2015 was ₹60 and ₹100 in 2025. What is the percentage increase in the price of petrol?

Solution:
Increase in petrol price = ₹100 – ₹60 = ₹40
Percentage increase = $\frac{40}{60}$ × 100% = $\frac{4}{6}$ × 100% = $\frac{2}{3}$ × 100% = $\frac{200}{3}$ % = 66.66%
Therefore, the petrol price increased by approximately 66.67%.

6. Samson bought a car for ₹4,40,000 after getting a 15% discount from the car dealer. What was the original price of the car?
Solution:
Let the marked price of car be y.
Discount percentage = 15%
Discount = 15% of y = $\frac{15}{100}$ × y = 0.15y
Sales price = Marked price – Discount = y – 0.15y = 0.85y
0.85y = 4,40,000
y = $\frac{4,40,000}{0.85}$ = ₹5,17,647
Therefore, the original price of the car was approximately ₹5,17,647.

7. 1600 people voted in an election and the winner got 500 votes. What percent of the total votes did the winner get? Can you guess the minimum number of candidates who stood for the election?
Solution:
Total people voted = 1600
Votes winner got = 500
Percentage of votes received by winner = $\frac{500}{1600}$ × 100% = 31.25%

8. The price of l kg of rice was ₹ 38 in 2024. It is ₹42 in 2025. What is the rate of inflation? (Inflation is the percentage increase in prices.)
Solution:
Increase in price = ₹42 – ₹38 = ₹4
Rate of inflation = $\frac{4}{38}$ × 100% = 10.52%

9. A number increased by 20% becomes 90. What is the number?
Solution:
Let the number be y
According to question,
y + 20% of y = 90
y + $\frac{20}{100}$ × y = 90
y + 0.2y = 90
1.2 y = 90
y = $\frac{90}{1.2}$ = 75.
Therefore, the number is 75.

10. A milkman sold two buffaloes for ₹80,000 each. On one of them, he made a profit of 5% and on the other a loss of 10%. Find his overall profit or loss.
Solution:
S.P of 1st buffalo = ₹80,000
Profit = 5%
C.P of 1st buffalo = $\frac{100}{100 + 5}$ × 80,000 = $\frac{100}{105}$ × 80,000 = ₹76,190
S.P of 2nd buffalo = ₹80,000
Loss = 10%
C.P of 2nd buffalo = $\frac{100}{100 - 10}$ × 80,000 = $\frac{100}{90}$ × 80,000 = ₹88,889
Total S.P = ₹80,000 + ₹80,000 = ₹1,60,000
Total C.P = ₹76,190 + ₹88,889 = ₹1,65,079
Loss = ₹1,65,079 – ₹1,60,000 = 5,079
Loss percentage = $\frac{5,079}{1,65,079}$ × 100% = 3%

11. The population of elephants in a national park increased by 5% in the last decade. If the population of the elephants last decade is p, the population now is
(i) p × 0.5
(ii) p × 0.05
(iii) p × 1.5
(iv) p × 1.05
(v) p + 1.50
Solution:
New population = p + 5% of p
= p + $\frac{5}{100}$ × p = p + 0.05p = 1.05p
Therefore,
(iv) p × 1.05 is the correct option.

12. Which of the following statement(s) mean the same as — “The demand for cameras has fallen by 85% in the last decade”?
(i) The demand now is 85% of the demand a decade ago.
(ii) The demand a decade ago was 85% of the demand now.
(iii) The demand now is 15% of the demand a decade ago.
(iv) The demand a decade ago was 15% of the demand now.
(v) The demand a decade ago was 185% of the demand now.
(vi) The demand now is 185% of the demand a decade ago.
Solution:
(iii) The demand now is 15% of the demand a decade ago.

