simple and compound interest Model Questions & Answers, Practice Test for ssc cgl tier 1 2023

Answer: (d)

Rate = ${{(7/3-1)}×100}/10$=13.33%

Answer: (b)

(1 + r) = $1+1/40=41/40$

∴ Amount = $1600 × 41/40 × 41/40 =1681$

∴ Compound interest = Rs. 1681 – Rs. 1600 = Rs. 81

Answer: (d)

Let principal = Rs. 100

Amount after two years = 100 × $(11/10)^2$ = Rs.121

∴ Compound interest for second year

= Rs. 121 – Rs. 110 = Rs. 11

But actual compound interest for second year

= Rs. 132 (i.e. 12 times of Rs. 11)

∴ Principal = 12 × Rs. 100 = Rs. 1200

Answer: (c)

The general formula for such question is :

A = P $(1 + {{r_1}/{100}})(1 + {{r_2}/{100}})(1 + {{r_3}/{100}})$

where A = Amount, P = Principal and $r_1 , r_2 , r_3$ are the rates of interest for different years.

2249.52 = P $(1 + {{3}/{100}}) (1 + {{4}/{100}})(1 + {{5}/{100}})$

or, 2249.52 = P(1.03) (1.04) (1.05)

∴ P = ${2249.52}/{1.03 × 1.04 × 1.05}$ = Rs.2000

Answer: (c)

C.I = $2000[(1+8/{100×4})^{4×9/12}-1]$

P=2000, R=8% p.a., t=9months=$9/12$year

C.I =$2000[(1+8/{100×4})^{4×9/12}-1](n=4)$

=$2000[(102/100)^3-1]$ =Rs. 122.

∴ the compound interest is Rs.122