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

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Class 8 Maths Ganita Prakash Part 2 Chapter 1 Solutions

Ganita Prakash Class 8 Chapter 1 Solutions Fractions in Disguise

Class 8 Maths Ganita Prakash Part 2 Chapter 1 Fractions in Disguise Solutions Question Answer

1.1 Fractions as Percentages

Expressing Fractions as Percentages

NCERT In-Text Questions (Pages 2-3)

Question 1.
Can you tell what percentage of the colour was made using yellow?

Solution:
It is clear that the part of the colour made using yellow is from \(\frac {75}{100}\) to \(\frac {100}{100}\)
Thus, the yellow part is \(\frac{100}{100}-\frac{75}{100}=\frac{25}{100}\)
Hence, 25% of the colour was made using yellow.

Question 2.
Try completing Method 3 by filling in the boxes.

Solution:

Figure it Out (Pages 3-4)

Question 1.
Express the following fractions as percentages.
(i) \(\frac {3}{5}\)
(ii) \(\frac {7}{14}\)
(iii) \(\frac {9}{20}\)
(iv) \(\frac {72}{150}\)
(v) \(\frac {1}{3}\)
(vi) \(\frac {5}{11}\)
Solution:

Question 2.
Nandini has 25 marbles, of which 15 are white. What percentage of her marbles are white?
(i) 10%
(ii) 15%
(iii) 25%
(iv) 60%
(v) 40%
(vi) None of these
Solution:
We are given that 15 out of 25 marbles are white, that is \(\frac {15}{25}\) marbles are white.
⇒ \(\frac{15}{25}=\frac{15 \times 4}{25 \times 4}=\frac{60}{100}\) = 60%
Hence, the correct answer is (iv) 60%.

Question 3.
In a school, 15 of the 80 students come to school by walking. What percentage of the students come by walking?
Solution:
We are given that 15 out of 80 students come to school by walking, that is \(\frac {15}{80}\) students come to school by walking.
⇒ \(\frac{15}{80}=\frac{15 \times 100}{80 \times 100}=\left(\frac{75}{4}\right) \times \frac{1}{100}=\frac{75}{4} \%\) or 18.75%
Hence, 18.75% students come to school by walking.

Question 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:
By observing the given positions of A, B, C, and D, we can align them with the given percentages as follows:
A = 38%
B = 55%
C = 72%
D = 93%

Question 5.
Pairs of quantities are shown below. Identify and write appropriate symbols ‘>’, ‘<’, ‘=’ in the blanks. Try to do it without calculations.
(i) 50% ___ 5%
(ii) \(\frac {5}{10}\) ___ 50%
(iii) \(\frac {3}{11}\) ___ 61%
(iv) 30% ___ \(\frac {1}{3}\)
Solution:
(i) Since 50 > 5
⇒ 50% > 5%

(ii) Since \(\frac{5}{10}=\frac{50}{100}\)
⇒ \(\frac {5}{10}\) = 50%

(iii) Since \(\frac{3}{10}=\frac{30}{100}<\frac{61}{100}\) and \(\frac{3}{10}>\frac{3}{11}\),
we have \(\frac {3}{11}\) < 61%.

(iv) Since \(\frac {1}{3}\) = 0.333… = \(\frac{33.33 \ldots}{100}>\frac{30}{100}\),
we have 30% < \(\frac {1}{3}\)

1.2 Percentage of Some Quantity

Free-hand Computations

NCERT In-Text Questions (Pages 7-8)

Question 1.
Try to calculate (without using pen and paper) the indicated percentages of the values shown in the table below. Write your answers in the table.

Solution:

Question 2.
How did you find these values? Discuss the methods with the class. Do you find anything interesting in the table?
Solution:
Do it yourself.

Question 3.
Using this understanding, mentally calculate how much 40% of the values in the table above would be.
Solution:

Question 4.
Using this observation, mentally calculate how much 15% of the values in the table would be.
Solution:
The 15% of the values in the table would be the sum of the values of 10% and 5%.

Question 5.
Suppose you have to mentally calculate the following percentages of some value: 75%, 90%, 70%, 55%. How would you do it? Discuss.
Solution:
Do it yourself.

The FDP Trio — Fractions, Decimals, and Percentages

NCERT In-Text Questions (Page 8)

Question 1.
Similarly, to find 10% of a quantity, what decimal value should be multiplied?
Solution:
Since 10% = \(\frac {10}{100}\) = 0.1, to find 10% of a quantity, the decimal value that needs to be multiplied is 0.1.

Question 2.
Complete the following table:

Solution:

Percentages Greater than 100

NCERT In-Text Questions (Pages 11-12)

Question 1.
On Days 5 and 6, his sales were ₹ 7800 and ₹ 9550, respectively. Calculate the percentage of the target achieved on these days.

Solution:
Day 5: Target = ₹ 5000, Sales = ₹ 7800
Percentage ol target achieved = (\(\frac {7800}{5000}\) × 100)% = 156%
Thus, he achieved 156% of his target on Day 5.
Day 6: Target = ₹ 5000, Sales = ₹ 9550
Percentage of target achieved = (\(\frac {9550}{5000}\) × 100)% = 191%
Thus, he achieved 191% of his target on Day 6.

