TA Exam Maths Topic Compound Interest: Formulas and Sample Problems

In this article we will share TA Exam Maths Topic Compound Interest: Formulas and Sample Problems. s Territorial Army 2 2017 Exam Mock Test Series TA Exam Maths Topic Compound...

Ashutosh
July 21, 2017
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In this article we will share TA Exam Maths Topic Compound Interest: Formulas and Sample Problems.

Territorial Army 2 2017 Exam Mock Test Series

TA Exam Maths Topic Compound Interest: Formulas and Sample Problems

Let Principal = P, Rate = R% per annum, Time = n years.

Q. A sum of money amounts to Rs.6690 after 3 years and to Rs.10,035 after 6 years on compound interest.find the sum.

A) 4360

B) 4460

C) 4560

D) 4660

Let the sum be Rs.P.then P(1+R/100)^3=6690…(i) and P(1+R/100)^6=10035…(ii) On dividing,we get (1+R/100)^3=10025/6690=3/2. Substituting this value in (i),we get: P*(3/2)=6690 or P=(6690*2/3)=4460
Hence,the sum is rs.4460.

Q. Find compound interest on Rs. 8000 at 15% per annum for 2 years 4 months, compounded annually
A) 2109
B) 3109
C) 4109
D) 6109

Time = 2 years 4 months = 2(4/12) years = 2(1/3) years. Amount = Rs’. [8000 X (1+(15/100))^2 X (1+ (1/3)*15)/100)] =Rs. [8000 * (23/20) * (23/20) * (21/20)] = Rs. 11109. .
:. C.I. = Rs. (11109 – 8000) = Rs. 3109.

The population of a town was 3600 three years back. It is 4800 right now. What will be the population three years down the line, if the rate of growth of population has been constant over the years and has been compounding annually?

A) Rs.600

B) Rs,6400

C) Rs.6500

D) Rs.6600

The population grew from 3600 to 4800 in 3 years. That is a growth of 1200 on 3600 during three year span.

Therefore, the rate of growth for three years has been constant.

The rate of growth during the next three years will also be the same.

Therefore, the population will grow from 4800 by = 1600

Hence, the population three years from now will be 4800 + 1600 = 6400

In what time will Rs. 1000 become Rs. 1331 at 10% per annum compounded annually?

A) 1 years

B) 2years

C) 3years

D) 4years

Principal = Rs. 1000; Amount = Rs. 1331; Rate = 10% p.a. Let the time be n years. Then,

[ 1000 (1+ (10/100))^n ] = 1331 or (11/10)^n = (1331/1000) = (11/10)^3
n = 3 years

Q. What is the difference between the compound interests on Rs. 5000 for 1 1/2 years at 4% per annum compounded yearly and half-yearly?

A) 2.04

B) 3.04

C) 4.04

D) 5.04

compounded yearly=rs.[5000*(1+4/100)(1+1/2*4/100)] = Rs. 5304. C.I. when interest is compounded half-yearly=rs.5000(1+2/100)^3 = Rs. 5306.04 Difference = Rs. (5306.04 – 5304) = Rs. 2.04