SSC Selection Post 2020 PYQ — Question 62

The correct answer is option 1 i.e. ₹4,680.

Time (T₁) =  $2\frac{2}{3} = \frac{8}{3}$ years
Rate (R₁) = 15% p.a.
Simple Interest (SI₁) = $\frac{P \times T_1 \times R_1}{100} = \frac{P \times \frac{8}{3} \times 15}{100}$
Time (T₂) =  $4\frac{1}{4} = \frac{17}{4}$years
Rate (R₂) = 12% p.a.
Simple Interest (SI₂) = $\frac{P \times T_2 \times R_2}{100} = \frac{P \times \frac{17}{4} \times 12}{100}$
$SI_2 - SI_1 = 514.80$
$SI_1 = \frac{P \times 8 \times 15}{3 \times 100} = \frac{P \times 8 \times 5}{100} = \frac{40P}{100} = \frac{2P}{5}$
$SI_2 = \frac{P \times 17 \times 12}{4 \times 100} = \frac{P \times 17 \times 3}{100} = \frac{51P}{100}$
$\frac{51P}{100} - \frac{2P}{5} = 514.80$
$\frac{51P}{100} - \frac{40P}{100} = 514.80$
$\frac{11P}{100} = 514.80$
$P = \frac{51480}{11}$ =  ₹4680