NCERT Solutions for Class 8 Maths Chapter 8 - Comparing Quantities

Exercise: 8.1

1. Find the ratio of the following:

(a) Speed of a cycle $15$km per hour to the speed of scooter $30$km per hour.

Ans: Speed of a cycle $ = $ $15$km

Speed of scooter $ = $$30$km

Ratio of the speed of a cycle to the speed of scooter 

$ = \dfrac{{15}}{{30}}$

$ = \dfrac{1}{2}$ 

The required ratio is $1:2$.

(b) $5$m to $10$km.

Ans:  $5$m to $10$km.

Since 1km $ = $$1000$ m

$ \Rightarrow $$\dfrac{{5\;{\text{m}}}}{{10\;{\text{km}}}}\; = \,\dfrac{5}{{10}} \times \dfrac{1}{{1000}}$

$ = \dfrac{1}{{2000}}$

$ \Rightarrow 1:2000$

The required ratio is $1:2000$.

(c) $50$ paise to Rs $5$.

Ans:  $50$ paise to Rs $5$

Since Rs 1 $ = $ $100$ paise      

$ \Rightarrow \dfrac{{50paise}}{{Rs5}}$                     

$ = \dfrac{{50}}{5} \times \dfrac{1}{{100}}$               

$ = \dfrac{1}{{10}}$                    

$ \Rightarrow 1:10$ 

The required ratio is $1:10$.

2. Convert the following ratios to percentages.

(a) $3:4$

Ans: $3:4$ $ = $ $\dfrac{3}{4}$

$ \Rightarrow $$\dfrac{3}{4} \times \dfrac{{100}}{{100}}$                        

$ = $$0.75 \times 100\% $ 

$ = 75\% $

The required ratio to percentage is $75\% $

(b) $2:3$

Ans: $2:3$ $ = $ $\dfrac{2}{3}$

$ = \dfrac{2}{3} \times \dfrac{{100}}{{100}}$

$ = \dfrac{{200}}{3}\% $ 

$ = 66\dfrac{2}{3}\% $

The required ratio to percentage is $66\dfrac{2}{3}\% $

3. $72\% $ of  $25$ students are good in mathematics. How many are not good in mathematics?

Ans: Total number of students $ = $ 25 .

Percentage of students are good in mathematics  $ = $$72\% $

Percentage of students who are not good in mathematics  $ = $ $(100 - 72)\% $

$ \Rightarrow 28\% $

$\therefore $ Number of students who are not good in mathematics $ = $$28\%  \times 25$

$ \Rightarrow \dfrac{{28}}{{100}} \times 25$

$ \Rightarrow \dfrac{{28}}{4}$

$ \Rightarrow 7$

Students are not good in mathematics $ = 7$

4. A football team won $10$ matches out of the total number of matches they played. If their win percentage was $40$, then how many matches did they play in all ?

Ans: The total number of matches won by the football team $ = 10$.

Percentage of team $ = 40\% $

The total number of matches played by the team $ = ?$

The total number of matches played by the team 

$ \Rightarrow 40\%  \times x = 10$

$ \Rightarrow \dfrac{{40}}{{100}} \times x = 10$

$ \Rightarrow x = 10 \times \dfrac{{100}}{{40}}$

$ \Rightarrow x = \dfrac{{100}}{4}$

$ \Rightarrow x = 25$

The total number of matches played by the team $ = 25$.

5. If Chameli had Rs .$600$ left after spending 75% of her money, how much did she have in the beginning?

Ans:Chameli’s money after spend $ = 600$

Percentage of money after spend $ = 75\% $

Beginning amount of chameli $ = ?$

Percentage of beginning amount 

$ = \left( {100{\text{ }} - {\text{ }}75} \right)\% \qquad $

$ = 25\% $

Beginning amount of chameli

$ \Rightarrow 25\%  \times x = 600$

$ \Rightarrow \left( {\dfrac{{25}}{{100}}} \right) \times x = 600$

$ \Rightarrow \left( {\dfrac{1}{4}} \right) \times x = 600$

$ \Rightarrow x = 600 \times 4$

$ \Rightarrow x = 2400$

Beginning amount of chameli $ = 2400$.

6.  If $60\% $ people in city like cricket,$30\% $like football and the remaining like other games, then what per cent of the people like other games? If the total number of people are $50$ lakh, find the exact number who like each type of game.

Ans: Total number of people in city $ = 50$lakh

Percentage of people like cricket $ = 60\% $

Percentage of people like football $ = 30\% $

Percentage of people like other games $ = ?$

Number of people like each type of game $ = ?$

Percentage of people like other games 

$ = \left( {100 - 60 - 30} \right)\% $ 

$ = \left( {100 - 90} \right)\% $

$ = 10\% $

Percentage of people like other games$ = 10\% $

Number of people like cricket 

$ = \left( {60\%  \times 50} \right)$

$ = \left( {\dfrac{{60}}{{100}} \times 50} \right)$

$ = \dfrac{{60}}{2}$

$ = 30$lakh

Number of people like cricket$ = 30$lakh

Number of people like football

$ = \left( {30\%  \times 50} \right)$

$ = \left( {\dfrac{{30}}{{100}} \times 50} \right)$

$ = \dfrac{{30}}{2}$

$ = 15$lakh

Number of people like football$ = 15$lakh

Number of people like other games

$ = \left( {10\%  \times 50} \right)$

$ = \left( {\dfrac{{10}}{{100}} \times 50} \right)$

$ = \dfrac{{10}}{2}$

$ = 5$lakh.

