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

Answer: (b)

Let the principal be = Rs.100

∴ Simple interest

= ${100×8×6}/100$ = Rs.48

∴ Amount (100 + 48) = Rs.148

∴ When the amount is = Rs.148, the principal = Rs.100

∴ When amount = Rs.28046, the principal

= $100/48$ × 28046 = Rs.18950

∴ Simple interest = (Rs.28046 – 18950) = Rs.9096

Answer: (a)

Amt=P$(1+10/100)^2$

$14520=P(110/100)^2$ or

P=$14520×100/110×100/110=12,000$

Answer: (a)

A = P$(1+ r/{100})^n = 4000(1+5/{100})^2$

= 4000 × ${21}/{20} = {21}/{20}$ = Rs.4410

Answer: (c)

Difference = ${\text"Sum"×r^2(300+r)}/(100)^3$

$={8000×2.5×2.5(300+2.5)}/{100×100×100}$

$={8×25×25×3025}/{100×100×100}=121/8$ = Rs. 15.125

Answer: (a)

C. I $[(1 + r/{100})^T - 1]$

4676.25=14500$[(1 + {r/{100}})^2-1]$

⇒${4676.25}/{14500} = (1 + {r}/{100})^2$ - 1

⇒${4676.25}/{14500} + 1 = (1 + {r}/{100})^2$

⇒${4676.25 × 14500}/{14500} = (1 + {r}/{100})^2$

⇒$√{{19176.25}/{14500}} = 1+r/{100}$⇒$√{1.3225}$ = 1+$r/{100}$

⇒$√{{13225}/{10000}}$= 1+$r/{100}$

⇒${115}/{100} = 1 + {r/{100}}⇒{r/{100}} = {115}/{100}-1$

⇒$r/{100} = {115 - 100}/{100}⇒{r/{100}} = {115}/{100}$

⇒r = 15%