Question 2.
Complete the table below. Mark the approximate locations in the following diagram.

Solution:

Figure it Out (Pages 12-14)

Estimate first before making any computations to solve the following questions. Try different methods, including mental computations.

Question 1.
Find the missing numbers. The first problem has been worked out.


Solution:
(ii) The whole (100%) is divided into 10 parts.
So, 1 part = \(\frac{100 \%}{10}=\frac{100}{10 \times 100}=\frac{10}{100}\) = 10%
Also, 100% corresponds to 90.
Therefore, 6 parts = 60% of 90
= 0.6 × 90
= 54
Therefore, the correct answer is

(iii) The whole (100%) is divided into 4 parts.
So, 1 part = \(\frac{100 \%}{4}=\frac{100}{4 \times 100}=\frac{25}{100}\) = 25%
Also, 100% corresponds to 140.
Therefore, 3 parts = 75% of 140
= 0.75 × 140
= 105
Therefore, the correct answer is

Question 2.
Find the value of the following and also draw their bar models.
(i) 25% of 160
(ii) 16% of 250
(iii) 62% of 360
(iv) 140% of 40
(v) 1% of 1 hour
(vi) 7% of 10 kg
Solution:
(i) 25% of 160 = \(\frac {25}{100}\) × 160
= \(\frac {1}{4}\) × 160
= 40

(ii) 16% of 250 = \(\frac {16}{100}\) × 250
= 0.16 × 250
= 40

(iii) 62% of 360 = \(\frac {62}{100}\) × 360
= 0.62 × 360
= 223.2

(iv) 140% of 40 = \(\frac {140}{100}\) × 40
= 1.4 × 40
= 56

(v) 1% of 1 hour (60 minutes) = \(\frac {1}{100}\) × 60 min = 0.6 minutes or 36 seconds

(vi) 7% of 10 kg = \(\frac {7}{100}\) × 10 = 0.7 kg or 700 g

Question 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:
Total orange paint = 60 ml
Red paint is \(\frac {3}{4}\) of the total orange paint.
So, red paint used = \(\frac {3}{4}\) × 60 ml
= 3 × 15
= 45 ml

Question 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 (Same percentage of a larger number is greater).
(ii) 37% of 148 < 73% of 148 (A higher percentage of the same number is greater).
(iii) 29% of 43 < 92% of 110 (Both the percentage and the number are smaller on the left).
(iv) 30% of 40 < 40% of 50 (Both the percentage and the number are smaller on the left).
(v) We can estimate that 45% of 200 is between 80 and 100, and 10% of 490 is between 40 (10% of 400) and 50 (10% of 500).
So, 45% of 200 > 10% of 490.
(vi) Since 24% of 80 > 24% of 64 (Same percentage of a larger number is greater).
So, 30% of 80 > 24% of 64

Question 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) If 30% of k is 70, then
60% of k = Double of 30% of k
= 2 × 70
= 140
90% of k = Three times 30% of k
= 3 × 70
= 210
120% of k = Four times 30% of k
= 4 × 70
= 280

(ii) If 100% of m is 215, then
10% of m = One tenth of 100% of m
= \(\frac {1}{10}\) × 215
= 21.5
1% of m = One tenth of 10% of m
= \(\frac {1}{10}\) × 21.5
= 2.15
6% of m = Six times 1% of m
= 2.15 × 6
= 12.9

(iii) If 90% of n is 270, then
9% of n = One tenth of 90% of n
= \(\frac {1}{10}\) × 270
= 27
18% of n = Double of 9% of n
= 2 × 27
= 54
100% of n = 300
Since 90% of n is 270
⇒ 10% of n is \(\frac {270}{9}\) = 30
⇒ 100% of n is 30 × 10 = 300

(iv) Do it yourself.

Question 6.
Fill in the blanks:
(i) 3 is ___ % of 300.
(ii) ___ is 40% of 4.
(iii) 40 is 80% of ___.
Solution:
(i) Let the unknown percentage be x.
According to the question,
3 = \(\frac {x}{100}\) × 300
Simplifying, 3 = 3x
⇒ x = 1
Thus, 3 is 1% of 300.

(ii) Let the unknown value be y.
According to the question,
y = \(\frac {40}{100}\) × 4
Simplifying, y = 0.4 × 4 = 1.6
⇒ y = 1.6
Thus, 1.6 is 40% of 4.

(iii) Let the unknown number be z.
According to the question,
40 = \(\frac {80}{100}\) × z
Simplifying, 40 = 0.8z
⇒ z = 50
Thus, 40 is 80% of 50.

Question 7.
Is 10% of a day longer than 1% of a week? Create such questions and challenge your peers.
Solution:
10% of a day: Since 1 day = 24 hours
∴ 10% of 24 hours = 0.1 × 24 = 2.4 hours
1% of a week: Since 1 week = 7 days
= 7 × 24 hours
= 168 hours
∴ 1% of 168 hours = 0.01 × 168 = 1.68 hours
Therefore, 10% of a day (2.4 hours) is longer than 1% of a week (1.68 hours).
Create such questions by yourself.