Number of people like other games$ = 5$lakh.

Exercise 8.2

1. A man got a $10\% $ increase in his salary. If his new salary is Rs $1,54,000$.Find his original salary.

Ans: Percentage of increment in salary $ = 10\% $

New salary $ = 1,54,000$

Original salary $ = x$

Formula:

Original salary $ + $Increment $ = $New salary. 

$ \Rightarrow x + 10\%  \times x = 1,54,000$

$ \Rightarrow x\left( {1 + 10\% } \right) = 1,54,000$

$ \Rightarrow x\left( {1 + \dfrac{{10}}{{100}}} \right) = 1,54,000$

$ \Rightarrow x\left( {\dfrac{{100 + 10}}{{100}}} \right) = 1,54,000$

$ \Rightarrow x\left( {\dfrac{{110}}{{100}}} \right) = 1,54,000$

$ \Rightarrow x = 1,54,000 \times \left( {\dfrac{{100}}{{110}}} \right)$

$ \Rightarrow x = 1400 \times 100$

$ \Rightarrow x = 140000$

The original salary $ = 1,40,000$.

2. On Sunday $845$ people went to the Zoo. On Monday only $169$ people went. What is the per cent decrease in the people visiting the zoo on Monday?

Ans: People went to zoo in Sunday $ = 845$

People went to zoo in Monday $ = 169$

Percentage decrease in number of people visit in Monday $ = ?$

Formula:

Percentage decrease in number of people visit in Monday $ = \left( {\dfrac{{Decrease{\text{ }}in{\text{ }}number{\text{ }}of{\text{ }}people{\text{ }}visit{\text{ }}in{\text{ }}Monday}}{{People{\text{ }}went{\text{ }}to{\text{ }}zoo{\text{ }}in{\text{ }}Sunday}}} \right) \times 100\% $

Decrease in number of people visit in Monday

$ = 845 - 169$

$ = 676$

Decrease in number of people visit in Monday$ = 676$.

Percentage decrease in number of people visit in Monday

$ = \left( {\dfrac{{676}}{{845}} \times 100} \right)\% $

$ = 0.8 \times 100\% $

$ = 80\% $

Percentage decrease in number of people visit in Monday$ = 80\% $

3. A shopkeeper buys $80$ articles for Rs $2,400$ and sells them for a profit of $16\% $. Find the selling price of one article.

Ans: Total number of articles$ = 80$

Articles price$ = 2400$

Percentage of profit $ = 16\% $

Selling price of one article $ = ?$

Formula:

Profit percent $ = \left( {\dfrac{{profit}}{{C.P}}} \right) \times 100$

Selling price of one article $ = $ C.P. $ + $ Profit

Cost of one article 

$ = \left( {\dfrac{{Articles{\text{ }}price}}{{Total{\text{ }}number{\text{ }}of{\text{ }}articles}}} \right)$

$ = \left( {\dfrac{{2400}}{{80}}} \right)$

$ = 30$

Cost of one article $ = 30$

Profit percent $ = \left( {\dfrac{{profit}}{{C.P}}} \right) \times 100$

$16 = \dfrac{{Profit}}{{30}} \times 100$
Profit $ = \dfrac{{16 \times 30}}{{100}}$

Profit $ = 0.16 \times 30$

Profit $ = 4.80$

Selling price of one article $ = $ C.P. $ + $ Profit

$ = $Rs $\left( {30 + 4.80} \right)$

$ = $Rs$34.80$

4. The cost of an article was Rs $15,500$. Rs $450$ were spent on its repairs. If it is sold for a profit of $15\% $, find the selling price of the article.

Ans:  Cost of an article \[ = \]Rs. \[15,500\]

Spent for repair = Rs. $450$    

Profit precentag  $ = 15\% $  

Selling price of one article  $ = ?$

Formula:

Profit percent $ = \left( {\dfrac{{profit}}{{C.P}}} \right) \times 100$

Selling price of one article $ = $ C.P. $ + $ Profit

Total cost of an article $ = $ Cost of an article $ + $ Spent for repair

$ = $Rs. $15500$ $ + $Rs. $450$

$ = $Rs $15950$ 

Profit percent $ = \left( {\dfrac{{profit}}{{C.P}}} \right) \times 100$

$15 = \left( {\dfrac{{profit}}{{15950}}} \right) \times 100$

Profit $ = \dfrac{{15 \times 15950}}{{100}}$

Profit $ = 15 \times 159.5$

Profit $ = 2392.5$

Selling price of one article $ = $ C.P. $ + $ Profit

$ = $ Rs $2392.5$

$ = $Rs$18342.5$

Selling price of one article $ = $Rs$18342.5$

5. A VCR and TV were bought for Rs $8,000$ each. The shopkeeper made a loss of $4\% $ on the VCR and a profit of $8\% $ on the TV. Find the gain or loss percent on the whole transaction.

Ans: Cost of VCR and TV $ = 8,000$

Loss Percentage in VCR $ = 4\% $

Profit percentage in TV $ = 8\% $

Whole gain and loss $ = ?$

Loss percentage  $ = 4\% $

if C.P. is Rs.$100$, then S.P. is Rs.$96$.

when C.P. is Rs. $8000$

S.P $ = \left( {\dfrac{{96}}{{100}} \times 8000} \right)$

S.P $ = 96 \times 80$

S.P $ = $Rs. $7680$

Profit percentage$ = 8\% $

if C.P. is Rs.$100$, then S.P. is Rs.$108$.

when C.P. is Rs. $8000$

S.P $ = \left( {\dfrac{{108}}{{100}} \times 8000} \right)$

S.P $ = 108 \times 80$

Selling price $ = $Rs. $8640$

Total selling price $ = $ Rs. $7680 + $ Rs $8640$

$ = $ Rs.$16320$

Total cost price $ = $ Rs. $8000$$ + $ Rs.$8000$

$ = $ Rs. $16000$

Since total selling price greater than total cost price .

