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

Answer: (d)

Let P = 100

$A_1/A_2={100+T_1R_1}/{100+T_2R_2}$

=${100+3×12}/{100+10×24}$

=$136/340=2/5$

∴ $A_1 : A_2$ = 2 : 5

Answer: (c)

R = ${S.I × 100}/{P × T} = {10230 × 100}/{27500 × 3}$ = 12.4%p.q

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

= $27500 [(1 + {{12.4}/{100}})^3 -1] = 27500 [({112.4}/{100})^3 -1]$

=27500$[{112.4 × 112.4 × 11.24 - 100 × 100 × 100}/{100 × 100 × 100}]$

=27500$[{1420034.624 - 1000000}/{1000000}]$

= 27500$[{420034.624}/{1000000}]$

= 2700 × 0.42 = Rs. 11550

Answer: (c)

SI = Rs.(12710 – 10250)= Rs.2460

time = ${\text"\text"S.I"." × 100}/\text"Principal×Rate"$ =${2460×100}/{10250×4}$ = 6 years

Answer: (e)

Simple interest

$\text"principle×time×rate"/100$=${16500×4×16}/100$=Rs.10560

Answer: (b)

Amount = $25000×{(1+20/100)^2×(1+10/100)^1$

=$25000×(6/5)^2×11/10=39600$

∴ Compound interest = 39600 – 25000 = 14600.