Question 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: Fodder given = 2 units, Fodder eaten = 1 unit.
Percentage of fodder eaten = \(\frac {1}{2}\) × 100 = 50%
Day 2: Fodder given = 3 units, Fodder eaten = 2 units.
Percentage of fodder eaten = \(\frac {2}{3}\) × 100 ~ 66.67%
Day 3: Fodder given = 4 units, Fodder eaten = 3 units.
Percentage of fodder eaten = \(\frac {3}{4}\) × 100 = 75%
Day 99: Fodder given = 100 units, Fodder eaten = 99 units.
Percentage of fodder eaten = \(\frac {99}{100}\) × 100 = 99%
In general, the percentage of fodder eaten is \(\left(\frac{n}{n+1} \times 100\right) \%\), where n is the day number.
As n increases, \(\frac{n}{n+1}\) gets closer to 1, so the percentage of fodder eaten approaches 100%, but it never reaches 100% exactly.

Question 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:
Let the total number of days required to pick berries from 100% of the plantation be x.
Since 20% of the plantation is completed in 18 days, the time taken is directly proportional to the area covered.
\(\frac{20 \%}{18 \text { days }}=\frac{100 \%}{x \text { days }}\)
⇒ x = \(\frac{18 \times 100}{20}\)
⇒ x = 18 × 5 = 90 days
Therefore, the workers will take 90 days to pick coffee berries from the entire plantation.
Reason for the assumption: The assumption that the rate of work remains the same is necessary because time is proportional to work only when the speed of working and the number of workers do not change. If the workers slow down due to fatigue, speed up, or if the number of workers changes, the time taken will no longer be directly proportional to the area covered.

Question 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 last?

Solution:
Total time = 90 minutes
Warm-up time = 10% of 90
= 0.10 × 90
= 9 minutes
Play time = 80% of 90
= 0.80 × 90
= 72 minutes
Cool down time = 10% of 90
= 0.10 × 90
= 9 minutes

Question 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:
Estimated world population in 2025 = 8.2 billion (approx.)
90% of 8.2 billion = 0.90 × 8.2 = 7.38 billion
Thus, approximately 7.38 billion people live in the Northern Hemisphere based on the 2025 world population.

Question 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:
The ingredients for halwa for 4 people are given in the following proportions:
Rava = 40%, Sugar = 40%, and Ghee = 20%.
(i) When the number of people increases, the total quantity of the ingredients increases, but the proportion of each ingredient remains the same.
Therefore, for 8 people, the proportions will be:
Rava = 40%, Sugar = 40%, and Ghee = 20%.

(ii) Total weight of the ingredients = 2 kg
Rava = 40% of 2 kg = \(\frac {40}{100}\) × 2 = 0.8 kg (or 800 g)
Sugar = 40% of 2 kg = \(\frac {40}{100}\) × 2 = 0.8 kg (or 800 g)
Ghee = 20% of 2 kg = \(\frac {20}{100}\) × 2 = 0.4 kg (or 400 g)
Therefore, in 2 kg of ingredients, there are 800 g rava, 800 g sugar, and 400 g ghee.

1.3 Using Percentages

To Compare Proportions

NCERT In-Text Questions (Page 15)

Question 1.
Complete this table by calculating the percentages to answer the questions:

Solution:

Question 2.
Check if the percentages of each product add up to 100.
Solution:
DEF: 66 + 20 + 8 + 6 = 100
Zacni: 68 + 16 + 10 + 6 = 100
Thus, the percentages of each product add up to 100.

Profit and Loss

NCERT In-Text Questions (Pages 17-18)

Question 1.
Find the profit percentage of the wholesaler and the manufacturer.
Solution:
Manufacturer:
Cost Price (CP) = ₹ 230, Selling Price (SP) = ₹ 253
Profit = SP – CP
= 253 – 230
= ₹ 23
Profit percentage = \(\frac {Profit}{CP}\) × 100
= \(\frac {23}{230}\) × 100
= 10%
Therefore, the profit percentage of the manufacturer is 10%.
Wholesaler:
Cost Price (CP) = ₹ 253, Selling Price (SP) = ₹ 300
Profit = SP – CP
= 300 – 253
= ₹ 47
Profit percentage = \(\frac {47}{253}\) × 100 ~ 18.58%
Therefore, the profit percentage of the wholesaler is 18.58%.

Question 2.
Shambhavi owns a stationery shop. She procures 200-page notebooks at ₹ 36 per book. She sells them with a profit margin of 20%. Find the selling price.
Solution:
Cost Price (CP) of one notebook = ₹ 36
Profit = 20% of CP
= \(\frac {20}{100}\) × 36 = ₹ 7.20
Selling Price (SP) = CP + Profit
= 36 + 7.20
= ₹ 43.20
Therefore, the selling price of one notebook is ₹ 43,20.

