Fractions In Disguise - NCERT Solutions Class 8 Maths (Ganita Prakash)

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Fractions In Disguise – NCERT Solutions Class 8 Maths (Ganita Prakash) includes all the questions with solutions given in the NCERT Class 8 Maths (Ganita Prakash).

NCERT Solutions Class 8

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Fractions In Disguise – NCERT Solutions

Q.1: 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 {tex}({CP})={/tex} ₹ 36 per book
Profit margin {tex}=20 \%{/tex}

Step 1: Calculate the profit amount Profit {tex}=20 \%{/tex} of CP
Profit {tex}=(\frac {20 }{ 100}) \times 36{/tex}
Profit {tex}= 0.20 \times 36{/tex}
Profit {tex}={/tex} ₹ 7.20
Step 2: Calculate Selling Price
Selling Price {tex}(S P)=C P+{/tex} Profit
{tex}S P=36+7.20 {/tex}
SP = ₹ 43.20

Q.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:

Part 1: Finding Selling Price
Given:

{tex}{MRP}=₹ 1800{/tex}
Discount {tex}=35 \%{/tex}

Step 1: Calculate discount amount Discount amount {tex}=35 \%{/tex} of 1800
Discount amount {tex}=(35 / 100) \times 1800{/tex}
Discount amount {tex}=0.35 \times 1800{/tex}
Discount amount {tex}=₹ 630{/tex}
Step 2: Calculate Selling Price
SP = MRP – Discount
SP = 1800 – 630
SP = ₹ 1170
Part 2: Finding Profit Percentage
Given:

Cost Price (CP) = ₹ 900
Selling Price (SP) = ₹ 1170

Step 1: Calculate profit Profit = SP – CP Profit = 1170 – 900 Profit = ₹ 270
Step 2: Calculate profit percentage Profit percentage {tex}=({/tex}Profit {tex}/ {CP}) \times 100 \%{/tex}
Profit percentage {tex}=(270 / 900) \times 100 \%{/tex}
Profit percentage {tex}=0.30 \times 100 \%{/tex}
Profit percentage = 30%
Therefore, the selling price is ₹ 1170, and the shopkeeper makes a profit of {tex}30 \%{/tex}.

Q.3: Surya wants to use a deep orange colour to capture the sunset. He mixes some red paint and yellow paint to make this colour. The red paint makes up {tex}\frac 34{/tex} of this mixture. What percentage of the colour is made with red?

Solution:

{tex}\frac{3}{4}{/tex} is 3 out of every 4.
That is, 6 out of every 8 (equivalent fraction).
That is, 30 out of every 40.
That is, 75 out of every 100.
{tex}\frac{3}{4}=\frac{6}{8}=\frac{30}{40}=\frac{75}{100}{/tex}
This means {tex}75 \%{/tex}.

Q.4: Surya won some prize money in a contest. He wants to save {tex}\frac{2}{5}{/tex} of the money to purchase a new canvas. Express this quantity as a percentage.

Solution:

Try to understand the different methods for solving this problem, as shown below.
Method 1:
{tex} \frac{2}{5}=\frac{20}{50}=\frac{40}{100} {/tex}
{tex} =40 \% . {/tex}
Method 2:
{tex} \frac{2}{5}=\frac{x}{100} {/tex}
{tex} x=\frac{2}{5} \times 100=40 {/tex}.

Q.5: Given a percentage, can you express it as a fraction? For example, express 24% as a fraction.

Solution:

Since a percentage is a fraction, {tex}24 \%{/tex} is the same as {tex}\frac{24}{100}{/tex}.
We can find other equivalent forms of {tex}\frac{24}{100}=\frac{12}{50}=\frac{6}{25}=\frac{48}{200}{/tex}.
In general, we can say that a percentage, {tex}z \%{/tex}, can be expressed by any of the fractions that are equivalent to {tex}\frac{z}{100}{/tex}.

Q.6: Express the {tex}\frac 35{/tex} fraction as a percentage.

Solution:

To convert a fraction to percentage, multiply it by 100.
{tex}\frac 35=(\frac 35) \times 100 \% {/tex}
{tex} =\frac {(3 \times 100) }{ 5 \% }{/tex}
{tex} =\frac {300 }{ 5 \% }{/tex}
{tex} =60 \% {/tex}
Therefore, {tex}\frac 3 5=60 \%{/tex}.

Q.7: Express the {tex}\frac {7}{14}{/tex} fraction as a percentage.

Solution:

First, let’s simplify the fraction:
{tex}\frac { 7}{ 14}= \frac 1 2 {/tex}
Now, converting to percentage:
{tex}\frac 1 2=(\frac 12) \times 100 \% {/tex}
{tex} =\frac {100 }{ 2 \% }{/tex}
{tex} =50 \% {/tex}
Therefore, {tex}\frac {7 }{ 14}=50 \%{/tex}.

Q.8: Express the {tex}\frac {9}{20}{/tex} fraction as a percentage.

Solution:

Converting to percentage:
{tex}\frac { 9}{20}=(\frac {9 }{ 20}) \times 100 \% {/tex}
{tex} =\frac {(9 \times 100) }{ 20 \% }{/tex}
{tex} =\frac {900 }{ 20 \%} {/tex}
{tex} =45 \% {/tex}
Therefore, {tex}\frac {9 }{ 20}=45 \%{/tex}.

Q.9: Express the {tex}\frac {72}{150}{/tex} fraction as a percentage.

Solution:

Converting to percentage:
{tex}\frac { 72 }{ 150}=(\frac {72 }{ 150}) \times 100 \% {/tex}
{tex} =\frac {(72 \times 100) }{ 150 \% }{/tex}
{tex} =\frac {7200 }{ 150 \% }{/tex}
{tex} =48 \% {/tex}
Therefore, {tex}\frac {72 }{ 150}=48 \%{/tex}.

Q.10: Express the {tex}\frac 13{/tex} fraction as a percentage.

Solution:

Converting to percentage:
{tex}\frac 1 3=(\frac 1 3) \times 100 \% {/tex}
{tex} =\frac {100 }{ 3 \%} {/tex}
{tex} =33.33 \% \text { (or } 33 \frac{1}{3} \% \text { ) } {/tex}
Therefore, {tex}\frac 13{/tex} = 33.33%

Q.11: Express the {tex}\frac {5}{11}{/tex} fraction as a percentage.

Solution:

Converting to percentage:
{tex}\frac {5 }{11}=(\frac {5 }{ 11}) \times 100 \% {/tex}
{tex} =\frac {500 }{ 11 \%} {/tex}
{tex}=45.45 \% \text { (approx.)}{/tex}
Therefore, {tex}\frac {5}{11}\approx 45.45 \%{/tex}.

Q.12: Nandini has 25 marbles, of which 15 are white. What percentage of her marbles are white?

10%
15%
25%
60%
40%
None of these

Solution:

Total marbles {tex}=25{/tex}
White marbles = 15
Fraction of white marbles {tex}=\frac {15 }{ 25}=\frac 3 5{/tex}
Converting to percentage:
{tex}(\frac 3 5) \times 100 \%=\frac {(3 \times 100) }{ 5 \%}{/tex} {tex}=\frac {300 }{ 5 \%}=60 \% {/tex}
Therefore, the correct answer is (iv) 60%.

Q.13: In a school, 15 of the 80 students come to school by walking. What percentage of the students come by walking?

Solution:

Total students {tex}=80{/tex}
Students who walk {tex}=15{/tex}
Fraction of students who walk = {tex}\frac {15}{80}{/tex}
Simplifying: {tex}\frac {15}{80}{/tex} {tex}=\frac {3 }{ 16}{/tex}
Converting to percentage:
{tex}(\frac {3 }{ 16}) \times 100 \%=\frac {(3 \times 100) }{ 16 \%}{/tex} {tex}=\frac {300 }{ 16 \%}=18.75 \% {/tex}
Therefore, {tex}18.75 \%{/tex} of students come to school by walking.

Q.14: 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:

Based on the positions shown in the diagram:

Person A is very close to the start {tex}\rightarrow{/tex} approximately {tex}{2 0} \boldsymbol{\%}{/tex} completed
Person B is about one-third through the race {tex}\rightarrow{/tex} approximately {tex}{3 8 \%}{/tex} completed
Person C is more than halfway {tex}\rightarrow{/tex} approximately {tex}{5 5 \%}{/tex} completed
Person D is close to the finish {tex}\rightarrow{/tex} approximately {tex}{7 2 \%}{/tex} completed

Matching:

{tex}{A} \rightarrow 20 \%{/tex}
{tex}{B} \rightarrow 38 \%{/tex}
{tex}{C} \rightarrow 55 \%{/tex}
{tex}{D} \rightarrow 72 \%{/tex}

Q.15: Pairs of quantities are shown below. Identify and write appropriate symbols ‘>’, ‘<’, ‘=’ in the blanks. Try to do it without calculations.

50% ________ 5%
{tex}\frac{5}{10} {/tex} ________ 50%
{tex}\frac{3}{11}{/tex} ________ 61%
{tex}30 \%{/tex} ________ {tex}\frac{1}{3}{/tex}

Solution:

{tex}50 \%{/tex} ________ 5%
50% > 5%
({tex}50 \%{/tex} means 50 out of 100, while {tex}5 \%{/tex} means 5 out of 100. Clearly 50 is greater than 5.)
{tex}\frac {5}{1 0}{/tex} ________ 50%
{tex}\frac {5}{1 0}{/tex} = 50%
({tex}\frac {5}{1 0}{/tex} {tex}=\frac 1 2=50 \%{/tex})
{tex}\frac{{3}}{{1 1}}{/tex} ________ 61%
{tex}\frac{{3}}{{1 1}}{/tex} < 61%
{tex}(\frac {3}{11} \approx 27.27 \%{/tex}, which is less than {tex}61 \%){/tex}
30% ________ {tex}\frac 13{/tex}
30% < {tex}\frac 13{/tex}
{tex}(30 \%=\frac {30 }{ 100}=0.3{/tex}, while {tex}\frac 1 3 \approx 0.333{/tex} or {tex}33.33 \%){/tex}

Q.16: Madhu and Madhav each ate biscuits of a different variety. Madhu’s biscuits had 25% sugar, while Madhav’s had 35% sugar. Can you tell who ate more sugar?

Solution:

As we just saw, percentages represent fractional quantities or proportions. It would be inappropriate to compare just the percentages when they are referring to different quantities or wholes. That is, if they both had 100 g of biscuits, then clearly Madhav ate more sugar- 35 g (35% of 100 g is 35 g per 100 g) vs. Madhu’s 25 g (25% of 100 g is 25 g per 100 g)

Q.17: We can find {tex}50 \%{/tex} of a value by multiplying {tex}\frac{1}{2}{/tex} with the value. Will multiplying the value by 0.5 also give the answer for {tex}50 \%{/tex} of the value?