Profit $ = $ Rs. $16320$$ - $ Rs. $16000$

$ = $ Rs. $320$

Profit percent $ = \left( {\dfrac{{profit}}{{C.P}}} \right) \times 100$

$ = \left( {\dfrac{{320}}{{16000}}} \right) \times 100$

$ = \dfrac{{320}}{{160}}$

$ = 2\% $

The gain percentage of shopkeeper in whole transaction$ = 2\% $

6. During a sale, a shop offered a discount of $10\% $ on the marked prices of all the items. What would a customer have to pay for a pair of jeans marked at Rs $1450$ and two shirts marked at Rs $850$ each?

Ans: Discount percentage $ = 10\% $

Price of pair jeans $ = $ Rs.$1450$

Price of shirt $ = $ Rs. $850$

Formula:

Discount $ = $ Marked price $ - $ Sale price.

One shirt price $ = $ Rs. $850$

Two shirt price = $2 \times $ Rs. $850$

$ \Rightarrow $ Rs. $1,700$

Total marked price = Rs. $\left( {1,450 + 1,700} \right)$

= Rs. $3,150$

Discount = Rs. $\left( {10\%  \times 3150} \right)$

= Rs. $\left( {\dfrac{{10}}{{100}} \times 3150} \right)$

Discount = Rs. $315$.

Discount $ = $ Total Marked price $ - $ Sale price

$ \Rightarrow $ Rs $315$ = Rs $3150$ − Sale price

$ \Rightarrow $Sale price $ = $ Rs $\left( {3150 - 315} \right)$

$ \Rightarrow $Sale price $ = $ Rs $2835$

Customer paid $ = $ Rs $2835$.

7. A milkman sold two of his buffaloes for Rs $20,000$ each. On one he made a gain of $5\% $ and on the other a loss of $10\% $. Find his overall gain or loss.

(Hint: Find CP of each).

Ans: Buffaloes $ = $ Rs. $20,000$

Gain percentage $ = 5\% $

Loss percentage $ = 10\% $

Overall gain or loss $ = ?$

Gain  percentage $ = 5\% $

if C.P. is Rs.$100$, then S.P. is Rs. $105$.

when C.P. is Rs. $20,000$

S.P $ = \left( {\dfrac{{100}}{{105}} \times 20,000} \right)$

S.P $ = 100 \times 190.47$

Selling price $ = $Rs. $19047$

Loss percentage $ = 10\% $

if C.P. is Rs.$100$, then S.P. is Rs. $90$.

when C.P. is Rs. $20,000$

S.P $ = \left( {\dfrac{{100}}{{90}} \times 20,000} \right)$

S.P $ = 100 \times 222.222$

Selling price $ = $Rs. $22222.22$.

Total selling price $ = $ Rs. $20,000 + $ Rs $20,000$

$ = $ Rs. $40,000$

Total cost price $ = $ Rs. $19047.62$$ + $ Rs. $22222.22$

$ = $ Rs. $41269.84$

Since total selling price less than total cost price .

Loss $ = $ Rs. $41269.84$$ - $ Rs. $40,000$

$ = $ Rs. $1269.84$

The overall loss of milkman $ = $ Rs. $1269.84$.

8. The price of a TV is Rs $13,000$. The sales tax charged on it is at the rate of $12\% $. Find the amount that Vinod will have to pay if he buys it.

Ans: Price of TV $ = $ Rs. $13,000$.

Sales tax percentage $ = 12\% $

Vinod have to pay $ = ?$

if  Rs.$100$, then Tax to be pay is Rs. $12$.

when  Rs. $13,000$

Tax to be pay $ = \left( {\dfrac{{12}}{{100}} \times 13,000} \right)$

Tax to be pay $ = 12 \times 130$

Tax to be pay $ = $Rs. $1,560$.

Vinod have to pay $ = $ price of TV $ + $ Tax to be pay

$ = $ Rs. $13,000$$ + $ Rs. $1560$

$ = $ Rs. $14560$

Vinod have to pay$ = $ Rs. $14560$.

9. Arun bought a pair of skates at a sale where the discount given was $20\% $. If the amount he pays is Rs $1,600$. find the marked price.

Ans: Discount in skates $ = 20\% $

Total amount $ = 1,600$

Marked price $ = x$

Formula:

Discount percent\[\]$ = \left( {\dfrac{{Discount}}{{Marked{\text{ }}price}}} \right) \times 100$ 

Discount percent $ = \left( {\dfrac{{Discount}}{{Marked{\text{ }}price}}} \right) \times 100$

$ \Rightarrow 20 = \left( {\dfrac{{Discount}}{x}} \right) \times 100$

Discount $ = \dfrac{{20 \times x}}{{100}}$

Discount $ = \dfrac{{1 \times x}}{5}$

Discount $ = $ Marked price $ - $ Total amount

$ \Rightarrow \dfrac{{1 \times x}}{5}$$ = x - 1600$

$ \Rightarrow 1600 = x - \dfrac{1}{5}x$

$ \Rightarrow 1600 = \dfrac{{5x - x}}{5}$

$ \Rightarrow 1600 = \dfrac{{4x}}{5}$

$ \Rightarrow \dfrac{{1600 \times 5}}{4} = x$

$ \Rightarrow x = 400 \times 5$

$ \Rightarrow x = 2000$

Marked price $ = 2000$.

