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

IMPORTANT quantitative aptitude EXERCISES

model 1 basic compound interest using formula Shortcuts »

Click to Read...

model 1 basic compound interest using formula Online Quiz

Click to Start..

Recently Added Subject & Categories For All Competitive Exams

New 100+ Compound Interest MCQ with Answers PDF for IBPS

Compound Interest verbal ability questions and answers solutions with PDF for IBPS RRB PO. Aptitude Objective MCQ Practice Exercises all competitive exams

02-Jul-2024 by Careericons

Continue Reading »

100+ Mixture and Alligation MCQ Questions PDF for IBPS

Most importantly Mixture and Alligation multiple choice questions and answers with PDF for IBPS RRB PO. Aptitude MCQ Practice Exercises all Bank Exams

02-Jul-2024 by Careericons

Continue Reading »

IBPS Profit and Loss Questions Solved Problems with PDF

Most important Profit and Loss multiple choice questions and answers with PDF for IBPS RRB PO. 100+ Aptitude MCQ Practice Exercises all competitive exams

28-Jun-2024 by Careericons

Continue Reading »

100+ Average Aptitude Questions Answers solutions MCQ PDF

New Average multiple choice questions and answers with PDF for IBPS RRB PO. 100+ Quantitative Aptitude MCQ Practice Exercises all competitive exams

28-Jun-2024 by Careericons

Continue Reading »