Solution:

Yes, since {tex}\frac{1}{2}=0.5{/tex}.
{tex} 50 \%=\frac{50}{100}=\frac{1}{2}=\frac{0.5}{1}=0.5 . {/tex}

Q.18: The maximum marks in a test are 75. If students score 80% or above in the test, they get an A grade. How much should Zubin score at least to get an A grade?

Solution:

We can find 80% of 75 in different ways, using our understanding of fraction and decimal multiplication, as well as of proportionality.
Fraction Multiplication {tex}\rightarrow \frac{80}{100} \times 75{/tex} {tex} =\frac{4}{5} \times 75=60 . {/tex}
Decimal Multiplication {tex}\rightarrow 0.8 \times 75=60{/tex}.
Proportional Reasoning {tex}\rightarrow{/tex} Out of 100, the minimum mark is 80.
Out of 75, it is {tex} \frac{75 \times 80}{100}=60 \text {.}{/tex}

Q.19: To prepare a particular millet kanji (porridge), suppose the ratio of millet to water to be mixed for boiling is 2:7. What percentage does the millet constitute in this mixture? If 500 ml of the mixture is to be made, how much millet should be used?

Solution:

This situation can be modelled as shown in the bar model on the right side.

The ratio of millet to the volume of the mixture is 2:9. In other words, in one unit of the mixture, millet occupies {tex}\frac{2}{9}{/tex} units and water occupies {tex}\frac{7}{9}{/tex} units.

Q.20: A cyclist cycles from Delhi to Agra and completes 40% of the journey. If he has covered 92 km, how many more kilometres does he have to travel to reach Agra?

Solution:

Let us first try to model this situation by a bar model.

Q.21: Kishanlal recently opened a garment shop. He aims to achieve a daily sales of at least ₹ 5000. The sales on the first 2 days were ₹ 2000 and ₹ 3500. What percentage of his target did he achieve?

Solution:

The percentage target achieved is visualised below.

% of target:
{tex} \frac{2000}{5000} \times 100=40 \% {/tex}
{tex} 40 \%=\frac{40}{100}=\frac{2}{5}=0.4 {/tex}

% of target:
{tex} \frac{3500}{5000} \times 100=70 \% {/tex}
{tex} 70 \%=\frac{70}{100}=\frac{7}{10}=0.7 {/tex}
It is 40% on Day 1 and 70% on Day 2.
Another way of saying it is- he was 60% short of his target on Day 1 and 30% short of his target on Day 2.

Q.22: A farmer harvested 260 kg of wheat last year. This year, they harvested 650 kg of wheat. What percentage of last year’s harvest is this year’s harvest?

Solution:

This year’s harvest {tex}=\frac{650}{260} \times 100=250 \%{/tex} of last year’s harvest.
{tex}250 \%{/tex} indicates that it is 2.5 times the original value.

Q.23: Find the missing numbers.

Solution:

Given: {tex}20 \%{/tex} of a number {tex}=75{/tex}
Let the number be x.
{tex} 20 \% \text { of } x=75 {/tex}
{tex} (\frac {20 }{ 100}) \times x=75 {/tex}
{tex} x=75 \times(\frac {100 }{ 20}) {/tex}
{tex} x=75 \times 5 {/tex}
{tex} x=375 {/tex}
Therefore, {tex}100 \%{/tex} of the number {tex}={3 7 5}{/tex}.

Q.24: Find the missing numbers.

Solution:

Given: 60% of a number {tex}=90{/tex}
Let the number be x.
{tex} 60 \% \text { of } x=90 {/tex}
{tex} (\frac {60 }{100}) \times x=90 {/tex}
{tex} x=90 \times(\frac {100 }{ 60}) {/tex}
{tex} x=90 \times(\frac {10 }{ 6}) {/tex}
{tex} x=\frac {900 }{ 6 }{/tex}
{tex} x=150 {/tex}
Therefore, {tex}100 \%{/tex} of the number {tex}=150{/tex}.

Q.25: Find the missing numbers

Solution:

The question needs to specify what value is obtained. Assuming we need to find {tex}100 \%{/tex} of 140:
{tex} 100 \% \text { of } 140={1 4 0} {/tex}
If a different percentage is implied, please provide the complete information.

Q.26: Find the value of 25% of 160 and also draw their bar models.

Solution:

{tex}25 \%{/tex} of {tex}160=(\frac {25 }{ 100}) \times 160{/tex}
{tex} =(\frac 1 4) \times 160 {/tex}
{tex} =\frac {160 }{ 4} {/tex}
{tex} =40 {/tex}

Q.27: Find the value of 16% of 250 and also draw their bar models.

Solution:

{tex}16 \% \text { of } 250=(\frac {16 }{ 100}) \times 250 {/tex}
{tex} =\frac {(16 \times 250) }{ 100 }{/tex}
{tex} =\frac {4000 }{ 100 }{/tex}
{tex} =40 {/tex}

Q.28: Find the value of 62% of 360 and also draw their bar models.

Solution:

{tex}62 \%{/tex} of {tex}360=(\frac {62 }{ 100}) \times 360{/tex}
{tex} =\frac {(62 \times 360) }{ 100 }{/tex}
{tex} =\frac {22,320 }{ 100 }{/tex}
{tex} =223.2 {/tex}

Q.29: Find the value of 140% of 40 and also draw their bar models.

Solution:

{tex}140 \%{/tex} of {tex}40=(\frac {140 }{ 100}) \times 40{/tex}
{tex} =\frac {(140 \times 40) }{ 100 }{/tex}
{tex} =\frac {5600 }{ 100 }{/tex}
{tex} =56 {/tex}

Q.30: Find the value of 1% of 1 hour and also draw their bar models.

Solution:

1 hour {tex}=60{/tex} minutes
{tex} 1 \% \text { of } 60 \text { minutes }=(\frac {1 }{ 100}) \times 60 {/tex}
{tex} =\frac {60 }{100 }{/tex}
{tex} =0.6 \text { minutes } {/tex}
{tex} =0.6 \times 60 \text { seconds } {/tex}
{tex} =36 \text { seconds}{/tex}
Therefore, {tex}1 \%{/tex} of 1 hour = {tex}{0 . 6}{/tex} minutes or {tex}{3 6}{/tex} seconds.

Q.31: Find the value of 7% of 10 kg and also draw their bar models.

Solution:

{tex}7 \%{/tex} of {tex}10 {~kg}=(\frac {7 }{ 100}) \times 10{/tex}
{tex} =\frac {70 }{ 100 }{/tex}
{tex} =0.7 {~kg} {/tex}
{tex} =700 \text { grams}{/tex}
Therefore, 7% of 10 kg = {tex}{0 . 7 ~ k g}{/tex} or {tex}{7 0 0}{/tex} grams.

Q.32: Surya made 60 ml of deep orange paint, how much red paint did he use if red paint made up {tex}\frac{3}{4}{/tex} of the deep orange paint?

Solution:

Total deep orange paint {tex}=60 {ml}{/tex}
Red paint {tex}={/tex} {tex}\frac 34{/tex} of total paint
Red paint {tex}=(\frac 3 4) \times 60 {ml}{/tex}
{tex} =\frac {(3 \times 60) }{ 4} {ml} {/tex}
{tex} =\frac {180 }{ 4 }{ml} {/tex}
{tex} =45 \ {ml} {/tex}
Therefore, Surya used {tex}{4 5 ~ m l}{/tex} of red paint.

Q.33: 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.

{tex}50 \% \text { of } 510 \ \square \ 50 \% \text { of } 515 {/tex}
{tex}37 \% \text { of } 148\ \square \ 73 \% \text { of } 148 {/tex}

Solution:

{tex}50 \%{/tex} of {tex}510=(\frac {50 }{ 100}) \times 510=255{/tex}
{tex} 50 \% \text { of } 515=(\frac {50 }{ 100}) \times 515=257.5 {/tex}
{tex} 255<257.5 {/tex}
Therefore, 50% of 510 < 50% of 515
Since the base value (148) is the same, we can directly compare the percentages.
{tex} 37 \%<73 \% {/tex}
Therefore, 37% of {tex}148<73 \%{/tex} of 148

Q.34: 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.

{tex}29 \% \text { of } 43 \ \square \ 92 \% \text { of } 110 {/tex}
{tex}30 \% \text { of } 40 \ \square \ 40 \% \text { of } 50 {/tex}

Solution:

{tex}29 \%{/tex} of {tex}43 \approx 0.29 \times 43 \approx 12.47{/tex}
{tex}92 \%{/tex} of {tex}110=0.92 \times 110=101.2{/tex}
12.47 < 101.2
Therefore, 29% of {tex}43<92 \%{/tex} of 110
30% of 40 {tex}=(\frac {30 }{ 100}) \times 40=12{/tex}
{tex}40 \%{/tex} of {tex}50=(\frac {40 }{ 100}) \times 50=20{/tex}
12 < 20
Therefore, {tex}30 \%{/tex} of {tex}40<40 \%{/tex} of 50

Q.35: 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.

{tex}45 \%{/tex} of 200 {tex}\square{/tex} 10% of 490
{tex}30 \%{/tex} of 80 {tex}\square{/tex} {tex}24 \%{/tex} of 64

Solution:

45% of {tex}200=(\frac {45 }{ 100}) \times 200=90{/tex}
{tex} 10 \% \text { of } 490=(\frac {10 }{ 100}) \times 490=49 {/tex}
{tex} 90>49 {/tex}
Therefore, 45% of {tex}200>10 \%{/tex} of 490
{tex}30 \%{/tex} of {tex}80=(\frac {30 }{ 100}) \times 80=24{/tex}
{tex} 24 \% \text { of } 64=(\frac {24 }{ 100}) \times 64=15.36 {/tex}
{tex} 24>15.36 {/tex}
Therefore, {tex}30 \%{/tex} of {tex}80>24 \%{/tex} of 64

Q.36: 30% of k is 70, 60% of k is ________, 90% of k is ________, 120% of k is ________.

Solution:

Given: 30% of {tex}{k}=70{/tex}
First, let’s find k:
{tex} (\frac {30 }{ 100}) \times k=70 {/tex}
{tex} k=70 \times(\frac {100 }{ 30}) {/tex}
{tex} k=\frac {7000 }{ 30 }{/tex}
{tex} k=\frac {700 }{ 3}=233.33 {/tex}
Now finding the required values:
60% of {tex}{k}=60 \%{/tex} is double of 30%
So, {tex}60 \%{/tex} of {tex}{k}=2 \times 70=140{/tex}
{tex}{9 0 \%}{/tex} of {tex}{k}=90 \%{/tex} is three times of {tex}30 \%{/tex}
So, {tex}90 \%{/tex} of {tex}{k}=3 \times 70=210{/tex}
120% of {tex}{k}=120 \%{/tex} is four times of 30%
So, {tex}120 \%{/tex} of {tex}{k}=4 \times 70={2 8 0}{/tex}

Q.37: 100% of m is 215, 10% of m is ________, 1% of m is ________, 6% of m is ________.

