Question

Joe invested an amount of $\$5000$ divided in two different schemes A and B at the simple interest rate of 5% p.a. and 3% p.a. respectively. If the total amount of simple interest earned in 2 years be $\$380$, then

• A. The amount invested in Scheme A is $\$3000$
• B. The amount invested in Scheme B is $\$2000$
• C. The amount invested in Scheme A is $\$2000$
• D. The amount invested in Scheme B is $\$3000$

Answer 1

Answer: (C), (D)

The amount invested in Scheme B is $\$3000$

Total amount invested $=\$5000$

Let the sum invested in Scheme A be $\$x$ and that in Scheme B be $\$(5000-x).$

Simple interest earned on Scheme A $=\frac{x\times 5\times 2}{100}$
$=\frac{10x}{100}$

Simple interest earned on Scheme B $=\frac{(5000-x)\times 3\times 2}{100}$
$=\frac{6(5000-x)}{100}$

Total simple interest earned $=\text{Simple interest earned on Scheme A}+\text{Simple interest earned on Scheme B}$
$=\frac{10x}{100}+\frac{6(5000-x)}{100}$

According to question total interest earned is $\$380$ So,

$\frac{10x}{100}+\frac{6(5000-x)}{100}=380$
$\Rightarrow \frac{10x+6(5000-x)}{100}=380$
$\Rightarrow \frac{10x+30000-6x}{100}=380$
$\Rightarrow \frac{(10x-6x)+30000}{100}=380$

$\Rightarrow \frac{4x+30000}{100}=380$

Cross multiplying, we get
$4x+30000=380\times 100$
$\Rightarrow 4x=38000-30000$
$\Rightarrow 4x=8000$

Dividing both the sides by $4,$ we get
$\frac{4x}{4}=\frac{8000}{4}$

Simplify and cancel common factors, we get
$x=\frac{4\times 2000}{4}$
$\Rightarrow x=2000$
Hence, amount invested in scheme A is $\$2000$
$\therefore 5000-x=5000-2000=3000$

So, the amount invested in Scheme B is $\$3000$