# NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities: In our daily life, we compare many things like his bottle is filled up to two-thirds of its height, his mark is 66% of my mark, my marks increased by 15% compared to the last test, etc. In this chapter, you will learn to compare quantities using ratios, percentage. Also, you will learn concepts of profit, loss and simple interest. In CBSE NCERT solutions for class 7 maths chapter 8 Comparing Quantities, you will get questions related to the above concepts which are useful in our daily life for the basics for calculations in our life. Ratio and percentage are two things which are useful in comparing quantities.

Ratio: The ratio tells how many times one number contains another. The unit must be the same to find the ratio of two quantities. If it is not same, you first convert it into the same unit then find the ratio.

Example- Find the ratio of 500 meters and 1 Km.

First, we have to convert it to the same unit. 1Km=1000m. Now the ratio is 500:1000 or 1:2. In solutions of NCERT for class 7 maths chapter 8 comparing quantities, there are many problems where you will be using the above concept to find the ratio of quantities.

Percentage: It simply means numbers or amount in each hundred. In other words, percentages are number or ration that represents a fraction of 100. It is represented by '%'. 5% means 5 out of 100. That is 5/100= 0.05. Let's see an example to calculate the percentage

Example: Out of 50 students there are 20 boys in the class then what is the percentage of girls in the class?

Solution:

$\text{Percentage of girls in the class }= \frac{30}{50}\times 100= 60 \%$

60% of students are girls. That is 50-20=30 are girls.

There are 24 questions in 4 exercises of this chapter. You will get solutions to all these questions in CBSE NCERT solutions for class 7 maths chapter 8 comparing quantities explained in a detailed manner. You will get NCERT solutions from class 6 to 12 by clicking on the above link.

## The main and subheadings of NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities are listed below:

8.1 Introduction

8.2 Equivalent Ratios

8.3 Percentage – Another Way Of Comparing Quantities

8.3.1 Meaning Of Percentage

8.3.2 Converting Fractional Numbers To Percentage

8.3.3 Converting Decimals To Percentage

8.3.4 Converting Percentages To Fractions Or Decimals

8.3.5 Fun With Estimation

8.4 Use Of Percentages

8.4.1 Interpreting Percentages

8.4.2 Converting Percentages To “How Many”

8.4.3 Ratios To Percents

8.4.4 Increase Or Decrease As a percent

8.5 Prices Related To An Item Or Buying And Selling

8.5.1 Profit Or Loss As A Percentage

8.6 Charge Given On Borrowed Money Or Simple Interest

8.6.1 Interest For Multiple Years

## Solutions of NCERT for class 7 maths chapter 8 comparing quantities-Exercise: 8.1

Question:1 Find the ratio of:

(a) Rs 5 to 50 paise
(b) 15 kg to 210 g

(c) 9 m to 27 cm
(d) 30 days to 36 hours

(a) Rs 5 to 50 paise
First, convert the given quantities into the same units,

So,  Rs 5 = $5 \times 100$ = 500 paise
Now, Ratio = $\frac{500}{50} = 10:1$

(b) 15 kg to 210 g
Converting the given quantities into the same unit.
So, 15 kg = $15 \times 1000$ = 15,000 g
Therefore, Required ratio
$\\=\frac{15000}{210}=500:7$

(c) 9 m to 27 cm
First, convert meter into centimetre
So,  9m = $9 \times 100 = 900 cm$
Therefore, the required ratio
$\frac{900}{27}=100:3$

(d) 30 days to 36 hours
Converting the given quantities into the same unit.
So, 30 days = $30 \times 24hr = 720 hrs$
Therefore, the ratio
$\frac{720}{36} = 20:1$

Given that,
6 student use 3 computers
So, 1 computer is used by $\frac{6}{3} =2$ students.

Therefore, for 24 students no. of computer required = $\frac{24}{2} =12$ computers

Hence the required number of computer is 12.

(i) How many people are there per $km^{2}$ in both these States?

(ii) Which State is less populated?

