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

Answer: (d)

Suppose at the rate of 12% p.a. he borrowed Rs. x.

36480 – 30000 = ${x × 12 × 2}/{100} + {(30000 - x) × 10 × 2}/{100}$

6480 = ${24x}/{100} + {600000 – 20x}/{100}$

6480 × 100 = 24x + 600000 – 20x

648000 – 600000

4x = 648000 – 600000 = 48000

x = ${48000}/4$ = Rs.12000

Answer: (a)

Let x be lent at 5% and (1200 – x) at 4%

Then we have, ${x×5×2}/100+{(1200-x)×4×2}/100=106$

⇒ x = 500.

Answer: (d)

P = Rs. 12500. N = 2 years, Rate = 10% . When interest is payable yearly

A = $12500{(1+10/100)}^2$ = Rs. 15125

When interest is payable half yearly

A = $12500{(1+5/100)}^4$ = Rs. 15193.82

Difference = Rs. (15193.82 – 15125)

= Rs. 68.82

Answer: (b)

Nanoo's interest for an year at 10% compounded half yearly

= $400{(1+5/100)}^2-400$

=${400×21×21}/{20×20}-400$

= Rs.441 – 400 = Rs. 41

Meenu's interest at simple interest

= ${400 ×10×1}/100$ = Rs. 40

Thus, Nanoo paid 41 – 40 = Rs. 1 more

Answer: (c)

Simple interest

= ${4000×5×2}/{100}$ = Rs.400

Compound interest

= $4000(1+5/100)^2 – 4000$

=${4000×105×105}/{100×100}-4000$

= 4410 – 4000 = Rs.410

∴Difference = 410 – 400

= Rs.10