MCQ
The difference between the interest obtained for $Rs. 1000$ at $12\%$ per annum for $3$ years and that for $Rs. 1500$ at $8\%$ per annum for $1\frac{1}{2}$ years is:
- A$Rs. 360$
- B$Rs. 300$
- ✓$Rs. 180$
- D$Rs. 200$
✓
Answer
Correct option: C.
$Rs. 180$
It is given that,
$\operatorname{Sum}\left( P _1\right)= Rs. 1000$
Rate $\left(R_1\right)=12 \%$
Time $\left(T_1\right)=3$ years
$\text{I}_1=\frac{\text{P}_1\ \times\ \text{R}_1\ \times\ \text{T}_1}{100}$
$=\frac{1000\ \times\ 12\ \times\ 3}{100}$
$=\text{Rs. }360\ ...(\text{i})$
$\operatorname{Sum}\left(P_2\right)= Rs. 1500$
Rate $\left(R_2\right)=8 \%$
Time $\left(T_2\right)=1 \frac{1}{2}$ year $=\frac{3}{2}$ year
$I_2=\frac{P_2 \times R_2 \times T_2}{100}$
$=\frac{1500 \times 8 \times 3}{100 \times 2}$
$=\text { Rs. } 180 \ldots \text { (ii) }$
Subtracting $(ii)$ from $(i),$ we get
$I_2-I_1=\text { Rs. } 360-\text { Rs. } 180$
$=\text { Rs. } 180$
Hence, the correct option is $(c).$
$\operatorname{Sum}\left( P _1\right)= Rs. 1000$
Rate $\left(R_1\right)=12 \%$
Time $\left(T_1\right)=3$ years
$\text{I}_1=\frac{\text{P}_1\ \times\ \text{R}_1\ \times\ \text{T}_1}{100}$
$=\frac{1000\ \times\ 12\ \times\ 3}{100}$
$=\text{Rs. }360\ ...(\text{i})$
$\operatorname{Sum}\left(P_2\right)= Rs. 1500$
Rate $\left(R_2\right)=8 \%$
Time $\left(T_2\right)=1 \frac{1}{2}$ year $=\frac{3}{2}$ year
$I_2=\frac{P_2 \times R_2 \times T_2}{100}$
$=\frac{1500 \times 8 \times 3}{100 \times 2}$
$=\text { Rs. } 180 \ldots \text { (ii) }$
Subtracting $(ii)$ from $(i),$ we get
$I_2-I_1=\text { Rs. } 360-\text { Rs. } 180$
$=\text { Rs. } 180$
Hence, the correct option is $(c).$
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