Given that,
Population of Rajasthan = 570 lakhs and population of UP = 1660 lakhs
Area of Rajasthan =$3\: lakh \: km^{2}$
and area of UP =$2\: lakh \: km^{2}$

Now,

(a) Number of peoples per $km^2$
In Rajsthan, = $\frac{population}{area}$
$=\frac{570}{3}= 190$

In UP,
$\frac{population}{area} = \frac{1160}{2}$
= 580

(b) Rajasthan is less populated as we can see that the number of people in per $km^2$ is less

## CBSE NCERT solutions for class 7 maths chapter 8 comparing quantities topic 8.3.1 meaning of percentage

 Height Number of  Children In Fraction In Percentage 110 cm 22 120 cm 25 128 cm 32 130 cm 21 Total 100

 Height No. of children In fraction In percentage 110 22 22/100 =0.22 22% 120 25 25/100 = 0.25 25% 128 32 32/100 =0.32 32% 130 21 21/100 =0.21 21% total 100 01 100%

Size 2 : 20        Size 3 : 30       Size 4 : 28      Size 5 : 14       Size 6 : 8

Write this information in tabular form as done earlier and find the Percentage of each shoe size available in the shop.

 Size No. of shoes Fraction Percentage 2 20 $\frac{20}{100}=0.2$ $20 \%$ 3 30 $\frac{30}{100}=0.3$ $30 \%$ 4 28 $\frac{28}{100}=0.28$ $28 \%$ 5 14 $\frac{14}{100}=0.14$ $14 \%$ 6 8 $\frac{8}{100}=0.08$ $8 \%$ Total 100

Solutions of NCERT for class 7 maths chapter 8 comparing quantities topic 8.3.1 subtopic when total is not hundred

 Colour Number Fraction Denominator Hundred In Percentage Green Blue Red Total

Fill the table and find the percentage of chips of each colour.

 Colour Number Fraction Denominator hundred In percentage Green 4 4/10=0.4 40/100 40% Blue 3 3/10=0.3 30/100 30% Red 3 3/10=0.3 30/100 30% Total 10 1 100/100 100%

We have,
Number of gold bangles = 20
Number of silver bangles = 10
Total bangles = 30

Therefore, the percentage of gold bangles
$\frac{20}{30}\times 100 = 66.66\%$

and the percentage of the silver bangles
$\frac{10}{30}\times 100 = 33.33\%$

Solutions of NCERT for class 7 maths chapter 8 comparing quantities topic 8.3.3 converting decimals to percentage

Question:1(a) Convert the following to per cents:

$(a)\frac{12}{16}$

$(a)\frac{12}{16}$
To convert into a percentage,
$\frac{12\times 100}{16}=\frac{1200}{16}\%=75\%$

Question:(b) Convert the following to per cents:

3.5

(b) 3.5
To convert into a percentage,
$\frac{35\times 100}{10\times 10}=\frac{3500}{10}\%=350\%$

Question: Convert the following to per cents:

$(c)\frac{49}{50}$

$(c)\frac{49}{50}$

To convert into a percentage,
$\frac{49\times 100}{50}=\frac{4900}{50}\%=98\%$

Question:1(d) Convert the following to per cents:

$(d)\frac{2}{2}$

$(d)\frac{2}{2}$
To convert into a percentage,
$\frac{2\times 100}{2}=\frac{200}{2}\%=100\%$

Question:(e) Convert the following to per cents:

0.05

(e) 0.05
To convert into a percentage,
$\frac{5\times 100}{100}=\frac{500}{100}\%=5\%$

We have a total number of student = 32
and the absent student = 8

Therefore, the percentage of absent students
$=\frac{8}{32}\times 100 =25\%$

We have a total  radio = 25
and the radios that are out of order  = 16

Therefore, the percentage of out of order radios
$=\frac{16}{25}\times 100 =64\%$

Given that,
The shop has total items =500
Defective item = 5

Therefore, the percentage of the defective item
$\Rightarrow \frac{5}{500}\times 100 = \frac{1}{100}\times 100 =1\%$

Given that,
The total no. of voters =120
Voted yes = 90

Therefore, the percentage of voted yes
$\Rightarrow \frac{90}{120}\times 100 = \frac{9}{12}\times 100= \frac{3}{4}\times 100=75\%$

CBSE NCERT solutions for class 7 maths chapter 8 comparing quantities topic 8.3.4 converting percentage to fractions or decimals

Question:1(i)   35% + _______ = 100%

Let the blank space be X
therefore, 35% + X = 100%
So, X = 100% - 35%
= 65%

Question:1(ii)  64% + 20% +________ % = 100%

Let the blank space be X
therefore,
64% + 20% +  X = 100%
So, X = 100% - 64% - 20%
= 16%

Question:1(iii)  45% = 100% – _________ %

Let the blank space be X
therefore,
45%  = 100% - X
So, X = 100% - 45%
= 55%

Question:1(iv)  70% = ______% – 30%

Let the blank space be X
therefore,
70%  = X - 30 %
So, X = 70%- 30%
= 100%

Given that,
65 % of students in a class have a bicycle.
So, the remaining percent of students have no bicycle
= 100% - 65%
= 35%

Hence 35% of students have no bicycle.