Solution:

Given: 100% of m = 215
This means {tex}{m}=215{/tex}
{tex} {1 0 \%} \text { of } {m}=(\frac {10 }{ 100}) \times 215=\frac {215 }{ 10}={2 1 . 5} {/tex}
{tex} {1 \%} \text { of } {m}=(\frac {1 }{ 100}) \times 215=\frac {215 }{ 100}={2 . 1 5} {/tex}
{tex} {6 \%} \text { of } {m}=6 \times(1 \% \text { of } m)=6 \times 2.15={1 2 . 9} {/tex}

Q.38: 90% of n is 270, 9% of n is ________, 18% of n is ________, 100% of n is ________.

Solution:

Given: {tex}90 \%{/tex} of {tex}n=270{/tex}
9% of {tex}{n}={9 \%}{/tex} is one-tenth of 90%
So, {tex}9 \%{/tex} of {tex}n=\frac {270 }{ 10}=30{/tex}
18% of {tex}{n}=18 \%{/tex} is double of 9%
So, {tex}18 \%{/tex} of {tex}n=2 \times 30=60{/tex}
100% of {tex}n={/tex} First find {tex}n{/tex}:
{tex}(\frac {90 }{ 100}) \times n=270{/tex}
{tex}n=270 \times(\frac {100 }{ 90}){/tex}
{tex}{n}=\frac {27000 }{ 90}{/tex}
{tex}n=300{/tex}
Therefore, 100% of {tex}n={3 0 0}{/tex}

Q.39: 3 is ________% of 300.

Solution:

Let the percentage be {tex}{x} \%{/tex}.
{tex} x \% \text { of } 300=3 {/tex}
{tex} (\frac {x }{ 100}) \times 300=3 {/tex}
{tex} 3 x=3 {/tex}
{tex} x=\frac 33=1 {/tex}
Therefore, 3 is {tex}{1 \%}{/tex} of 300.

Q.40: ________ is 40% of 4.

Solution:

Let the number be y.
{tex} y=40 \% \text { of } 4 {/tex}
{tex} y=(\frac {40 }{ 100}) \times 4 {/tex}
{tex} y=\frac {160 }{ 100} {/tex}
{tex} y=1.6 {/tex}
Therefore, {tex}{1 . 6}{/tex} is {tex}40 \%{/tex} of 4.

Q.41: 40 is 80% of ________.

Solution:

Let the number be {tex}z{/tex}.
{tex} 40=80 \% \text { of } z {/tex}
{tex} 40=(\frac {80 }{ 100}) \times z {/tex}
{tex} 40=(\frac 4 5) \times z {/tex}
{tex} z=40 \times(\frac 5 4) {/tex}
{tex} z=\frac {200 }{ 4 }{/tex}
{tex} z=50 {/tex}
Therefore, 40 is {tex}80 \%{/tex} of 50.

Q.42: Is 10% of a day longer than 1% of a week?

Solution:

Let’s calculate both:
{tex}{1 0 \%}{/tex} of a day: 1 day {tex}=24{/tex} hours {tex}10 \%{/tex} of 24 hours {tex}=(\frac {10 }{ 100}) \times 24=2.4{/tex} hours
1% of a week: 1 week {tex}=7{/tex} days {tex}=7 \times 24=168{/tex} hours 1% of 168 hours {tex}=(\frac {1 }{ 100}) \times{/tex} 168 = 1.68 hours
Comparing: 2.4 hours > 1.68 hours
Therefore, Yes, 10% of a day (2.4 hours) is longer than 1% of a week (1.68 hours).

Q.43: 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:

Let’s calculate the percentage eaten each day:
Day 1: Ate 1 out of {tex}2=(\frac 1 2) \times 100 \%=50 \%{/tex}
Day 2: Ate 2 out of {tex}3=(\frac 2 3) \times 100 \%=66.67 \%{/tex}
Day 3: Ate 3 out of {tex}4=(\frac 3 4) \times 100 \%=75 \%{/tex}
Day 4: Ate 4 out of {tex}5=(\frac 4 5) \times 100 \%=80 \%{/tex}
Day 5: Ate 5 out of {tex}6=(\frac 5 6) \times 100 \%=83.33 \%{/tex}
…continuing this pattern…
Day 10: Ate 10 out of {tex}11=(\frac {10 }{ 11}) \times 100 \%=90.91 \%{/tex}
Day 20: Ate 20 out of {tex}21=(\frac {20 }{ 21}) \times 100 \%=95.24 \%{/tex}
Day 50: Ate 50 out of {tex}51=(\frac {50 }{ 51}) \times 100 \%=98.04 \%{/tex}
Day 99: Ate 99 out of {tex}100=(\frac {99 }{ 100}) \times 100 \%=99 \%{/tex}
Observation: The percentage of fodder eaten by the bull increases each day and approaches {tex}100 \%{/tex} as the days progress. The bull’s eating percentage follows the pattern: {tex}(\frac {n }{(n+1)}) \times 100 \%{/tex}, where {tex}n{/tex} is the day number. As {tex}n{/tex} increases, the percentage gets closer and closer to {tex}100 \%{/tex}, but never quite reaches it.

Q.44: 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:

Time taken for {tex}20 \%{/tex} of plantation {tex}=18{/tex} days
Rate of work remains constant

Method 1: Using Proportion
If 20% takes 18 days, then:

{tex}20 \% \rightarrow 18{/tex} days
{tex}100 \% \rightarrow{/tex} ?

Using proportion: {tex}20: 18:: 100: x{/tex}
{tex} x=\frac {(100 \times 18) }{ 20 }{/tex}
{tex} x=\frac {1800 }{ 20 }{/tex}
{tex} x=90 \text { days}{/tex}
Method 2: Using Logic
{tex}20 \%{/tex} of work takes 18 days {tex}100 \%=5 \times 20 \%{/tex}
So, {tex}100 \%{/tex} will take {tex}=5 \times 18=90{/tex} days
Why is the assumption necessary?
The assumption that the rate of work stays the same is necessary because:

Workers might get tired over time, reducing their efficiency
Different parts of the plantation might have varying difficulty levels
Weather conditions might change
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 {tex}{9 0}{/tex} days to complete the entire plantation.

Q.45: 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:

Total training time {tex}=90{/tex} minutes
The ratio is: Warm up : Play : Cool down {tex}=10 \%: 80 \%: 10 \%{/tex}
Warm up time: {tex}10 \%{/tex} of 90 minutes {tex}=(\frac {10 }{ 100}) \times 90=9{/tex} minutes
Play time: 80% of 90 minutes {tex}=(\frac {80 }{100}) \times 90=72{/tex} minutes
Cool down time: {tex}10 \%{/tex} of 90 minutes {tex}=(\frac {10 }{ 100}) \times 90=9{/tex} minutes
Verification: {tex}9+72+9=90{/tex} minutes
Therefore:

Warm up {tex}=9{/tex} minutes
Play {tex}=72{/tex} minutes
Cool down {tex}=9{/tex} minutes

Q.46: 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:

Current world population {tex}(2024) \approx 8{/tex} billion {tex}(8,000,000,000){/tex}
Population in Northern Hemisphere {tex}=90 \%{/tex} of world population
{tex} =(\frac {90 }{ 100}) \times 8,000,000,000 {/tex}
{tex} =(\frac {9 }{ 10}) \times 8,000,000,000 {/tex}
{tex} =\frac {72,000,000,000 }{ 10} {/tex}
{tex} =7,200,000,000 {/tex}
{tex} =7.2 \text { billion}{/tex}
Therefore, approximately {tex}{7 . 2}{/tex} billion people (or {tex}{7 , 2 0 0}{/tex} million people) live in the Northern Hemisphere.

Q.47: A recipe for the dish, halwa, for 4 people has the following ingredients in the given proportions- Rava: 40%, Sugar: 40%, and Ghee: 20%.

If you want to make halwa for 8 people, what is the proportion of each of the above ingredients?
If the total weight of the ingredients is 2 kg, how much rava, sugar and ghee are present?

Solution:

The proportions remain the same regardless of the number of people.
When we increase the quantity from 4 people to 8 people, we double the amounts, but the proportions (percentages) stay constant.
Therefore:
- Rava: 40%
- Sugar: 40%
- Ghee: 20%
The proportions do not change when scaling the recipe.
Total weight {tex}=2 {~kg}=2000{/tex} grams
Rava: {tex}40 \%{/tex} of {tex}2000 {~g}=(\frac {40 }{ 100}) \times 2000=800{/tex} grams {tex}=0.8 {~kg}{/tex}
Sugar: 40% of {tex}2000 {~g}=(\frac {40 }{ 100}) \times 2000=800{/tex} grams {tex}=0.8 {~kg}{/tex}
Ghee: 20% of {tex}2000 {~g}=(\frac {20 }{ 100}) \times 2000=400{/tex} grams {tex}=0.4 {~kg}{/tex}
Verification: {tex}800+800+400=2000{/tex} grams
Therefore:
- Rava {tex}=800{/tex} grams {tex}(0.8 {~kg}){/tex}
- Sugar {tex}=800{/tex} grams (0.8 kg)
- Ghee {tex}=400{/tex} grams {tex}(0.4 {~kg}){/tex}

Q.48: Eesha scored 42 marks out of 50 on an English test and 70 marks out of 80 in a Science test. Since she lost only 8 marks in English but 10 marks in Science, she thinks she has done better at English. Reema does not agree! She argues that since Eesha has scored more marks in Science, she has done better at Science. Vishu thinks we cannot compare the scores because the maximum marks are different. Who do you think is correct?

Solution:

If the maximum marks are the same, the comparison becomes easier, isn’t it? For these kinds of comparisons, we need to convert both values to percentages.
English score as a percentage {tex}=\frac{42}{50} \times 100=84 \%{/tex}
Science score as a percentage {tex}=\frac{70}{80} \times 100=87.5 \%{/tex}.
The Science score (as a percentage) is higher than the English score (as a percentage). So, we can conclude that Eesha has scored better on the Science test.

Q.49: Madhu and Madhav recently learnt about the importance of reading labels on processed food before purchase. They are at a shop to buy badam drink mix. They are looking at two products and wondering which has a larger share of badam. Can you figure it out? Which product uses a smaller proportion of food chemicals?

Solution:

It is easier to compare the proportions of the ingredients if we convert them into percentages. For example,
DEF’s sugar content as a percentage of total weight {tex}=\frac{99}{150} \times 100=66 \%{/tex}

Q.50: Do the following two statements mean the same thing?