Question 3.
She sells crayon boxes at ₹ 50 per box with a profit margin of 25%. How much did Shambhavi buy them from the wholesaler?
Solution:
Selling Price (SP) = ₹ 50
Profit = 25% of Cost Price (CP)
So, SP = CP + 25% of CP
⇒ 50 = \({CP}\left(1+\frac{25}{100}\right)\)
⇒ 50 = \(\frac {125}{100}\) × CP
⇒ CP = \(\frac{50 \times 100}{125}\)
⇒ CP = ₹ 40
Therefore, Shambhavi bought each crayon box for ₹ 40.

Question 4.
Could we have just calculated the loss percentage per kg instead? Would it be the same?
Solution:
Yes, we could have calculated the loss percentage per kg instead, and it would be the same, as
Cost Price (CP) per kg = ₹ 35
Selling Price (SP) per kg = 300 ÷ 10 = ₹ 30
So, loss = ₹ 5, and the loss percentage = \(\frac {5}{35}\) × 100 = 14.28%

Question 5.
Due to heavy rains, Snehal could not transport strawberries to Hyderabad from his farm in Panchgani. He sells some of his stock at ₹ 80 per kg with a 12% loss. What is the cost price?
Solution:
Selling Price (SP) per kg = ₹ 80, Loss = 12%
When there is a loss, SP = \(\frac{100-{loss} \%}{100} \times \mathrm{CP}\)
⇒ 80 = 88% of CP
⇒ CP = \(\frac{80 \times 100}{88}\)
⇒ CP = ₹ 90.91 (approximately)
Hence, the cost price of strawberries is ₹ 90.91 per kg (approximately).

Question 6.
A utensil store is offering a 35% discount on the cooker with an MRP of ₹ 1800. What is the selling price? If the cost price was ₹ 900, what is the percentage profit made after the sale?
Solution:
Marked Price (MRP) = ₹ 1800, discount = 35%
Discount = 35% of 1800
= \(\frac {35}{100}\) × 1800
= ₹ 630
Selling price (SP) = MRP – Discount
= 1800 – 630
= ₹ 1170
Next, cost price (CP) = ₹ 900, selling price (SP) = ₹ 1170
⇒ Profit = SP – CP
= 1170 – 900
= ₹ 270
Percentage profit = \(\frac {Profit}{CP}\) × 100
= \(\frac {270}{900}\) × 100
= 30%
Thus, selling price = ₹ 1170, percentage profit = 30%

Taxes

NCERT In-Text Questions (Page 19)

Question 1.
Check if the calculations are correct in the bill shown.

Solution:
1 bulb is for ₹ 150.
So, the price of 3 bulbs is 3 × 150 = ₹ 450
CGST is 9% of ₹ 450 = 0.09 × 450 = ₹ 40.50
SGST is 9% of ₹ 450 = 0.09 × 450 = ₹ 40.50
Total = 450 + 40.50 + 40.50 = ₹ 531.00
Hence, the calculations are correct in the bill.

Question 2.
You may share any bills you have at home with the class. Observe the different elements present in the bills. Are there any similarities or differences in these bills?
Solution:
Do it yourself.

Figure it Out (Pages 19-20)

Question 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:
Cost price (CP) = ₹ 75, Selling price (SP) = ₹ 110
Profit = SP – CP
= 110 – 75
= ₹ 35
Profit margin (with respect to cost) = \(\frac {35}{75}\) × 100 = 46.67%

Question 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 (CP) = ₹ 475
Required profit margin = 50%
Profit = 50% of 475
= \(\frac {50}{100}\) × 475
= ₹ 237.50
Selling price (SP) = CP + profit
= 475 + 237.50
= ₹ 712.50
Therefore, the selling price of the chair is ₹ 712.50.

Question 3.
The total sales of a company (also called revenue) were ₹ 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:
Total sales (Revenue) = ₹ 2.5 crore
Profit margin = 25%
Since, profit = Total sales (Revenue) × Profit margin
And, expenditure = Revenue – profit
∴ Profit = 2.5 × 25% = 0.625 crore
∴ Expenditure = 2.5 – 0.625 = 1.875 crore.

Question 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:
Original Price = ₹ 300, discount = 25%
Discounted amount = \(\frac {25}{100}\) × 300 = ₹ 75
Amount to be paid = 300 – 75 = ₹ 225
Therefore, Anwar will pay ₹ 225.

Question 5.
The petrol price in 2015 was ₹ 60 and ₹ 100 in 2025. What is the percentage increase in the price of petrol?
(i) 50%
(ii) 40%
(iii) 60%
(iv) 66.66%
(v) 140%
(vi) 160.66%
Solution:
Old Price (2015) = ₹ 60
New price (2025) = ₹ 100
Increase in price = 100 – 60 = ₹ 40
Percentage increase = \(\frac {40}{60}\) × 100 = 66.66%
Therefore, the correct option is (iv) 66.66%.

