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

Class 8 Maths Fractions in Disguise Solutions

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

NCERT (Page 2-3)

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

Solution:
We observe the given bar model shows
the equivalence between \(\frac{3}{4}\) and 75%.
Here, four block equally divides into 25%.
The percentage of grey colour in bar model = 25%

Question 2.
(i) Try completing this by filling the boxes.

Solution:
Here, in the ‘bar model’ that relates fractions to percentages. Since, the whole (100%) is divided into 5 equal parts, we can calculate the value of each part to fill in those boxes.

(ii) Yes, it is the same as finding \(\frac{2}{5}\) of 100.

Figure it Out (NCERT Page 3 – 4)

Question 1.
Express the following fractions as percentages.
(i) \(\frac{3}{5}\)
Solution:
The following fractions as percentage are as follows.
Let x be the required percentage.
We have, \(\frac{3}{5}=\frac{x}{100}\)
⇒ x = \(\frac{3}{5}\) × 100 = 60%

(ii) \(\frac{7}{14}\)
Solution:
Do same as part (i). Ans. 50%

(iii) \(\frac{9}{20}\)
Solution:
Do same as part (i). Ans. 45%.

(iv) \(\frac{72}{150}\)
Solution:
Do same as part (i). Ans. 48%

(iv) \(\frac{1}{3}\)
Solution:
Do same as part (i). Ans. 33\(\frac{1}{3}\)%

(iv) \(\frac{5}{11}\)
Solution:
Do same as part (i). Ans. 45\(\frac{5}{11}\)%

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:
(iv) Given, the total number of marbles = 25
and the number white marbles = 15
∴ The portion of white marbles = \(\frac{15}{25}\)
∴ The percentage of white marbles = \(\frac{15 \times 4}{25 \times 4}=\frac{60}{100}\) = 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:
Given, the total number of students in a school = 80 and the number of students who come to school by
walking = 15
∴ The portion of students who come to school by walking = \(\frac{15}{80}\)
.‘.The required percentage = \(\frac{15}{80}\) × 100
= \(\frac{150}{8}\)%
= \(\frac{75}{4}\)%
= 18\(\frac{3}{4}\)%

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:
The figure is a number line from start 0% to finish 100%. Here, Runner A is the closest to the start line, having completed the least distance. The lowest available option greater than 20% is 38%.
Runner B is slightly further along. The next available option is 55%.
Runner C is past the halfway point and closer to finish than start the next option is 72%.
Runner D is the closest to the finish line, having completed the most distance. The next option is 84%.
Hence, A → (iii), B → (i), C → (iv) and D → (v).

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% Solution: We have, 50% and 5%. 50% > 5%

(ii) \(\frac{5}{10}\) ____ 50%
Solution:
We have, \(\frac{5}{10}=\frac{5 \times 10}{10 \times 10}\)
= \(\frac{50}{100}\) = 50%
\(\frac{5}{10}\) = 50%

(iii) \(\frac{3}{11}\) _____ 61%
Solution:
We have, \(\frac{3}{11}=\frac{3}{11}\) × 100
= \(\frac{300}{11}\)
= 27\(\frac{3}{11}\)

(iv) 30% ____ \(\frac{1}{3}\)
Solution:
We have, \(\frac{1}{3}=\frac{1}{3}\) × 100
= 33\(\frac{1}{3}\) %
30% < 33\(\frac{1}{3}\)%

NCERT (Page 7-8)

Question 1.
(i) 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:
The complete table is as follows.

(ii) How did you find these values? Discuss the methods with the class. Do you find anything interesting in the table?
Solution:
To calculate the percentages mentally, one can use simple mental methods.
(a) To find 25%, divide the number by 4.
e.g. 25% of 100 = \(\frac{100}{2}\) = 25
(b) To find 10%, simply move the decimal point one place to the left.
e.g. 10% of 100 is 10.0 or 10.
(c) To find 20%, calculate 10% and then double the result e.g. 10% of 100 is 10, so 20 % of 100 is
2 × 10 = 20.
(d) To find 5%, calculate 10% and then halve the result e.g. 10% of 100 is 10, so 5% of 100 is \(\frac{10}{2}\) = 5.

Some interesting facts are as follows.
(a) Methods involve using simple fractions and decimal shifts for mental calculation.
(b) One interesting observation is that the values for 20% are exactly double the values for 10%.
(c) Another observation is that the values for 5% are exactly half of the values for 10%.
(d) The values for 25% are the sum of the values for 20% and 5%.

(iii) Using this understanding, mentally calculate how much 40% of the values in the table above would be.
Solution:
Since, 40% is exactly doceble of 20%.
So, 40% of the values 100, 200, 50, 80, 10, 35 and 287 are 40, 80, 20, 32, 4, 14 and 114.8

(iv) Using the observation, mentally calculate how much 15% of the values in the table would be.
Solution:
(a) We have, 15% of 100

Here, we observe that 15% = 10% + 5%
Therefore, the 15% values for each number in the given table are 15, 30, 7.5,12,1.5, 5.25 and 43.05.

Question 2.
Suppose you have to mentally calculate the following percentages of some value: 75%, 90%, 70%, 55%. How would you do it? Discuss.
Solution:
Mental calculation methods involve breaking down percentages into easier – to calculate components like 10%, 50% or 25% and then using addition, subtraction or multiplication to find the final values.
For 75% Calculate 25% (or \(\frac{1}{4}\)) and multiply it by 3, or
calculate (50% + 25%).
For 90% → Calculate 10% and subtract it from 100% of the value.
For 70% → Calculate 10% and multiply it by 7 or calculate (50% + 20%).
For 55% → Calculate 50% (i.e. half the value) and add 5% (i.e. half of 10%).