The population of this state in 1991 is 165% of that in 1961.
The population of this state has increased by 65% from 1961 to 1991.

Solution:

Yes, both mean the same. Suppose p is the population of the state in 1961 and q is the population of the state in 1991.
Statement A implies,
{tex} q=165 \% \text { of } p {/tex}
{tex} q=\frac{165}{100} \times p=1.65 p {/tex}
Statement B implies,
{tex} q=p+65 \% \text { of } p {/tex}
{tex} q=p+0.65 \times p=1.65 p {/tex}
In other words, the population of the state in 1991 is 1.65 times that in 1961.

Q.51: Find out the percentage profit Kishanlal made on this sweater.

Solution:

We shall consider the cost price to be 100% to find out the percentage profit made with reference to the cost price. The following rough diagram describes this situation.
The profit amount is ₹ 130.
The percentage profit is {tex}\frac{130}{300} \times 100=43.3 \%{/tex}.

Q.52: 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.

Solution:

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 {tex}\frac{50}{350} \times 100=14.28 \%{/tex}.

Q.53: Shyamala had procured decorative vases at ₹ 2650 per piece. One of the pieces was slightly damaged. She decides to sell it at a loss of 18%. How much will she get by selling this piece?

Solution:

Two methods of solving this are shown.
Method 1:
With respect to the buying price being 100%, the selling price is {tex}18 \%{/tex} less than the buying price. That is, the selling price would be {tex}82 \%{/tex}.
{tex} 82 \% \text { of } 2650=0.82 \times 2650=2173 . {/tex}
Method 2:
The loss amount is 18% of 2650.
That is, {tex}\frac {18}{100}\times 2650{/tex} = 477.
Reducing this from the buying price, 2650 – 477 = 2173.
The sale amount of the damaged vase would be ₹ 2173.

Q.54: If one deposits ₹ 6000 in the bank, what is the amount after 3 years?

Solution:

That depends on the choice of FD. There are two possibilities:

Option 1: The interest is paid out regularly (for example, every year).
The principal amount is returned after the maturity period.
Option 2: The interest gained every time (say after each year) is added back to the FD, thus increasing the principal amount for the subsequent period. After the maturity period, the entire amount is returned. This phenomenon is called compounding.

We can see that with compounding, the final amount is more.

Q.55: What percent is the total amount received with respect to the amount deposited in both the options?

Solution:

This can be calculated by finding {tex}\frac{\text { total amount received }}{\text { amount deposited }} \times 100{/tex}.
Without Compounding
{tex} \frac{7800}{6000} \times 100=130 \%=1.3 \text {.}{/tex}
In other words, the total amount
{tex} \text { received }= 6000 \times(1+0.1+0.1+0.1) {/tex}
{tex} = 6000 \times 1.3 {/tex}
The percentage gain over 3 years is {tex}30 \%{/tex}.
With Compounding
{tex} \frac{7986}{6000} \times 100=133.1 \%=1.331 {/tex}
In other words, the total amount received
{tex} =6000 \times 1.1 \times 1.1 \times 1.1 {/tex}
{tex} =6000 \times 1.331 {/tex}
The percentage gain over 3 years is {tex}33.1 \%{/tex}.

Q.56: What is the amount we get back if we invest ₹ 6000 at an interest rate of 10% p.a. for ‘t’ years?

Solution:

No Compounding
Here, the interest gained every term is paid back. Therefore, the principal amount for every term shall remain the same, and as a result, the interest gained every term also shall be the same.
The interest gained in 1 term is {tex}6000 \times 0.1{/tex}
The interest gained in 3 terms is {tex} 6000 \times 0.1 \times 3 {/tex}
The total amount at the end of an FD of 3 years is {tex} 6000+(6000 \times 0.1 \times 3) . {/tex}
The interest gained in 1 term is {tex}p \times r{/tex} ({tex}p{/tex} is the principal, {tex}r{/tex} is the rate of interest in percentage)
The interest gained in {tex}t{/tex} terms is {tex} p \times r \times t {/tex}
The total amount at the end of an FD of {tex}t{/tex} years is {tex} p+(p \times r \times t)=p+p r t {/tex}
{tex} =p(1+r t) . {/tex}

Q.57: A TV is bought at a price of ₹ 21,000. After 1 year, the value of the TV depreciates by 5%. Find the value of the TV after one year.

Solution:

The amount of reduction in the {tex}\text {value is } 5 \% \text { of } {/tex}21,000
{tex} =0.05 \times 21,000 {/tex} {tex} =1050 {/tex}.
The current value is {tex}21,000-1050{/tex}
= 19,950.
The value of the TV after 1 year will be {tex}95 \%{/tex} of the current value
{tex} =95 \% \text { of } 21,000 =0.95 \times 21,000 {/tex}
{tex} =19,950 . {/tex}
The value of the TV after 1 year will be ₹ 19,950.

Q.58: The population of a village was observed to be reducing by about 10% every decade. If the current population is 1250, what is the expected population after 3 decades?

Solution:

The population 1 decade later will be 0.9 times the population of the current decade.
Therefore, the population after 1 decade will be {tex}1250 \times 0.9{/tex}.
The population after 2 decades will be {tex}1250 \times 0.9 \times 0.9{/tex}.
The population after 3 decades will be {tex}1250 \times 0.9 \times 0.9 \times 0.9 =911.25{/tex}.
First {tex}\text {decade’s decrease } =0.1 \times 1250 {/tex} {tex} =125 {/tex}
Population after 1 decade {tex} =1250-125=1125 . {/tex}
Second decade’s decrease {tex} =0.1 \times 1125=112.5 \cong 112 . {/tex}
Population after 2 decades {tex} \text { = } 1125-112=1013 {/tex}
Third decade’s population decrease {tex} =0.1 \times 1013=101.3 \cong 101 . {/tex}
Population after 3 decades {tex} \text { = } 1013-101 \text { = } 912 . {/tex}
Rounding off, we can say that the expected population after 3 decades will be around 910.

Q.59: A bakery called Cakely is offering a 30% + 20% discount on all cakes. Another bakery called Cakify is offering a 50% discount on all cakes. Would you rather choose Cakely or Cakify if you want the cheaper cost?

Solution:

It seems that both the options should give the same benefit. Although mathematically {tex}30 \%+20 \%{/tex} is the same as {tex}50 \%{/tex}, the usage of {tex}30 \%+20 \%{/tex} in shopping means compounding.
Suppose you want to buy a cake worth ₹ 200.
Cakely’s {tex}30 \%+20 \% \rightarrow{/tex}
Applying the {tex}30 \%{/tex} discount {tex}\rightarrow{/tex} the price of cake is ₹ 200 – ₹ 60 = ₹ 140.
Applying the {tex}20 \%{/tex} discount on ₹ 140 {tex}\rightarrow{/tex} the price of cake is ₹ 140 – ₹ 28 = ₹ 112.
Cakify’s 50% {tex}\rightarrow{/tex} The 50% discount makes the price of the cake ₹ 100.

Q.60: After Surbhi bought cookware from the wholesaler, she kept a profit margin of 50% on all the products. To clear off the remaining stock, she thought she would offer a 50% discount and come out without any loss.

Do you think she didn’t make any loss?
If she had sold goods (originally) for ₹ 12,000 after discount, how much loss did she incur? What is the percentage loss?
What should have been the percentage discount offered so that she sold the goods at the price she had bought (i.e., no profit or loss)?

Solution:

Let us model the situation first.
Suppose the worth of the products she bought from the wholesaler is {tex}x{/tex}.
The worth corresponding to the selling price (with a {tex}50 \%{/tex} margin) is {tex}1.5 x{/tex}.
A {tex}50 \%{/tex} discount on this price will make the worth {tex}0.75 x{/tex}.

This means the selling price is {tex}\frac{3}{4}{/tex} of the price the goods were bought at, i.e., a {tex}25 \%{/tex} loss.
If she had sold goods worth ₹ 12,000,
{tex} 0.75 x=12,000 {/tex}
{tex} x=16,000 . {/tex}
She lost ₹ 4000.
To sell the goods at the same price, the discount offered should be
{tex} 1.5 x-d \times(1.5 x)=x {/tex}
{tex} d=\frac{1}{3}=0.33 {/tex}
The discount offered should have been {tex}33.33 \%{/tex}.

Q.61: 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:

Cost Price (CP) {tex}={/tex} ₹ 75
Selling Price (SP) = ₹ 110

Step 1: Calculate profit Profit {tex}=S P-C P{/tex}
Profit {tex}=110-75{/tex}
Profit {tex}=₹ 35{/tex}
Step 2: Calculate profit percentage with respect to cost Profit percentage = {tex}\frac {\text {Profit}}{\text {CP}}{/tex}{tex}\times 100 \%{/tex} Profit percentage {tex}=(\frac {35 }{ 75}) \times 100 \%{/tex}
Profit percentage {tex}=0.4667 \times 100 \%{/tex}
Profit percentage {tex}=46.67 \%{/tex} (approximately {tex}47 \%{/tex})
Therefore, the shopkeeper’s profit margin is {tex}46.67 \%{/tex}.

Q.62: 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:

Cost Price {tex}({CP})={/tex} ₹ 475
Profit margin {tex}=50 \%{/tex}

Step 1: Calculate profit amount Profit {tex}=50 \%{/tex} of CP Profit {tex}=(\frac {50 }{ 100}) \times 475{/tex} Profit {tex}= 0.50 \times 475{/tex} Profit {tex}=₹ 237.50{/tex}
Step 2: Calculate Selling Price {tex}S P=C P+{/tex} Profit
{tex}S P=475+237.50 {/tex}
SP = ₹ 712.50
Alternative Method: {tex}{SP}={CP} \times(1+{/tex} profit percentage)
{tex} {SP}=475 \times(1+0.50){/tex}
{tex}SP= 475 \times 1.50 {/tex}
{tex}{SP}=₹ \ 712.50{/tex}
Therefore, the carpenter should sell each chair at ₹ 712.50.

Q.63: 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:

Revenue (Total Sales) = ₹ 2.5 crore
Profit margin {tex}=25 \%{/tex}

Understanding: Profit margin of {tex}25 \%{/tex} means profit is {tex}25 \%{/tex} of the revenue.
Step 1: Calculate profit amount Profit {tex}=25 \%{/tex} of Revenue
Profit {tex}=(25 / 100) \times 2.5{/tex} crore
Profit {tex}=0.25 \times 2.5{/tex} crore
Profit {tex}=₹ 0.625{/tex} crore
Step 2: Calculate total expenditure Revenue {tex}={/tex} Expenditure + Profit {tex}2.5={/tex} Expenditure + 0.625
Expenditure {tex}=2.5-0.625{/tex}
Expenditure {tex}=₹ 1.875{/tex} crore
Converting to easier format: ₹ 1.875 crore {tex}={/tex} ₹ 1,87,50,000
Therefore, the total expenditure of the company last year was ₹ 1.875 crore or ₹ 1,87,50,000.

