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

Question :1

Rahul has borrowed Rs. 20,000 from two money tenders. On one he had to pay 8% p.a. S.I. and on the other amount he ple & Compound Interest C-81 had to pay 12% p.a. SI. After 3 years, he paid total interest of Rs.5760, find the amount borrowed at 12% p.a.

Answer: (c)

Let amount borrowed at 8% be x,

Let amount borrowed at 12% will be (20,000 – x)

${x×8×3}/100+{(20,000-x)×12×3}/100$=5760

On solving x = 8000 which is the amt. borrowed at 8%.

So, amt. borrowed at 12% = 20,000 – 8000

= 12,000

Question :2

Sonika invested an amount of Rs. 5800 for 2 years. At what rate of compound interest will she get an amount of Rs. 594.5 at the end of two years ?

Answer: (c)

594.5 = 5800 $[(1 + r/{100})^2 - 1] = {594.5}/{5800} = (1 + r/{100})^2 - 1$

0.1025 + 1 = $(1 + r/100)^2 $

1.1025 = ${(100 + r)^2}/{10000}$

1.1025 × 10000 = $(100 + r)^2$ .

11025 = $(100 + r)^2$ .

105 = 100 + r

r = 5%

Question :3

Ms. Sandhya deposits an amount of Rs. 31,400 to obtain a simple interest at the rate of 12 per cent per annum for 8 years. What total amount will Ms. Sandhya get at the end of 8 years?

Answer: (b)

Simple Interest = $\text"P×R×T"/100$

=${31400×8×12}/100$=Rs.30144

∴ Required amount = Rs.(31400 + 30144)

= Rs.61544

Question :4

Population of a town increases at a certain rate per cent per annum. Present population of the town is 3600 and in 5 years it becomes 4800. How much will it be in 10 years ?

Answer: (a)

Let rate of increase in population = r% p.a.

Then 4800=3600$(1+r/100)^5$

∴ $(1+r/100)^5=4800/3600=4/3$

Population in the next 5 years will become

4800 × $4/3$ = 6400.

Question :5

At a certain rate of simple interest, a certain sum doubles itself in 10 years. It will treble itself in years ________ .

Answer: (d)

It doubles in 10 yrs.

Then trebles in 20 yrs.

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