NCERT Solutions for Class 7 Maths Chapter 8 - Comparing Quantities

Exercise 8.1

1.Find the ratio of:

(a) Rs $\text{5}$ to $\text{50}$ paise

Ans: We know that to evaluate the ratio of two quantities, both the quantities must be in the same unit.

So, converting the given quantities into the same quantity

Converting Rs $\text{5}$ in paise,

Since Rs $\text{1 = 100}$ paise,

Therefore, Rs $\text{5 = 500}$ paise

Calculating the required ratio,

i.e., the ratio of $\text{500}$ paise to $\text{50}$ paise,

The ratio is $\text{=}\frac{\text{500}}{\text{50}}\text{ = }\frac{\text{10}}{\text{1}}\text{ = 10 : 1}$

The required ratio is $\text{10 : 1}$

(b) $\text{15}$ kg to $\text{210}$ g

Ans: We know that to evaluate the ratio of two quantities, both the quantities must be in the same unit.

So, converting the given quantities into the same quantity

Converting $\text{15}$ kg to grams

Since, $\text{1}$ kg $\text{= 1000}$g 

Therefore, $\text{15}$kg $\text{= 15 }\times \text{ 1000}$

             $\Rightarrow \text{15}$kg $\text{= 15000}$g 

Calculating the required ratio,

i.e., the ratio of $\text{15000}$g to $\text{210}$g

The ratio is $\text{= }\frac{\text{15000}}{\text{210}}\text{ = }\frac{\text{500}}{\text{7}}$

Thus, the required ratio is $\text{500 : 7}$

(c) $\text{9}$m to $\text{27}$cm

Ans: We know that to evaluate the ratio of two quantities, both the quantities must be in the same unit.

So, converting the given quantities into the same quantity

Converting $\text{9}$m to cm

Since, $\text{1}$m $\text{= 100}$cm

Therefore, $\text{9}$m $\text{= 9 }\times \text{ 100}$cm

             $\Rightarrow \text{9}$m $\text{= 900}$cm

Calculating the required ratio,

i.e., the ratio of $\text{900}$cm to $\text{27}$cm

The ratio is $\text{= }\frac{\text{900}}{\text{27}}\text{ = }\frac{\text{100}}{\text{3}}$

Thus, the required ratio is $\text{100 : 3}$

(d) $\text{30}$days to $\text{36}$hours

Ans: We know that to evaluate the ratio of two quantities, both the quantities must be in the same unit.

So, converting the given quantities into the same quantity

Converting $\text{30}$days to hours

Since, $\text{1}$day $\text{= 24}$hours

Therefore, $30$days $\text{= 30 }\times \text{ 24}$hours

             $\Rightarrow \text{30}$days $\text{= 720}$hours

Calculating the required ratio,

i.e., the ratio of $\text{720}$ hours to $36$ hours

The ratio is $\text{= }\frac{\text{720}}{\text{36}}\text{ = }\frac{\text{20}}{\text{1}}$

Thus, the required ratio is $\text{20 : 1}$.

2. In a computer lab, there are $\text{3}$ computers for every $\text{6}$ students. How many computers will be needed for $\text{24}$ students?

Ans: For this type of question, we use the unitary method,

Since, $\text{6}$ students need $\text{3}$ computers,

Therefore, by the unitary method,

$\text{1}$ student needs $\frac{\text{3}}{\text{6}}$ computers 

Therefore, $\text{24}$ students need $\frac{\text{3}}{\text{6}}\text{ }\times \text{ 24 = 12}$ computers,

Thus, $\text{24}$ students require $\text{12}$ computers.

3. Population of Rajasthan $\text{= 570}$ lakhs and population of U.P. $\text{= 1660}$ lakhs. Area of Rajasthan $\text{= 3}$ lakh km2 and area of U.P. $\text{= 2}$ lakh km2.

(i) How many people are there per km2 in both states?

Ans: The population of Rajasthan $\text{= 570}$ lakhs and the population of U.P. $\text{= 1660}$ lakhs.

Area of Rajasthan $=3$ lakh km2 and the area of U.P. $\text{= 2}$ lakh km2 

Since, the people present per km2 $=\frac{Population}{Area}$ 

So, the people in Rajasthan $=\frac{\text{570}}{\text{3}}=\text{190}$ people per km2

And, The people in U.P. $=\frac{\text{1660}}{\text{2}}=\text{830}$ people per km2 

(ii) Which state is less populated?

Ans: It is clear from part $\text{(i)}$, that Rajasthan is less populated.

