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

Question : 21 [SSC CGL Prelim 2000]

The compound interest on Rs.10,000 in 2 years at 4% per annum, the interest being compounded half-yearly, is :

a) Rs.828. 82

b) Rs.636.80

c) Rs.912. 86

d) Rs.824.32

Answer: (d)

Using Rule 1,

A = 10,000$(1 + 2/100)^4$

=10,000$(51/50)^4$ =10824.3216

Interest = 10,824.3216 - 10,000

= Rs.824.32

Question : 22 [SSC CGL Tier-I 2013]

A man borrows Rs.21000 at 10% compound interest. How much he has to pay annually at the end of each year, to settle his loan in two years ?

a) Rs.12300

b) Rs.12000

c) Rs.12200

d) Rs.12100

Answer: (d)

If each instalment be x, then Present worth of first instalment

= $x/{1 + 10/100} = {10x}/11$

= Present worth of second instalment

= $x/(1 + 10/100)^2 = 100/121x$

$10/11x + 100/121x$ = 21000

${110x + 100x}/121 = 21000$

210x = 21000 × 121

$x = {21000 × 121}/210$ = Rs.12100

Using Rule 9,
If a sum 'P' is borrowed at r% annual compound interest which is to be paid in 'n' equal annual installments including interest, then
(i) For n = 2, Each annual installment
= $p/{(100/{100 + r}) + (100/{100 + r})^2$
(ii) For n = 3, Each annual installment
= $p/{(100/{100 + r}) + (100/{100 + r})^2 + (100/{100 + r})^3$

Here, n = 2, p = Rs.21000, r = 10%

Each annual instalment

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

= $21000/{100/110 + (100/110)^2}$

= $21000/{100/110 + 10000/12100}$

= $21000/{10/11 + 100/121}$

= $21000/{110 + 100} × 121$

= $21000/210 × 121$ = 12100

Question : 23 [SSC CGL Prelim 2008]

A certain sum, invested at 4% per annum compound interest, compounded half yearly, amounts to Rs.7,803 at the end of one year. The sum is

a) Rs.7,700

b) Rs.7,000

c) Rs.7,500

d) Rs.7,200

Answer: (c)

Using Rule 1,

Let the sum be P.

As, the interest is compounded half-yearly,

R = 2%, T = 2 half years

A = P$(1 + R/100)^T$

7803 = P$(1 + 2/100)2$

7803 = $(1 + 1/50)^2$

7803 = P$× 51/50 × 51/50$

P = ${7803 × 50 × 50}/{51 × 51}$ = Rs.7500

Question : 24 [SSC CGL Prelim 2005]

In how many years will Rs.2,000 amounts to Rs.2,420 at 10% per annum compound interest?

a) 1$1/2$ years

b) 3 years

c) 2 years

d) 2$1/2$ years

Answer: (c)

Using Rule 1,

According to question,

2420 = 2000$(1 + 10/100)^t$

$2420/2000 = (11/10)^t$

or $(11/10)^t = 121/100$

or, $(11/10)^t = (11/10)^2$

t = 2 years

Question : 25 [SSC CGL Prelim 2005]

At what rate per cent per annum will Rs.2304 amount to Rs.2500 in 2 years at compound interest ?

a) 4$1/3$%

b) 4$1/2$%

c) 4$1/6$%

d) 4$1/5$%

Answer: (c)

Using Rule 1,

Let the rate per cent per annum be r. Then,

2500 = 2304$(1 + r/100)^2$

$(1 + r/100)^2 = 2500/2304 = (50/48)^2$

$1 + r/100 = 50/48 = 25/24$

$r/100 = 25/24 - 1 = 1/24$

r = $100/24 = 25/6 = 4{1}/6$%

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

Classification MCQ Test: IBPS RRB PO 2024 Scale-1 Prelims

Free 150+ Classification Verbal Reasoning Multiple choice questions and answers practice test series PDF, Free Quiz For IBPS RRB PO Officer Scale 1 Level 2024

28-Mar-2024 by Careericons

Continue Reading »

100+ Analogy Verbal Questions Answers PDF For IBPS RRB PO

Most Important Analogy Verbal Reasoning Multiple choice questions and answers practice test series PDF & Free Quiz For IBPS RRB PO Officer Scale 1 Level 2024

27-Mar-2024 by Careericons

Continue Reading »

IBPS RRB Clerk 2024 Seating Arrangement Questions Answers

Seating Arrangement: Verbal Reasoning Practice multiple choice questions and answers test & Quiz PDF for IBPS RRB (Assistant Multipurpose) Clerk Prelims 2024

26-Mar-2024 by Careericons

Continue Reading »

Top Aptitude Time & Distance Questions PDF: IBPS RRB Clerk

Important Aptitude Time and Distance multiple choice questions answers practice test PDF for IBPS RRB Clerk (Office Assistant Multipurpose) Prelims 2024 Exam

25-Mar-2024 by Careericons

Continue Reading »