We have,

A basket is full of apples, oranges and mangoes.
50% are apples, 30% are oranges.

So, the remaining percent are mangoes = 100% - 50% - 30%
= 20%

Solutions for NCERT class 7 maths chapter 8 comparing quantities topic 8.3.5 fun with estimation

Question:1(a) What per cent of this figure is shaded?

In the above figure, there are a total of 4 parts and out of which 3 parts are shaded
therefore, the percentage of the shaded part
$\Rightarrow \frac{3}{4}\times 100 = 75\%$

Question:1(b)  What per cent of this figure is shaded?

In the above figure,
$\Rightarrow \frac{1}{4} +\frac{1}{8}+\frac{1}{8} = \frac{1}{2}$

Therefore, the percentage of the shaded part
$\Rightarrow \frac{1}{2}\times 100 = 50\%$

Solutions of NCERT for class 7 maths chapter 8 comparing quantities topic 8.4.2 converting percentage to how many

Question:1(a) Find: 50% of 164

50% of 164
$\\\Rightarrow \frac{50}{100}\times 164\\ \Rightarrow\frac{1}{2}\times 164 = 82$

So 50% of 164 is 82, that is half of 164

Question:1(b) Find: 75% of 12

75% of 12
$\\\Rightarrow \frac{75}{100}\times 12\\ \Rightarrow\frac{3}{4}\times 12 = 9$

$(c) 12\tfrac{1}{2}\: ^{o}/_{o}\ of\ 64$

it means 12.5% of 64
Therefore,
$\\\Rightarrow \frac{12.5}{100}\times 64\\ \Rightarrow 0.125\times 64 = 8$

We have,
8% children of a class of 25 like getting wet in the rain

therefore, the number of children get wet
8 % of 25
$\frac{8}{100}\times 25 = 2$

Hence out of 25 only 2 children get wet in the rain

Question:1 9 is 25% of what number?

Let the number be X
therefore, 25% of X = 9
$\\\frac{25}{100}\times X = 9\\ X = \frac{900}{25}=36$

Question:2 75% of what number is 15?

Let the number be X
therefore, 75% of X = 15
$\\\Rightarrow \frac{75}{100}\times X = 15\\ \Rightarrow X = \frac{1500}{75}=20$

Solutions for class 7 maths chapter 8 comparing quantities-Exercise: 8.2

$(a)\frac{1}{8}$
$(b)\frac{5}{4}$
$(c)\frac{3}{40}$
$(d)\frac{2}{7}$

To convert the given fractional into the per cent we have to do multiplication in numerator and denominator by 100
Now,
$(a)\frac{1}{8}$
$\Rightarrow \frac{1}{8}\times100 = \frac{100}{8}\%=12.5\%$

$(b)\frac{5}{4}$
$\Rightarrow \frac{5}{4}\times100 = \frac{500}{4}\%=125\%$

$(c)\frac{3}{40}$
$\Rightarrow \frac{3}{40}\times 100 = \frac{300}{40}\%=7.5\%$

$(d)\frac{2}{7}$
$\Rightarrow \frac{2}{7}\times100 = \frac{200}{7}\%=28.57\%$

(a) 0.65
(b) 2.1
(c) 0.02
(d) 12.35

To convert the given fractional decimal into a per cent, multiply the denominator and numerator by 100.
So,
(a) 0.65
$0.65\times 100=65\%$

(b) 2.1
$2.1\times 100=210\%$

(c) 0.02
$0.02\times100=2\%$

(d) 12.35
$12.35\times100=1235\%$

We have,
The fraction of the coloured part = $\frac{1}{4}$
Therefore,
$\frac{1}{4}\times \frac{100}{100}=\frac{100}{4}\%=25\%$

In the given figure, the fraction of the coloured part is $3/6$
Therefore, Percentage of the coloured part
$=\frac{3\times 100}{5\times 100} = \frac{300}{5}\% = 60\%$