Question 3.
Find 10% of a quantity, what decimal value should be multiplied?
Solution:
We have,
10% = \(\frac{10}{100}=\frac{1}{10}\) = 0.1
Thus, to find 10% of a quantity, multiply the quantity by 0.1.

Question 4.
Complete the following table:

Solution:
(ii) We have, 100% = \(\frac{100}{100}\)
= 1 = 1.0

(iii) We have, 25% = \(\frac{25}{100}=\frac{1}{4}\)
= 0.25

(iv) We have, 75% = \(\frac{75}{100}=\frac{3}{4}\)
= 0.75

(v) We have, 10% = \(\frac{10}{100}=\frac{1}{10}\)
= 0.1

(vi) We have, 1% = \(\frac{1}{100}\)
= 0.01

(vii) We have, 5% = \(\frac{5}{100}\)
= 0.05

(viii) We have, 43% = \(\frac{43}{100}\)
= 0.43

The complete table is given below.

NCERT (Page 12)

Question 1.
Complete the table below. Mark the approximate locations in the following diagram.

Solution:
(i) (a) We have, 90% = \(\frac{90}{100}=\frac{9}{10}\)
= 0.9

(b) We have, 110% = \(\frac{110}{100}=\frac{11}{10}\)
= 1.1

(c) We have, 200% = \(\frac{200}{100}\)
= 2.0

(d) We have, 250% = \(\frac{250}{100}=\frac{25}{10}=\frac{5}{2}\)
= 2.5

(e) We have, 15% = \(\frac{15}{100}=\frac{3}{20}\)
= 0.15

(f) We have, 173% = \(\frac{173}{100}\)
= 1.73

(g) We have, 358% = \(\frac{358}{100}=\frac{179}{50}\)
= 358

(h) We have, 28.9% = \(\frac{1}{2}\)
= 0.289

(i) We have, 305% = \(\frac{1}{2}\)
= 3.05

(ii) The complete table is given below.

Figure it Out (NCERT Pg 12-14)

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

Solution:
(ii) From the figure there are 10 equal parts.
Since, total 10 parts = 100%.
So, each part = \(\frac{100}{10}\) = 10%
∵ 90 is the whole.
∴ 10% of90 = \(\frac{10}{100}\) × 90 = 9
Since, the value of 1 part is 9.
Then, the value of 6 parts = 9 × 6 = 54
Hence, the missing numbers are 10% and 54.

(iii) From the figure, there are 4 equal parts.
Since, total 4 parts = 100%
So, each part = \(\frac{100}{4}\) = 25%.
∵ 140 is the whole.
∴ 25% of 140 = \(\frac{25}{100}\) × 140 = 35
Since, the value of 1 part is 35.
Then, the value of 3 parts = 35 × 3 = 105
Hence, the missing numbers are 25% and 105.

Question 2.
Find the value of the following and also draw their bar models.
(i) 25% of 160
Solution:
We have, 25% of 160 = \(\frac{25}{100}\) × 160 = 5 × 8 = 40

(ii) 16% of 250
Solution:
Do same as part (i). Ans. 40

(iii) 62% of 360
Solution:
Do same as part (i). Ans. 223.2

(iv) 140% of 40
Solution:
Do same as part (i). Ans. 56

(v) 1% of 1 hour
Solution:
Do same as part (i). Ans. 0.6 min or 36 s

(vi) 7% of 10 kg
Solution:
Do same as part (i). Ans. 0.7 kg or 700 grams.

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:
Given, the total volume of deep orange paint = 60 mL
and the fraction of red paint = \(\frac{3}{4}\)
∴ The amount of red paint = Total volume × \(\frac{3}{4}\)
= 60 × \(\frac{3}{4}\)
= 15 × 3
= 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) We have, 50% of 510
= \(\frac{50}{100}\) × 510
= \(\frac{1}{2}\) × 510
= 255

and 50% of 515 = \(\frac{1}{2}\) × 515
= \(\frac{50}{100}\) × 515
= 257.5

(ii) Do same as part (i). Ans. 37% of 148 [<]73% of 148

(iii) Do same as part (i). Ans. 29% of 43 [<] 92% of 110

(iv) Do same as part (i). Ans. 30% of 40 [<] 40% of 50 (v) Do same as part (i). Ans. 45% of 200 [>] 10% of 490

(vi) Do same as part (i). Ans. 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 ____ and 120% of k is ____
Solution:
Given, 30% ofk is 70.
∵ 60% is double of that amount. .
∴ 60% of k = 2 x (30% of k) = 2 x 70 = 140

Now, 90% is triple of that amount.
∴ 90% of k = 3 x (30% of k) =3×70=210
Similarly, 120% of k = 4 x (30% of k) = 4 x 70 = 280
Therefore, 30% ofk is 70, 60% of k is 140, 90% of k is 210 and 120% of k is 280.

(ii) 100% of m is 215,10% of m is ____ 1% of m is ____ 6% of m is ____
Solution:
Do same as part (i). Ans. 21.5,2.15 and 12.9

(iii) 90% of n is 270, 9% of n is ____ 18% of n is ______ 100% of n is _____
Solution:
Do same as part (i). Ans. 27, 54 and 300.

