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# Solved Examples(Set 5) - Simple Interest

 21. In how many years, $₹150$ will produce the same simple interest at $6\%$ as $₹800$ produce in $2$ years at $4\dfrac{1}{2}\%~?$ A. $9$ B. $6$ C. $8$ D. $4$

Explanation:

## Solution 1

Assume, simple interest on $₹150$ at $6\%$ for $n$ years = simple interest on $₹800$ at $4\dfrac{1}{2}\%$ for $2$ years

$\dfrac{150×6×n}{100}=\dfrac{800×\dfrac{9}{2}×2}{100}\\\Rightarrow 150×6×n=800×\dfrac{9}{2}×2\\\Rightarrow n=8$

## Solution 2

SI on $₹800$ at $\dfrac{9}{2}\%$ for $2$ years
$\equiv$ SI on $₹800$ at $6\%$ for $\left(2×\dfrac{9}{12}=\dfrac{3}{2}\right)$ years
$\equiv$ SI on $₹150$ at $6\%$ for $\left(\dfrac{3}{2}×\dfrac{800}{150}=8\right)$ years

## Solution 3

Let required number of years be $n$

SI $\propto$ PRT

Therefore,
$150×6×n=800×\dfrac{9}{2}×2\\\Rightarrow n=8$

 22. A sum of $₹2500$ amounts to $₹3875$ in $4$ years at the rate of simple interest. What is the rate of interest? A. $13.75\%$ B. $12\%$ C. $6\%$ D. $12.25\%$

Explanation:

Simple interest $=3875-2500=1375$

$\text{R}=\dfrac{100×1375}{2500×4}=13.75$

 23. What is the interest due after $40$ days for $₹3200$ at $10\%$ A. $35.52$ B. $36.21$ C. $35.07$ D. $34$

Explanation:

Required simple interest
$=\dfrac{3200×10×\dfrac{40}{365} }{100}=35.07$
 24. What is the rate of interest at which $₹150$ becomes $₹220$ in $10$ years (assume simple interest)? A. $\dfrac{11}{3}\%$ B. $\dfrac{14}{3}\%$ C. $12\%$ D. $14\%$

Explanation:

Simple interest $=220-150=70$

$\text{R}=\dfrac{100×70}{150×10}=\dfrac{14}{3}$

 25. A person invested $₹2600$ by dividing into $3$ parts. The rate of interest per annum for the first part, second part and third part are $4\%,6\%$ and $8\%$ respectively. At the end of the year, he got the same simple interest in all three parts. What is the money invested at $4\%~?$ A. $₹1600$ B. $₹1200$ C. $₹800$ D. $₹2200$

Explanation:

Let the parts be $x,y$ and $z$

simple interest on $x$ at $4\%$ for $1$ year
= simple interest on $y$ at $6\%$ for $1$ year
= simple interest on z at $8\%$ for $1$ year

$\dfrac{x×4×1}{100}=\dfrac{y×6×1}{100}=\dfrac{z×8×1}{100}\\\Rightarrow 2x=3y=4z$

$x:y=3:2=6:4\\x:z=2:1=6:3$

$x:y:z=6:4:3$

Since the total amount is $2600,$
$x=2600×\dfrac{6}{6+4+3}=1200$