Question 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:
Selling Price (after discount) = ₹ 4,40,000
Discount = 15%
So, ₹ 4,40,000 is 85% of the original price.
Original Price = \(\frac{4,40,000 \times 100}{85}\) = ₹ 5,17,647 (approximately)

Question 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 votes = 1600, votes obtained by winner = 500
Percentage of votes = \(\frac {500}{1600}\) × 100 = 31.25%
Since no one else got 500 or more votes, at least 4 candidates must have contested.
Thus, the winner got 31.25% of the votes, and the minimum number of candidates that must have contested is 4.

Question 8.
The price of 1 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:
Price in 2024 = ₹ 38, price in 2025 = ₹ 42
Increase in price = 42 – 38 = ₹ 4
Inflation rate = \(\frac {4}{38}\) × 100 ~ 10.53%

Question 9.
A number increased by 20% becomes 90. What is the number?
Solution:
Let the number be x.
Increase = 20% of x = 0.2x
New number after increase, x + 0.2x = 1.2x
Therefore, 1.2x = 90 (Given)
⇒ x = 75
Hence, the required number is 75.

Question 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:
Selling price of each buffalo = ₹ 80,000
The first buffalo was sold at 5% profit, which means the selling price is 105% of the cost price.
Therefore, CP = \(\frac{80,000 \times 100}{105}\) ~ ₹ 76,190
The second buffalo was sold at a 10% loss, which means the selling price is 90% of the cost price.
Therefore, CP = \(\frac{80,000 \times 100}{90}\) ~ ₹ 88,889
Total CP ~ 76,190 + 88,889 = ₹ 1,65,079
Total SP = 80,000 + 80,000 = ₹ 1,60,000
Loss = 1,65,079 – 1,60,000 = ₹ 5,079
Hence, his overall loss is approximately ₹ 5,079.

Question 11.
The population of elephants in a national park increased by 5% in the last decade. If the population of the elephants over the 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:
Population last decade = p
Increase = 5%, that is 0.05 of p.
New population = p + 0.05p = 1.05p
Hence, the correct option is (iv) p × 1.05.

Question 12.
Which of the following statements means 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:
A fall of 85% means only 15% remains.
Hence, the correct option is (iii) The demand now is 15% of the demand a decade ago.

Growth and Compounding

Figure it Out (Pages 22-23)

Question 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:
Without compounding:
Interest each year = 10% of 20000
= 0.1 × 20000
= ₹ 2000
For 2 years, total interest = 2 × 2000 = ₹ 4000
Total amount after 2 years = 20000 + 4000 = ₹ 24,000
With compounding annually:
Year 1: Interest = 10% of 20000 = ₹ 2000
Amount after Year 1 = 20000 + 2000 = ₹ 22,000
Year 2: Interest = 10% of 22000 = 0.1 × 22000 = ₹ 2200
Amount after Year 2 = 22000 + 2200 = ₹ 24,200
Difference = ₹ 24,200 – ₹ 24,000 = ₹ 200 extra with compounding.
Percentage gain (Without Compounding):
Total gain = 10% + 10% = 20%
The amount becomes 120% of the principal.
Percentage gain (With Compounding):
After 1 year, the principal becomes 110%
After 2 years, the principal becomes 110% of 110% = 110% × 110% = 121%
So, total gain = 21%
The amount becomes 121% of the principal.
Thus, with compounding, one gets ₹ 200 or 1% more on the principal amount of ₹ 20,000.

Question 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:
Without compounding:
Interest per year = 5% of 20000
= 0.05 × 20000
= ₹ 1000
For 4 years, total interest = 4 × 1000 = ₹ 4000
Total amount after 4 years = 20000 + 4000 = ₹ 24,000
With compounding annually:
Year 1: Principal = ₹ 20,000, interest = 5% of ₹ 20,000 = ₹ 1000
After Year 1, amount = ₹ 20,000 + ₹ 1000 = ₹ 21,000
Year 2: Principal = ₹ 21,000, interest = 5% of ₹ 21,000
= 0.05 × 21000
= ₹ 1050
After Year 2, total amount = ₹ 21,000 + ₹ 1050 = ₹ 22,050
Year 3: Principal = ₹ 22,050, interest = 5% of ₹ 22,050
= 0.05 × ₹ 22,050
= ₹ 1102.5
After Year 3, total amount = ₹ 22,050 + ₹ 1102.5 = ₹ 23,152.50
Year 4: Principal = ₹ 23,152.5, interest = 5% of ₹ 23,152.50
= 0.05 × ₹ 23,152.5
= ₹ 1157.625
After Year 4, total amount = ₹ 23,152.5 + ₹ 1157.625 = ₹ 24,310.125
Thus, if one deposits ₹ 20,000 for 4 years, he/she would get approximately ₹ 310 more when deposited with compounding.

Questions 3.
Do you observe anything interesting in the solutions of the two questions above? Share and discuss.
Solution:
Do it yourself.