Figure it Out (Page 22 – 23)

1. Bank of Yahapur offers an interest of 10% p.a. Compare how much one gets if they deposit ₹20,000 for a period of 2 years with compounding and without compounding annually.
Solution:
Principal = ₹20,000
Interest = 10% p.a.
Time = 2 years
(i) Without compounding:
Amount = p × (1 + rt)
= 20,000 × (1 + $\frac{10}{100}$ × 2)
= 20,000 × (1 + 0.1 × 2)
= 20,000 × (1 + 0.2)
= 20,000 × 1.2 = ₹24,000.
(ii) With compounding:
Amount = p × (1 + r)^t
= 20,000 × (1 + $\frac{10}{100}$ )²
= 20,000 × (1 + 0.1)²
= 20,000 × (1.1)²
= 20,000 × 1.1 × 1.1 = ₹24,200.
Without compounding: ₹24,000
With compounding: ₹24,200
Difference = ₹24,200 – ₹24,000 = ₹200.
Therefore, with compounding, you earn ₹200 more.

2. Bank of Wahapur offers an interest of 5% p.a. Compare how much one gets if one deposits ₹20,000 for a period of 4 years with compounding and without compounding annually.
Solution:
(i) With compounding:
Amount = p × (1 + rt)
= 20,000 × (1 + $\frac{5}{100}$ × 2)
= 20,000 × (1 + 0.05 × 4)
= 20,000 × (1 + 0.2)
= 20,000 × 1.2 = ₹24,000.
(ii) With compounding:
Amount = p × (1 + r)^t
= 20,000 × (1 + $\frac{5}{100}$ )⁴
= 20,000 × (1 + 0.05)⁴
= 20,000 × (1.05)⁴
= 20,000 × 1.05 × 1.05 × 1.05 × 1.05 = ₹24,310.125
Therefore, with compounding, the final amount is more.

3. Do you observe anything interesting in the solutions of the two questions above? Share and discuss.
Solution:
The interest earned with compounding is always greater than the interest earned without compounding.

Figure it Out (Page 24)

4. Jasmine invests amount ‘p’ for 4 years at an interest of 6% p.a. Which of the following expression(s) describe the total amount she will get after 4 years when compounding is not done?

Solution:
Total amount = p + (p × r × t)
= p + (p × $\frac{6}{100}$ × 4)
= p + (p × 0.06 × 4)
Therefore,
(vii) p+ (p × 0.06 × 4) is the correct expression.

5. The post office offers an interest of 7% p.a. How much interest would one get if one invests ₹50,000 for 3 years without compounding? How much more would one get if it was compounded?
Solution:
(i) Without compounding:
Amount = p × (1 + rt)
= 50,000 × (1 + $\frac{7}{100}$ × 3)
= 50,000 × (1 + 0.07 × 3)
= 50,000 × (1 + 0.21)
= 50,000 × (1.21) = ₹60,500.
(ii) With compounding:
Amount = p × (1 + r)^t
= 50,000 × (1 + $\frac{7}{100}$ )³
= 50,000 × (1 + 0.07)³
= 50,000 × (1.07)³
= 50,000 × 1.07 × 1.07 × 1.07 = ₹61252.15
Extra interest due to compounding = 11,252.15 − 10,500 = ₹752.15.

6. Giridhar borrows a loan of ₹12,500 at 12% per annum for 3 years without compounding and Raghava borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?
Solution:
(i) Giridhar’s interest without compounding:
Interest = 12,500 × $\frac{12}{100}$ × 3 = 12,500 × 0.12 × 3 = ₹4,500.
(ii) Raghava’s interest with compounding:
Amount = p × (1 + r)^t
= 12,500 × (1 + $\frac{10}{100}$ )³
= 12,500 × (1 + 0.1)³
= 12,500 × (1.1)³
= 12,500 × 1.1 × 1.1 × 1.1 = ₹16637.5
Interest = ₹16637.5 – ₹12,500 = ₹4137.5
Difference = ₹4,500 – ₹4137.5 = ₹ 362.50
Therefore, Giridhar paid ₹ 362.50 more than Raghava.