In the given figure, the fraction of the coloured part is $3/8$
Therefore, Percentage of the coloured part
$=\frac{3\times 100}{8\times 100}=\frac{300}{8}\% = 37.5\%$

Here per cent implies for ($\% \rightarrow \frac{1}{100}$)

Therefore,
(a) 15% of 250
$15\times \frac{1}{100}\times 250 =\frac{75}{2}$
= 37.5

(b) 1$\%$ of 1 hour
= 1 $\%$ of 60 minutes
$=\frac{1}{100}\times 60 =\frac{3}{5}\ min$
$=\frac{3}{5}\times 60 = 36\ sec$

(c) 20$\%$  of Rs 2500
$=\frac{20}{100}\times 2500 =Rs\ 500$

(d) 75$\%$ of 1 kg
We know that 1kg = 1000 gm, therefore,
$\frac{75}{100}\times 1000g=750g$
=0.75 kg

Question:5 Find the whole quantity if

(a) 5% of it is 600.
(b) 12% of it is Rs 1080.
(c) 40% of it is 500 km.
(d) 70% of it is 14 minutes.
(e) 8% of it is 40 litres.

(a) Let the whole quantity be X
Therefore, 5 % of X = 600
$\\\Rightarrow \frac{5}{100}\times X=600\\ \Rightarrow X = \frac{600\time 100}{5}=12000$
Thus the required whole quantity is 12000

(b) 12 % of X = Rs 1080
$\\\Rightarrow \frac{12}{100}\times X=1080\\ \Rightarrow X = \frac{1080\time 100}{12}=9000$

(c) 40 % of X = 500 km
$\\\Rightarrow \frac{40}{100}\times X=500\\ \Rightarrow X = \frac{50000\time 100}{40}=1250\ km$

(d) 70% of X = 14 minutes
$\\\Rightarrow \frac{70}{100}\times X=14\ min\\ \Rightarrow X = \frac{1400\time 100}{70}=20\ min$

(e) 8 % of X = 40 litre
$\\\Rightarrow \frac{8}{100}\times X=40\ L\\ \Rightarrow X = \frac{4000}{8}=500\ L$

(a) 25%
(b) 150%
(c) 20%
(d) 5%

(a) 25 %
$\Rightarrow \frac{25}{100} =0.25 = \frac{1}{4}$

(b) 150%
$\Rightarrow \frac{150}{100} =1.5 = \frac{3}{2}$

(c) 20%
$\Rightarrow \frac{20}{100} =0.2 = \frac{1}{5}$

(d) 5%
$\Rightarrow \frac{5}{100} =0.05 = \frac{1}{20}$

Given that,
30 % are female and
40 % are males
Total percentage of females and males
= 30 % + 40% = 70%
Therefore, Percentage of children
= (100% -70%) = 30 %

Given that,
Total number of voters = 15000
Percentage of voters who voted = 60%
Therefore, remaining 40% voters didn't vote

Thus, the number of people who did not vote
$40\%\ of\ 15000$
$=\frac{40}{100}\times 15000$
$=6000$

Given that,
10 % of her salary = 4000
Let her total salary be X
Therefore,
10% of X = 4000
$\Rightarrow \frac{10}{100}\times X = 4000$
$X = \frac{4000}{10}\times 100$
X = Rs 40,000

Given that,
Total match played by the cricket team = 20
and, a match won = 25%

According to question,
Number of winning match = 25 % of 20
$\\\Rightarrow \frac{25}{100}\times 20=5\\$

hence they won total 5 matches out of 20 match

CBSE NCERT solutions for class 7 maths chapter 8 comparing quantities topic 8.4.3 ratio of percents

We have 15 sweets.
Manu wants 20% and Sonu wants 80% of the sweets respectively.

Therefore, 20% of 15
= $\frac{20}{100}\times 15 =3$

and 80% of 15 sweets
$\frac{80}{100}\times 15 =12$

Hence give 3 sweets to Manu and 12 sweets to Sonu.

We have,
angles of a triangle are in the ratio 2 : 3: 4.
Let the angle be $2x,3x\and\ 4x$ respectively

And the sum of all the angle
$9x =180$ [by angle sum property]
so,
$x =\frac{180}{9}=20$

now, the angles are
40, 60 and 80

Solutions for NCERT class 7 maths chapter 8 comparing quantities topic 8.4.4 increase or decrease as per cent

Question:1 Find Percentage of increase or decrease:

– Price of shirt decreased from Rs 280 to Rs 210.

