SSC Selection Post 2020 PYQ — Question 58

The correct answer is option 4 i.e. ₹1,600.

 

R = $\frac{8}{2} = 4\%$  per half-year

n = 2

$CI = P \left(1 + \frac{r}{100}\right)^n - P = P \left[ \left(1 + \frac{4}{100} \right)^2 - 1 \right]$

$CI = P \left[ (1.04)^2 - 1 \right] = P(1.0816 - 1) = P \cdot 0.0816$

$CI = 652.80 = P \cdot 0.0816 \Rightarrow P = \frac{652.80}{0.0816} = ₹8000$

$SI = \frac{P \cdot R \cdot T}{100} = \frac{8000 \cdot 8 \cdot 2.5}{100}$

$SI = \frac{160000}{100} = ₹1600$