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S.chand mathematics solution class 8 chapter 9 simple interest and compound interest

Exercise 9 (A)


QUESTION 1

What will be the simple interest for 1 year 4 months on a sum of 25800 at the rate of 14% p.a. ?

Sol :

We know \text{I}=\dfrac{\text{P}\times \text{R} \times \text{T}}{100}

\text{T}=1+\dfrac{4}{12} =\dfrac{12+4}{12}=\dfrac{16}{12} =\dfrac{4}{3}

\text{I}=\dfrac{25800\times 14 \times \dfrac{4}{3}}{100}

=\dfrac{258\times 14 \times 4}{3}

= 4816

 


QUESTION 2

What will 5000 amount to at 8% p.a. for the period from april 6 , 2017 to june 18 , 2017 ?

Sol :

Lets find time period (T)

\begin{matrix}dd&mm&yy\\18&06&2017\\06&04&2017\\\hline 12&02&0000\\\hline\end{matrix} 12 days and 2 months ( may month have 31 days)

12 days and 2 month = 73 days

\text{I}=\dfrac{\text{P}\times \text{R} \times \text{T}}{100}

\text{I}=\dfrac{5000\times 8 \times \dfrac{73}{365}}{100}

=\dfrac{8000}{100}

= 80

Amount = Principal + Interest

= 5000 + 80

= 5080

 


QUESTION 3

A sum of 3200 becomes 3456 in two years at a certain rate of simple interest . What is the rate of interest per annum ?

Sol :

T = 2 year , R = ? , P = 3200

After adding simple interest to principal = 3456

S.I = 3456 – 3200 = 256

256=\dfrac{3200\times \text{R}\times 2}{100}

\dfrac{25600}{3200\times 2}=\text{R}

R = 4

R = 4 % per annum

 


QUESTION 4
Find the principal on which a simple interest of 56 will be obtained after 9 months at the rate of 3\dfrac{2}{3} per annum ?

Sol :

\text{I}=\dfrac{\text{P}\times \text{R} \times \text{T}}{100}

56=\dfrac{\text{P}\times 3\dfrac{2}{3}\times \dfrac{9}{12}}{100}

5600=\text{P}\times \dfrac{11}{3}\times \dfrac{3}{4}

5600=\text{P}\times \dfrac{11}{4}

\text{P}=5600\times \dfrac{4}{11}

P = 2036.36

 


QUESTION 5

In what time will 1860 amount to 2641.20 at simple interest of 12% per annum ?

Sol :

T = ? , P = 1860 , R = 12 % per annum

S.I = 2641.20 – 1860 = 781.2

\text{I}=\dfrac{\text{P}\times \text{R} \times \text{T}}{100}

781.2=\dfrac{1860\times 12\times \text{T}}{100}

\text{T}=\dfrac{781.2\times 100}{1860\times 12}

=\dfrac{78120}{22320}=\dfrac{7}{2}

=3\dfrac{1}{2} years

 


QUESTION 6

Firoz invested a sum of money at an annual simple interest rate of 10% . At the end of 4 years , the amount received by Firoz was 7700 . What was the sum invested ?

Sol :

P = ? , R = 10% per annum , T = 4 years , I = ?

Let the principal be x

\text{I}=\dfrac{\text{P}\times \text{R} \times \text{T}}{100}

\text{I}=\dfrac{x \times 10 \times 4}{100}

\text{I}=\dfrac{2x}{5}

 

Amount = Principal + Interest

7700 = x + \dfrac{2x}{5}

7700=\dfrac{7x}{5}

x=\dfrac{7700\times 5}{7}

x = 5500

Which means Principal = 5500 , Firoz invested 5500 rupees

 


QUESTION 7
A person deposited 400 for 2 years , 550 for 4 years and 1200 for 6 years at the same rate of simple interest . If he received a total interest of 1020 . What is the rate of interest per annum ?

Sol :

Let the rate of interest be R per annum

According to question ,

\therefore \dfrac{400\times 2\times \text{R}}{100}+\dfrac{500\times 4\times \text{R}}{100} + \dfrac{1200\times 6\times \text{R}}{100} = 1020

8R + 22R + 72R = 1020

102R = 1020

\text{R}=\dfrac{1020}{102}

R = 10% per annum

 


QUESTION 8
The simple interest on a certain sum for 8 months at 4% is 129 less than the simple interest on the same sum for 15 months at 5% per annum . What is the sum ?

Sol :

Let the Principal be P

T1 = 8 \text{months} = \dfrac{8}{12}\text{year}

 

T2 = 15 months = 12 months + 1 month

= 1 year + 1 month

T2 =1+\dfrac{1}{12}

T2 =\dfrac{13}{12}

 

According to question

\dfrac{\text{P}\times 5 \times 15}{100\times 12}-\dfrac{\text{P}\times 4 \times 8}{100\times 12}=129

\dfrac{43P}{1200}=129

P = 3600

So , the sum is 3600

 


QUESTION 9
The simple interest on a sum of 5 years is one-fourth of the sum . What is the rate of interest per annum ?

Sol :

T = 5 years , R = ? , \text{I}=\dfrac{1}{4}P , P = P

\dfrac{1}{4}\text{P}=\dfrac{\text{P}\times \text{R}\times 5}{100}

\dfrac{1}{4}=\dfrac{\text{R}}{20}

R = 5 % per annum

 


QUESTION 10

Simple interest on a certain sum is \dfrac{16}{25} of the sum . Find the time (in years) . If the rate per cent and time are equal ?

Sol :

\text{I}=\dfrac{\text{P}\times \text{R} \times \text{T}}{100}

\dfrac{16}{25}P=\dfrac{P \times x \times x}{100}

\dfrac{16\times 100}{25}={x}^2

x2 = 64

x = 8

Which means R = 8 ,  T = 8 years

 


QUESTION 11

At a certain rate of simple interest , a certain sum of money becomes double of itself in 10 years . In what time will it becomes treble of itself ?

Sol :

Let Principal be P and Rate of interest be R

Amount = 2P (given)

Amount = Principal + Interest

2P = P + I

I = P

And Time period (T) = 10 years

So , \text{I}=\dfrac{\text{P}\times \text{R} \times \text{T}}{100}

P=\dfrac{P\times R \times 10}{100}

R = 10 % per annum

Now , the second case where Amount become triple

Amount = 3P (given)

Amount = Principal + Interest

3P = P + I

I = 2P

Putting this in formula we get

2P=\dfrac{P\times 10 \times T}{100}

2=\dfrac{T}{10}

T = 20 years

 


QUESTION 12

A part of 1500 was lent at 10% per annum and the rest at 7% per annum simple interest . The total interest earned in 3 years was 396 . What was the sum lent at 10% p.a. ?

Sol :

Here , we have two cases

Case 1 : Let the principal be P and R = 10 and T = 3

Case 2 : Let the principal be ( 1500 – P ) , R = 7 and T = 3

Then adding interest of both the cases , which is equal to total interest

\dfrac{P\times 10 \times 3}{100}+\dfrac{(1500-P)\times 7 \times 3}{100}=396

\dfrac{30P+(1500\times 21)-21P}{100}=396

30P + 31500 – 21P = 39600

9P = 8100

P = 900

 


 

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