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 three 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. Here you will get solutions to three exercises of this chapter.

Exercise:8.1

Exercise:8.2

Exercise:8.3

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

Here are the NCERT solutions for class 7 maths chapter 8 Comparing Quantities:

 

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

Answer:

(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

Question:2 In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students? 

Answer:

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.

Question:3 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}.

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

           (ii) Which State is less populated?

Answer:

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

Question:1 Find the Percentage of children of different heights for the following data.

            Height    

    Number of  Children

   In Fraction

     In Percentage

110 cm

22

 

 

120 cm

25

 

 

128 cm

32

 

 

130 cm

21

 

 

Total

100

 

 

 

 

 

 

 

 

 

 

 

 

 

Answer:

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%

 

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

Question:1 A collection of 10 chips with different colours is given.

  

Colour

Number

Fraction

Denominator Hundred

In Percentage

Green

 

 

 

 

Blue

 

 

 

 

Red

 

 

 

 

Total

 

 

 

 

 

 

 

 

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

Answer:

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%

Question:2 Mala has a collection of bangles. She has 20 gold bangles and 10 silver bangles. What is the percentage of bangles of each type? Can you put it in the tabular form as done in the above example?

Answer:

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}

 

Answer:

(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

Answer:

(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}

 

Answer:

(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}

Answer:

(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

Answer:

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

Question:2(i)  Out of 32 students, 8 are absent. What per cent of the students are absent?

Answer:

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\%

 

Question:2(ii)  There are 25 radios, 16 of them are out of order. What per cent of radios are out of order?

Answer:

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\%

 

Question:2(iii) A shop has 500 items, out of which 5 are defective. What per cent are defective?

Answer:

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\%

 

Question:2(iv) There are 120 voters, 90 of them voted yes. What per cent voted yes?

Answer:

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%

Answer:

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

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

Answer:

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

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

Answer:

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

Question:1(iv)  70% = ______% – 30%

Answer:

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

Question:2  If 65% of students in a class have a bicycle, what per cent of the student do not have bicycles?

Answer:

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.

Question:3 We have a basket full of apples, oranges and mangoes. If 50% are apples, 30% are oranges, then what per cent are mangoes?

Answer:

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?

            

         You can make some more figures yourself and ask your friends to estimate the shaded parts.

Answer:

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?

            

  You can make some more figures yourself and ask your friends to estimate the shaded parts.

Answer:

In the above figure, 
The total shaded part = 
\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

Answer:

  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

Answer:

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

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

Answer:

(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

Question:2 8% children of a class of 25 like getting wet in the rain. How many children like getting wet in the rain.

Answer:

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?

Answer:

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?

Answer:

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

Question:1 Convert the given fractional numbers to per cents.

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

Answer:

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\%

Question:2 Convert the given decimal fractions to per cents.

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

Answer:

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\%

Question:3(i) Estimate what part of the figure is coloured and hence find the per cent which is coloured.

          

Answer:

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

 

Question:3(ii) Estimate what part of the figure is coloured and hence find the per cent which is coloured.

         

Answer:

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\%

 

Question:3(iii) Estimate what part of the figures is coloured and hence find the per cent which is coloured.

         
 

Answer:

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\%

Question:4 Find: (a) 15\% of 250
         (b) 1\% of 1 hour
         (c) 20\%  of Rs 2500
         (d) 75\% of 1 kg

Answer:

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.

Answer:

(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

Question:7 In a city, 30% are females, 40% are males and remaining are children. What per cent are children?

Answer:

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 %

Question:8 Out of 15,000 voters in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?

Answer:

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

Question:9 Meeta saves Rs 4000 from her salary. If this is 10% of her salary. What is her salary?

Answer:

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

Question:10  A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?

Answer:

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

Question:1 Divide 15 sweets between Manu and Sonu so that they get 20 % and 80 % of them respectively.

Answer:

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.

Question:2 If the angles of a triangle are in the ratio 2 : 3 : 4. Find the value of each angle.

Answer:

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.

Answer:

(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%

Question:2 My mother says, in her childhood petrol was Rs 1 a litre. It is Rs 52 per litre today. By what Percentage has the price gone up?

Answer:

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

Question:1 A shopkeeper bought a chair for Rs 375 and sold it for Rs 400. Find the gain Percentage.

Answer:

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

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

Question:2 Cost of an item is Rs 50. It was sold with a profit of 12%. Find the selling price.

Answer:

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

Question:3  An article was sold for Rs 250 with a profit of 5%. What was its cost price?

Answer:

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

Question:4 An item was sold for Rs 540 at a loss of 5%. What was its cost price?

Answer:

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

Question:1 Rs 10,000 is invested at 5% interest rate p.a. Find the interest at the end of one year.

Answer:

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

Question:2 Rs 3,500 is given at 7% p.a. rate of interest. Find the interest which will be received at the end of two years.

Answer:

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

Question:3 Rs 6,050 is borrowed at 6.5% rate of interest p.a.. Find the interest and the amount to be paid at the end of 3 years.

Answer:

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

Question:4 Rs 7,000 is borrowed at 3.5% rate of interest p.a. borrowed for 2 years. Find the amount to be paid at the end of the second year.

Answer:

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

Question:1 You have Rs 2,400 in your account and the interest rate is 5%. After how many years would you earn Rs 240 as interest.

Answer:

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

Question:2 On a certain sum the interest paid after 3 years is Rs 450 at 5% rate of interest per annum. Find the sum.

Answer:

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

Question:1(a) Tell what is the profit or loss in the following transaction. Also find profit per cent or loss per cent in each case.

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

Answer:

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\%

Question:1(b) Tell what is the profit or loss in the following transaction. Also find profit per cent or loss per cent in each case.

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

Answer:

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\%

Question:1(c) Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent in each case.

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

Answer:

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\%

Question:1(d) Tell what is the profit or loss in the following transaction. Also find profit per cent or loss per cent in each case.

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

Answer:

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

Answer:

(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

Answer:

(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

Answer:

  (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

Answer:

(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\%

Question:3 The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.

Answer:

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\%

Question:4 Arun bought a car for Rs 3,50,000. The next year, the price went upto Rs 3,70,000. What was the Percentage of price increase?

Answer:

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\%

Question:5 I buy a T.V. for Rs 10,000 and sell it at a profit of 20%. How much money do I get for it?

Answer:

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

Question:6 Juhi sells a washing machine for Rs 13,500. She loses 20% in the bargain. What was the price at which she bought it?

Answer:

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

Question:7(i) Chalk contains calcium, carbon and oxygen in the ratio 10:3:12. Find the percentage of carbon in chalk.

Answer:

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\%

Question:7(ii) If in a stick of chalk, carbon is 3g, what is the weight of the chalk stick?

Answer:

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

Question:8 Amina buys a book for Rs 275 and sells it at a loss of 15%. How much does she sell it for?

Answer:

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 

Question:9(a) Find the amount to be paid at the end of 3 years in this case:

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

Answer:

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

Question:9(b) Find the amount to be paid at the end of 3 years in this case:

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

Answer:

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

Question:10 What rate gives Rs 280 as interest on a sum of Rs 56,000 in 2 years?

Answer:

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\%

Question:11 If Meena gives an interest of Rs 45 for one year at 9% rate p.a. What is the sum she has borrowed?

Answer:

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 

Solutions of NCERT for class 7 maths

CBSE NCERT solutions for class 7 science

 

 

 

 

 

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!!!

 

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