model 4 difference in ci & si Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 6 EXERCISES

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The following question based on compound interest topic of quantitative aptitude

Questions : On what sum does the difference between the compound interest and the simple interest for 3 years at 10% is 31 ?

(a) 1000

(b) 1500

(c) 1100

(d) 1200

The correct answers to the above question in:

Answer: (a)

Let the sum be x

r = 10%, n = 3 years

S.I. = ${x × r × n}/100$

S.I.= ${x × 10 × 3}/100 = 3/10x$

C.I.= $[(1 + r/100)^n - 1]x$

= $[(1 + 10/100)^3 - 1]x$

= $[(11/10)^3 - 1]x$

$(1331/1000 - 1)x = 331/1000x$

$331/1000x - 3/10x$ = 31

or $({331 - 300})/1000x = 31$

or $31/1000x$ = 31

or x = 1000

Sum = Rs.1000

Using Rule 6,

Here, C.I. - S.I. = Rs.31, R = 10%, T = 3 years, P = ?

C.I. - S.I. = P × $(R/100)^2 × (3 + R/100)$

31 = P × $(10/100)^2(3 + 10/100)$

31 = P × $1/100 × 31/10$ ⇒ P = Rs.1000

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Read more difference in ci and si Based Quantitative Aptitude Questions and Answers

Question : 1

On a certain sum of money, the difference between the compound interest for a year, payable half-yearly, and the simple interest for a year is 180. If the rate of interest in both the cases is 10%, then the sum is

a) 54,000

b) 60,000

c) 62,000

d) 72,000

Answer: (d)

If the interest is compounded half yearly,

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

= P$[(1 + 5/100)^2 - 1]$

= P$[(21/20)^2 - 1] = {41P}/400$

S.I. = ${P × R × T}/100 = {P × 10}/100 = P/10$

${41P}/400 - P/10$ = 180

${41P - 40P}/400$ = 180

$P/400 = 180$

P = Rs.72000

Using Rule 6,

Here, C.I. - S.I. = Rs.180

Interest is compounded half yearly

R = $10/5$ = 5%, T = 2 years

C.I. - S.I. = P$(R/100)^2$

180 = P$(5/100)^2$

P = 180 × 20 × 20 ⇒ P = Rs.72000

Question : 2

The difference between compound interest and simple interest of a sum for 2 years at 8 per cent is 768. The sum is

a) 1,70,000

b) 1,00,000

c) 1,20,000

d) 1,10,000

Answer: (c)

Using Rule 6,

Simple Approach

Sum = (CI - SI)$(100/r)^2$

= 768 × $(100/8)^2$ = Rs.1,20,000

Question : 3

The difference between the compound interest and simple interest on 10,000 for 2 years is 25. The rate of interest per annum is

a) 12%

b) 5%

c) 10%

d) 7%

Answer: (b)

Using Rule 6,

Difference = ${PR}^2/10000$

25 = ${10000 × R^2}/10000$ ⇒ R = 5%

Question : 4

On what sum of money will the difference between S.I and C.I for 2 years at 5% per annum be equal to 25 ?

a) 9000

b) 10,000

c) 9,500

d) 10,500

Answer: (b)

Using Rule 6,

Difference = ${PR^2}/10000$

25 = ${P × 5 × 5}/10000$

P = Rs.10000

Question : 5

The difference between the compound interest and simple interest for the amount 5,000 in 2 years is 32. The rate of interest is

a) 12%

b) 5%

c) 10%

d) 8%

Answer: (d)

Using Rule 6,

Difference of 2 years

= ${p × r^2}/10000$

32 = ${5000 × r^2}/10000$

$r^2 = {32 × 10000}/5000$ = 64

r = $√{64}$ = 8%

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