(iv) Make 2 more such questions and challenge your peers.
Solution:
Do 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 x be the missing number.
∴ 3 = \(\frac{x}{100}\) × 300
⇒ 3 = 3x
⇒ x = 1

(ii) Let y be the missing number.
∴ y = \(\frac{40}{100}\) × 4 = \(\frac{4}{10}\) × 4 = \(\frac{16}{10}\) = 1.6

(iii) Let z be the missing number.
∴ 40 = \(\frac{40}{100}\) × z
⇒ 40 = \(\frac{4}{10}\) × 4
⇒ z = 40 × \(\frac{5}{4}\)
= 10× 5 = 50

Question 7.
(i) Is 10% of a day longer than 1% of a week?
Solution:
Yes
We have, 10% of a day = \(\frac{10}{100}\) × 24 h [∵ 1 day = 24 h]
= 0.1 × 24 = 2.4h
and 1% of a week = \(\frac{1}{100}\) × 1 week
Now, a week has 7 days.
So, 7 days = 7 × 24 h = 168 h
∴ 1% of a week = \(\frac{1}{100}\) × 168 = 1.68 h
Since, 2.4 h > 1.68 h
Therefore, 10% of a day is longer than 1% of a week,

(ii) Create such questions and challenge your peers.
Solution:
Do 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:
The percentage is calculated by dividing the amount of fodder eaten by the amount given and multiplying by 100.
In every instance described, the bull eats exactly 1 unit less than the amount of fodder given.
The observations is that the percentage of fodder eaten is not constant, it changes slightly each day.

Hence, the percentage eaten increases each day, approaching 100% as the amount of fodder given increases.

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:
Given, the workers take 18 days to complete 20% of the work.
We assume that the rate of work stays the same.
Let x be the total number of days to complete the work. According to the question,
20% of x = 18
⇒ \(\frac{20}{100}\) × x = 18
⇒ x = \(\frac{18 \times 100}{20}\) = 18 × 5 = 90 days

Why is the assumption necessary?
The assumption that the rate of work stays the same is necessary because
(i) workers might get tired over time, reducing their efficiency
(ii) different parts of the plantation might have varying difficulty levels.
(iii) weather conditions might change.
(iv) some workers might take breaks or leave.
Without this assumption, we cannot accurately predict the total time needed, as the rate could vary significantly. Therefore, workers will take 90 days to complete the entire plantation.

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 min. How long should each activity be done?
Solution:
Given, the warm-up is 10% of 90 min of training session.
∴ Warm-up duration = 10% of 90 min
= \(\frac{10}{100}\) × 90 = 9 min 100

Play time is 80% of 90 min of training session.
Play duration = 80% × 90 min
= \(\frac{80}{100}\) × 90 = 72 min

and the cool down is 10% of 90 min of training session.
Cool down duration = 10% × 90 min
= \(\frac{10}{100}\) × 90 = 9 min
Therefore, the warm-up should be 9 min. play should be 72 min and the cool down should be 9 min.

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:
The question requires “this year’s worldwide population”.
As of the estimated world population is approximately 8.28 billion people.
∵ 90% of the world’s population lives in the Northern hemisphere.
∴ Population in Northern hemisphere = 90% × Total world population 90
= \(\frac{90}{100}\) × 8.28 billion
= 7.45 billion
Therefore, the approximate number of people living in the Northern hemisphere is 7.45 billion people.

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?
Solution:
The proportion of ingredients in a recipe is a ratio and does not change based on the quantity being prepared. Doubling the number of people requires doubling the absolute amount of each ingredient, but their relative proportion stays constant.
Therefore, the proportion of each ingredient remains the same as for 8 people – Rava 40%, Sugar 40% and Ghee 20%.

(ii) If the total weight of the ingredients is 2 kg, how much rava, sugar and ghee are present?
Solution:
Given, total weight of ingredients = 2 kg
∴ Weight of Rava = 40% of 2 kg = \(\frac{40}{100}\) × 2 = 0.8 kg
Weight of Sugar = 40% of 2 kg = \(\frac{40}{100}\) × 2 = 0.8 kg
and weight of Ghee = 20% of 2 kg = \(\frac{20}{100}\) × 2 = 0.4 kg

NCERT (Page 15)

Question 1.

Nutritional Facts
Ingreadients (DEF)
Sugar	99g
Milk solids	30g
Badam powder	12g
Food chemicals	9g
Total weight	150g

Nutritional Information

Nutritional Information
Ingreadients (Zacni)
Sugar	272g
Milk solids	64g
Badam powder	40g
Food chemicals	24g
Total weight	400g

(i) Complete this table by calculating the percentages to answer the questions:

Maybe they should call it ‘Sugar drink mix’ Ya! It is very important to know your contents,
Answer:
Given, the total weight for DEF = 150 g
∴ Percentage of milk solids = \(\frac{30}{150}\) × 100 = 20%
Percentage of badam powder = \(\frac{12}{150}\) × 100 = 8%
Percentage of food chemicals = \(\frac{9}{150}\) × 100 = 6%

Again, the total weight for zacni = 400 g
∴ Percentage of sugar = \(\frac{272}{400}\) × 100 = 68%
Percentage of milk solids = \(\frac{64}{400}\) × 100 = 16%
Percentage of badam powder = \(\frac{40}{400}\) × 100 = 10%
Percentage of food chemicals = \(\frac{24}{400}\) × 100 = 6%

The completed table is as follows.

(ii) Check if the percentages of each product add upto 100.
Answer:
Sum of ingreadients percentages for DEF = 66% + 20% + 8% + 6% = 100%
Sum of ingreadients percentages for zacni = 68% + 16% + 10% + 6% = 100%
Yes, the percentages for each product add up to 100%.

NCERT (Page 17)

Question 1.
(i) 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:
Given, cost price of one notebook (CP) = ₹ 36
and profit margin = 20%
Now, Profit amount = \(\frac{\text { Profit percentage × Cost price }}{100}\)
= \(\frac{20}{100}\) × 36
= ₹ 7.20 100
∴ The selling price (SP) = Cost price + Profit amount = 36 + 720 = ₹ 4320
Therefore, the selling price is ₹ 43.20.

(ii) She sells crayon boxes at ₹ 50 per box with a profit margin of 25%. How much did Shambhavi buy them from the wholesaler?
Solution:
Given, selling price (SP) of crayon boxes = ₹ 50 and profit margin = 25%
∴ Selling price = Cost price + Profit amount
⇒ 50 = CP + \(\frac{25}{100}\) × CP
⇒ 50 = CP (1 + 0.25)
⇒ 50 = 1.25 × CP
⇒ CP = \(\frac{50}{1.25}\) = ₹ 40
Therefore, Shambhavi bought the crayon boxes from the wholesaler for ₹ 40 per box.