7. Consider an amount ₹1000. If this grows at 10% p.a., how long will it take to double when compounding is done vs. when compounding is not done? Is compounding an example of exponential growth and not-compounding an example of linear growth?
Solution:
Let the initial amount be ₹1000 and the rate = 10% per annum.
(i) Without compounding:
Amount = p × (1 + rt)
2000 = 1000 × (1 + $\frac{10}{100}$ t)
$\frac{2000}{1000}$ = (1 + 0.10t)
2 = 1 + 0.10t
0.10t = 2 – 1
t = 10 years
(ii) With compounding:
Amount = p × (1 + r)^t
2000 = 1000(1 + $\frac{10}{100}$ )^t
$\frac{2000}{1000}$ = (1 + 0.1 )^t
2 = (1.1 )^t
Taking logs or estimating: t ≈ 7.27 years

8. The population of a city is rising by about 3% every year. If the current population is 1.5 crore, what is the expected population after 3 years?
Solution:
Population after 3 years = 1.5 × (1 + $\frac{3}{100}$ )³ crore
= 1.5 × (1 + 0.03)³ crore
= 1.5 × (1.03)³ crore
= 1.5 × 1.03 × 1.03 × 1.03 crore
= 1.639 crore

9. In a laboratory, the number of bacteria in a certain experiment increases at the rate of 2.5% per hour. Find the number of bacteria at the end of 2 hours if the initial count is 5,06,000.
Solution:
Total bacteria after 2 hours = 5,06,000 × (1 + $\frac{25}{1000}$ )²
= 5,06,000 × (1 + 0.025)²
= 5,06,000 × (1.025)²
= 5,06,000 × 1.025 × 1.025
= 5,31,616

Try it Out (Page 28 – 30)

1. The population of Bengaluru in 2025 is about 250% of its population in 2000. If the population in 2000 was 50 lakhs, what is the population in 2025?
Solution:
Population in 2000 = 50 lakh
Population in 2025 = 50 lakh + 250% of 50 lakh
= 50 lakh + 2.5 × 50 lakh
= 50 lakh + 125 lakh
= 175 lakh or 1.75 crore.

2. The population of the world in 2025 is about 8.2 billion. The populations of some countries in 2025 are given. Match them with their approximate percentage share of the worldwide population.
[Hint: Writing these numbers in the standard form and estimating can help].

Solution:
Germany = $\frac{83}{8200}$ × 100% = $\frac{83}{82}$ = 1.01% ≈ 1%

India = $\frac{1.46}{8.2}$ × 100% = $\frac{1460}{8200}$ × 100% = $\frac{1460}{82}$ = 17.8% ≈ 18%

Bangladesh = $\frac{175}{8200}$ × 100% = $\frac{175}{82}$ ≈ 2%

USA = $\frac{347}{8200}$ × 100% = $\frac{347}{82}$ ≈ 4%

3. The price of a mobile phone is ₹8,250. A GST of 18% is added to the price. Which of the following gives the final price of the phone including the GST?

Solution:
Mobile phone price = ₹8,250
GST = 18%
Final price including the GST = ₹8,250 + 18% of ₹8,250
= 8,250 + $\frac{18}{100}$ × 8250
= 8250 + 8250 × 0.18
= 8250 × (1 + 0.18)
= 8250 × 1.18
Therefore, correct options are:
(v) 8250 × 1.18
(vi) 8250 + 8250 × 0.18

4. The monthly percentage change in population (compared to the previous month) of mice in a lab is given: Month 1 change was +5%, Month 2 change was –2%, and Month 3 change was –3%. Which of the following statement(s) are true? The initial population is p.
(i) The population after three months was p × 0.05 × 0.02 × 0.03.
(ii) The population after three months was p × 1.05 × 0.98 × 0.97.
(iii) The population after three months was p + 0.05 – 0.02 – 0.03.
(iv) The population after three months was p.
(v) The population after three months was more than p.
(vi) The population after three months was less than p.
Solution:
Population after 3 months = p × (1 + $\frac{5}{100}$ ) (1 – $\frac{2}{100}$ ) (1 – $\frac{3}{100}$ )
= p × (1 + 0.05) × (1 – 0.02) × (1 – 0.03)
= p × 1.05 × 0.98 × 0.97
= 0.9987p
Therefore, the correct option are:
(ii) The population after three months was p × 1.05 × 0.98 × 0.97.
(vi) The population after three months was less than p.