Exercise 8.2

1. Convert the given fractional numbers to percent: 

(a) $\frac{\text{1}}{\text{8}}$

Ans: A fractional number is given, $\frac{1}{8}$

$\frac{1}{8} =\frac{1}{8} \times 100 \%$

$=\frac{25}{2} \%=12.5 \%$

$\Rightarrow  \frac{1}{8}=12.5 \%$

(b) $\frac{\text{5}}{\text{4}}$

Ans: A fractional number is given, $\frac{\text{5}}{\text{4}}$

$\frac{5}{4}=\frac{5}{4} \times 100 \%$

$=5 \times 25 \%$

$=125 \%$

$\Rightarrow  \frac{5}{4}=125 \%$

(c) $\frac{\text{3}}{\text{40}}$

Ans: A fractional number is given, $\frac{\text{3}}{\text{40}}$

$\frac{3}{40}=\frac{3}{40} \times 100 \%$

$\quad=\frac{15}{2} \%$

$=7.5 \%$

$\Rightarrow \frac{3}{40}=7.5 \%$

(d) $\frac{\text{2}}{\text{7}}$

Ans: A fractional number is given, $\frac{\text{2}}{\text{7}}$

$\frac{2}{7}=\frac{2}{7} \times 100 \% $

$=\frac{200}{7} \%$

$=28 \frac{4}{7} \%$

$\Rightarrow  \frac{2}{7}=28 \frac{4}{7} \%$

2. Convert the given decimal fractions to percents:

(a) $\text{0}\text{.65}$

Ans: A decimal fraction is given, needed to convert it into percentage from

$\Rightarrow 0.65=\frac{65}{100}$

$\Rightarrow 0.65=65 \%$

(b) $\text{2}\text{.1}$

Ans: A decimal fraction is given, needed to convert it into percentage from

$\Rightarrow 2.1=\frac{21}{10} \times 100 \%$

$\Rightarrow 2.1=210 \%$

(c) $\text{0}\text{.02}$

Ans: A decimal fraction is given, needed to convert it into percentage from

$\Rightarrow 0.02=\frac{2}{100}$

$\Rightarrow 0.02=2 \%$

(d) $\text{12}\text{.35}$

Ans: A decimal fraction is given, needed to convert it into percentage from

$\Rightarrow 12.35=\frac{1235}{100} \times 100 \%$

$\Rightarrow 12.35=1235 \%$

3. Estimate what part of the figure is colored and hence find the percent which is colored:

(i)   

Ans: It is clear from the figure that the colored part is $\text{= }\frac{\text{1}}{\text{4}}$

Therefore, the required percentage of colored parts,

 is $=\frac{1}{4} \times \frac{25}{25} $

$=\frac{25}{100} $

$=25 \%$

(ii)  

Ans: It is clear from the figure that the colored part is $\text{= }\frac{\text{3}}{\text{5}}$

Therefore, the required percentage of colored parts,

 is $\text{= }\frac{\text{3}}{\text{5}}\text{  }\!\!\times\!\!\text{  }\frac{\text{20}}{\text{20}}$

     $\text{= }\frac{\text{60}}{\text{100}}$

     $\text{= 60 }\%$

(iii)  

Ans: It is clear from the figure that the colored part is $\text{= }\frac{\text{3}}{\text{8}}$

Therefore, the required percentage of colored parts,

 is $\text{= }\frac{\text{3}}{\text{8}}\text{  }\!\!\times\!\!\text{  }\frac{\text{100}}{\text{100}}$

     $\text{= }\frac{\text{300}}{\text{8}}\text{ }\%$

     $\text{= 37}\text{.5 }\%$

4. Find: 

(a) $\text{15 }\%$ of $\text{250}$

Ans: Needed to find the required percentage of the given number $\text{250}$ i.e.,

$\text{15 }\%\text{  }$ of $\text{250}$ $\text{= }\frac{\text{15}}{\text{100}}\text{  }\!\!\times\!\!\text{  250}$

                      $\text{= 15  }\!\!\times\!\!\text{  2}\text{.5}$

                        $\text{= 37}\text{.5}$

$\Rightarrow \text{ 15 }\%\text{  }$ of $\text{250}$$\text{= 37}\text{.5}$

(b). $\text{1 }\%$ of $\text{1}$ hour

Ans: Needed to find the required percentage i.e.,

$\text{1 }\%$ of $\text{1}$ hour $\text{= 1 }\%$ of $\text{60}$ minutes

                      $\text{= 1 }\%$ of $\text{60  }\!\!\times\!\!\text{  60}$ seconds

                      $\text{= }\frac{\text{1}}{\text{100}}\text{  }\!\!\times\!\!\text{  60  }\!\!\times\!\!\text{  60}$ seconds

                      $\text{= 36}$ seconds

$\Rightarrow \text{1 }\%$ of $\text{1}$ hour $\text{= 36}$ seconds

(c). $\text{20 }\%$ of Rs$\text{2500}$

Ans: Needed to find the required percentage of the given currency Rs$\text{2500}$ i.e.,

$\text{20 }\%$ of Rs$\text{2500}$ $\text{= }\frac{\text{20}}{\text{100}}\text{  }\!\!\times\!\!\text{  2500}$

                           $\text{= 20  }\!\!\times\!\!\text{  25}$

                           $\text{=}$ Rs $\text{500}$

$\Rightarrow \text{20 }\%$ of Rs$\text{2500}$$\text{=}$Rs$\text{500}$

(d). $\text{75 }\%$ of $\text{1}$ kg

Ans: Needed to find the required percentage of the given quantity $\text{1}$ kg i.e.,

$\text{75 }\%$ of $\text{1}$ kg $\text{= }\frac{75}{\text{100}}\text{  }\!\!\times\!\!\text{  1}$ kg

