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# Solved Examples(Set 10) - Simple Interest

 46. A four years NSC certificate was purchased for $₹500$ with $₹1000$ being the maturity value. Find the rate of simple interest. A. $16\%$ B. $24\%$ C. $22\%$ D. $25\%$

Explanation:

P $=500$
T $=4$
SI $=1000-500=500$

$\text{R}=\dfrac{100×500}{500×4}=25$

 47. If simple interest on a certain sum of money for $8$ years at $4\%$ per annum is same as the simple interest on $₹560$ for $8$ years at the rate of $12\%$ per annum then the sum of money is A. $₹1680$ B. $₹1820$ C. $₹1040$ D. $₹1120$

Explanation:

## Solution 1

Let the sum of money be $x$

$\dfrac{x×4×8}{100 }=\dfrac{560×12×8}{100 }\\\Rightarrow x×4×8=560×12×8\\\Rightarrow x×4=560×12\\\Rightarrow x=560×3=1680$

## Solution 2

simple interest on $₹560$ for $8$ years at $12\%$
$=\dfrac{560×12×8}{100}=537.6$

That is, simple interest on the required sum for $8$ years at $4\%$ is $537.6$

Required sum $=\dfrac{100×537.6}{4×8}=1680$

## Solution 3

Let the sum of money be $x$

SI $\propto$ PR (because T is constant here)

Therefore, $4x=560×12$
$\Rightarrow x=1680$

 48. If a sum of money trebles itself in $40$ years, what is the rate of interest? A. $4\%$ B. $5\%$ C. None of these D. $6\%$

Explanation:

## Solution 1

Let the sum of money be $x$
$x$ becomes $3x$ in $40$ years.

Simple interest $=3x-x=2x$

R $=\dfrac{100×2x}{x×40}=5$

## Solution 2

Refer formula

R $=\dfrac{100(3-1)}{40}=5$ Abhinas B
2015-01-26 02:03:43
A person invests money in 3 different schemes for 6 years, 10 years and 12 years at 10%,12%,and 15% simple interest. At the completion of each scheme he gets the same interest. what is the ratio of his investment ? 0 0 reply Raj
2015-02-09 19:31:19
Let x, y and z be the amounts he invests in the three schemes respectively.
Let simple interest in each case be I

I = x*6*10/100  => x = 100I/(6*10)
I = y*10*12/100 => y = 100I/(10*12)
I = z*12*15/100 => y = 100I/(12*15)

Required ratio = x : y : z
= 100I/(6*10) : 100I/(10*12) : 100I/(12*15)
= 1/60 : 1/120 : 1/180
= 6 : 3 : 2 0 0