5. A shopkeeper initially set the price of a product with a 35% profit margin. Due to poor sales, he decided to offer a 30% discount on the selling price. Will he make a profit or a loss? Give reasons for your answer.
Solution:
Let the CP be ₹100.
Profit = 35%
Initial SP = 100 + 35% of 100 = 100 + 35 = ₹135
Discount percentage = 30%
Discount = 30% of 135 = 0.30 × 135 = ₹40.5
Final SP = 135 − 40.5 = ₹94.5
Loss = 100 − 94.5 = ₹5.5
Therfore, the shopkeeper makes a loss.
Reason: Even though he first added a 35% profit, the 30% discount is applied on the higher price (₹135), which reduces the price more significantly. As a result, the final price becomes less than the cost price, leading to a loss.

6. What percentage of area is occupied by the region marked ‘E’ in the figure?

Solution:
Total area of the figure = 12 × 12 = 144 sq. units
The bottom-left square has $\frac{1}{4}$ of the total area.
Area of bottom-left square =
Area of region E = 18 sq. units
Percentage of region maked E = $\frac{18}{144}$ × 100% = 12.25%

7. What is 5% of 40? What is 40% of 5? What is 25% of 12? What is 12% of 25? What is 15% of 60? What is 60% of 15? What do you notice?
Can you make a general statement and justify it using algebra, comparing x% of y and y% of x?
Solution:
5% of 40 = $\frac{5}{100}$ × 40 = 2
40% of 5 = $\frac{40}{100}$ × 5 = 2
25% of 12 = $\frac{25}{100}$ × 12 = 3
12% of 25 = $\frac{12}{100}$ × 25 = 3
15% of 60 = $\frac{15}{100}$ × 60 = 9
60% of 15 = $\frac{60}{100}$ × 15 = 9
Observation: x% of y = y% of x
Justification: Let x and y be any numbers.
x% of y = $\frac{x}{100}$ × y
y% of x = $\frac{y}{100}$ × x
Since multiplication is commutative, $\frac{x}{100}$ × y = $\frac{y}{100}$ × x.

8. A school is organising an excursion for its students. 40% of them are Grade 8 students and the rest are Grade 9 students. Among these Grade 8 students, 60% are girls.
[Hint: Drawing a rough diagram can help].
(i) What percentage of the students going to the excursion are Grade 8 girls?
(ii) If the total number of students going to the excursion is 160, how many of them are Grade 8 girls?
Solution:
Let the total students be 100.
Number of Grade 8 students = 40% of 100 = $\frac{40}{100}$ × 100 = 40
Number of Grade 9 students = 60% of 100 = $\frac{60}{100}$ × 100 = 60
(i) Percentage of Grade 8 girls = 60% of 40 = $\frac{60}{100}$ × 40 = 24%
(ii) Number of Grade 8 girls = 24% of 160
= $\frac{24}{100}$ × 160 = 38.4
Therefore, there are about 38 Grade 8 girls.

9. A shopkeeper sells pencils at a price such that the selling price of 3 pencils is equal to the cost of 5 pencils. Does he make a profit or a loss? What is his profit or loss percentage?
Solution:
Let the CP of 1 pencil = ₹1
CP of 5 pencils = 5 × 1 = ₹5
SP of 3 pencils = ₹5
SP of 1 pencil = ₹ $\frac{5}{3}$
Profit per pencil = $\frac{5}{3}$ – 1 = $\frac{5 - 3}{3}$ = ₹ $\frac{2}{3}$
Profit % = $\frac{2 / 3}{1}$ × 100% = $\frac{2}{3}$ × 100 = $\frac{200}{3}$ = 66 $\frac{2}{3}$ %
Therefore, the shopkeeper makes a profit of 66 $\frac{2}{3}$ %.