10. I purchased a hair-dryer for Rs $5,400$ including $8\% $ VAT. Find the price before VAT was added.

Ans: Hair-dryer rate include VAT $ = 5,400$

Tax percentage $ = 8\% $

Rate before VAT $ = ?$

VAT$ = 8\% $

If VAT without Rs.$100$, then price is Rs. $108$

when  Rs. $5400$

Rate before VAT $ = \left( {\dfrac{{100}}{{108}} \times 5400} \right)$

Rate before VAT $ = 100 \times 50$

Rate before VAT $ = $ Rs. $5000$.

Exercise 8.3

1. Calculate the amount and compound interest on

(a) Rs $10800$ for $3$ years at \[12\dfrac{1}{2}\% \] per annum compounded annually.

(b) Rs \[18000\] for\[2\dfrac{1}{2}\] years at \[10\% \]  per annum compounded annually.

(c) Rs \[62500\] for \[1\dfrac{1}{2}\] years at \[8\% \]  per annum compounded half yearly.

(d) Rs \[8000\] for \[1\] year at \[9\% \]  per annum compound half yearly.

(You could use the year by year calculation using SI formula to verify)

(e) Rs \[10000\] for \[1\] year at \[8\% \] per annum compounded half yearly.

Ans:

(a) Principal (P) \[ = \] Rs. \[10,800\].

Rate (R) = \[12\dfrac{1}{2}\% \]

\[ \Rightarrow \dfrac{{24 + 1}}{2}\% \]

\[ \Rightarrow \dfrac{{25}}{2}\% \] (annual)

Number of years n \[ = 3\]

Formula :

Amount ,A \[ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}\]

Compound interest \[ = \] A \[ - \] P

A \[ = 10,800{\left( {1 + \dfrac{{25}}{{2 \times 100}}} \right)^3}\]

A \[ = 10,800{\left( {1 + \dfrac{{25}}{{200}}} \right)^3}\]

A  \[ = 10,800{\left( {\dfrac{{200 + 25}}{{200}}} \right)^3}\]

A \[ = 10,800{\left( {\dfrac{{225}}{{200}}} \right)^3}\]

A\[ = 10,800{\left( {\dfrac{9}{8}} \right)^3}\]

A \[ = 10800{\left( {1.125} \right)^3}\]

A \[ = 10800 \times 1.423828\]

A \[ = 15377.34\] (approximately)

Compound interest \[ = \] A \[ - \] P

\[ \Rightarrow \] Rs. \[\left( {15377.34 - 10800} \right)\]

\[ \Rightarrow \] Rs. \[4577.34\].

(b) Principal (P) \[ = \] Rs. \[18,000\].

Rate (R) = \[10\% \] (annual)

Number of years n \[ = 2\dfrac{1}{2}\] Years (2 Years and 6 month)

The amount for 2 years and 6 months can be calculated by first calculating the amount for 2 years using the compound interest formula, and then calculating the simple interest for 6 months on the amount obtained at the end of 2 years.

Firstly, the amount for 2 years has to be calculated.

Formula :

Amount ,A \[ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}\]

Compound interest \[ = \] A \[ - \] P

A \[ = 18000{\left( {1 + \dfrac{{10}}{{100}}} \right)^2}\]

A \[ = 18,000{\left( {1 + \dfrac{1}{{10}}} \right)^2}\]

A  \[ = 18,000{\left( {\dfrac{{10 + 1}}{{10}}} \right)^2}\]

A \[ = 18,000{\left( {\dfrac{{11}}{{10}}} \right)^2}\]

A \[ = 18,000{\left( {1.1} \right)^2}\]

A \[ = 18,000 \times 1.21\]

A \[ = 21780\] (approximately)

By taking Rs. \[21,780\] as principal.

The S.I for next 6 month will be calculated.

i.e \[\dfrac{1}{2}\] year

S.I \[ = \] Rs \[\left( {\dfrac{{21,780 \times \dfrac{1}{2} \times 10}}{{100}}} \right)\]

S.I \[ = \] Rs\[\left( {\dfrac{{21,780 \times 5}}{{100}}} \right)\]

S.I \[ = \] Rs \[\left( {\dfrac{{21,78}}{2}} \right)\]

S.I \[ = \] Rs \[1089\].

Interest for first two years \[ = \] Rs. \[\left( {21780 - 18000} \right)\]

\[ = \] Rs. \[3780\].

And interest for next half years\[ = \] Rs. \[1089\].

Total compound interest \[ = \] Rs. \[3780\]\[ + \]\[1089\]

\[ = \] Rs. \[4869\].

Compound interest \[ = \] A \[ - \] P

A \[ = \] P\[ + \] Compound interest

\[ \Rightarrow \] Rs. \[\left( {18000 + 4869} \right)\]

\[ \Rightarrow \] Rs. \[22,869\].

Amount \[ = \] Rs. \[22,869\].

(c) Principal (P) \[ = \] Rs. \[62,500\].

Rate (R) = \[8\% \] per annum or \[4\% \]per half year.

