model 1 basic compound interest using formula Practice Questions Answers Test with Solutions & More Shortcuts

Question : 26 [SSC SAS 2010]

At what percent per annum will Rs.3000/- amounts to Rs.3993/- in 3 years if the interest is compounded annually?

a) 13%

b) 9%

c) 11%

d) 10%

Answer: (d)

Using Rule 1,
If A = Amount, P = Principal, r = Rate of Compound Interest (C.I.), n = no. of years then,
A=P$(1 + r/100)^n$, C.I. = A - P
C.I. = P$[(1 + r/100)^n - 1]$

P = Rs.3000, A = Rs.3993, n = 3 years

A = P$(1 + r/100)^n$

$(1 + r/100)^n = A/P$

$(1 + r/100)^3 = 3993/3000 = 1331/1000$

$(1 + r/100)^3 = (11/10)^3$

1 + $r/100 = 11/10$

$r/100 = 11/10$ - 1

$r/100 = 1/10 ⇒ r = 100/10$ = 10%

Question : 27 [SSC CGL Prelim 2008]

A certain sum of money yields Rs.1261 as compound interest for 3 years at 5% per annum. The sum is

a) Rs.8000

b) Rs.9000

c) Rs.7500

d) Rs.8400

Answer: (a)

Let the principal be Rs.x. Now,

C.I. = P$[(1 + R/100)^T - 1]$

1261 = $x[(1 + 5/100)^3 - 1]$

1261 = $x(9261/8000 - 1)$

1261 = $x({9261 - 8000}/8000)$

= ${1261x}/8000$

$x = {1261 × 8000}/1261$ = Rs.8000

Question : 28 [SSC CPO S.I.2004]

The principal, which will amount to Rs.270.40 in 2 years at the rate of 4% per annum compound interest, is

a) Rs.220

b) Rs.200

c) Rs.250

d) Rs.225

Answer: (c)

Using Rule 1,

Let the principal be Rs.P.

270.40 = P $(1 + 4/100)^2$

270.40 = P $(1 + 0.04)^2$

P = ${270.40}/{1.04 × 1.04}$ = Rs.250

Question : 29 [SSC CGL Prelim 2008]

In what time will Rs.10,000 amount to Rs.13310 at 20% per annum compounded half yearly?

a) 3 years

b) 1$1/2$ years

c) 2$1/2$ years

d) 2 years

Answer: (b)

Using Rule 1,
If A = Amount, P = Principal, r = Rate of Compound Interest (C.I.), n = no. of years then,
A=P$(1 + r/100)^n$, C.I. = A - P
C.I. = P$[(1 + r/100)^n - 1]$

Using Rule 2,
Compound interest is calculated on four basis:

RateTime(n)
Annuallyr%t years
Half–yearly
(Semi-annually
$r/2$%t × 2 years
Quarterly$r/4$%t × 4 years
Monthly$r/12$%t × 12 years

The rate of interest is compounded half yearly,

r = 10% per half year

Let time = $T/2$ years = half years

According to the question,

Amount = P$(1 + R/100)^t$

13310 = 10000$(1 + 10/100)^T$

$13310/10000 = (11/10)^T$

$(11/10)^T = 1331/1000 = (11/10)^3$

T = 3 half years =1$1/2$ years

Question : 30 [SSC MTS 2013]

A man saves Rs.2000 at the end of each year and invests the money at 5% compound interest. At the end of 3 years he will have :

a) Rs.2205

b) Rs.4305

c) Rs.4205

d) Rs.6305

Answer: (d)

Using Rule 1,

Amount = $2000(1 + 5/100)^2 + 2000(1 + 5/100)$

= 2000 × $(21/20)^2 + 2000(21/20)$

= 2000 × $21/20 × 41/20$ = Rs.4305

Required amount

= 4305 + 2000 = Rs.6305

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