                     $\text{= 0}\text{.75}$kg

$\Rightarrow \text{75 }\%$ of $\text{1}$ kg $\text{= 0}\text{.75}$kg

5. Find the whole quantity if: 

(a). $\text{5 }\%$ of it is $\text{600}$

Ans: Let the required whole quantity be $\text{x}$

Therefore, $\text{5 }\%$ of $\text{x}$ $\text{= 600}$

             $\Rightarrow \text{ }\frac{\text{5}}{\text{100}}\text{  }\!\!\times\!\!\text{  x = 600}$

             $\Rightarrow \text{ x = }\frac{\text{600  }\!\!\times\!\!\text{  100}}{\text{5}}$

             $\Rightarrow \text{ x = 12000}$

(b). $\text{12 }\%$ of it is Rs$\text{1080}$

Ans: Let the required whole quantity be $\text{x}$

Therefore, $\text{12 }\%$ of $\text{x}$ $\text{= Rs 1080}$

             $\Rightarrow \text{ }\frac{\text{12}}{\text{100}}\text{  }\!\!\times\!\!\text{  x = 1080}$

             $\Rightarrow \text{ x = }\frac{\text{1080  }\!\!\times\!\!\text{  100}}{\text{12}}$

             $\Rightarrow \text{ x = Rs 9000}$

(c). $\text{40 }\%$ of it is $\text{500}$ km

Ans: Let the required whole quantity be $\text{x}$

Therefore, $\text{40 }\%$ of $\text{x}$ $\text{= 500}$km

             $\Rightarrow \text{ }\frac{\text{40}}{\text{100}}\text{  }\!\!\times\!\!\text{  x = 500}$

             $\Rightarrow \text{ x = }\frac{\text{500  }\!\!\times\!\!\text{  100}}{\text{40}}$

             $\Rightarrow \text{ x = 1250}$km

(d). $\text{70 }\%$ of it is $\text{14}$ minutes

Ans: Let the required whole quantity be $\text{x}$

Therefore, $\text{70 }\%$ of $\text{x}$ $\text{= 14}$minutes

             $\Rightarrow \text{ }\frac{\text{70}}{\text{100}}\text{  }\!\!\times\!\!\text{  x = 14}$

             $\Rightarrow \text{ x = }\frac{\text{14  }\!\!\times\!\!\text{  100}}{\text{70}}$

             $\Rightarrow \text{ x = 20}$ minutes

(e). $\text{8 }\%$ of it is $\text{40}$ liters

Ans: Let the required whole quantity be $\text{x}$

Therefore, $\text{8 }\%$ of $\text{x}$ $\text{= 40}$liters

             $\Rightarrow \text{ }\frac{\text{8}}{\text{100}}\text{  }\!\!\times\!\!\text{  x = 40}$

             $\Rightarrow \text{ x = }\frac{\text{40  }\!\!\times\!\!\text{  100}}{\text{8}}$

             $\Rightarrow \text{ x = 500}$ liters

6. Convert given percents to decimal fraction and also fraction to simplest form:

(a). $\text{25 }\%$

Ans: We have given a percent $\text{25 }\%$ 

Fraction form$\text{= }\frac{\text{25}}{\text{100}}$

Simplest fractional form $\text{= }\frac{\text{1}}{\text{4}}$

Decimal form $\text{= 0}\text{.25}$

(b). $\text{150 }\%$

Ans: We have given a percent $\text{150 }\%$ 

Fraction form $\text{= }\frac{\text{150}}{\text{100}}$

Simplest fractional form $\text{= }\frac{\text{3}}{\text{2}}$

Decimal form $\text{= 1}\text{.5}$

(c). $\text{20 }\%$

Ans: We have given a percent $\text{20 }\%$ 

Fraction form $\text{= }\frac{\text{20}}{\text{100}}$

Simplest fractional form $\text{= }\frac{\text{1}}{\text{5}}$

Decimal form $\text{= 0}\text{.2}$

(d). $\text{5 }\%$

Ans: We have given a percent $\text{5 }\%$ 

Fraction form $\text{= }\frac{\text{5}}{\text{100}}$

Simplest fractional form $\text{= }\frac{\text{1}}{\text{20}}$

Decimal form $\text{= 0}\text{.05}$

7. In a city, $\text{30 }\%$ are females, $\text{40 }\%$ are males and remaining are children. What percent are children?

Ans:  Let the percentage of children be $\text{x  }\%$ 

It is given that the percentage of females and males are $\text{30 }\%$ and $\text{40 }\%$ respectively.

And, the total percentage $\text{= 100 }\%\text{  =}$ Percentage of males and Percentage of females and Percentage of children

$\Rightarrow \text{ 100 }\%\text{  = 30 }\%\text{  + 40 }\%\text{  + x }\%$

$\Rightarrow \text{ 100 }\%\text{  = 70 }\%\text{  + x }\%$

$\Rightarrow \text{ x }\%\text{   = 100 }\%\text{  - 70 }\%\text{  }$

$\Rightarrow \text{ x }\%\text{  = 30 }\%$

Thus $\text{30 }\%\text{  }$is the population of children in the city.

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

Ans: The total number of voters $\text{= 15,000}$

The percentage of people who voted $\text{= 60 }\%$

So, the percentage of people who didn’t vote $\text{= 100 }\%\text{  - 60 }\%$ 

 $\text{= 40 }\%$

And, the number of actual candidates, who didn’t vote $=40 \%$ of $\text{15000}$ 

 $\text{= 6000}$

Thus, $\text{6,000}$ people out of $\text{15,000}$ did not vote.

