Solved Examples(Set 10) - Simple Interest

Answer: Option D

Explanation:

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

$\text{R}=\dfrac{100×500}{500×4}=25$

Answer: Option A

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, $4x=560×12$
$\Rightarrow x=1680$

Answer: Option B

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 $=\dfrac{100(3-1)}{40}=5$