Question 2.
The rice stock in Raghu’s provision store is getting old. He had purchased the rice at ₹ 35 per kg. To clear his stock, he sells 10 kg rice for ₹ 300. Find out the percentage loss.
The amount Raghu had paid towards buying the 10 kg rice Is ₹ 350. He sold It for ₹ 300.
The loss is ₹ 350 – ₹ 300 = ₹ 50
The percentage loss is \(\frac{50}{350}\) × 100 = 14.28%

Could we have just calculated the loss percentage per kg Instead? Would It be the same?
Solution:
Given, cost price per kg = ₹ 35
Selling price per kg = ₹ 300 + 10 = ₹ 30
Loss per kg = ₹ 35 – ₹ 30 = 5
Percentage of loss per kg = \(\frac{5}{35}\) × 100% = 1428%
Hence, the percentage loss remains the same whether it is calculated per kg or calculated for the total quantity because both cost price and loss change is in the same proportion.

NCERT (Page 18)

Question 1.
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:
Given, selling price (SP) of the strawberries = ₹ 80 per kg
and loss percentage = 12%
∴ The selling price would be (100 – 12)% i.e. 88%.
∴ SP = 88% of Cost price
⇒ 80 = \(\frac{88}{100}\) × Cost price (CP)
⇒ CP = \(\frac{8000}{88}\)
= ₹ 90.91 per kg
Hence, the cost price is approximately ₹ 90.91 per kg.

Question 2.
A utensil store is offering a 35% discount on the cooker with an MRP ₹ 1800. What is the selling price? If the cost price was ₹ 900, what is the percentage profit made after the sale?
Solution:
Given, MRP of a cooker = ₹ 1800
and discount percentage = 35%
∴ Discount amount = 35% of 1800 = \(\frac{35}{100}\) × 1800
= 35 × 18
= ₹ 630
Now, selling price (SP) = MRP – Discount
= 1800 – 630 = ₹ 1170
Since, the cost price of cooker is ₹ 900.
∴ Profit amount = 1170 – 900 = ₹ 270
Therefore, the profit percentage = \(\frac{\text { Profit }}{\text { Cost Price }}\) × 100
= \(\frac{270}{900}\) × 100 = 30%
Hence, the selling price of cooker is ₹ 1170 and the profit percentage made after the sale is 30%.

NCERT (Page 19)

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

Solution:
The sub total is calculated by multiplying the quantity by the price per item.
∴ Sub total = Quantity × Price per item
= 3 × 150.00
= ₹ 450.00
The sub total amount in the bill is correct.
Since, the CGST (Central goods and Services Tax) is 9% of the subtotal.
CGST = 9% × Sub total
= \(\frac{9}{100}\) × 450.00
= ₹ 40.50 100
The CGST in the bill is correct.

Since, SGST (State Goods and Services Tax) is also 9% of the sub-total.
∴ SGST = 9% × Sub total
= \(\frac{9}{100}\) × 450.00
= 40.50
The SGST amount in the bill is correct.
Now, total amount = Sub total + CGST + SGST
= 450.00 + 4050 + 40.50
= ₹ 531.00
The total amount in the bill is correct.
Hence, all calculations in the bill are correct.

(ii) 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:
(ii) Do yourself

Figure it Out (NCERT Pg 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:
Given, die cost price (CP) of geometry box = ₹ 75
and selling price (SP) = ₹ 110
∴ Profit amount = SP – CP = 110 – 75 = ₹ 35
Now, profit margin = \(\frac{\text { Profit }}{\text { Cost price }}\) × 100
= \(\frac{35}{75}\) × 100
= 46.67%
Hence, the shopkeeper’s profit margin with respect to the cost is approximately 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:
Given, the cost price of materials = ₹ 475
and the desired profit margin = 50%
∴ Profit amount = \(\frac{50}{100}\) × 475 = ₹ 237.50 100
Now, selling price = Cost price + Profit amount = 475 + 237.50 = 71250
Hence, the chair should be sold for ₹ 712.50.

Question 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:
Given, the total sales of a company last year = ₹ 2.5 crore
and profit margin = 25%
∴ Profit amount = 25% of 2.5 crore
= \(\frac{25}{100}\) × 2.5 = 0.625 crore
Now, expenditure = Revenue – Profit
= 25 crore – 0.625 crore = 1.875 crore
Hence, the total expenditure of the company last year was ₹ 1.875 crore or ₹ 18750000.

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:
Given, the original price of a shirt = ₹ 300
and discount percentage = 25%
∴ Discounted amount = 25% × 300
= \(\frac{25}{100}\) × 300
= ₹ 75
Now, final price = Original price – Discounted amount = 300 – 75 = ₹ 225
Hence, Anwar will have to pay ₹ 225 to buy the shirt.

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:
(iv) Given, the initial price of petrol in 2015 = ₹ 60
and the final price of petrol in 2025 = ₹ 100
Now, increase in the price of petrol
= Price of petrol in 2025 – price of petrol in 2015 = 100 – 60 = ₹ 40
∴ Percentage increase = \(\frac{\text { Increase in the price }}{\text { Price in } 2015}\) × 100
= \(\frac{40}{60}\) × 100
= \(\frac{2}{3}\) × 100
= 66. 60%
Hence, the calculated percentage increase is approximately 66.66%.