Q.64: 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:

Original Price (Marked Price) = ₹ 300
Discount = 25%

Method 1:
Step 1: Calculate discount amount
Discount {tex}=25 \%{/tex} of 300
Discount {tex}=(25 / 100) \times{/tex} 300
Discount {tex}=0.25 \times 300{/tex}
Discount {tex}=₹ 75{/tex}
Step 2: Calculate selling price Selling Price {tex}={/tex} Original Price – Discount SP {tex}=300- 75{/tex}
SP = ₹ 225
Method 2: {tex}S P={/tex} Original Price {tex}\times(1-{/tex} discount percentage)
{tex} S P=300 \times(1-0.25) {/tex}
{tex}S P= 300 \times 0.75 {/tex}
SP = ₹ 225
Therefore, Anwar will have to pay ₹ 225 for the shirt.

Q.65: The petrol price in 2015 was ₹ 60 and ₹ 100 in 2025. What is the percentage increase in the price of petrol?

50%
40%
60%
66.66%
140%
160.66%

Solution:

Given:

Price in 2015 = ₹ 60
Price in {tex}2025=₹ \ 100{/tex}

Step 1: Calculate increase in price Increase = Price in 2025 – Price in 2015
Increase = 100 – 60 Increase = ₹ 40
Step 2: Calculate percentage increase Percentage increase = (Increase/Original Price{tex}) \times 100 \%{/tex}
Percentage increase {tex}=(40 / 60) \times 100 \%{/tex}
Percentage increase {tex}=(2 / 3) \times 100 \%{/tex}
Percentage increase {tex}=66.67 \%{/tex} (approximately {tex}66.66 \%{/tex})
Therefore, the correct answer is (iv) 66.66%.

Q.66: 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:

Given:

Price paid after discount {tex}({SP})=₹ 4,40,000{/tex}
Discount {tex}=15 \%{/tex}

Understanding: If there’s a {tex}15 \%{/tex} discount, Samson paid {tex}85 \%{/tex} of the original price.
Step 1: Set up the equation {tex}85 \%{/tex} of Original Price
{tex}=₹ 4,40,000(85 / 100) \times{/tex} Original Price
{tex}=4,40,0000.85 \times{/tex} Original Price
{tex}=4,40,000{/tex}
Step 2: Calculate original price
Original Price {tex}=4,40,000 / 0.85{/tex}
Original Price {tex}= 4,40,000 \div 0.85{/tex}
Original Price {tex}={/tex} ₹ 5,17,647.06 (approximately ₹ 5,17,647)
Verification: Discount {tex}=15 \%{/tex} of 5,17,647 {tex}=0.15 \times 5,17,647=₹ 77,647{/tex}
Price after discount {tex}=5,17,647-77,647=₹ 4,40,000 \checkmark{/tex}
Therefore, the original price of the car was approximately ₹ 5,17,647.

Q.67: 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:

Part 1: Percentage of votes won
Given:

Total votes {tex}=1600{/tex}
Winner’s votes {tex}=500{/tex}

Percentage of votes {tex}=({/tex}Winner’s votes/Total votes{tex}) \times 100 \%{/tex}
Percentage of votes {tex}= (500 / 1600) \times 100 \%{/tex}
Percentage of votes {tex}=(5 / 16) \times 100 \%{/tex}
Percentage of votes {tex}={/tex} 31.25%
Part 2: Minimum number of candidates
The winner got 500 votes out of 1600. Remaining votes {tex}=1600-500=1100{/tex} votes
For the winner to win with 500 votes, no other candidate should have more than 500 votes.
If there was only 1 other candidate, that candidate would have 1100 votes (more than the winner) – so the winner wouldn’t win.
If there were 2 other candidates sharing 1100 votes equally: Each would get 550 votes (still more than 500) – winner wouldn’t win.
If there were 3 other candidates: Maximum any could get (if votes distributed unevenly) could be more than 500.
For the winner to definitely win with 500 votes, we need at least 3 candidates total (winner +2 others), where the other 1100 votes are distributed such that no single candidate gets more than 500.
Therefore, the winner got {tex}31.25 \%{/tex} of total votes, and the minimum number of candidates who stood for the election is {tex}{3}{/tex} candidates.

Q.68: 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:

Price in 2024 = ₹ 38 per kg
Price in 2025 = ₹ 42 per kg

Step 1: Calculate increase in price Increase = Price in 2025 – Price in 2024 Increase = 42 – 38 Increase = ₹ 4
Step 2: Calculate rate of inflation
Rate of inflation {tex}=({/tex}Increase/Original Price{tex}) \times 100 \%{/tex}
Rate of inflation {tex}=(4 / 38) \times 100 \%{/tex}
Rate of inflation {tex}=0.1053 \times 100 \%{/tex}
Rate of inflation = 10.53% (approximately)
Therefore, the rate of inflation is approximately {tex}10.53 \%{/tex}.

Q.69: A number increased by 20% becomes 90. What is the number?

Solution:

Given:

A number increased by {tex}20 \%=90{/tex}

Let the original number be x.
Step 1: Set up the equation Number {tex}+20 \%{/tex} of Number {tex}=90 {x}+(20 / 100) \times {x}{/tex}
{tex}=90 {x}+ 0.20 x=90{/tex}
{tex}1.20 x=90{/tex}
Step 2: Solve for x
{tex}{x}=90 / 1.20{/tex}
{tex} {x}=90 \div 1.20 {/tex}
{tex}{x}=75{/tex}
Verification: 20% of 75 {tex}=(20 / 100) \times 75=15{/tex}
{tex}75+15=90{/tex}
Therefore, the number is 75.

Q.70: 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:

Given:

Selling price of each buffalo = ₹ 80,000
Buffalo 1: Profit of {tex}5 \%{/tex}
Buffalo 2: Loss of {tex}10 \%{/tex}

For Buffalo 1 (5% profit):
{tex} S P=C P+5 \% \text { of } C P \ 80,000{/tex} {tex}=C P \times(1+0.05) \ 80,000{/tex}
{tex}=C P \times 1.05 \ C P_1{/tex} {tex}=\frac {80,000 }{ 1.05} {/tex}
{tex} C P_1=₹ \ 76,190.48 {/tex}
For Buffalo 2 (10% loss):
{tex} S P=C P-10 \% \text { of } C P{/tex}
80,000 {tex}=C P \times(1-0.10) {/tex}
80,000 {tex}=C P \times 0.90 C P_2\ =\frac {80,000 }{ 0.90} {/tex}
{tex} C P_2=₹ \ 88,888.89 {/tex}
Overall calculation:
Total Cost Price {tex}={CP}_1+{CP}_2{/tex}
Total {tex}{CP}=76,190.48+88,888.89{/tex}
Total {tex}{CP}={/tex} ₹ 1,65,079.37
Total Selling Price {tex}=80,000+80,000{/tex}
Total SP {tex}=₹ \ 1,60,000{/tex}
Overall Loss {tex}={/tex} Total CP – Total SP
Overall Loss {tex}=1,65,079.37-1,60,000{/tex}
Overall Loss = ₹ 5,079.37
Percentage Loss {tex}=(5,079.37 / 1,65,079.37) \times 100 \%{/tex}
Percentage Loss {tex}=3.08 \%{/tex}
Therefore, the milkman incurred an overall loss of approximately ₹ 5,079 or 3.08%.

Q.71: 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

p {tex}\times{/tex} 0.5
p {tex}\times{/tex} 0.05
p {tex}\times{/tex} 1.5
p {tex}\times{/tex} 1.05
p + 1.50

Solution:

Given:

Original population {tex}={p}{/tex}
Increase {tex}=5 \%{/tex}

Step 1: Calculate new population
New population {tex}={/tex} Original {tex}+5 \%{/tex} of Original
New population {tex}=p+(5 / 100) \times p{/tex}
New population {tex}=p+0.05 p{/tex}
New population {tex}=1.05 p{/tex}
New population {tex}={p} \times 1.05{/tex}
Therefore, the correct answer is (iv) {tex}{p} \times {1 . 0 5}{/tex}.

Q.72: Which of the following statement(s) mean the same as – “The demand for cameras has fallen by 85% in the last decade”?

The demand now is 85% of the demand a decade ago.
The demand a decade ago was 85% of the demand now.
The demand now is 15% of the demand a decade ago.
The demand a decade ago was 15% of the demand now.
The demand a decade ago was 185% of the demand now.
The demand now is 185% of the demand a decade ago.

Solution:

Let’s analyze what “demand has fallen by {tex}85 \%{/tex}” means:
If original demand (a decade ago) {tex}=100 \%{/tex}
Fallen by {tex}85 \%{/tex} means the demand decreased by {tex}85 \%{/tex}
Current demand {tex}=100 \%-85 \%=15 \%{/tex}
So, current demand is {tex}15 \%{/tex} of what it was a decade ago.
Checking each statement:

The demand now is {tex}85 \%{/tex} of the demand a decade ago.
- This would mean demand fell by only {tex}15 \%{/tex}. Incorrect
The demand a decade ago was {tex}85 \%{/tex} of the demand now.
- This would mean demand increased. Incorrect
The demand now is {tex}15 \%{/tex} of the demand a decade ago.
- If demand fell by {tex}85 \%{/tex}, then {tex}100 \%-85 \%=15 \%{/tex} remains. Correct ✓
The demand a decade ago was {tex}15 \%{/tex} of the demand now.
- This would mean demand increased significantly. Incorrect
The demand a decade ago was {tex}185 \%{/tex} of the demand now.
- Let’s check: If current demand {tex}=15 \%{/tex}, then original {tex}=100 \%{/tex}
- {tex}100 \%=(100 / 15) \times 15 \%=6.67 \times 15 \% \approx 667 \%{/tex} of current
- We can also think: If now {tex}=15{/tex}, then ago {tex}=100{/tex}
- So ago {tex}=(100 / 15) \times{/tex} now {tex}\approx 6.67 \times{/tex} now
- Wait, let’s recalculate: If now is {tex}15 \%{/tex} of ago, then ago {tex}=(100 / 15) \times{/tex} now = 6.67 {tex}\times{/tex} now
- This doesn’t match {tex}185 \%{/tex}. Incorrect
Actually, let me recalculate ({tex}v{/tex}): If current demand {tex}=x{/tex}, and it is {tex}15 \%{/tex} of original demand {tex}({d}){/tex} Then {tex}{x}=0.15 {~d}{/tex} So {tex}{d}={x} / 0.15=(100 / 15) {x}=6.67 {x}{/tex}
This means original demand was 667% of current demand, not 185%. Incorrect
The demand now is {tex}185 \%{/tex} of the demand a decade ago.
- This would mean demand increased by {tex}85 \%{/tex}. Incorrect
Therefore, only statement (iii) is correct.

