simple and compound interest Model Questions & Answers, Practice Test for ibps clerk prelims 2023

Answer: (d)

R = ${12000 × 100}/{40000 × 3 }[R = {S.I × 100}/{P × T}]$ = 10%

CI = P$ [(1 + {{r}/{100}})^t -1] = 40000[(1 + {10}/{100})^3 -1]$

= 40000$[({{11}/{10}})^3 -1]$ = 40000$[{1331 - 1000}/{1000}]$

=40000$[{31}/{1000}]$ = 40 × 331 = Rs.13240

Answer: (d)

Principal = ${\text"S.I"×100}/\text"R×T"={57200×100}/{11×8}$

= 65000

Answer: (b)

SI=${P×Y×2}/100={2YP}/100$

SI=${P×2Y×(2+X)}/100={2Y(2+X)}/100$

$5×{2YP}/100=2YP{(2+X)}/100$

5 = 2 + x

5 – 2 = x = 3.

Answer: (b)

Suppose the rate of interest = r% and the sum = Rs. A

Now, A + ${A + r × 4}/100$ = 600;

or, A + ${Ar}/{25}$ = 600

or, A $[1 + {r/{25}}]$ = 600 ...(1)

And, A + ${A × r × 6}/{100}$ = 650;

or, A $[1 + {{3r}/{50}}]$ = 650...(2)

Dividing (1) by (2), we have

${1 + {r/25}}/{1 + {3r}/{50}} = {600}/{650};$ or, ${(25 + r) × 2}/{50 + 3r} = {12}/{13}$

or, ${(25 + r) × 2}/{50 × 3r} = {12}/{13}$

or, (50 + 2r) × 13 = (50 + 3r) × 12

or, 650 + 26r = 600 + 36r; or, 10r = 50

∴ r = 5%

Answer: (b)

Suppose Rs. x was lent at 6% per annum.

Thus, ${x × 6 × 5}/{100} + {(7000 - x) × 4 × 5}/{100} = 1600$

${3x}/{10} + {7000 - x}/{5}$ = 1600

or, ${3x + 14,000 - 2x}/{10}$ = 1600

∴ x = 16000 - 14000 = Rs. 2000