– Marks in a test increased from 20 to 30.

(i) The decrease in price is 280-210=70

Percentage of decrease
$=\frac{70}{280}\times 100$
= 25%

(ii) increase of mark is 30-20=10

Percentage of increase
$=\frac{10}{30}\times 100$
= 33.33%

Total increase in petrol = 52 -1 = Rs51

Therefore, the percentage of price increase
$\Rightarrow \frac{51}{52}\times 100 = 98.07\%$

Solutions of NCERT for class 7 maths chapter 8 comparing quantities topic 8.5.1 profit or loss as a percentage

We have,
SP = 400
CP =375
So,  SP - CP = 25

SInce SP > CP
Therefore, Gain(%)
$\Rightarrow \frac{25}{375}\times 100 = 6.6\%$

We have,
CP = Rs 50
Profit  = 12 %
SP = ?

Therefore,
$SP = CP(1+\frac{profit\%}{100})$
Put the values in the above equation, we get

$\\= 50(1+\frac{12}{100})\\ =50 \times 1.12\\ = Rs\ 56$

We have,
SP = Rs 250
Profit = 5%
CP =?

Therefore,
$SP = CP (1+\frac{profit\%}{100})$
put the values in the above equation, we get

$\\\Rightarrow 250 = CP (1+\frac{5}{100})\\ \Rightarrow 250 = CP\times \frac{21}{20}\\$
$CP = \frac{250\times 20}{21} = 238.095$

We have,
SP = Rs 540
Loss = 5%
CP =?

Therefore,
$SP = CP(1-\frac{loss\%}{100})$
put the values in the above equation, we get

$\\\Rightarrow 540 = CP(1-\frac{5}{100})\\ \Rightarrow540 = CP\times \frac{19}{20}$

$\Rightarrow CP = \frac{540 \times 20}{19} = Rs\ 568.42$

CBSE NCERT solutions for class 7 maths chapter 8 comparing quantities topic 8.6.1 interest for multiple years

Given that,
Principal = 10,000
Rate= 5% p.a
T = 1 year

Therefore,
$Interest = \frac{P\times R\times T}{100}$
Substituting the values in the above formula, we get
$= \frac{10,000\times 5\times 1}{100}$
$= Rs\ 500$

Given that,
Principal = 3,500
Rate= 7% p.a
T = 2 year

Therefore,
$Interest = \frac{P\times R\times T}{100}$
Substituting the values in the above formula, we get
$= \frac{3,500\times 7\times 2}{100}$
$= Rs\ 490$

Given that,
Principal = 6050
Rate= 6.5% p.a
T = 3 year

Therefore,
$Interest = \frac{P\times R\times T}{100}$
Substituting the values in the above formula, we get
$= \frac{6050\times 6.5\times 3}{100}$
$= Rs\ 1179.75$

So, the total amount to be paid = Rs 1179.75  + Rs 6050 =Rs 7229.75

Given that,
Principal = 7,000
Rate= 3.5% p.a
T = 2 year

Therefore,
$Interest = \frac{P\times R\times T}{100}$
Substituting the values in the above formula, we get
$= \frac{7000\times 3.5\times 2}{100}$
$= Rs\ 490$

So, the total amount to be paid = Rs 490  + Rs 7000 =Rs 7490

Solutions of NCERT for class 7 maths chapter 8 comparing quantities topic 8.6.1 interest for multiple years

Given that,
Principal = 2,400
Rate= 5% p.a
T = ?
Interest = Rs 240

Therefore,
$Interest = \frac{P\times R\times T}{100}$
Substituting the values in the above formula, we get
$240= \frac{2,400\times 7\times T}{100}$
$T = \frac{10}{5} = 2\ year$

Given,
Simple Interest, S.I.= $Rs. \ 450$
Time, $T = 3\ yrs$
Rate, $R= 5\%$
We know,
$P= S.I \times \frac{100}{r} \times t$
$\\ =450 \times \frac{100}{3} \times 5 \\ = 3000$

Therefore, the sum is $Rs.\ 3000$

Solutions of NCERT for class 7 maths chapter 8 comparing quantities-Exercise: 8.3

Gardening shears bought for Rs 250 and sold for Rs 325.

Given that,
Selling price =  Rs 325
Cost price = Rs 250
Since SP > CP
Therefore, profit = SP-CP = Rs 75

And, Profit %
$\Rightarrow \frac{75}{250}\times 100 = 30\%$

A refrigerator bought for Rs 12,000 and sold at Rs 13,500.