With Compounding

NCERT In-Text Questions (Page 24)

Question 1.
Suppose we want to know the expression/formula to find the total interest amount gained at the end of the maturity period. What would be the formula for each of the two options?
Solution:
Without compounding:
The total amount after t years is p(1 + rt), where p is the principal and r is the rate of interest per annum.
So, the total interest earned is p(1 + rt) – p = prt
Hence, the total interest earned at the end of the maturity period is prt.
With compounding annually:
The total amount after t years is p(1 + r)^t, where p is the principal and r is the rate of interest per annum.
So, the total interest earned is p(1 + r)^t – p = p[(1 + r)^t – 1]
Hence, the total interest earned at the end of the maturity period is p[(1 + r)^t – 1].

Figure it Out (Page 24)

Question 4.
Jasmine invests an amount ‘p’ for 4 years at an interest of 6% p.a. Which of the following expressions describes the total amount she will get after 4 years when compounding is not done?
(i) p × 6 × 4
(ii) p × 0.6 × 4
(iii) p × \(\frac {0.6}{100}\) × 4
(iv) p × \(\frac {0.06}{100}\) × 4
(v) p × 1.6 × 4
(vi) p × 1.06 × 4
(vii) p + (p × 0.06 × 4)
Solution:
Principal (p) = p
Rate (r) = 6% = 0.06
Time (t) = 4 years
Since compounding is not done,
Simple Interest: p × r × t = p × 0.06 × 4
Total Amount after 4 years = p + (p × 0.06 × 4)
Hence, the correct option is (vii).

Question 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 were compounded?
Solution:
Given: Principal (p) = ₹ 50,000
Rate of interest (r) = 7% = 0.07
Time (t) = 3 years
Interest without compounding = p × r × t
= 50,000 × 0.07 × 3
= ₹ 10,500
Interest with compounding = p(1 + r)^t – p
= 50,000(1.07)³ – 50,000
= 50,000 × 1.225043 – 50,000
~ ₹ 11,252
Extra interest due to compounding = 11,252 – 10,500 = ₹ 752

Question 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:
For both Giridhar and Raghava,
Principal (p) = ₹ 12,500
Time (t) = 3 years
For Giridhar (Without compounding):
Rate (r) = 12% = 0.12
Interest = p × r × t
= 12,500 × 0.12 × 3
= ₹ 4500
For Raghava (Compounded annually):
Rate (r) = 10% = 0.10
Interest = p(1 + r)^t – p
= 12,500(1.10)³ – 12,500
= 12,500 × 1.331 – 12,500
= 16,637.50 – 12,500
= ₹ 4137.50
Difference in interest = 4500 – 4137.50 = ₹ 362.50
Hence, Giridhar pays more interest of ₹ 362.50.

Question 7.
Consider an amount of ₹ 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:
Given: p = ₹ 1000, r = 10% = 0.10
Without compounding:
Let the amount double in t years.
Amount doubles when total interest = ₹ 1000.
Interest = p × r × t
⇒ 1000 = 1000 × 0.10 × t
⇒ t = 10 years
With compounding:
Let the amount double in x years, then we have
2p = p(1 + 0.1)^x
⇒ 2p = p(1.1)^x
⇒ 2 = (1.1)^x
To find the value of x by the trial and error method,
(1.1)⁵ = 1.61 < 2
(1.1)⁷ = 1.95 < 2
(1.1)⁸ = 2.14 > 2
So, the amount doubles in about 7.5 years.
Yes, compounding is an example of exponential growth (growth on growth), while not-compounding is a linear growth (fixed growth based on the original principal).

Question 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:
Present population (P0) = 1.5 crore
Rate of increase (r) = 3% = 0.03 per year
Time (t) = 3 years
Population increases every year on the increased population, so compounding applies.
Let the final population be (P).
P = P0(1 + r)^t
= 1.5(1 + 0.03)³
= 1.5(1.03)³ [∵ (1.03)³ = 1.093]
= 1.5 × 1.093
= 1.6395 crore
Hence, the expected population after 3 years is approximately 1.64 crore.

Question 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:
Initial number of bacteria (N₀) = 5,06,000
Rate of increase (r) = 2.5% = 0.025 per hour
Time (t) = 2 hours
Since the number of bacteria increases every hour on the increased count, compounding applies.
Let the number of bacteria after 2 hours be N.
N = N₀ (1 + r)^t
= 5,06,000(1 + 0.025)²
= 5,06,000(1.025)² [∵ (1.025)² = 1.050625]
= 5,06,000 × 1.050625
= 5,31,616.25
Hence, the number of bacteria after 2 hours is ~ 5,31,616.

Tricky Percentages

NCERT In-Text Questions (Pages 25-27)

Question 1.
You have won a contest. The organisers offer you two options to choose from:
Option A: You deposit ₹ 100, and you get back ₹ 300.
Option B: You deposit ₹ 1000, and you get back ₹ 1500.
What is the percentage gain each option gives? You can choose any option only once. Which option would you choose? Why?
Solution:
Option A:
Deposit = ₹ 100, Amount received = ₹ 300
Gain = 300 – 100 = ₹ 200
Percentage Gain = \(\frac {200}{100}\) × 100 = 200%
Option B:
Deposit = ₹ 1000, Amount received = ₹ 1500
Gain = 1500 – 1000 = ₹ 500
Percentage Gain = \(\frac {500}{1000}\) × 100 = 50%

