myhelper: S.chand mathematics solution class 8 chapter 9 simple interest and compound interest Exercise 9 (A)

Exercise 9 (A)

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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

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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

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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

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QUESTION 4
Find the principal on which a simple interest of 55 will be obtained after 9 months at the rate of $3\dfrac{2}{3}$ per annum ?

Sol :

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

$55=\dfrac{\text{P}\times 3\dfrac{2}{3}\times \dfrac{9}{12}}{100}$

$5500=\text{P}\times \dfrac{11}{3}\times \dfrac{3}{4}$

$5500=\text{P}\times \dfrac{11}{4}$

$\text{P}=5500\times \dfrac{4}{11}$

P = 2000

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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

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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

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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 ,

$\therefore \dfrac{400\times 2\times \text{R}}{100}+\dfrac{550\times 4\times \text{R}}{100} + \dfrac{1200\times 6\times \text{R}}{100} = 1020$

8R + 22R + 72R = 1020

102R = 1020

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

R = 10% per annum

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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

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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

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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

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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

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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

S.chand mathematics solution class 8 chapter 9 simple interest and compound interest Exercise 9 (A)

Exercise 9 (A)

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