10. The bus fares were increased by 3% last year and by 4% this year. What is the overall percentage price increase in the last 2 years?
Solution:
Let initial fare be ₹100.
After 3% increase: 100 × 1.03 = ₹103
After 4% increase: 103 × 1.04 = ₹107.12
Increase = 107.12 − 100 = 7.12
Percentage increase = $\frac{7.12}{100}$ × 100% = 7.12%

11. If the length of a rectangle is increased by 10% and the area is unchanged, by what percentage (exactly) does the breadth decrease by?
Solution:
Length = L
Breadth = B
Area = L × B
New length = 1.10L
New breadth = B′
Since area is unchanged, L × B = 1.10L × B′
B′ = $\frac{L \times B}{1.10 L}$ = $\frac{B}{1.10}$ = $\frac{100 \times B}{110}$ = $\frac{10 B}{11}$
Decrease in breadth = B − B′ = B – $\frac{10 B}{11}$ = $\frac{11 B - 10 B}{11}$ = $\frac{B}{11}$
Percentage decrease = $\frac{1 / 11}{1}$ × 100% = $\frac{1}{11}$ × 100% = $\frac{100}{11}$ % = 9 $\frac{1}{11}$ %.
Therefore, the breadth decreases by 9111% (exactly).

12. The percentage of ingredients in a 65 g chips packet is shown in the picture. Find out the weight each ingredient makes up in this packet.

Solution:
Potato = 70% of 65 g = 0.70 × 65 = 45.5 g
Veg oil = 24% of 65 g = 0.24 × 65 = 15.6 g
Salt = 3% of 65 g = 0.03 × 65 = 1.95 g
Spice = 3% of 65 g = 0.03 × 65 = 1.95 g
Verification: 45.5 + 15.6 + 1.95 + 1.95 = 65.05 g

13. Three shops sell the same items at the same price. The shops offer deals as follows:
Shop A: “Buy 1 and get 1 free”
Shop B: “Buy 2 and get 1 free”
Shop C: “Buy 3 and get 1 free”
Answer the following:
(i) If the price of one item is ₹100, what is the effective price per item in each shop? Arrange the shops from cheapest to costliest.
(ii) For each shop, calculate the percentage discount on the items.
[Hint: Compare the free items to the total items you receive.]
(iii) Suppose you need 4 items. Which shop would you choose? Why?
Solution:
(i) Effective price per item in Shop A = $\frac{100}{2}$ = ₹50
Effective price per item at Shop B = $\frac{200}{3}$ = ₹66 $\frac{2}{3}$
Effective price per item at Shop C = $\frac{300}{4}$ = ₹75
Shop A < Shop B < Shop C (cheapest to costliest)
(ii) Discount at Shop A = $\frac{1}{2}$ × 100% = 50%
Discount at Shop B = $\frac{1}{3}$ × 100% = $\frac{100}{3}$ = 33.3%
Discount at Shop C = $\frac{1}{4}$ × 100% = 25%
(iii) I need 4 items, so I would choose Shop A. I would pay for 2 items and get 2 items free.

14. In a room of 100 people, 99% are left-handed. How many left-handed people have to leave the room to bring that percentage down to 98%?
Solution:
Total people = 100
Left-handed persons = 99% of 100 = 99
Right-handed persons = 1
Let x left-handed people leave.
Left-handed remaining = 99 − x
Total people remaining = 100 − x
$\frac{99 - x}{100 - x}$ = 98%
$\frac{99 - x}{100 - x}$ = $\frac{98}{100}$
100(99 – x) = 98(100 – x)
9900 – 100x = 9800 – 98x
100x – 98x = 9900 – 9800
2x = 100
x = 50
Therefore, 50 left-handed people must leave.

15. Look at the following graph.

Based on the graph, which of the following statement(s) are valid?
(i) People in their twenties are the most computer-literate among all age groups.
(ii) Women lag behind in the ability to use computers across age groups.
(iii) There are more people in their twenties than teenagers.
(iv) More than a quarter of people in their thirties can use computers.
(v) Less than 1 in 10 aged 60 and above can use computers.
(vi) Half of the people in their twenties can use computers.
Solution:
(i) True
(ii) True
(iii) True
(iv) False
(v) True
(vi) False

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