Question 6.
Samson bought a car for ₹ 440000 after getting a 15% discount from the car dealer. What was the original price of the car?
Solution:
Given, price of car after 15% discount (SP) = ₹ 440000
If a discount of 15% is given, the samson pays
(100 – 15)% i.e. 85% of the original price.
Let the original price be ₹ x.
So, 85% of original price = SP 85
⇒ \(\frac{85}{100}\) × x = 440000
⇒ 85x = 44000000
⇒ x = \(\frac{44000000}{2}\) = 517647.05
Hence, the original price of the car was approximately ₹ 517647.05.

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:
Given, the total number of people who voted in an election = 1600
and number of votes winner got in an election = 500
∴ Percentage of winner voted = \(\frac{500}{1600}\) × 100 = \(\frac{5}{16}\) × 100
= 31.25%
Since, the winner has 500 votes.
For the winner to actually win, each of the other candidates must have less than 500 votes.
∴ Total remaining votes = 1600 – 500 = 1100
If there was only one other candidates, that candidates would have received 1100 votes, which is more than 500, meaning the first candidates would not be the winner.
If there were two other candidates, they could
have received 550 votes each (\(\frac{1100}{2}\) = 550), still more than the winner.
If there were three other candidates, they could have received approximately 366.67 each [\(\frac{1100}{3}\) = 366.67 ]
Therefore, at least four candidates are required for a person with 500 votes to be the winner.

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:
Given, the initial price of rice in 2024 (P_In) = ₹ 38
and the final price of rice in 2025 (P_Fi) = ₹ 42
∴ The change in price = P_Fi – P_In
= 42 – 38 = ₹ 4
So, the rate of inflation
= \(\frac{\text { Change in price }}{P_{\mathrm{In}}}\) × 100 = \(\frac{4}{38}\) × 100
= 10526%
= 10.53%
Hence, the rate of inflation is approximately 10.53%.

Question 9.
A number increased by 20% becomes 90. What is the number?
Solution:
Let the original number be x.
According to the question,
x + 20% of x = 90
⇒ x + \(\frac{20}{100}\) × x = 90
⇒ x + 0.20x = 90
⇒ 1.20x = 90
⇒ x = \(\frac{90}{1.20}\) = 75
Hence, the number is 75.

Question 10.
A milkman sold two buffaloes for ₹ 80000 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:
Given, the selling price of each buffalo = ₹ 80000,
profit percentage on first buffalo = 5%
and loss percentage on second buffalo = 10%
Let the cost price of first buffalo be ₹ x.
∴ SP = CP + 5% of x = x + \(\frac{5}{100}\) × x
= x + 0.05 = 1.05x
∴ 1.05x = 80000
⇒ x = \(\frac{80000}{1.05}\)
= 76190.48
Let the cost price of second buffalo be ₹ y.
SP = CP – 10% of y = y – 0.1y
⇒ 80000 = 0.9y
⇒ y = 88888.89
Total cost price = 76190.48 + 88888.89
= ₹ 16507937
and total selling price = 80000 + 80000
= ₹ 160000
Since, total SP is less than the total CP, so there is an overall loss.
∴ Loss = CP – SP
= 165079.37 – 160000
= ₹ 5079.37

Question 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 × 05
(ii)p × 0.05
(iii) p × 1.5
(iv) p × 1.05
(v) p + 1.50
Solution:
(iv) If the population of the elephants last decade is p.
New population after an increase of 5%
= p + 5% of p = p + \(\frac{5}{100}\) × p
= p + 0.05p
= p(1 + 0.05)
= p × 1.05

Question 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) If the demand for cameras has fallen by 85% in the last decade.
∴ The current demand is (100% – 85%) i.e. 15% of the demand a decade ago.
Therefore,
Now, to express the demand a decade ago as a percentage of the present demand, we use the formula.
Percentage = \(\left(\frac{\text { Earlier demand }}{\text { Present demand }}\right)\) × 100
= \(\left(\frac{100}{15}\right)\) × 100
= 666.67% ≈ 6675
This implies that the demand a decade ago was approximately 667% of the present demand and not 185%.
Hence, only statement (iii) is correct.

Figure it Out (NCERT Pg 22-23)

Question 1.
Bank of Yahapur offers an interest of 10% per annum Compare how much one gets if they deposit ₹ 20000 for a period of 2 yr with compounding and without compounding annually.
Solution:
Given, deposit amount = ₹ 20000, number of years = 2 and interest = 10% per annum
The principal amount is returned after the maturity period without compounding annually.

The principal amount is returned after the maturity period with compounding annually.

Hence, with compounding one gets ₹ 200 more (24200 – 24000) compared to without compounding.

Question 2.
Bank of Wahapur offers an interest of 5% per annum Compare how much one gets if one deposits ₹ 20000 for a period of 4 yr with compounding and without compounding annually.
Solution:
Do same as Question 1.

Question 3.
Do you observe anything interesting in the solutions of the two questions above? Share and discuss.
Solution:
An interesting observation is that the total amount after 2 yr without compounding is the same as the total amount after 4 yr without compounding.

NCERT (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:
If p is the principal, r is the rate of interest in decimal form and t is the time period.
∴ The expression or formula to find the total interest amount gained at the end of the maturity period for without compounding (simple interest) is
SI = p × r × t
and the total amount at the end of an FD of t yr = p + prt
For compounding (compound interest):
The total amount in the FD after t yr = p × (1 + r)^t

Figure it Out (NCERT Pg 24)

Question 1.
Jasmine invests amount pfor 4 yr at an interest of 6% per annum. Which of the following expression(s) describe the total amount she will get after 4 yr, 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:
(vii) Given, principal amount = p,
number of years (f) = 4 yr and rate of interest (r) = 6%
∴ The total amount without compounding
= p + prt = p + p × \(\frac{6}{100}\) × 4
= p + (p × 0.06 × 4)