Given that,
CP = Rs 12000
SP = Rs 13500

SInce SP > CP
Therefore, Profit = SP - CP = 1500

and, Profit %
$\Rightarrow \frac{1500}{12000}\times 100 = 12.5\%$

A cupboard bought for Rs 2,500 and sold at Rs 3,000.

We have,
CP = Rs 2500
SP = Rs 3000

Since SP > CP
Therefore, Profit = SP - CP = 500

and, Profit %
$\Rightarrow \frac{500}{2500}\times 100 = 20\%$

A skirt bought for Rs 250 and sold at Rs 150.

We have,
SP = Rs 150
CP = Rs 250

Since CP > SP
Therefore, Loss = CP-SP = Rs 100

and, Loss%
$\Rightarrow \frac{100}{250}\times 100= 40\%$

Question:2(a) Convert each part of the ratio to percentage:

3 : 1

(a) 3: 1
The total sum of the ratio is 3 + 1 = 4
Therefore, the percentage of the first part
$\frac{3}{4}\times 100 = 75\%$

Percentage of the second part
$=\frac{1}{4}\times 100 = 25\%$

Question:2(b) Convert each part of the ratio to percentage:

2 : 3: 5

(b) 2 : 3: 5
Sum of the ratio part = 2 + 3 + 5 = 10
Therefore, the percentage of the first part
$=\frac{2}{10}\times 100 = 20\%$

the percentage of the second part
$=\frac{3}{10}\times 100 = 30\%$

the percentage of the third part
$=\frac{5}{10}\times 100 = 50\%$

Question:2(c) Convert each part of the ratio to percentage:

1:4

(c) 1:4
Sum of the ratio part = 4 + 1 = 5
therefore, the percentage of the first part
$=\frac{1}{5}\times 100 = 20\%$

the percentage of the second part
$=\frac{4}{5}\times 100 = 80\%$

Question:2(d) Convert each part of the ratio to percentage:

1 : 2 : 5

(d) 1 : 2 : 5

Sum of the ratio part = 1 + 2 + 5 = 8
therefore, the percentage of the first part
$=\frac{1}{8}\times 100 = 12.5\%$

the percentage of the second part
$=\frac{2}{8}\times 100 = 25\%$

the percentage of the third part
$=\frac{5}{8}\times 100 = 62.5\%$

Given that,
Initial population  = 25,000
Final population = 24,500

Total decrement = 1000  (25000 - 24500)
Therefore, percentage in decrease
$\Rightarrow \frac{500}{25000}\times100 = 2\%$

Given that,
Original price (OP) = 3,50,000
Increased price (IP) = 3,70,000

Therefore, increase in price = OP - IP = 20,000

Thus, percentage of the increase price
$=\frac{20,000}{3,50,000}\times 100 = 5.71\%$

Here we have,
CP = Rs 10,000
Profit = $20\%$
SP =?

We know that,
$SP =CP(1+\frac{profit}{100})$...........(i)

By substituting the values in eq (i), we get

$SP =10,000(1+\frac{20}{100})$
$=10,000\times (\frac{6}{5})$
= Rs 12,000

We have,
Sp of the washing m/c = Rs 13,500
Loss(%) = 20%
CP = ?

We know that,
$SP = CP (1-\frac{Loss}{100})$
Putting the values in the above equation we get;

$\\13,500= CP (1-\frac{20}{100})\\ 13500 = CP (1-\frac{1}{5}) = CP \times (\frac{4}{5})$
Therefore,
$CP = 13,500\times \frac{5}{4} = Rs\ 16875$

We have,
The ratio of calcium, carbon and oxygen in the chalk = 10 : 3: 12
Now, Sum of the ratio = 10 + 3 + 12 = 25

Therefore, the percentage of carbon in the chalk
$\Rightarrow \frac{3}{25}\times 100 =12\%$

We have the weight of the carbon = 3g
Therefore, the weight of the chalk
$\Rightarrow \frac{3}{3}\times 25g = 25g$

Hence the weight of the chalk is 25g

Here we have,
CP of the book = Rs 275
Loss = 15%

We know that,
$SP = CP (1-\frac{Loss}{100}) = 275(1-\frac{15}{100})$
$= 275\times (\frac{85}{100})$
$= Rs\ 233.75$

Hence the required selling price is Rs 233.75

Principal = Rs 1,200 at 12% p.a.