Number of years n \[ = 1\dfrac{1}{2}\] years

Formula :

Amount ,A \[ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}\]

Compound interest \[ = \] A \[ - \] P

A \[ = 62,500{\left( {1 + \dfrac{4}{{100}}} \right)^3}\]

A \[ = 62,500{\left( {1 + \dfrac{1}{{25}}} \right)^3}\]

A  \[ = 62,500{\left( {\dfrac{{25 + 1}}{{25}}} \right)^3}\]

A \[ = 62,500{\left( {\dfrac{{26}}{{25}}} \right)^3}\]

A \[ = 62,500{\left( {1.04} \right)^3}\]

A \[ = 62,500\left( {1.1248} \right)\]

A \[ = 70,304\] (approximately)

Compound interest \[ = \] A \[ - \] P

\[ \Rightarrow \] Rs. \[\left( {70304 - 62500} \right)\]

\[ \Rightarrow \] Rs. \[7,804\].

(d) Principal (P) \[ = \] Rs. \[8,000\].

Rate (R) = \[9\% \] (annual) or \[\dfrac{9}{2}\% \] (half year)

Number of years n \[ = 1\]

There will be 2 half years in 1 year.

Formula :

Amount ,A \[ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}\]

Compound interest \[ = \] A \[ - \] P

A \[ = 8,000{\left( {1 + \dfrac{9}{{200}}} \right)^2}\]

A \[ = 8,000{\left( {1 + 0.045} \right)^2}\]

A  \[ = 8,000{\left( {1.045} \right)^2}\]

A \[ = 8,000 \times 1.092\]

A \[ = 8,736.2\] (approximately)

Compound interest \[ = \] A \[ - \] P

\[ \Rightarrow \] Rs. \[\left( {8736.2 - 8000} \right)\]

\[ \Rightarrow \] Rs. \[736.20\].

(e) Principal (P) \[ = \] Rs. \[10,000\].

Rate (R) = \[8\% \] per annum or 4% per half year.

Number of years n \[ = 1\] year

There will be 2 half years in 1 year.

Formula :

Amount ,A \[ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}\]

Compound interest \[ = \] A \[ - \] P

A \[ = 10,000{\left( {1 + \dfrac{4}{{100}}} \right)^2}\]

A \[ = 10,000{\left( {1 + \dfrac{1}{{25}}} \right)^2}\]

A  \[ = 10,000{\left( {\dfrac{{25 + 1}}{{25}}} \right)^2}\]

A \[ = 10,000{\left( {\dfrac{{26}}{{25}}} \right)^2}\]

A \[ = 10,000{\left( {1.04} \right)^2}\]

A \[ = 10,000 \times 1.0816\]

A \[ = 10,816\] (approximately)

Compound interest \[ = \] A \[ - \] P

\[ \Rightarrow \] Rs. \[\left( {10816 - 10000} \right)\]

\[ \Rightarrow \] Rs. \[816\].

2. Kamala borrowed Rs \[26400\] from a Bank to buy a scooter at a rate of \[15\% \]p.a. compounded yearly. What amount will she pay at the end of \[2\] years and \[4\] months  to clear the loan?

(Hint: Find A for \[2\] years with interest is compounded yearly and then find SI on 

the\[2\] nd year amount for \[\dfrac{4}{{12}}\]years.)

Ans: Principal (P) \[ = \] Rs. \[26,400\]

Rate (R) = \[15\% \] per annum 

Number of years n  =  \[2\dfrac{4}{{12}}\] year

Formula :

Amount ,A \[ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}\]

Compound interest \[ = \] A \[ - \] P

The amount for 2 years and 4 months can be calculated by first calculating the amount for 2 years using the compound interest formula, and then calculating the simple interest for 4 months on the amount obtained at the end of 2 years.

Firstly, the amount for 2 years has to be calculated.

A \[ = 26,400{\left( {1 + \dfrac{{15}}{{100}}} \right)^2}\]

A \[ = 26,400{\left( {1 + \dfrac{3}{{20}}} \right)^2}\]

A  \[ = 26,400{\left( {\dfrac{{20 + 3}}{{20}}} \right)^2}\]

A \[ = 26,400{\left( {\dfrac{{23}}{{20}}} \right)^2}\]

A \[ = 26,400{\left( {1.15} \right)^2}\]

A \[ = 26,400 \times 1.3225\]

A \[ = 34,914\] (approximately)

By taking Rs. \[34,914\] as principal.

The S.I for next \[\dfrac{1}{3}\]years will be calculated.

S.I \[ = \] Rs \[\left( {\dfrac{{34914 \times \dfrac{1}{3} \times 15}}{{100}}} \right)\]

S.I \[ = \] Rs \[\left( {\dfrac{{34914 \times 0.333 \times 15}}{{100}}} \right)\]

S.I \[ = \] Rs \[\left( {\dfrac{{174569.999}}{{100}}} \right)\]

S.I \[ = \] Rs \[1745.70\].

Interest for first two years \[ = \] Rs. \[\left( {34914 - 26400} \right)\]

\[ = \] Rs. \[8514\].

And interest for next \[\dfrac{1}{3}\] years\[ = \] Rs. \[1,745.70\].

Compound interest \[ \= \] Rs. \[\left( {8514 + 1745.70} \right)\]

\[ \Rightarrow \] Rs. \[10,259.70\].

Amount \[ = \] P \[ + \] C.I

\[ \Rightarrow \] Rs. \[\left( {26,400 + 10,259.70} \right)\]

\[ \Rightarrow \] Rs. \[36,659.70\].