Question 2.
A provision store is offering a stock clearance sale. Customers can choose one of the two options—20% discount or ₹ 50 discount—for any purchase above ₹ 150. Which option would you choose if you want to:
(i) buy items worth ₹ 180
(ii) buy items worth ₹ 225
(iii) buy items worth ₹ 300
Solution:
(i) Purchase worth ₹ 180
Option A (20% discount):
20% of 180 = 0.20 × 180 = ₹ 36
Option B (₹ 50 discount): ₹ 50
Better option: (B) ₹ 50 discount (greater savings)

(ii) Purchase worth ₹ 225
Option A (20% discount):
20% of 225 = 0.20 × 225 = ₹ 45
Option B (₹ 50 discount): ₹ 50
Better option: (B) ₹ 50 discount (greater savings)

(iii) Purchase worth ₹ 300
Option A (20% discount):
20% of 300 = 0.20 × 300 = ₹ 60
Option B (₹ 50 discount): ₹ 50
Better option: (A) ₹ 60 (greater savings)

Question 3.
Ariba and Arun have some marbles. Ariba says, “The number of marbles with me is 120% of the marbles Arun has”. What would be an appropriate statement Arun could make comparing the number of marbles he has with Ariba’s?
Solution:
Ariba says her marbles are 120% of Arun’s marbles.
Let the number of marbles with Arun be 100.
Then, the number of marbles with Ariba = 120% of 100 = 120
Arun has 100 marbles, while Ariba has 120 marbles.
So, Arun’s marbles as a percentage of Ariba’s marbles = \(\frac {100}{120}\) × 100
= \(\frac {5}{6}\) × 100
= 83\(\frac {1}{3}\)%
Appropriate Statement: Arun could say, “The number of marbles with me is 83\(\frac {1}{3}\)% of Ariba’s marbles.”

Figure it Out (Pages 28-30)

Question 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 lakhs
Population percentage in 2025 ~ 250% of the population of 2000
So, population in 2025 = \(\frac {250}{100}\) × 50 lakhs
= 2.5 × 50,00,000
= 1,25,00,000
Hence, the population in 2025 is 1.25 crore.

Question 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:
Population as a percentage of the world population (8.2 billion):
Germany = \(\frac {83,000,000}{8,200,000,000}\) ~ 0.0101 ~ 1%
India = \(\frac {1,460,000,000}{8,200,000,000}\) ~ 0.178 ~ 18%
Bangladesh = \(\frac {175,000,000}{8,200,000,000}\) ~ 0.0213 ~ 2%
USA = \(\frac {347,000,000}{8,200,000,000}\) ~ 0.0423 ~ 4%

Question 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?
(i) 8250 + 18
(ii) 8250 + 1800
(iii) 8250 + \(\frac {18}{100}\)
(iv) 8250 × 18
(v) 8250 × 1.18
(vi) 8250 + 8250 × 0.18
(vii) 1.8 × 8250
Solution:
Final Price = Price + (Price × GST)
= 8250 + 8250 × 0.18
= 8250 × (1 + 0.18)
= 8250 × 1.18
Hence, the correct options are option (v) 8250 × 1.18 and (vi) 8250 + 8250 × 0.18.

Question 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 statements 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:
The initial population is p.
After Month 1: p × 1.05
After Month 2: p × 1.05 × 0.98
After Month 3: p × 1.05 × 0.98 × 0.97
And, p × 1.05 × 0.98 × 0.97 < p
Hence, the correct options 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.

Question 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 cost price be C, the marked price be M, and the selling price after discount be S.
Profit margin: M = C + 0.35C = 1.35C
Discount 30%: S = M – 0.3M
= 0.7M
= 0.7 × 1.35C
= 0.945C
Comparing S with C: 0.945C < C → Loss
Now, Loss % = \(\frac{\mathrm{C}-\mathrm{S}}{\mathrm{C}} \times 100\)
= \(\frac{C-0.945 C}{C} \times 100\)
= 5.5%

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

Solution:
The region has occupied one-eighth of the figure.
\(\frac{1}{8}=\frac{100}{8} \%\) = 12.5%

Question 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
In every pair, x% of y = y% of x.
Let x and y be any two numbers.
x% of y = \(\frac{x}{100} \times y=\frac{x y}{100}\)
y% of x = \(\frac{y}{100} \times x=\frac{y x}{100}\)
Since multiplication is commutative (xy = yx),
x% of y = y% of x

Question 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:
(i) Grade 8 girls = 60% of 40%
= \(\frac {60}{100}\) × 40%
= 24%
∴ 24% of the students are Grade 8 girls.

(ii) Total students = 160
24% of 160 = \(\frac {24}{100}\) × 160 = 38.4
Since the number of students must be a whole number, there are 38 students in Grade 8, girls.