9. Meeta saves Rs $\text{400}$ from her salary. If this is $\text{10 }\%$ of her salary. What is her salary?

Ans:  Let $\text{x}$ be the salary of Meeta.

Since $\text{10 }\%$ of her salary $\text{= Rs 400}$

$\Rightarrow \text{ 10 }\%\text{  of x = 400}$

$\Rightarrow \text{ 10 }\%\text{   }\!\!\times\!\!\text{  x = 400}$

$\Rightarrow \text{ }\frac{\text{10}}{\text{100}}\text{x = 400}$

$\Rightarrow \text{ x = Rs 4,000}$

Therefore, the salary of Meeta is $\text{Rs 4,000}$.

10. A local cricket team played $\text{20}$ matches in one season. It won $\text{25 }\%$ of them. How many matches did they win?

Ans: The local cricket team played $\text{20}$ matches.

They won $\text{25 }\%$ of matches out of $\text{20}$ 

Therefore, the number of matches the cricket team won $\text{= 25  }\%\text{  of  20}$

 $\text{= }\frac{\text{25}}{\text{100}}\text{  }\!\!\times\!\!\text{  20}$

 $\text{= 5}$ matches

So, the local cricket team won $\text{5}$ matches out of $\text{20}$.

Exercise 8.3

1. Tell what is the profit or loss in the following transactions. Also, find the profit per cent or loss per cent in each case.

(a). Gardening shears bought for $\text{Rs 250}$ and sold for $\text{Rs 325}$.

Ans: Here, the Cost price of the gardening shears $\text{= Rs 250}$

Also, the Selling price of gardening shears $\text{= Rs 325}$

Since, the Selling Price (S.P.) is greater than Cost Price (C.P.)

Therefore, here we have is the profit.

$\therefore $ Profit $\text{=}$ S.P. $-$ C.P. $\text{= 325-250}$

$\Rightarrow $ Profit $\text{=}$ $\text{Rs 75}$

So, the profit percentage $\text{= }\frac{\text{Profit}}{\text{C}\text{.P}}\text{  }\!\!\times\!\!\text{  100}$

$\Rightarrow $ Profit$\text{  }\%\text{  = }\frac{\text{75}}{\text{250}}\text{  }\!\!\times\!\!\text{  100}$

$\Rightarrow $ Profit $\text{ }\%\text{  = 30 }\%$

(b). A refrigerator bought for $\text{Rs 12,000}$ and sold at $\text{Rs 13,500}$.

Ans: Here, the Cost price of the refrigerator $\text{= Rs 12,000}$

Also, the Selling price of the refrigerator $\text{= Rs 13,500}$

Since, the Selling Price (S.P.) is greater than Cost Price (C.P.)

Therefore, here we have is the profit.

$\therefore $ Profit $\text{=}$ S.P. $-$ C.P. $\text{= 13,500 - 12,000}$

$\Rightarrow $ Profit $\text{=}$ $\text{Rs 1,500}$

So, the profit percentage $\text{= }\frac{\text{Profit}}{\text{C}\text{.P}}\text{  }\!\!\times\!\!\text{  100}$

$\Rightarrow $ Profit $\text{ }\%\text{  = }\frac{\text{1500}}{12000}\text{  }\!\!\times\!\!\text{  100}$

$\Rightarrow $ Profit $\text{ }\%\text{  = 12}\text{.5 }\%$

(c). A cupboard bought for $\text{Rs 2,500}$ and sold at $\text{Rs 3,000}$.

Ans: Here, the Cost price of the cupboard $\text{= Rs 2,500}$

Also, the Selling price of the cupboard $\text{= Rs 3,000}$

Since, Selling Price (S.P.) is greater than Cost Price (C.P.)

Therefore, here we have is the profit.

$\therefore $ Profit $\text{=}$ S.P. $-$ C.P. $\text{= 3,000 - 2,500 }$

$\Rightarrow $ Profit $\text{=}$ $\text{Rs 500}$

So, the profit percentage $\text{= }\frac{\text{Profit}}{\text{C}\text{.P}}\text{  }\!\!\times\!\!\text{  100}$

$\Rightarrow $ Profit $\text{ }\%\text{  = }\frac{\text{500}}{\text{2500}}\text{  }\!\!\times\!\!\text{  100}$

$\Rightarrow $ Profit $\text{ }\%\text{  = 20 }\%$

(d). A skirt bought for $\text{Rs 250}$ and sold at $\text{Rs 150}$.

Ans: Here, the Cost price of the skirt $\text{= Rs 250}$

Also, the Selling price of the skirt $\text{= Rs 150}$

Since, Selling Price (S.P.) is lower than Cost Price (C.P.)

Therefore, here we have is the loss.