Question 2.
The post office offers an interest of 7% per annum. How much interest would one get if one invests ₹ 50000 for 3 yr without compounding? How much more would one get, if it was compounded?
Solution:
Given, principal amount (p) = ₹ 50000,
rate of interest (r) = 7% and time period (f) = 3 yr
Interest without compounding = p × r × t
= 50000 × \(\frac{7}{100}\) × 3
= 500 × 21
= ₹ 10500 100

Interest with compounding = p(1 – r)^t – p
= 50000 \(\left(1+\frac{7}{100}\right)^3\) – 50000
= 50000 × (1.07)³ – 50000
= 50000 × 1.225043 – 50000
= 61252.15 – 50000
= ₹ 11252.15

∴ Difference = Interest with compounding – Interest without compounding
= 11252.15 – 10500
= ₹ 752.15

Question 3.
Girldhar borrows a loan of ₹ 12500 at 12% per annum for 3 yr 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:
ForGiridhar
Principal amount (p) = ₹ 12500,
rate of interest (r) = 12% and time period (f) = 3 yr
∴ Interest without compounding = prt
= 12500 × \(\frac{12}{100}\) × 3
= 125 × 36
= ₹ 4500

For Raghava
Principal amount (p) = ₹ 12500,
rate of interest (r) = 10% and time period (f) = 3 yr
∴ Interest with compounding = p(1 + r)^t – p
= 12500 (1 + \(\frac{10}{100}\))³ – 12500
= 12500 (1.1)³ – 12500
= 12500 × 1.331 – 12500
= 16637.50 – 12500
= ₹ 4137.50
Therefore, Giridhar pays ₹ 4500 interest and Raghava pays ₹ 4137.50 interest.
So, difference = 4500 – 4137.50 = ₹ 362.50 .
Hence, Giridhar pays more interest by ₹ 362.50.

Question 4.
Consider an amount ₹ 1000. If this grows at 10% per annum, 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, principal amount (p) = ₹ 1000
and rate of interest (r) = 10%

For without compounding
If the principal amount is double then
2p = p + prt
⇒ 2 × 1000 = 1000 + 1000 × \(\frac{10}{100}\) × t
⇒ 2000 – 1000 = 100t
⇒ t = \(\frac{1000}{100}\) = 10yr

For compounding
The total amount 2p = p (1 + r)^t
⇒ 2 = (1 + \(\frac{10}{100}\))^t
⇒ 2 = (1 + 0.1)^t
⇒ 2 = (1.1)^t
⇒ (1.1)⁷ = (1.1)^t [∵ 2 = (1.1)⁷ (approx)]
On comparing the powers, we get ⇒ t = 7 yr (approx)
It will take 10 yr for the amount to double, when compounding is not done.
It will take approximately 7 yr for the amount to double, when compounding is done.
Yes, compounding is an example of exponential growth and not-compounding is an example of linear growth.

Question 5.
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 yr?
Solution:
Given, principal amount (p) = 1.5 crore = 15000000,
rate of interest (r) = 3%
and time period (f) = 3yr
Since, population involved compound growth, where the population increases by a constant percentage each year.
∴ The expected population after 3 yr = p(1 + r)^t
= 15000000 (1 + \(\frac{3}{100}\))³
= 15000000 (1 + 0.03)³
= 15000000 (1.03)³
= 15000000 × 1.092727
= 16390905
Hence, the expected population after 3 yr is 16390905.

Question 6.
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 h if the initial count is 506000.
Solution:
Given, number of bacteria (principal)
(p) = ₹ 506000,
rate of increase per hour (r) = ₹ 2.5%
and number of period (f) = 2 h
Since, bacteria involved compounding growth.
So, the number of bacteria at the end of 2 h
= p(1 + r)^t
= 506000(1 + \(\frac{2.5}{100}\))²
= 506000 (1 + 0.025)²
= 506000 (1.025)²
= 506000 × 1.050625
= 531626.25
Hence, the number of bacteria at the end of 2 h is approximately 531626.

NCERT (Page 26)

Question 1.
A provision store is offering a stock clearance sale. Customers can choose one of the two options 20% s 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:
Customers can choose one of the two options.
20% discount or ₹ 50 discount for any purchase above ₹ 150.
(i) The discount amount is calculated as 20% of the purchase value.
∴ Discount = 20% of 180
= \(\frac{20}{100}\) × 180
= 2 × 18
= ₹ 36
So, the amount paid would be 180 – 36 = ₹ 144
∵ The 20% discount is ₹ 36 and the fixed discount is ₹ 50.
∴ ₹ 50 discount is better.

(ii) Given, the purchase amount = ₹ 225
The 20% discount is calculated as
20% of 225 = \(\frac{20}{100}\) × 225 = 45
∵ The 20% discount is ₹ 45 and the fixed discount is ₹ 50.
₹ 50 discount is better.

(iii) Given, the purchase amount = ₹ 300
The 20% discount is calculated as
20% of300 = \(\frac{20}{100}\) × 300 = 60
∵ The 20% discount is ₹ 60 and the fixed discount is ₹ 50.
∴ 20% discount is better.

NCERT (Page 27)

Question 1.
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:
Let Abe the number of marbles Ariba has andl/be the number of marbles Arun has.
According to the question,
A = 120% of U = \(\frac{120}{100}\) × U = 1.2 × U
⇒ U = \(\frac{1}{2}\) × A = 0.8333 × A = \(\frac{83.33}{100}\) × A
⇒ U = 8333% × A
Hence, Arun could state that he has approximately 83.33% of the marbles Ariba has.

Figure it Out (NCERT Pg 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:
Given, the population of Bengaluru in 2025 is about 250% of its population in 2000.
∴ Population in 2025
= 250% × Population in 2000
= \(\frac{250}{100}\) × Population in 2000
Since, the population in 2000 was 50 lakhs.
Population in 2025 = \(\frac{250}{100}\) × 50 lakhs
= 25 x 50 lakhs
= 125 lakhs
Hence, the population in 2025 is 125 lakhs.