Given that,
Principal (PA)= Rs 1,200 at 12% p.a. (Rate of interest)
and T = 3 year
We know that,
$Interest = \frac{P\times R\times T}{100}$
Now, putting the values in the above equation, we get
$\Rightarrow \frac{1200\times 12\times 3}{100} = Rs\ 432$

So, Amount = PA + PI = 1200 + 432 = Rs 1632

(b) Principal = Rs 7,500 at 5% p.a.

Given that,
Principal (PA)= Rs 7,500 at 5% p.a. (Rate of interest)
and T = 3 year
We know that,
$Interest = \frac{P\times R\times T}{100}$
Now, putting the values in the above equation, we get
$\Rightarrow \frac{7,500\times 5\times 3}{100} = Rs\ 1125$

So, Amount = PA + PI = 7,500 + 1125 = Rs 8625

Given that,
Principal = Rs 56,000
Interest = Rs 280
Time = 2 year

Rate= ?

Therefore,
$Rate = \frac{100\times I}{P\times T}$
$= \frac{100\times 280}{56,000\times 2} = 0.25\%$

Given that,
Principal =?
Interest = Rs 45
Time = 1 year

Rate= 9% p.a

Therefore,
$Principal = \frac{100\times I}{R\times T}$
$= \frac{100\times 45}{9\times 1} = Rs\ 500$

Hence she borrowed total Rs 500

## NCERT Solutions for Class 7 Maths - Chapter-wise

 Chapter No. Chapter Name Chapter 1 Solutions of NCERT for class 7 maths chapter 1 Integers Chapter 2 CBSE NCERT solutions for class 7 maths chapter 2 Fractions and Decimals Chapter 3 NCERT solutions for class 7 maths chapter 3 Data Handling Chapter 4 Solutions of NCERT for class 7 maths chapter 4 Simple Equations Chapter 5 CBSE NCERT solutions for class 7 maths chapter 5 Lines and Angles Chapter 6 NCERT solutions for class 7 maths chapter 6 The Triangle and its Properties Chapter 7 Solutions of NCERT for class 7 maths chapter 7 Congruence of Triangles Chapter 8 NCERT solutions for class 7 maths chapter 8 comparing quantities Chapter 9 CBSE NCERT solutions for class 7 maths chapter 9 Rational Numbers Chapter 10 NCERT solutions for class 7 maths chapter 10 Practical Geometry Chapter 11 Solutions of NCERT for class 7 maths chapter 11 Perimeter and Area Chapter 12 CBSE NCERT solutions for class 7 maths chapter 12 Algebraic Expressions Chapter 13 NCERT solutions for class 7 maths chapter 13 Exponents and Powers Chapter 14 Solutions of NCERT for class 7 maths chapter 14 Symmetry

## NCERT Solutions for Class 7- Subject-wise

There are two additional topics profit and loss percentage and simple interest in this chapter. The last exercise of NCERT solutions for class 7 maths chapter 8 Comparing Quantities has all 11 questions related to these two topics.

Profit and Loss Percentage:- If the selling price (SP) > cost price (CP) then there is a profit P= SP-CP.  If selling price (SP) is < cost price (CP) then there is a loss L= CP-SP.

$profit\ per\ cent=\frac{SP-CP}{CP}\times100\%=\frac{profit}{CP}\times100\%$
$loss\ per\ cent=\frac{CP-SP}{CP}\times100\%=\frac{loss}{CP}\times100\%$

Simple interest: If we deposit 100 rupees for the interest of 10% for a year then you will get 100+10% of 100=100+10=110 rupees after a year. Here 100 rupees is known as the principal or sum (P), 10% is the rate percent per annum (R), Then the interest we are getting (I) after 1 year is (If we are borrowing money then the interest is to be paid)

$I=\frac{PR}{100}=\frac{100\times10}{100}=10$

The amount (A) you will receive total amount =  Principle amount +Interest =100+10=110

Suppose if you are borrowing or taking a loan for more than one year (say T years) then the interest to be paid after T years is

$I=\frac{PRT}{100}$

There are practice questions given after every topic to get a better understanding of the concept. In CBSE NCERT solutions for class 7 maths chapter 8 comparing quantities, you will also get solutions to practice questions given after every topic. Solve all the questions and examples to understand the concept and score well in the exam.

Happy learning!!!