Q.73: 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:

Given:

Principal {tex}(P)=₹\ 20,000{/tex}
Rate of interest {tex}(r)=10 \%{/tex} p.a. {tex}=0.10{/tex}
Time period {tex}({t})=2{/tex} years

Without Compounding (Simple Interest):
Interest for Year 1 {tex}={P} \times {r}=20,000 \times 0.10={/tex} ₹ 2,000
Interest for Year 2 {tex}={P} \times {r}= 20,000 \times 0.10=₹ 2,000{/tex}
Total Interest {tex}=2,000+2,000=₹\ 4,000{/tex}
Total Amount {tex}={/tex} Principal + Total Interest
Total Amount {tex}=20,000+4,000=₹ 24,000{/tex}
Alternative Formula: Amount {tex}={P}(1+{rt}){/tex}
Amount {tex}=20,000(1+0.10 \times 2){/tex}
Amount {tex}= 20,000(1+0.20){/tex}
Amount {tex}=20,000 \times 1.20{/tex}
Amount {tex}=₹ 24,000{/tex}
With Compounding:
Year 1: Amount {tex}=P \times(1+r)=20,000 \times 1.10=₹ 22,000{/tex}
Year 2: Amount {tex}=22,000 \times(1 +{r}){/tex} {tex}=22,000 \times 1.10=₹ 24,200{/tex}
Comparison:

Without compounding: ₹ 24,000
With compounding: ₹ 24,200
Difference: ₹ 24,200 – ₹ 24,000 = ₹ 200

Therefore, with compounding, one gets ₹ 200 more than without compounding.

Q.74: 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:

Given:

Principal {tex}(P)=₹ 20,000{/tex}
Rate of interest {tex}(r)=5 \%{/tex} p.a. {tex}=0.05{/tex}
Time {tex}\operatorname{period}({t})=4{/tex} years

Without Compounding:
Interest per year {tex}={P} \times {r}=20,000 \times 0.05=₹ 1,000{/tex}
Total Interest for 4 years {tex}=1,000 \times{/tex} 4 = ₹ 4,000
Total Amount {tex}=P+{/tex} Total Interest
Amount {tex}=20,000+4,000=₹ \ 24,000{/tex}
Using Formula: Amount {tex}={P}(1+{rt}){/tex}
Amount {tex}=20,000(1+0.05 \times 4){/tex}
Amount {tex}= 20,000(1+0.20){/tex}
Amount {tex}=20,000 \times 1.20{/tex}
Amount {tex}=₹ 24,000{/tex}
With Compounding:
Year 1: {tex}20,000 \times 1.05=₹ 21,000{/tex}
Year 2: {tex}21,000 \times 1.05=₹ 22,050{/tex}
Year 3: {tex}22,050 \times 1.05=₹ 23,152.50{/tex}
Year {tex}4: 23,152.50 \times 1.05{/tex} {tex}=₹ 24,310.125 \approx ₹ 24,310.13{/tex}
Using Formula: Amount {tex}={P}(1+{r})^ {t}{/tex}
Amount {tex}=20,000(1.05)^4{/tex}
Amount {tex}=20,000 \times{/tex} 1.21550625
Amount {tex}={/tex} ₹ {tex}24,310.125 \approx{/tex} ₹ {tex}24,310.13{/tex}

Q.75: 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?

{tex}p \times 6 \times 4{/tex}
{tex}p \times 0.6 \times 4{/tex}
{tex}p \times \frac{0.6}{100} \times 4{/tex}
{tex}p \times \frac{0.06}{100} \times 4{/tex}
{tex}p \times 1.6 \times 4{/tex}
{tex}p \times 1.06 \times 4{/tex}
{tex}p+(p \times 0.06 \times 4){/tex}

Solution:

When compounding is not done, we use simple interest.
Formula: Amount {tex}={/tex} Principal + Simple Interest
Amount {tex}={P}+({P} \times {r} \times {t}){/tex}
Amount {tex}={P} +({P} \times 0.06 \times 4){/tex}
Amount {tex}={P}(1+0.06 \times 4){/tex}
Amount {tex}={P}(1+0.24){/tex}
Amount {tex}={P} \times 1.24{/tex}
Now let’s check each option:

{tex}{p} \times 6 \times 4=24 {p}{/tex} Incorrect
{tex}{p} \times 0.6 \times 4=2.4 {p}{/tex} Incorrect
{tex}{p} \times 0.6 / 100 \times 4={p} \times 0.006 \times 4=0.024 {p}{/tex} Incorrect
{tex}{p} \times 0.06 / 100 \times 4={p} \times 0.0006 \times 4=0.0024 {p}{/tex} Incorrect
{tex}{p} \times 1.6 \times 4=6.4 {p}{/tex} Incorrect
{tex}{p} \times 1.06 \times 4=4.24 {p}{/tex} Incorrect (This would be for compound interest formula applied incorrectly)
{tex}p+(p \times 0.06 \times 4)=p+0.24 p=1.24 p{/tex} Correct

Therefore, only option (vii) {tex}{p}+({p} \times {0 . 0 6} \times {4}){/tex} is correct.

Q.76: 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:

Given:

Principal {tex}(P)=₹ \ 50,000{/tex}
Rate {tex}(r)=7 \%{/tex} p.a. {tex}=0.07{/tex}
Time {tex}({t})=3{/tex} years

Without Compounding (Simple Interest):
Simple Interest {tex}={P} \times {r} \times {t}\ {/tex}
{tex}{\text {SI}}=50,000 \times 0.07 \times 3\ {/tex}
{tex} {\text {SI}}=50,000 \times 0.21{/tex}
{tex}{\text {SI}}=₹ \ 10,500{/tex}
Total Amount {tex}={P}+{SI}=50,000+10,500=₹ \ 60,500{/tex}
With Compounding:
Amount {tex}=P(1+r)^ t{/tex} Amount {tex}=50,000(1.07)^3{/tex}
Amount {tex}=50,000 \times 1.225043{/tex}
Amount = ₹ 61,252.15
Compound Interest {tex}={/tex} Amount – Principal
{tex}{CI}=61,252.15-50,000{/tex}
{tex} {CI}=₹ 11,252.15{/tex}
Difference: Extra interest with compounding {tex}={CI}-{SI}{/tex}
Extra interest {tex}=11,252.15{/tex} – 10,500
Extra interest = ₹ 752.15
Therefore, without compounding, one would get ₹ 10,500 as interest. With compounding, one would get ₹ 752.15 more, making the total interest ₹ 11,252.15.

Q.77: 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:

Given:

Principal for both {tex}=₹ 12,500{/tex}
Time period for both {tex}=3{/tex} years

Giridhar (12% without compounding):
Simple Interest {tex}={P} \times {r} \times {t}{/tex}
{tex} {SI}=12,500 \times 0.12 \times 3 {/tex}
{tex}{SI}=12,500 \times 0.36{/tex}
{tex}{SI}=₹\ 4,500{/tex}
Raghava (10% with compounding):
Amount {tex}=P(1+r)^ t{/tex}
Amount {tex}=12,500(1.10)^3{/tex}
Amount {tex}=12,500 \times 1.331{/tex}
Amount {tex}={/tex} ₹ 16,637.50
Compound Interest {tex}={/tex} Amount – Principal
{tex}{CI}=16,637.50-12,500 {/tex}
{tex}{CI}=₹ 4,137.50{/tex}
Comparison: Giridhar’s interest {tex}={/tex} ₹ 4,500
Raghava’s interest {tex}={/tex} ₹ 4,137.50
Difference {tex}=4,500-4,137.50=₹ 362.50{/tex}
Therefore, Giridhar pays more interest by ₹ 362.50.

Q.78: 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:

Given:

Principal {tex}(P)=₹ 1,000{/tex}
Rate (r) = 10% p.a. = 0.10
Target amount = ₹ 2,000 (double)

Without Compounding (Simple Interest):
Amount {tex}=P(1+r t)\ 2,000{/tex}
{tex}=1,000(1+0.10 \times t) 2=1+0.10 t{/tex}
{tex}t=1 / 0.10 t=10{/tex} years
With Compounding:
Amount {tex}=P(1+r)^t\ 2,000=1,000(1.10)^t \ 2=(1.10)^t{/tex}
Taking logarithm on both sides: {tex}\log (2)={t} \times \log (1.10){/tex}
{tex} {t}=\log (2) / \log (1.10){/tex}
{tex}t= 0.3010 / 0.0414 t \approx 7.27{/tex} years
So it takes between 7 and 8 years, approximately 7.27 years.

Q.79: 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:

Given:

Current population {tex}({P})=1.5{/tex} crore {tex}=1,50,00,000{/tex}
Growth rate {tex}(r)=3 \%{/tex} per year {tex}=0.03{/tex}
Time period {tex}({t})=3{/tex} years

Population growth follows compounding (exponential growth).
Formula: Population after t years {tex}={P}(1+{r})^{{t}}{/tex}
Calculation: Population after 3 years {tex}=1.5{/tex} crore {tex}\times(1.03)^3=1.5{/tex} crore {tex}\times 1.092727{/tex} = 1.639 crore (approximately)
Detailed calculation: After Year 1: {tex}1.5 \times 1.03=1.545{/tex} crore
After Year 2: {tex}1.545 \times 1.03=1.59135{/tex} crore
After Year 3: {tex}1.59135 \times 1.03=1.6391{/tex} crore
Converting: 1.6391 crore {tex}=1,63,91,000{/tex}
Therefore, the expected population after 3 years is approximately 1.64 crore or 1,63,91,000.

Q.80: 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:

Given:

Initial count {tex}(P)=5,06,000{/tex}
Growth rate {tex}(r)=2.5 \%{/tex} per hour {tex}=0.025{/tex}
Time period {tex}({t})=2{/tex} hours

Bacterial growth follows compounding (exponential growth).
Formula: Number of bacteria after {tex}t{/tex} hours {tex}=P(1+r)^t{/tex}
Calculation: Number after 2 hours {tex}=5,06,000 \times(1.025)^2=5,06,000 \times 1.050625={/tex} 5,31,616.25
Since we cannot have fractional bacteria, we round to the nearest whole number.
Detailed calculation: After 1 hour: 5,06,000 {tex}\times 1.025=5,18,650{/tex}
After 2 hours: {tex}5,18,650 \times 1.025=5,31,616.25{/tex}
Therefore, the number of bacteria at the end of 2 hours is approximately {tex}5,31,616{/tex}.

