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

Answer: (b)

Suppose principle is P

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

1414.40 = P $[(1 + {8/{100}})^2 - 1]$

1414.40 = P [1.1664 – 1]

P = ${1414.40}/{0.1664}$ = Rs. 8500

So A = P + SI = 8500 + 1414.40 = Rs. 9914.40

Answer: (e)

Amount = Principal $(1+\text"Rate"/100)^\text"Time"$

=$5000(1+8/100)^2=5000(1+2/25)^2$

=$5000×{27/25}×{27/25}$=5832Rs.

∴ CI=Rs.(5832-5000)=832Rs.

Answer: (a)

In 2 years, Rs. 1 will become $(1+{15/100})^2$ times of itself

= $(115/100)^2$ times of itself = $13225/10000$ times of itself

∴ Increase = $13225/10000-1=3225/10000= 32.25%$

Answer: (a)

Required Amount = 24000 $(1+{14×8}/100)$

= 24000 × $212/100$ = Rs.50880

Answer: (b)

Compound Interest

= $3080[(1+7/100)^3-1]$

=$3080[(107/100)^3-1]$

=$3080×[{1225043-1000000}/1000000]$

$3080×225043/1000000$

= Rs.693 (approximate)