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:
Given, the world population in 2025 is approximately 8.2 billion.
Now, we will calculate the percentage share for each country.
(i) For Germany The percentage share for Germany

(ii) For India Given, the population of India in 2025 = 1.46 billion
∴ The percentage share for India Population of India
= \(\frac{\text { Population of India }}{\text { Population of the world }}\) × 100
= \(\frac{1.46 \text { billion }}{8.2 \text { billion }}\) × 100 ≈ 18%

(iii) For Bangladesh Given, the population of Bangladesh in 2025 = 175 million
∴ The percentage share for Bangladesh
= \(\frac{\text { Population of Bangladesh }}{\text { Population of the world }}\) × 100
= \(\frac{175 \text { million }}{8.2 \text { billion }}}\) × 100
≈ 2%

(iv) For USA Given, the population of USA in 2025
= 347 million
∴ The percentage share for USA
= \(\frac{\text { Population of USA }}{\text { Population of the world }}\) × 100
= \(\frac{347 \text { million }}{8.2 \text { billion }}\) × 100
≈ 4%

Question 3.
The price of a mobile phone is ₹ 8250. 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:
(v) and (vi) Given, the price of a mobile phone = ₹ 8250
Since, the GST amount is 18% of the original price.
∴ GST amount = 18% of 8250 = \(\frac{18}{100}\) × 8250
= 0.18 × 8250
So, the final price of a mobile phone including the GST = 8250 + 0.18 × 8250
= 8250 (1 + 0.18)
= 8250 × 1.18

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 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:
(ii) and (vi)
Given, the initial population is p.
∵ The positive and negative sign shows that population is increasing and decreasing, respectively.
∴ The final population after three months
= p × (1 + 0.05) (1 – 0.02) (1 – 0.03)
= p × 1.05 × 0.98 × 0.97
Now, the overall factor = 1.05 × 0.98 × 0.97 = 0.99897
Since, the overall factor is less than 1, the final population is less than the initial population p.
Hence, the true statements are (ii) and (vi).

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 (CP) of the product be x.
∵ The shopkeeper set the initial price (selling price SP,) with a 35% profit margin.
SP₁ = x + 35% of x
= x + \(\frac{35}{100}\) × 100
= x (1 + 035)
= 1.35 × x ,..(i)

Now, 30% discount was offered on the selling price (SP,).
∴ SP₂ = SP₁ – 30% of SP₁
= SP₁ – \(\frac{30}{100}\) × SP₁
= SP₁ (1 – 0.30)
= SP₁ × 0.70
= 1.35 × x × 0.70
[from Eq. (i)]
⇒ SP₂ = 0.945x
Since, 0.945x < x

Hence, the shopkeeper will make a loss.

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

Solution:
First of all, we count the dots along each side of big square, there are 8 dots.
So, there are 8 equal small squares along each side.
∴ The area of big square = 8 × 8 = 64 sq units

Similarly, the total area of square D and E = 4 × 4 = 16
A diagonal cuts a square, it divides the area into two equal part, so each triangle D and E has half the area of the 4×4 square.
Therefore, area of E = \(\frac{16}{2}\) = 8
The required percentage = \(\frac{\text { Area of } E}{\text { Total area }}\) × 100
= \(\frac{8}{64}\) × 100 = 125%
Hence, region E occupies 12.5% of the total area.

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:
We have, 5% of 40 = \(\frac{5}{100}\) × 40
= 2
We have, 40% of 5 = \(\frac{4}{100}\) × 5
= 2
We have, 12% of 25 = \(\frac{12}{100}\) × 25
= 3
We have, 25% of 12 = \(\frac{25}{100}\) × 12
= 3
We have, 15% of 60 = \(\frac{15}{100}\) × 60
= \(\frac{900}{100}\) = 9
We have, 60% of 15 = \(\frac{60}{100}\) × 15
= \(\frac{900}{100}\) = 9

Observation – The result for x% ofy is the same as the result for y% of x. x% of y = y% of x
⇒ \(\frac{x}{100}\) × y = \(\frac{y}{100}\) × x
⇒ \(\frac{x y}{100}=\frac{x y}{100}\)

Hence proved.

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:
Let the total percentage of students be 100%.
Here, grade 8 students make up 40% of the total of those grade 8 students, 60% are girls.

(i) Percentage of grade 8 girls = 40% × 60% = \(\frac{40}{100} \times \frac{60}{100}\)
= 0.24 = 24%
(ii) Given, the total number of students =160
Since, 24% of the total students are grade 8 girls.
∴ Number of grade 8 girls = 24% of 160
= \(\frac{24}{100}\) × 160
= 38.4
Hence, the number of grade 8 girls is 38.

Question 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:
Given, the selling price of 3 pencils is equal to the cost of 5 pencils.
∴ 3 SP = 5 CP
⇒ SP = \(\frac{5}{3}\) CP Since, \(\frac{5}{3}\) = 167 > 1 so SP > CP.
Therefore, the shopkeeper makes a profit.
Now, profit percentage = \(\frac{\text { Profit }}{\mathrm{CP}}\) × 100

= 66.67%
Hence, the shopkeeper makes a profit of approximately 66.67% or 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 yr?
Solution:
Let the original price be P.
After a 3% increase, the new price
= P × (1 + 3%)
= P × (1 + 0.03)
= 1.03P
Now, the second increases of 4% is applied to the price after the first year.
The price after the second year = 1.03P × (1 + 4%)
= 1.03P × (1+0.04)
= 1.03P × 1.04
= 1.0712 P
∴ The overall percentage increase = \(\frac{1.0712 P-P}{P}\) × 100
= \(\frac{0.0712 P}{P}\) × 100
= 0.0712 × 100
= 7.12%
Hence, the overall percentage price increase in the last 2 yr 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 initial length be l and the initial breadth be b.
The initial area (A) = l × b … (i)
Since, the new length l_new is increased by 10% of the original length.
l_new = l + 10% of l = l + \(\frac{10}{100}\) × l
= 1 + 0.1l
= 1.1l …(ii)