3. Fabina borrows Rs $12,500$ at $12\% $ per annum for $3$years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?

Ans: Fabina and Radha borrowed money (P) $ = 12,500$

Percentage per annum for Fabina (R) $ = 12\% $

Percentage per annum for Radha (R) $ = 10\% $

Years (T)$ = 3$

Interest paid by Fabina $ = \dfrac{{P \times R \times T}}{{100}}$

$ = $ Rs. $\left( {\dfrac{{12500 \times 12 \times 3}}{{100}}} \right)$

$ = $ Rs. $125 \times 12 \times 3$

$ = $ Rs. $4,500$

Amount paid by Radha at the end of 3 years

A $ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$

A $ = 12500{\left( {1 + \dfrac{{10}}{{100}}} \right)^3}$

A $ = 12500{\left( {1 + \dfrac{1}{{10}}} \right)^3}$

A $ = 12500{\left( {1 + 0.1} \right)^3}$

A $ = 12500{\left( {1.1} \right)^3}$

A = $12500 \times 1.331$

A = $16637.5$

Compound interest \[ = \] A \[ - \] P

\[ \Rightarrow \] Rs. \[\left( {16637.50 - 12500} \right)\]

\[ \Rightarrow \] Rs. \[4,137.50\].

The interest paid by Fabina is Rs. $4,500$ and by Radha is Rs. \[4,137.50\].

$\therefore $Fabina pays more interest.

\[ \Rightarrow \] Rs. $4,500 - $ Rs. \[4,137.50\].

\[ \Rightarrow \] Rs. $362.50$

$\therefore $ Fabina will have to pay  Rs. $362.50$ more.

4. I borrowed Rs $12,000$ from Jamshed at $6\% $ per annum simple interest for $2$ years. Had I borrowed this sum at $6\% $ per annum compound interest, what extra amount would I have to pay?

Ans: Principle (P) $ = $ Rs $12,000$

Rate $ = $ $6\% $ per annum

Time (T) $ = $$2$ years.

Formula :

Compound interest \[ = \] A \[ - \] P

S.I $ = \dfrac{{P \times R \times T}}{{100}}$

$ = $ Rs. $\left( {\dfrac{{12,000 \times 6 \times 2}}{{100}}} \right)$

$ = $ Rs. $120 \times 6 \times 2$

$ = $ Rs. $1,440$

Compound interest \[ = \] A \[ - \] P

Find amount A,

A $ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$

A $ = 12,000{\left( {1 + \dfrac{6}{{100}}} \right)^2}$

A $ = 12,000{\left( {1 + \dfrac{3}{{50}}} \right)^2}$

A $ = 12,000{\left( {1 + 0.06} \right)^2}$

A $ = 12,000{\left( {1.06} \right)^2}$

A = $12,000 \times 1.1236$

A = $13,483.20$

Compound interest \[ = \] A \[ - \] P

\[ \Rightarrow \] Rs. \[\left( {13483.20 - 12000} \right)\]

\[ \Rightarrow \] Rs. \[1,483.20\].

Difference in interests = C.I. $ - $S.I

\[ \Rightarrow \] Rs. \[\left( {1,483.20 - 1,440} \right)\]

\[ \Rightarrow \] Rs. \[43.20\].

The extra amount to be paid $ = $ Rs. \[43.20\].

5. Vasudevan invested Rs $60,000$ at an interest rate of $12\% $ per annum compounded half yearly. What amount would he get

(i) after $6$ months?

(ii) after $1$ year?

Ans: (i)Given :

Principal (P) $ = $ Rs. $60,000$

Rate $ = $$12\% $per annum or $6\% $ per half year.

Number n $ = $ $6$ months or $1$ half year.

Formula :

A $ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$

A $ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$

A $ = 60,000{\left( {1 + \dfrac{6}{{100}}} \right)^1}$

A $ = 60,000{\left( {\dfrac{{100 + 6}}{{100}}} \right)^{}}$

A $ = 60,000{\left( {\dfrac{{106}}{{100}}} \right)^{}}$

A $ = 60,000{\left( {1.06} \right)^{}}$

A = $63,600$

(ii) There are 2 half years in 1 year.

Number n $ = 2$ 

A $ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$

A $ = 60,000{\left( {1 + \dfrac{6}{{100}}} \right)^2}$

A $ = 60,000{\left( {\dfrac{{100 + 6}}{{100}}} \right)^2}$

A $ = 60,000{\left( {\dfrac{{106}}{{100}}} \right)^2}$

A $ = 60,000{\left( {1.06} \right)^2}$

A $ = 60,000 \times 1.1236$

A = $67416$.

6. Arif took a loan of Rs $80,000$ from a bank. If the rate of interest is $10\% $ per annum, find the difference in amounts he would be paying after $1\dfrac{1}{2}$ years if the interest is

(i) Compounded annually.

(ii) Compounded half yearly.

Ans:

(i) Principal (P) $ = $ Rs. $80,000$

Rate $ = $ $10\% $per annum.

Number n $ = $$1\dfrac{1}{2}$years.

Formula:

A $ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$

S.I $ = \dfrac{{P \times R \times T}}{{100}}$

The amount for 1 year and 6 months can be calculated by first calculating the amount for 1 year using the compound interest formula, and then calculating the simple interest for 6 months on the amount obtained at the end of 1 year.Firstly, the amount or 1 year has to be calculated.

