SSC CHSL 2022 PYQ — Question 74

The correct answer is Option 4 i.e Rs. 5,250.

$P = {\frac{X}{(1 + {R \over 100})^2}} + {\frac{X}{(1 + {R \over 100})}}$

where, P = Principal and X = Instalment

According to the values given in the question,

The sum borrowed(P), 

$P = {\frac{3025}{(1 + {10 \over 100})^2}} + {\frac{3025}{(1 + {10\over 100})}}$

⇒ $P = {\frac{3025 \times 100}{11 \times 11}} + {\frac{3025 \times 10}{11}}$ 

⇒ P = 2500 + 2750

⇒ P = 5250

Hence, the sum borrowed is Rs. 5,250.