Q.81: 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:

Population in {tex}2000=50{/tex} lakhs
Population in {tex}2025=250 \%{/tex} of population in 2000

Calculation: Population in 2025 {tex}=250 \%{/tex} of 50 lakhs {tex}=(\frac {250 }{ 100}) \times 50{/tex} lakhs {tex}=2.5 \times{/tex} 50 lakhs = 125 lakhs
Converting: 125 lakhs {tex}=1.25{/tex} crore {tex}=1,25,00,000{/tex}
Understanding: 250% means 2.5 times the original value.
Therefore, the population of Bengaluru in 2025 is 125 lakhs or 1.25 crore.

Q.82: 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

Solution:

World population {tex}=8.2{/tex} billion {tex}=8,200{/tex} million
For Germany (83 million): Percentage {tex}=(83 / 8200) \times 100 \%{/tex} {tex}=0.0101 \times 100 \%= 1.01 \% \approx 1 \%{/tex}
For India ({tex}{1 . 4 6}{/tex} billion {tex}{= 1 , 4 6 0}{/tex} million): Percentage {tex}=(1,460 / 8,200) \times 100 \%{/tex} {tex}= 0.178 \times 100 \%{/tex} {tex}=17.8 \% \approx 18 \%{/tex}
For Bangladesh (175 million): Percentage {tex}=(175 / 8,200) \times 100 \%{/tex} {tex}=0.0213 \times 100 \% =2.13 \% \approx 2 \%{/tex}
For USA (347 million): Percentage {tex}=(347 / 8,200) \times 100 \%{/tex} {tex}=0.0423 \times 100 \%={/tex} 4.23% {tex}\approx{/tex} Approximately 4

Q.83: 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?

{tex}8250+18{/tex}
{tex}8250+1800{/tex}
{tex}8250+\frac{18}{100}{/tex}
{tex}8250 \times 18{/tex}
{tex}8250 \times 1.18{/tex}
{tex}8250+8250 \times 0.18{/tex}
{tex}1.8 \times 8250{/tex}

Solution:

Given:

Price = ₹ 8,250
{tex}{GST}=18 \%{/tex}

Calculating final price: GST amount {tex}=18 \%{/tex} of 8,250
GST amount {tex}=(18 / 100) \times{/tex} 8,250
GST amount {tex}=0.18 \times 8,250{/tex}
GST amount {tex}=₹ 1,485{/tex}
Final price {tex}={/tex} Price + GST amount
Final price {tex}=8,250+1,485{/tex}
Final price {tex}=₹ 9,735{/tex}
Alternative method: Final price {tex}={/tex} Price {tex}\times(1+{/tex} GST rate)
Final price {tex}=8,250 \times(1+{/tex} 0.18 )
Final price {tex}=8,250 \times 1.18{/tex}
Final price {tex}=₹ 9,735{/tex}
Now checking each option:

{tex}8250+18=8,268{/tex} Incorrect
{tex}8250+1800=10,050{/tex} Incorrect
{tex}8250+18 / 100=8250+0.18=8,250.18{/tex} Incorrect
{tex}8250 \times 18=1,48,500{/tex} Incorrect
{tex}8250 \times 1.18=9,735{/tex} Correct
{tex}8250+8250 \times 0.18=8250+1485=9,735{/tex} Correct
{tex}1.8 \times 8250=14,850{/tex} Incorrect

Therefore, options (v) {tex}{8 2 5 0} \times {1 . 1 8}{/tex} and (vi) {tex}{8 2 5 0}+{8 2 5 0} \times {0 . 1 8}{/tex} are correct.

Q.84: 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.

The population after three months was p {tex}\times{/tex} 0.05 {tex}\times{/tex} 0.02 {tex}\times{/tex} 0.03.
The population after three months was p {tex}\times{/tex} 1.05 {tex}\times{/tex} 0.98 {tex}\times{/tex} 0.97.
The population after three months was p + 0.05 – 0.02 – 0.03.
The population after three months was p.
The population after three months was more than p.
The population after three months was less than p.

Solution:

Given:

Initial population {tex}={p}{/tex}
Month {tex}1:+5 \%{/tex} change
Month 2: -2% change
Month 3: {tex}-3 \%{/tex} change

Calculating population after each month:
After Month 1: Population increases by {tex}5 \%={p} \times(1+0.05)={p} \times 1.05{/tex}
After Month 2: Population decreases by {tex}2 \%={p} \times 1.05 \times(1-0.02){/tex} {tex}={p} \times 1.05 \times 0.98{/tex}
After Month 3: Population decreases by {tex}3 \%={p} \times 1.05 \times 0.98 \times(1-0.03){/tex} {tex}={p} \times 1.05 \times 0.98 \times 0.97{/tex}
Calculating the final value: {tex}={p} \times 1.05 \times 0.98 \times 0.97={p} \times 0.998{/tex} (approximately) {tex}={/tex} 0.998p
Since {tex}0.998<1{/tex}, the final population is less than {tex}p{/tex}.
Checking each statement:

{tex}{p} \times 0.05 \times 0.02 \times 0.03={p} \times 0.00003{/tex} Incorrect – This doesn’t represent percentage changes correctly
{tex}{p} \times 1.05 \times 0.98 \times 0.97=0.998 {p}{/tex} Correct
{tex}{p}+0.05-0.02-0.03={p}{/tex} Incorrect – Percentage changes multiply, not add
The population after three months was {tex}{p} .0 .998 {p} \neq {p}{/tex} Incorrect
The population after three months was more than p. 0.998 p < p Incorrect
The population after three months was less than {tex}{p} .0 .998 {p}<{p}{/tex} Correct

Therefore, statements (ii) and (vi) are true.

Q.85: 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 of the product be ₹ 100 (for easy calculation).
Step 1: Calculate initial selling price with {tex}{3 5 \%}{/tex} profit
Selling Price {tex}={CP} \times(1+{/tex} profit percentage)
{tex} {SP}=100 \times(1+0.35) {/tex}
{tex}{SP}=100 \times 1.35 {/tex}
S.P = ₹ 135
Step 2: Calculate selling price after 30% discount
Discount {tex}=30 \%{/tex} of SP
Discount {tex}=0.30 \times 135=₹ 40.50{/tex}
Final Selling Price {tex}={SP}-{/tex} Discount
Final {tex}{SP}=135-40.50={/tex} ₹ 94.50
Alternative calculation: Final SP {tex}={SP} \times{/tex} (1- discount percentage)
Final SP {tex}=135 \times{/tex} (1-0.30)
Final SP=135 {tex}\times{/tex} 0.70
Final SP = ₹ 94.50
Step 3: Compare with cost price
Cost Price {tex}={/tex} ₹ 100
Final Selling Price {tex}={/tex} ₹ 94.50
Since Final SP < CP, the shopkeeper makes a loss.
Loss {tex}={CP}-{/tex} FinalSP
{tex}=100-94.50=₹ 5.50{/tex}
Loss Percentage {tex}=({/tex}Loss{tex}/{/tex}CP{tex}) \times 100 \%{/tex}
Loss Percentage {tex}=(5.50 / 100) \times 100 \%{/tex}
Loss Percentage {tex}=5.5 \%{/tex}
Verification using general formula: If {tex}C P=x{/tex}, then:

SP with {tex}35 \%{/tex} profit {tex}=1.35 {x}{/tex}
After {tex}30 \%{/tex} discount {tex}=1.35 x \times 0.70=0.945 x{/tex}

Since {tex}0.945<1{/tex}, there’s a loss of {tex}(1-0.945) \times 100 \%=5.5 \%{/tex}
Therefore, the shopkeeper will make a loss of {tex}{5 . 5} \boldsymbol{\%}{/tex}.
Reason: Although he initially added a {tex}35 \%{/tex} profit margin, the {tex}30 \%{/tex} discount is calculated on the increased selling price (not the cost price), which results in a larger absolute discount amount that exceeds the original profit.

Q.86: What percentage of area is occupied by the region marked ‘E’ in the figure?

Solution:

Without seeing the exact figure, I’ll explain the general approach:
Step 1: Identify the total area

The total figure represents 100%

Step 2: Determine what fraction region E occupies

Count the number of equal parts in the figure
Count how many parts region E occupies

Step 3: Calculate percentage
Percentage {tex}=({/tex}Area of {tex}{E} /{/tex} Total Area{tex}) \times 100 \%{/tex}
Example: If the figure is divided into equal parts:

If total parts {tex}=10{/tex} and E occupies 2 parts
Percentage of {tex}E=(2 / 10) \times 100 \%=20 \%{/tex}

Q.87: 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:

Calculating each:
{tex}5 \%{/tex} of {tex}40=(5 / 100) \times 40=0.05 \times 40=2{/tex}
{tex}40 \%{/tex} of {tex}5=(40 / 100) \times 5=0.40 \times 5=2{/tex}
{tex}25 \%{/tex} of {tex}12=(25 / 100) \times 12=0.25 \times 12=3{/tex}
{tex}12 \%{/tex} of {tex}25=(12 / 100) \times 25=0.12 \times 25=3{/tex}
{tex}15 \%{/tex} of {tex}60=(15 / 100) \times 60=0.15 \times 60=9{/tex}
{tex}60 \%{/tex} of {tex}15=(60 / 100) \times 15=0.60 \times 15=9{/tex}
Observation: In each pair, the results are equal!

{tex}5 \%{/tex} of {tex}40=40 \%{/tex} of 5
{tex}25 \%{/tex} of {tex}12=12 \%{/tex} of 25
{tex}15 \%{/tex} of {tex}60=60 \%{/tex} of 15

General Statement: {tex}{x \%}{/tex} of {tex}{y}={y \%}{/tex} of {tex}{x}{/tex} for any values of x and y.
Algebraic Justification:
Let’s prove that {tex}{x} \%{/tex} of y equals {tex}{y} \%{/tex} of x.
{tex}x \%{/tex} of {tex}y=(x / 100) \times y=x y / 100{/tex}
{tex}y \%{/tex} of {tex}x=(y / 100) \times x=y x / 100{/tex}
Since multiplication is commutative {tex}(x y=y x){/tex}:
{tex}x y / 100=y x / 100{/tex}
Therefore, {tex}{x} \%{/tex} of {tex}{y}={y} \%{/tex} of {tex}{x}{/tex}
This is a beautiful mathematical property that shows the symmetry in percentage calculations!

Q.88: 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.

What percentage of the students going to the excursion are Grade 8 girls?
If the total number of students going to the excursion is 160, how many of them are Grade 8 girls?