A $ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$

A $ = 80,000{\left( {1 + \dfrac{{10}}{{100}}} \right)^1}$

A $ = 80,000\left( {\dfrac{{100 + 10}}{{100}}} \right)$

A $ = 80,000\left( {\dfrac{{110}}{{100}}} \right)$

A $ = 80,000\left( {1.1} \right)$

A = $88,000$

By taking Rs. \[88,000\] as principal.

The S.I for next \[\dfrac{1}{2}\]years will be calculated.

S.I $ = \dfrac{{P \times R \times T}}{{100}}$

S.I \[ = \] Rs \[\left( {\dfrac{{88000 \times 10 \times \dfrac{1}{2}}}{{100}}} \right)\]

S.I \[ = \] Rs \[8800 \times 1 \times 0.5\]

S.I \[ = \] Rs \[4,400\].

Interest for first years \[ = \] Rs. \[\left( {88000 - 80000} \right)\]

\[ = \] Rs. \[8000\].

And interest for next \[\dfrac{1}{2}\] years\[ = \] Rs. \[4,400\].

Total Compound interest 

\[ \Rightarrow \] Rs. \[\left( {8000 + 4400} \right)\]

\[ \Rightarrow \] Rs. \[12400\].

Amount \[ = \] P \[ + \] C.I

\[ \Rightarrow \] Rs. \[\left( {80000 + 12400} \right)\]

\[ \Rightarrow \] Rs. \[92,400\].

(ii) Given :

The interest is compounded half yearly.

Rate $ = 10\% $ per annum or $5\% $ per half year.

Number n  $ = 1\dfrac{1}{2}$  years

Formula:

A $ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$

A $ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$

A $ = 80,000{\left( {1 + \dfrac{5}{{100}}} \right)^3}$

A $ = 80,000{\left( {\dfrac{{100 + 5}}{{100}}} \right)^3}$

A $ = 80,000{\left( {\dfrac{{105}}{{100}}} \right)^3}$

A $ = 80,000{\left( {1.05} \right)^3}$

A = $80,000 \times 1.157625$

A =$92,610$.

Difference in amount 

= Rs. $\left( {92,610 - 92,400} \right)$

= Rs. $210$.

7. Maria invested Rs \[8,000\] in a business. She would be paid interest at $5\% $ per annum compounded annually. Find.

(i) The amount credited against her name at the end of the second year

The interest for the $3$rd year.

Ans:

(i) Priniciple (p)$ = $Rs$8000$

Rate(R)             $ = $$5\% $ per annum

Number(n)      $ = $$2$years

Formula:

A $ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$

A $ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$

A $ = 8,000{\left( {1 + \dfrac{5}{{100}}} \right)^2}$

A $ = 8,000{\left( {\dfrac{{100 + 5}}{{100}}} \right)^2}$

A $ = 8,000{\left( {\dfrac{{105}}{{100}}} \right)^2}$

A $ = 8,000{\left( {1.05} \right)^2}$

A = $8,000 \times 1.1025$

A = $8820$

(ii) The interest for the next one year, i.e. the third year, has to be calculated. By taking Rs $8820$ as principal, the S.I. for the next year will be calculated

S.I$ = $Rs$\left( {\dfrac{{8820 \times 5 \times 1}}{{100}}} \right)$

S.I$ = $Rs $\left( {\dfrac{{44100}}{{100}}} \right)$

S.I$ = $Rs $441$

8. Find the amount and the compound interest on Rs $10,000$ for years at $10\% $ per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?

Ans: Principle (p)$ = $Rs $10,000$

Rate(R)             $ = $ $10\% $ per annum or $5\% $ per half year.

Number(n)      $ = $ $1\dfrac{1}{2}$years.

There will be $3$ half years in$1\dfrac{1}{2}$years.

Formula:

A $ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$

A $ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$

A $ = 10,000{\left( {1 + \dfrac{5}{{100}}} \right)^3}$

A $ = 10,000{\left( {\dfrac{{100 + 5}}{{100}}} \right)^3}$

A $ = 10,000{\left( {\dfrac{{105}}{{100}}} \right)^3}$

A $ = 10,000{\left( {1.05} \right)^3}$

A = $10,000 \times 1.157625$

A = $11576.25$

Compound interest \[ = \] A \[ - \] P

\[ \Rightarrow \] Rs. \[\left( {131576.25 - 10000} \right)\]

\[ \Rightarrow \] Rs. \[1,576.25\].

The amount for 1 year and $6$ months can be calculated by first calculating the amount for  1 year using the compound interest formula, and then calculating the simple interest for $6$ months on the amount obtained at the end of 1  year.The amount for the first year has to be calculated first.

A $ = 10,000{\left( {1 + \dfrac{{10}}{{100}}} \right)^1}$

A $ = 10,000\left( {\dfrac{{100 + 10}}{{100}}} \right)$

A $ = 10,000\left( {\dfrac{{110}}{{100}}} \right)$

A $ = 10,000\left( {1.1} \right)$

A = $11,000$

By taking Rs. \[11,000\] as principal.

The S.I for next \[\dfrac{1}{2}\]years will be calculated.

S.I $ = \dfrac{{P \times R \times T}}{{100}}$

S.I \[ = \] Rs \[\left( {\dfrac{{11000 \times 10 \times \dfrac{1}{2}}}{{100}}} \right)\]

S.I \[ = \] Rs \[1100 \times 1 \times 0.5\]

S.I \[ = \] Rs \[550\].

Interest for first years \[ = \] Rs. \[\left( {11000 - 10000} \right)\]

\[ = \] Rs. \[1000\].