Question 9.
A shopkeeper sells pencils at a price such that the price of 3 pencils equals the cost of 5 pencils. Does he make a profit or a loss? What is his profit or loss percentage?
Solution:
Let the cost price of 1 pencil be ₹ 1
Then, the cost price of 5 pencils = ₹ 5
Given that selling price of 3 pencils = cost price of 5 pencils = ₹ 5
So, selling price of 1 pencil = ₹ \(\frac {5}{3}\)
Since SP > CP, there is a profit.
Profit per pencil = \(\frac {5}{3}\) – 1 = \(\frac {2}{3}\)
Profit percentage = \(\frac{\frac{2}{3}}{1} \times 100=\frac{200}{3} \%=66 \frac{2}{3} \%\)
Hence, the shopkeeper makes a profit of 66\(\frac {2}{3}\)%.

Question 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 the original bus fare be ₹ 100.
After a 3% increase last year:
New fare = 100 + 3 = ₹ 103
After a further 4% increase this year:
New fare = 103 + 4% of 103
= 103 + 4.12
= ₹ 107.12
Overall increase in 2 years = 107.12 – 100 = ₹ 7.12
∴ Overall percentage increase = \(\frac {7.12}{100}\) × 100 = 7.12%
Hence, the overall percentage price increase in the last 2 years is 7.12%.

Question 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:
Let the original length and breadth be L and B, respectively.
Original area = LB
Length increased by 10%
⇒ New length = 1.1L
Since area is unchanged, 1.1L × B’ = LB
B’ = \(\frac{\mathrm{LB}}{1.1 \mathrm{~L}}=\frac{\mathrm{B}}{1.1}=\frac{10}{11} \mathrm{~B}\)
Decrease in breadth = \(\mathrm{B}-\frac{10}{11} \mathrm{~B}=\frac{1}{11} \mathrm{~B}\)
Percentage decrease = \(\frac{\frac{1}{11} \mathrm{~B}}{\mathrm{~B}} \times 100=\frac{100}{11} \%=9 \frac{1}{11} \%\)
Hence, the breadth decreases by 9\(\frac {1}{11}\)%

Question 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:
Weight of the packet = 65 g
Potato = 70% of 65 g
= \(\frac {70}{100}\) × 65
= 45.5 g
Vegetable oil = 24% of 65 g
= \(\frac {24}{100}\) × 65
= 15.6 g
Salt = 3% of 65
= \(\frac {3}{100}\) × 65
= 1.95 g
Spices = 3% of 65
= \(\frac {3}{100}\) × 65
= 1.95 g
Hence, the weights of the ingredients:
Potato = 45.5 g, Vegetable oil = 15.6 g, Salt = 1.95 g, and Spices = 1.95 g.

Question 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) The price of one item is ₹ 100.
Shop A: Buy 1 and get 1 free → Pay for 1 and get 2
Total cost = ₹ 100 for 2 items
Effective price per item = \(\frac {100}{2}\) = ₹ 50
Shop B: Buy 2 and get 1 free → Pay for 2 and get 3
Total cost = ₹ 200 for 3 items
Effective price per item = \(\frac {200}{3}\) ~ ₹ 66.67
Shop C: Buy 3 and get 1 free → Pay for 3 and get 4
Total cost = ₹ 300 for 4 items
Effective price per item = \(\frac {300}{4}\) = ₹ 75
Cheapest to costliest: Shop A, Shop B, Shop C

(ii) Percentage discount
Shop A: Free items = 1, total items = 2
Discount = \(\frac {1}{2}\) × 100 = 50%
Shop B: Free items = 1, total items = 3
Discount = \(\frac {1}{3}\) × 100 = 33\(\frac {1}{3}\)%
Shop C: Free items = 1, total items = 4
Discount = \(\frac {1}{4}\) × 100 = 25%

(iii) If 4 items are needed
Shop A: Buy 2 items + 2 items free → Pay ₹ 200 (for 4 items)
Shop B: Buy 3 items + 1 item free → Pay ₹ 300
Shop C: Buy 3 items + 1 item free → Pay ₹ 300
Hence, Shop A should be chosen as it gives 4 items for the least cost (₹ 200).

Question 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:
Initially, left-handed people = 99
Right-handed people = 1
Let the left-handed people leave the room.
Then remaining left-handed people = 99 – x
Remaining people = 100 – x
Given the new percentage of left-handed people = 98%
\(\frac{99-x}{100-x}=\frac{98}{100}\)
⇒ 99 – x = 98 – 0.98x
⇒ 1 = 0.02x
⇒ x = 50
Hence, 50 left-handed people must leave the room.

Question 15.
Look at the following graph.

Based on the graph, which of the following statements are valid?
(i) People in their twenties are the most computer-literate among all age groups.
(ii) Women lag 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:
Based on the given graph, the following are the valid statements:
(i) People in their twenties are the most computer-literate among all age groups.
Looking at the “Twenties” category, both the female (26%) and the male (37%) bars are the longest compared to those of any other age group.

(ii) Women lag in the ability to use computers across age groups.
In every single category (from Children to Seniors), the dark blue bar (Female) is shorter than the yellow bar (Male).

(v) Less than 1 in 10 aged 60 and above can use computers.
Seniors (60+) show that 2% for females and 4% for males can use computers, and “1 in 10” is 10%.