$\therefore $ Loss $\text{=}$ C.P. $\text{-}$ S.P.$\text{= 250 - 150}$

$\Rightarrow $ Loss $\text{=}$ $\text{Rs 100}$

So, the loss percentage $\text{= }\frac{\text{Loss}}{\text{C}\text{.P}}\text{  }\!\!\times\!\!\text{  100}$

$\Rightarrow $ Loss$\text{ }\%\text{  = }\frac{\text{100}}{\text{250}}\text{  }\!\!\times\!\!\text{  100}$

$\Rightarrow $ Loss$\text{ }\%\text{  = 40 }\%$

2. Convert each part of the ratio to percentage:

(a). $\text{3 : 1}$

Ans: The given ratio is $\text{3 : 1}$

The total of the parts of the given ratio is $\text{3 + 1 = 4}$

So, the fractional part of the given ratio is $\frac{\text{3}}{\text{4}}\text{ : }\frac{\text{1}}{\text{4}}$

Therefore, the percentage of the parts is $\text{= }\frac{\text{3}}{\text{4}}\text{  }\!\!\times\!\!\text{  100 : }\frac{\text{1}}{\text{4}}\text{  }\!\!\times\!\!\text{  100}$

                                                                 $\text{= 75 }\%\text{  : 25 }\%$

Thus, the percentage of the parts of given ratio is $\text{= 75 }\%\text{  : 25 }\%$

(b). $\text{2 : 3 : 5}$

Ans: The given ratio is $\text{2 : 3 : 5}$

The total of the parts of the given ratio is $\text{2 + 3 + 5 = 10 }$

So, the fractional part of the given ratio is $\frac{\text{2}}{\text{10}}\text{ : }\frac{\text{3}}{\text{10}}\text{ : }\frac{\text{5}}{\text{10}}$

Therefore, the percentage of the parts is $\text{= }\frac{\text{2}}{\text{10}}\text{  }\!\!\times\!\!\text{  100 : }\frac{\text{3}}{\text{10}}\text{  }\!\!\times\!\!\text{  100 : }\frac{\text{5}}{\text{10}}\text{  }\!\!\times\!\!\text{  100}$

                                                                 $\text{= 20 }\%\text{  : 30 }\%\text{  : 50 }\%$

Thus, the percentage of the parts of given ratio is $\text{= 20 }\%\text{  : 30 }\%\text{  : 50 }\%$

(c). $\text{1 : 4}$

Ans: The given ratio is $\text{1 : 4}$

The total of the parts of the given ratio is $\text{1 + 4 = 5}$

So, the fractional part of the given ratio is $\frac{\text{1}}{\text{5}}\text{ : }\frac{\text{4}}{\text{5}}$

Therefore, the percentage of the parts is $\text{= }\frac{\text{1}}{\text{5}}\text{  }\!\!\times\!\!\text{  100 : }\frac{\text{4}}{\text{5}}\text{  }\!\!\times\!\!\text{  100}$

                                                                 $\text{= 20 }\%\text{  : 80 }\%$

Thus, the percentage of the parts of given ratio is $\text{= 20 }\%\text{  : 80 }\%$

(d). $\text{1 : 2 : 5}$

Ans: The given ratio is $\text{1 : 2 : 5}$

The total of the parts of the given ratio is $\text{1 + 2 + 5 = 8}$

So, the fractional part of the given ratio is $\frac{\text{1}}{\text{8}}\text{ : }\frac{\text{2}}{\text{8}}\text{ : }\frac{\text{5}}{\text{8}}$

Therefore, the percentage of the parts is $\text{= }\frac{\text{1}}{\text{8}}\text{  }\!\!\times\!\!\text{  100 : }\frac{\text{2}}{\text{8}}\text{  }\!\!\times\!\!\text{  100 : }\frac{\text{5}}{\text{8}}\text{  }\!\!\times\!\!\text{  100}$

                                                                 $\text{= 12}\text{.5 }\%\text{  : 25 }\%\text{  : 62}\text{.5 }\%$

Thus, the percentage of the parts of given ratio is $\text{= 12}\text{.5 }\%\text{  : 25 }\%\text{  : 62}\text{.5 }\%$

3. The population of a city decreased from $\text{25,000}$ to $\text{24,500}$. Find the percentage decrease.

Ans: The population of a city got decreased from $\text{25,000}$ to $24,500$

The decreased population $\text{= 25,000 - 24,500 = 500}$

Decreased Percentage $\text{= }\frac{\text{Decreased Population}}{\text{Original Population }}\text{  }\!\!\times\!\!\text{  100}$

                                     $\text{= }\frac{500}{25000}\text{ }\times \text{ 100}$

                                     $\text{= 2 }\%$

Thus, the required percentage decrease is $2%$.

4. Arun bought a car for $\text{Rs 3,50,000}$. The next year, the price went up to $\text{Rs 3,70,000}$. What was the percentage of price increase?

Ans: The price of the car got increased from $\text{Rs 3,50,000}$ to $\text{Rs 3,70,000}$

So, the change in the price of the car $\text{= 3,70,000 - 3,50,000}$ $\text{= Rs 20,000}$

Therefore, Increased percentage $\text{= }\frac{\text{Change in Amount}}{\text{Original Amount}}\text{  }\!\!\times\!\!\text{  100}$

                                                     $\text{= }\frac{20000}{350000}\text{ }\times \text{ 100}$

                                                      $\text{= 5}\frac{\text{5}}{\text{7}}\text{ }\%$

Thus, the percentage of the increased price is $\text{ 5}\frac{\text{5}}{\text{7}}\text{ }\%$

5. I buy a T.V. for $\text{Rs 10,000}$ and sell it at a profit of $\text{20 }\%$. How much money do I get for it?

Ans: The money I got by selling the T.V. is known as the selling price.