Total Compound interest 

\[ \Rightarrow \] Rs. \[\left( {1000 + 550} \right)\]

\[ \Rightarrow \] Rs. \[1550\].

Therefore, the interest would be more when compounded half yearly than the interest when compounded annually.

9. Find the amount which Ram will get on Rs $4096$, he gave it for $18$ months at $12\dfrac{1}{2}\% $ per annum, interest being compounded half yearly.

Ans:  Principle (p)$ = $Rs $10,000$

Rate(R)             $ = $ $12\dfrac{1}{2}\% $ per annum or $\dfrac{{25}}{4}\% $ per half year.

Number(n)      $ = $ $18$ months.

There will be $3$ half years in $18$ months.

Formula:

A $ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$

A $ = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$

A $ = 4096{\left( {1 + \dfrac{{25}}{{400}}} \right)^3}$

A $ = 4096{\left( {1 + \dfrac{1}{{16}}} \right)^3}$

A $ = 4096{\left( {\dfrac{{16 + 1}}{{16}}} \right)^3}$

A $ = 4096{\left( {\dfrac{{17}}{{16}}} \right)^3}$

A $ = 4096{\left( {1.0625} \right)^3}$

 A = $4096 \times 1.1994628$

 A = $4913$

Amount = $4913$.

10. The population of a place increased to $54000$ in $2003$ at a rate of $5\% $ per annum

(i) find the population in $2001$

(ii) what would be its population in $2005$?

Ans:  Population in the year $2003$$ = 54,000$

(i) $54000$ $ = $population in 2001 $ \times $ ${\left( {1 + \dfrac{5}{{100}}} \right)^2}$

$54000$$ = $population in 2001 $ \times $${\left( {\dfrac{{100 + 5}}{{100}}} \right)^2}$

$54000$$ = $population in 2001 $ \times $${\left( {\dfrac{{105}}{{100}}} \right)^2}$

$54000$$ = $population in 2001 $ \times $${\left( {1.05} \right)^2}$

$54000$$ = $population in 2001 $ \times $$1.1025$

$\dfrac{{54000}}{{1.1025}}$$ = $population in 2001 

population in 2001 $ = 48979.591$

(ii) population in 2005

population in 2001 $ = $$54000{\left( {1 + \dfrac{5}{{100}}} \right)^2}$

population in 2001 $ = 54000$${\left( {\dfrac{{100 + 5}}{{100}}} \right)^2}$

population in 2001  $ = 54000{\left( {\dfrac{{105}}{{100}}} \right)^2}$

population in 2001 $ = 5400{\left( {1.05} \right)^2}$

population in 2001 $ = 54000 \times 1.1025$

population in 2001 $ = 59.535$

11. In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5%  per hour. Find the bacteria at the end of 2 hours if the count was initially $5,06,000$.

Ans:  Initial count of bacteria $ = 5,06,000$

Bacteria at the end of $2$ hours

$ = 506000{\left( {1 + \dfrac{{2.5}}{{100}}} \right)^2}$

$ = 506000{\left( {1 + 0.025} \right)^2}$

$ = 506000{\left( {1.025} \right)^2}$

$ = 506000\left( {1.050625} \right)$

$ = 531616.25$(approximately)

The count of bacteria at the end of 2 hours$ = 531616.25$.

12. A scooter was bought at Rs $42,000$. Its value depreciated at the rate of $8\% $ per annum. Find its value after one year.

Ans:  Principal (P) $ = $ Rs $42,000$

Rate(R)             $ = $ $8\% $ per annum .

Number(n)      $ = $ $1$ year.

Formula:

S.I $ = \dfrac{{P \times R \times T}}{{100}}$

S.I \[ = \] Rs \[\left( {\dfrac{{42000 \times 8 \times 1}}{{100}}} \right)\]

S.I \[ = \] Rs \[8 \times 420\]

S.I \[ = \] Rs \[3360\].

Value after $1$ year \[ = \] Rs. \[\left( {42000 - 3360} \right)\]

\[ = \] Rs. \[38,640\].

Benefits of Referring To NCERT Solutions For Class 8 Maths Chapter 8- Comparing Quantities

The benefits of referring to the Class 8 NCERT Solutions for Comparing Quantities are as follows.

• The solutions to all the exercise questions in the chapter Comparing Quantities are given in the PDF. So students can refer to these solutions to understand how to present the answers in the exam in a step-wise method.

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• NCERT Solutions for Class 8 Maths Chapter 8  Comparing Quantities are available for free to download.

Important Questions for Practice

1. Express 10& and 35% as decimals.

2. Calculate the ratio of 7 m to 49 m.

3. 5. If 45% of 15 students like English, find out the number of students who dislike English.

4. A shopkeeper bought two watches for Rs. 8,000 each. After selling the watches, there was a loss of 4% on the 1st watch while a profit of 8% on the 2nd watch. Calculate the overall gain or loss per cent on the whole transaction made by the shopkeeper.

5. In Class 8, there are 120 girls. Calculate the total number of students if 12% of the total students are boys.

Conclusion

NCERT Solutions for Class 8 Maths Chapter 8 on Comparing Quantities by Vedantu are a helpful resource for students. This chapter likely covers important concepts related to comparing quantities, such as understanding ratios, percentages, and their applications in real-life situations. The solutions provided by Vedantu aim to simplify complex mathematical ideas, making them more accessible to students. One crucial section could be the practical application of comparing quantities, where students learn how these mathematical concepts are used in everyday scenarios. Overall, these solutions serve as a valuable tool for students to grasp and apply mathematical concepts effectively.