Solution:

Finding percentage of Grade 8 girls:
Grade 8 students {tex}=40 \%{/tex} of total students
Among Grade 8 students, girls {tex}=60 \%{/tex}
Grade 8 girls {tex}=60 \%{/tex} of (40% of total) {tex}=60 \%{/tex} of 40% {tex}=(60 / 100) \times(40 / 100){/tex} of total {tex}= 0.60 \times 0.40{/tex} of total {tex}=0.24{/tex} of total {tex}=24 \%{/tex} of total
Alternative method: If total students {tex}=100{/tex}
Grade 8 students {tex}=40 \%{/tex} of {tex}100=40{/tex}
Grade 8 girls {tex}=60 \%{/tex} of {tex}40=24{/tex}
Percentage {tex}=(24 / 100) \times 100 \%=24 \%{/tex}
Therefore, {tex}{2 4 \%}{/tex} of the students going to the excursion are Grade {tex}{8}{/tex} girls.
Finding number of Grade 8 girls when total {tex}=160{/tex}:
Grade 8 students = 40% of 160 {tex}=0.40 \times 160=64{/tex}
Grade 8 girls = 60% of 64 {tex}=0.60 \times 64=38.4{/tex}
The answer 38.4 suggests there might be rounding involved, but mathematically it works out. In practical terms, this would be approximately {tex}{3 8}{/tex} or {tex}{3 9}{/tex} students.
However, using exact calculation: {tex}{3 8 . 4}{/tex}, which we can round to {tex}{3 8}{/tex} students (assuming we round down) or the problem expects us to work with the exact percentage giving us the decimal answer.
Therefore, there are {tex}{3 8}{/tex} Grade {tex}{8}{/tex} girls (or precisely 38.4 as per calculation).

Q.89: 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: Selling price of 3 pencils {tex}={/tex} Cost price of 5 pencils
Let the cost price of one pencil be ₹ x.
Step 1: Set up the relationship
Cost price of 5 pencils {tex}=5 x{/tex}
Selling price of 3 pencils {tex}=5 x{/tex}
Therefore, selling price of 1 pencil {tex}=5 x / 3{/tex}
Step 2: Compare cost and selling price for 1 pencil
Cost price of 1 pencil {tex}=x{/tex}
Selling price of 1 pencil {tex}=5 x / 3{/tex}
Since {tex}5 x / 3>x{/tex} (because {tex}5 / 3>1{/tex}), the shopkeeper makes a profit.
Step 3: Calculate profit
Profit per pencil {tex}=S P-C P=5 x / 3-x{/tex} {tex}=5 x / 3-3 x / 3=2 x / 3{/tex}
Step 4: Calculate profit percentage
Profit% {tex}=({/tex}Profit{tex}/{/tex}CP){tex} \times 100 \%=(2 x / 3 \div x) \times 100 \%{/tex} {tex}=(2 x / 3 \times 1 / x) \times 100 \%{/tex} {tex}=(2 / 3) \times 100 \%=66.67 \%{/tex} (or {tex}66 \frac{2}{3} \%{/tex})
Alternative approach: If CP of 1 pencil {tex}=₹ 1{/tex}, then:

CP of 5 pencils = ₹ 5
SP of 3 pencils = ₹ 5
SP of 1 pencil {tex}={/tex} ₹ 5/3 = ₹ 1.67

Profit {tex}=1.67-1=₹ 0.67{/tex}
Profit {tex}\%=(0.67 / 1) \times 100 \%=66.67 \%{/tex}
Therefore, the shopkeeper makes a profit of {tex}{6 6 . 6 7 \%}{/tex} (or {tex}{6 6} \frac{{2}}{{3}} \boldsymbol{\%}{/tex}).

Q.90: 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 {tex}₹ 100{/tex}.
Step 1: Calculate fare after first year (3% increase)
Fare after Year {tex}1={/tex} Original fare {tex}\times(1+0.03)=100 \times 1.03=₹ 103{/tex}
Step 2: Calculate fare after second year (4% increase on Year 1 fare)
Fare after Year {tex}2={/tex} Fare after Year {tex}1 \times(1+0.04)=103 \times 1.04=₹ 107.12{/tex}
Step 3: Calculate overall percentage increase
Overall increase {tex}={/tex} Final fare – Original fare {tex}=107.12-100={/tex} ₹ 7.12
Overall percentage increase {tex}=(7.12 / 100) \times 100 \%=7.12 \%{/tex}
Alternative formula approach: Overall multiplier {tex}=1.03 \times 1.04=1.0712{/tex}
Overall increase {tex}=(1.0712-1) \times 100 \%=7.12 \%{/tex}
Therefore, the overall percentage price increase in the last 2 years is 7.12%.

Q.91: 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 original length {tex}={/tex} L
Let original breadth {tex}={/tex} B
Original area {tex}={L} \times {B}{/tex}
Step 1: Calculate new length
New length {tex}=L+10 \%{/tex} of L
{tex}=L \times(1+0.10)=1.1 L{/tex}
Step 2: Find new breadth (area remains same)
Original area {tex}={/tex} New area {tex}{L} \times {B}=1.1 {/tex}
{tex}{L} \times{/tex} (New breadth)
{tex} {B}=1.1 \times({/tex}New breadth)
New breadth = B/1.1
New breadth = (10/11)B
Step 3: Calculate decrease in breadth
Decrease = Original breadth – New breadth = B – (10/11)B = (11/11)B – (10/11)B = (1/11)B
Step 4: Calculate percentage decrease
Percentage decrease = (Decrease/Original breadth){tex} \times 100 \%=[(1 / 11) {B} \div {B}] \times 100 \%{/tex} {tex}=(1 / 11) \times 100 \%=9.09 \%{/tex} (or {tex}91 / 11 \%{/tex})
Exact answer: 100/11% = {tex}91 / 11 \%{/tex}
Therefore, the breadth decreases by exactly {tex}{9 1 / 1 1 \%}{/tex} or 9.09% (approximately).

Q.92: 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:

Total weight of packet {tex}=65 {~g}{/tex}
Assuming the percentages shown in the picture are: (Please refer to your textbook for exact percentages)
General formula: Weight of ingredient {tex}=({/tex} Percentage {tex}/ 100) \times{/tex} Total weight
Example calculation (using hypothetical percentages):
If the ingredients are:

Potato: {tex}60 \%{/tex}
Oil: 30%
Salt and spices: {tex}10 \%{/tex}

Then:

Weight of Potato {tex}=(60 / 100) \times 65{/tex} {tex}=0.60 \times 65=39 {~g}{/tex}
Weight of Oil {tex}=(30 / 100) \times 65{/tex} {tex}=0.30 \times 65=19.5 {~g}{/tex}
Weight of Salt and spices {tex}=(10 / 100) \times 65{/tex} {tex}=0.10 \times 65=6.5 {~g}{/tex}

Verification: {tex}39+19.5+6.5=65 {~g}{/tex}

Q.93: 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:

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.
For each shop, calculate the percentage discount on the items.
[Hint: Compare the free items to the total items you receive.]
Suppose you need 4 items. Which shop would you choose? Why?

Solution:

Price of one item {tex}=₹ 100{/tex}
Shop A: Buy 1 get 1 free
- Pay for 1 item, get 2 items total
- Amount paid = ₹ 100
- Items received {tex}=2{/tex}
- Effective price per item {tex}=100 / 2=₹ 50{/tex}
Shop B: Buy 2 get 1 free
- Pay for 2 items, get 3 items total
- Amount paid = ₹ 200
- Items received = 3
- Effective price per item = {tex}\frac {200 }{ 3}{/tex} = ₹ 66.67
Shop C: Buy {tex}{3}{/tex} get {tex}{1}{/tex} free
- Pay for 3 items, get 4 items total
- Amount paid = ₹ 300
- Items received {tex}=4{/tex}
- Effective price per item {tex}\frac {300}{4}{/tex} = ₹ 75
Arrangement from cheapest to costliest: Shop A (₹ 50) < Shop B (₹ 66.67) < Shop C (₹75)
Shop A: Original price for 2 items = ₹ 200 Amount paid = ₹ 100
Discount {tex}=200-100=₹ 100{/tex}
Percentage discount {tex}=(100 / 200) \times 100 \%=50 \%{/tex}
Shop B: Original price for 3 items = ₹ 300
Amount paid = ₹ 200
Discount {tex}=300-200=₹ 100{/tex}
Percentage discount {tex}=(100 / 300) \times 100 \%=33.33 \%{/tex}
Shop C: Original price for 4 items {tex}={/tex} ₹ 400
Amount paid {tex}={/tex} ₹ 300
Discount {tex}=400-300=₹ 100{/tex}
Percentage discount {tex}=(100 / 400) \times 100 \%=25 \%{/tex}
Summary:
- Shop A: 50% discount
- Shop B: 33.33% discount
- Shop C: 25% discount
Shop A:
Buy 1 get {tex}1 \rightarrow{/tex} To get 4 items, pay for 2
Cost {tex}=2 \times 100=2002 \times 100=₹ 200{/tex}
Shop B:
Buy 2 get {tex}1 \rightarrow{/tex} For 4 items, pay for 3
Cost {tex}=3 \times 100=3003 \times 100=₹ 300{/tex}
Shop C:
Buy 3 get {tex}1 \rightarrow{/tex} Get 4 items by paying for 3
Cost {tex}=3 \times 100=3003 \times 100=₹ 300{/tex}
Best choice: Shop A because it gives 4 items for the lowest cost of ₹ 200.

Q.94: 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 people = 99
Right-handed people = 1
Let x left-handed people leave the room.
Then,
Left-handed remaining {tex}=99-{x}{/tex}
Total people remaining {tex}=100-{x}{/tex}
We want left-handed people to be {tex}98 \%{/tex} of the remaining people:
{tex} \frac{99-x}{100-x} =\frac{98}{100} {/tex}
{tex} 100(99-x) =98(100-x) {/tex}
{tex} 9900-100 x =9800-98 x {/tex}
{tex} 100 =2 x {/tex}
{tex} x =50 {/tex}
Therefore, 50 left-handed people must leave the room.

Q.95: Look at the following graph.

Based on the graph, which of the following statement(s) are valid?

People in their twenties are the most computer-literate among all age groups.
Women lag behind in the ability to use computers across age groups.
There are more people in their twenties than teenagers.
More than a quarter of people in their thirties can use computers.
Less than 1 in 10 aged 60 and above can use computers.
Half of the people in their twenties can use computers.

Solution:

People in their twenties are the most computer-literate among all age groups.
Teenagers {tex}=24 \%+29 \%=53 \%{/tex}
Twenties {tex}=26 \%+37 \%=63 \%{/tex} (highest)
This statement is true.
Women lag behind in the ability to use computers across age groups.
In every age group, the percentage for females is lower than that for males.
This statement is true.
There are more people in their twenties than teenagers.
The graph shows computer usage, not population size.
This statement is false.
More than a quarter of people in their thirties can use computers.
Thirties {tex}=14 \%+25 \%=39 \%{/tex}, which is more than {tex}25 \%{/tex}.
This statement is true.
Less than 1 in 10 aged 60 and above can use computers.
Seniors {tex}=2 \%+4 \%=6 \%{/tex}, which is less than {tex}10 \%{/tex}.
This statement is true.
Half of the people in their twenties can use computers.
Twenties {tex}=63 \%{/tex}, not exactly 50%.
This statement is false.