Since, I bought it at $\text{Rs 10,000}$, therefore

The Cost Price of T.V. $\text{= Rs 10,000}$

We know that, S.P. = C.P. $\text{+}$ Profit           …… (1)

As, I sell it at $\text{20 }\%$ profit.

Then, the Profit $\text{= 20 }\%$ of C.P.

$\Rightarrow $ Profit $\text{= }\frac{\text{20}}{\text{100}}\text{  }\!\!\times\!\!\text{  10000}$

$\Rightarrow $ Profit $\text{= Rs 2,000}$

Now, from equation (1) we get,

S.P. $\text{=}$ C.P. $\text{+}$ Profit

$\Rightarrow \text{ S}\text{.P}\text{. = 10,000 + 2,000}$

$\Rightarrow \text{ S}\text{.P}\text{. = Rs 12,000}$

Hence, I sold the T.V. at a price of $\text{Rs 12,000}$

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

Ans: Juhi sells the washing machine for $\text{Rs 13,500}% 

Therefore, the selling price of the washing machine $\text{= Rs 13,500}$

As, she loses $\text{20 }\%$ in the bargain,

Therefore, the loss percent $\text{= 20 }\%$ 

Loss $\text{= 20 }\%\text{  of C}\text{.P}\text{.}$

$\Rightarrow $ Loss $\text{= }\frac{\text{20}}{\text{100}}\text{  }\!\!\times\!\!\text{  C}\text{.P}\text{.}$

$\Rightarrow $ Loss $\text{= }\frac{\text{C}\text{.P}\text{.}}{\text{5}}$

We know that, S.P. = C.P. $-$ Loss 
$\Rightarrow \text{ S}\text{.P}\text{. = C}\text{.P}\text{. - }\frac{\text{C}\text{.P}\text{.}}{\text{5}}$

$\Rightarrow \text{ 13500 = }\frac{\text{4 C}\text{.P}\text{.}}{\text{5}}$

$\Rightarrow$  C.P. $\text{= }\frac{\text{13500  }\!\!\times\!\!\text{ 5}}{\text{4}}$

$\Rightarrow \text{ C}\text{.P}\text{. = Rs 16,875}$

Hence, the price at which Juhi bought the washing machine is $\text{Rs 16,875}$.

7. (i) Chalk contains Calcium, Carbon, and Oxygen in the ratio $\text{10 : 3 : 12}$. Find the percentage of Carbon in chalk.

Ans: We have given the ratio of the component present in a Chalk, i.e., the ratio of Calcium, Carbon, and Oxygen as $\text{10 : 3 : 12}$ 

Therefore, the total parts $\text{10 + 3 +12 = 25}$

Thus, the part of the carbon $\text{= }\frac{\text{3}}{\text{25}}$

$\Rightarrow $ The percentage of Carbon in the chalk is $\text{= }\frac{\text{3}}{\text{25}}\text{  }\!\!\times\!\!\text{  100  = 12 }\%$

(ii) If in a stick of chalk, Carbon is $\text{3}$ g, what is the weight of the chalk stick?

Ans: $\text{3}$g Carbon is present in the chalk.

Let the weight of the chalk is $\text{x}$ g.

Then from part (i) $\text{12 }\%$ of $\text{x = 3}$

$\Rightarrow \text{ }\frac{\text{12}}{\text{100}}\text{  }\!\!\times\!\!\text{  x = 3}$

$\Rightarrow \text{ x = }\frac{\text{3  }\!\!\times\!\!\text{  100}}{\text{12}}$

 $\Rightarrow \text{ x = 25}$g

Thus, the weight of the chalk is $\text{25 g}$.

8. Amina buys a book for $\text{Rs 275}$ and sells it at a loss of $\text{15 }\%$. How much does she sell it for?

Ans: Let the Selling Price (S.P.) of the book is $\text{Rs x}$.

Since, Amina bought the book for $\text{Rs 275}$ 

Therefore, the Cost Price (C.P.) of the book is $\text{Rs 275}$

She losses $\text{15 }\%$ while selling the book

Lost Percent $\text{= 15 }\%$ and,

$\Rightarrow \text{ }$Loss $\text{= 15 }\%$ of C.P

$\Rightarrow \text{ }$Loss $\text{= 15 }\%$ of $\text{Rs 275}$

$\Rightarrow \text{ }$Loss $\text{= }\frac{\text{15}}{\text{100}}\text{  }\!\!\times\!\!\text{  275}$

$\Rightarrow \text{ }$Loss $\text{= Rs 41}\text{.25}$ 

We know that, S.P. = C.P. $-$ Loss   

$\Rightarrow $ S.P. $\text{= 275 - 41}\text{.25}$

$\Rightarrow $ S.P. $\text{= Rs 233}\text{.75}$

Thus, Amina sells her book at $\text{Rs 233}\text{.75}$

9. Find the amount to be paid at the end of $\text{3}$ years in each case:

(a) Principal $\text{= Rs 1,200}$ at $\text{12 }\%$ p.a.

Ans: We know that Amount $\text{=}$ Principal $\text{+}$Simple Interest (S.I.)