Since, the area remains unchanged.
∴ A_new = A
Now, A_new = l_new × b_new
⇒ A = 1.1l × b_new
⇒ l × b = 1.1l × b_new
⇒ b_new = \(\frac{b}{1.1}=\frac{10}{11}\) = b

Now, the decrease in breadth = b – b_new
= b – \(\frac{10}{11}\)b = \(\frac{11 b-10 b}{11}=\frac{b}{11}\)

∴ The percentage decrease = \(\frac{1}{2}\) × 100
= \(\frac{\frac{b}{11}}{b}\) × 100 = \(\frac{100}{11}\)%

Hence, the breadth decreases by exactly \(\frac{110}{11}\)% or approximately 9.09%.

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:
Given, total weight of the packet = 65 g
and the percentage of potato = 70%
∴ Weight of potato = 70% of 65 g 70
= \(\frac {70}{100}\) × 65 = 455 g 100

The percentage of vegetable oil = 24%
∴ Weight of vegetable oil = 24% of 65 g
= \(\frac {24}{100}\) × 65 = 15.6 g

The percentage of salt = 3%
∴ Weight of salt = 3% of 65 g
= \(\frac {3}{100}\) × 65 = 1.95 g

The percentage of spices = 3%
∴ Weight of spices = 3% of 65 g
= \(\frac {3}{100}\) × 65 = 1.95 g 100
Hence, the weight of potato is 45.5 g, vegetable oil is 15.6 g, spices is 1.95 g and salt is 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) Given, the price of one item = ₹ 100
For Shop A Total items received = 2
∴ Effective Price = \(\frac{\text { Cost of } 1 \text { item }}{\text { Total items }}=\frac{100}{2}\) = 50

For Shop B Total items received = 3
∴ Effective Price = \(\frac{\text { Cost of } 2 \text { item }}{\text { Total items }}=\frac{200}{3}\) = 66.7

For Shop C Total items received = 4
∴ Effective Price = \(\frac{\text { Cost of } 3 \text { item }}{\text { Total items }}=\frac{300}{4}\) = 75

Shop A ₹ 50 per item,
Shop B ₹ 66.67 per item
and Shop C ₹ 75 per item
Therefore, the arrangement from cheapest to costliest is Shop A < Shop B < Shop C.

(ii) For Shop A 1 free item out of 2 total items
∴ Percentage discount = \(\frac{\text { Free items }}{\text { Total items }}\) × 100 = 50%
= \(\frac{1}{2}\) × 100 = 50%

For Shop B 1 free item out of 3 total items
∴ Percentage discount = \(\frac{\text { Free items }}{\text { Total items }}\) × 100 = 50%
= \(\frac{1}{3}\) × 100 = 50%

For Shop C 1 free item out of 4 total items
∴ Percentage discount = \(\frac{\text { Free items }}{\text { Total items }}\) × 100 = 50%
= \(\frac{1}{4}\) × 100 = 50%

∴ Shop A 50% discount,
Shop B 33.33% discount
and Shop C 25% discount.

(iii) For Shop A Buy 2, get 2 free
∴ Cost of 2 items = 2 × 100 = ₹ 200
For Shop B Cannot buy exactly 4 items with the deal (get 3 or 6)
For Shop C Buy 3, get 1 free.
∴ Cost of 3 items = 3 × 100 = ₹ 300
Hence, Shop A would be chosen because it provides 4 items for a total of Rs 200, which is the lowest total cost.

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:
Given, the number of people = 100
and the initial percentage of left-handed people = 99%
∴ The initial number of left-handed people
= 100 × \(\frac{99}{100}\) = 99

Let x be the number of left-handed people, who leave the room.
The number of left-handed people in the room = 99 – x
The total number of people remaining in the room
= 100 – x
The new percentage of left-handed people is 98%.
⇒ \(\frac{99-x}{100-x}=\frac{98}{100}\)
⇒ 99 – x = 0.98 (100 – x)
⇒ 99 – 98 = x – 0.98x
⇒ 1 = 0.02x
⇒ x = \(\frac{1}{0.02}\) = 50
Hence, 50 left-handed people have to leave the room to bring the percentage down to 98%.

Question 15.
Look at the following graph.
Ability to use computer by age gender (2023).
The ability to use computers is highest among those in their twenties and teenagers.

Note NSS Round 79, Comprehensive Annual Modular Survey,
National Statistics Office
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) The total computer literacy for people in their twenties = 26% + 37% = 63%
The total computer literacy for people in their teenagers
= 24% + 29% = 53%
The total computer literacy for people in their thirties
= 14% + 25% = 39%
Here, comparing there totals, people in their twenties have the highest combined percentage of computer literacy.
This statement is valid.

(ii) In all visible age group (children to seniors), the female percentage (black bar) is less than the male percentage (grey bar). So, this statement is valid.

(iii) The graph only shows the percentage of people within each group, who can use a computer, not the total number of people in each age group.
The population size cannot be inferred from this graph.
This statement is invalid.

(iv) We know that a quarter is 25%.
The total for people in their thirties
= 14% + 25% = 39%
So, 39% is more than 25%.
This statement is valid.

(v) We know that 1 in 10 is 10%.
The total for seniors (aged 60 and above)
= 2%+ 4% = 6%
So, 6% is less than 10%.
This statement is valid.

(vi) We know that half is 50%.
The total for people in their twenties
= 26% + 37% = 63%
So, 63% is more than 50%.
This statement is invalid.