And, S.I. $\text{= }\frac{\text{P }\!\!\times\!\!\text{ R }\!\!\times\!\!\text{ T}}{\text{100}}$

Here, P $\text{=}$Principal $\text{= Rs 1,200}$

R $\text{=}$ Rate $\text{= 12 }\%\text{  }$and,

T $\text{=}$ Time $\text{= 3}$ years

Therefore, S.I. $\text{= }\frac{\text{1200 }\!\!\times\!\!\text{ 12 }\!\!\times\!\!\text{ 3}}{\text{100}}$

$\Rightarrow \text{ S}\text{.I}\text{. = Rs 432}$

So, the Amount $\text{= 1200 + 432}$

$\Rightarrow $ Amount $\text{= Rs 1,632}$

(b) Principal $\text{= Rs 7,500}$ at $5%$ p.a.

Ans: We know that Amount $\text{=}$ Principal $\text{+}$Simple Interest (S.I.)

And, S.I. $\text{= }\frac{\text{P }\!\!\times\!\!\text{ R }\!\!\times\!\!\text{ T}}{\text{100}}$

Here, P $\text{=}$Principal $\text{= Rs 7,500}$

R $\text{=}$ Rate $\text{= 5 }\%\text{  }$ and,

T $\text{=}$ Time $\text{= 3}$ years

Therefore, S.I. $\text{= }\frac{\text{7500 }\!\!\times\!\!\text{ 5 }\!\!\times\!\!\text{ 3}}{\text{100}}$

$\Rightarrow \text{ S}\text{.I}\text{. = Rs 1,125}$

So, the Amount $\text{= 7,500 + 1,125}$

$\Rightarrow $ Amount $\text{= Rs 8,625}$

10. What rate gives $\text{Rs 280}$ as interest on a sum of $\text{Rs 56,000}$ in $\text{2}$ years?

Ans:  Given,

Simple Interest (S.I.) $\text{= Rs 280}$

Principal $\text{= Rs 56,000}$ and,

Time $\text{= 2}$ years 

Let $\text{R}$ be the required rate of interest.

As, S.I. $\text{= }\frac{\text{P }\!\!\times\!\!\text{ R }\!\!\times\!\!\text{ T}}{\text{100}}$

$\Rightarrow $ $\text{280 = }\frac{\text{56000 }\!\!\times\!\!\text{ R }\!\!\times\!\!\text{ 2}}{\text{100}}$

$\Rightarrow $ $\text{R = }\frac{\text{280 }\!\!\times\!\!\text{ 100}}{\text{56000 }\!\!\times\!\!\text{ 2}}$

$\Rightarrow $ \[R = 0.25\% \]

Thus, the required rate of interest is, \[R = 0.25\% \]

11. If Meena gives an interest of $\text{Rs 45}$ for one year at $\text{9 }\%$ rate p.a. What is the sum she has borrowed?

Ans: Let the sum / Principal borrowed by Meena is P

Given, Simple Interest, S.I. $\text{= Rs 45}$

Rate of Interest, R $\text{= 9 }\%\text{  p}\text{.a}\text{.}$ and,

Time, $\text{T = 1}$ year

As, S.I. $\text{= }\frac{\text{P }\!\!\times\!\!\text{ R }\!\!\times\!\!\text{ T}}{\text{100}}$

$\Rightarrow$ $\text{45 = }\frac{\text{P }\!\!\times\!\!\text{ 9 }\!\!\times\!\!\text{ 1}}{\text{100}}$

$\Rightarrow$ $\text{P = }\frac{\text{45 }\!\!\times\!\!\text{ 100}}{\text{9 }\!\!\times\!\!\text{ 1}}$

$\Rightarrow$ $\text{P = Rs 500}$

Thus, the sum Meena borrowed is, $\text{Rs 500}$.

NCERT Solutions for Class 7 Maths – Free PDF Download

Class 7 Maths Chapter 8 Includes:

Facts

• Various quantities (of the same kind) are compared using their ratios.

• If two fractions are equal, their ratios are equivalent.

• When two ratios are equal then the four quantities are in proportion.

• Percentages are numerators of fractions with denominator 100. The percentage is also a way of comparing quantities.

• To convert a percent into decimal, drop the sign of percent and then shift the decimal point two places to the left.

• To convert a fraction into percent, multiply the fraction by 100, and write % sign on the right of the number.

• Profit = SP – CP (when SP > CP).

• Loss = CP – SP (when CP > SP).

• Percent gain or loss is always calculated on CP.

• Money borrowed is called the principal.

• Simple interest = (P x R x T)/100

• Amount = Principal + Interest

Ratio

Comparison means to know “ how many times one quantity is of the other”, or to know – what part one quantity is of the other. It is called the ratio of quantities of the same kind and in the same units.

The ratio of two numbers ‘a’ and ‘b’ (b ≠ 0) is a/b and it is denoted by a:b. A ratio in the simplest form is also called the ratio in the lowest terms.

Note: To compare two quantities or to find the ratio of two quantities, their units must be the same.

Different ratios can also be compared by writing them as ‘like-fractions’.

Let us go through some examples to get a better understanding. Also, you can download and refer to the Exercise 8.1 in NCERT Solutions for class 7 chapter 11 from Vedantu for free, and prepare for your exams.

Example 1: Find the ratio of 5 km and 400 m.

Solution: The units of the two distances in the question are different.

First we need to convert the distances to the same unit.

5 km = 5 x 1000 m = 5000 m

Now, 5 km : 400 m = 5000 m : 400 m

= 5000 : 400 = 5000/200 : 400/200

= 25 : 2 (Since, the HCF of 5000 and 400 is 200)

Thus, the required ratio is 25:2.

Example 2: Find the ratio of 30 days and 36 hours.

Solution: Since the given durations have different units, we need to first convert them to the same unit.

30 days = 30 x 24 hours = 720 hours

Now, 30 days : 36 hours = 720 hours : 36 hours 

       = 720/36 : 36/36 (Since, the HCF of 720 and 36 is 36)

      =  20 : 1

Thus, the required ratio is 20:1.

Equivalent Ratios

We can compare various ratios by converting them to like fractions. If the like fractions are equal, then the given ratios are said to be equivalent.

Example: Are the ratios 2 : 3 and 3: 4 equivalent?

Solution: The corresponding fractions of the ratios (2:3 and 3:4) are 2/3 and 3/4.

Now, 2/3 = 2/3 x 4/4 = 8/12 (∴ LCM of 3 and 4 is 12)

and, 3/3 = 3/4 x 3/4 = 9/12

since, 8 < 9  

or, 8/12 < 9/12 or 2/3 < 3/4

Thus, 2:3 and 3:4 are not equivalent.     

Referring to the Exercise 8.2 in NCERT Solutions of Chapter 11 for class 7, you need to learn what percentage is and the conversion of fractional numbers to Percentage.

Percentage

Percentages are numerators of fractions having denominators as 100. Percentages are used in comparing results. As mentioned before, the percent is represented by the symbol % and it is expressed as parts of hundredths.

Example: 9 % means 9 out of 100. It is written as 9% = 9/100 = 0.09.

Percentage when Total is not Hundred.

In case, the total is not a hundred, then we need to convert the fraction into an equivalent fraction with denominator 100. For this purpose, we can use any method. Some methods are given below:

Method I: We multiply the given fraction by 100/100. [ ∵ 100/100 = 1], which does not change the value of the fraction.

Example: 12/20 = 12/20 x 100/100 = 3/5 x 100/100 = 60/100 = 60%

Method II: We can use the unitary method.

Example: For 20, the corresponding value is 8

For 1, the corresponding value is 8/20.

For 100, the corresponding value is (8/20  x 100) % = 40%

Method III: We multiply and divide the given fraction by a number such that the denominator becomes 100.

Example: 12/20 = 12/20 x 5/5 = 60/100 = 60%

Ratios To Percent

Sometimes, the parts of a whole quantity are given in the form of ratios. We can convert them into percentages. Let us look into an example to understand this well.

Example: The population of a state increased from 5,50,5000 to 6,05,000. What is the percentage increase in the population?

Solution:  Initial population = 5,50,000

Increased population = 6,05,000

∴  Increase in population = 6,05,000 – 5,50,000

= 55,000

∴ Percent increase = 55,000/5,50,000  x 100% = 10%

Note:  We convert the increase or decrease in a quantity as a percentage of the initial amount.

Profit Or Loss As A Percentage

We know that the buying price of any item is called its cost price (CP) and the price at which it is sold is called its selling price (SP).

Note

1. If CP < SP, then there is a profit in the transaction, and Profit = SP – CP.

2. If CP > SP, then there is a loss in the transaction and Loss = CP – SP.

3. Percent profit or percent loss is always calculated on the CP.

Simple Interest

The money borrowed is called the Principal (or sum borrowed). For using a bank's money for some time, the borrower has to pay some extra money to the bank. This extra money is called Interest. The total money paid back with interest is called Amount.

Amount = Principal + Interest

Interest is generally given in percent for a period of one year. Generally, the rate of interest is written in percent per year or per annum.

 So, Principal is denoted by P, Rate of interest by R, and Time by T.

Now,  simple interest, or Interest = (Principal x Rate x Time)/100, or I = PRT/100

Download and solve Exercise 8.3 from NCERT Solutions to chapter 8 for class 7. For further clarification, you can get in touch with any of the faculties at Vedantu.

Why should you Use Solutions Provided by Vedantu for Class 7 Maths?

Key Features of NCERT Solutions, These solutions are designed to help students achieve proficiency in their studies. They are crafted by experienced educators who excel in teaching Math. Some of the features include:

• Comprehensive explanations for each exercise and questions, promoting a deeper understanding of the subject.

• Clear and structured presentation for easy comprehension.

• Accurate answers aligned with the curriculum, boosting students' confidence in their knowledge.

• Visual aids like diagrams and illustrations to simplify complex concepts.

• Additional tips and insights to enhance students' performance.

• Chapter summaries for quick revision.

• Online accessibility and downloadable resources for flexible study and revision.

Conclusion

The NCERT Solutions for Class 7 Maths Chapter 8 - Comparing Quantities, provided by Vedantu, is a valuable tool for Class 7 students. It helps introduce Math concepts in an accessible manner. The provided solutions and explanations simplify complex ideas, making it easier for Class 7 Students to understand the material. By using Vedantu's resources, Students can develop a deeper understanding of NCERT concepts. These solutions are a helpful aid for Class 7 students, empowering them to excel in their studies and develop a genuine appreciation for Comparing Quantities.