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 : Find the difference between the compound interest and the simple interest on 32,000 at 10% p.a. for 4 years.

(a) 2501.20

(b) 2051.20

(c) 2025.20

(d) 2052.50

The correct answers to the above question in:

Answer: (b)

Using Rule 1,

S.I. = $\text"Principal × Time × Rate"/100$

= ${32000 × 4 × 10}/100$ = Rs.12800

C.I. = P$[(1 + R/100)^4 - 1]$

= 32000$[(1 + 10/100)^4 - 1]$

= 32000 $[(1.1)^4$ - 1]

= 32000 (1.4641 - 1)

= 32000 × 0.4641= Rs.14851.2

Required difference

= 14851.2 - 12800 = Rs.2051.2

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

Question : 1

If the difference between S.I. and C.I. for 2 years on a sum of money lent at 5% is 6, then the sum is

a) 2000

b) 2200

c) 2600

d) 2400

Answer: (d)

Using Rule 6,
The difference between C.I. and S.I. on a sum 'P' in 2 years at the rate of R% rate of compound interest will be
C.I - S.I. = P$(R/100)^2 = {S.I. × R}/200$
For 3 years, C.I. - S.I. = P$(R/100)^2 × (3 + R/100)$

Difference = ${Pr^2}/10000$

6 = ${P × 5 × 5}/10000$

P = 6 × 400 = Rs.2400

Question : 2

What sum will give 244 as the difference between simple interest and compound interest at 10% in 1 1 2 years compounded half yearly ?

a) 28,000

b) 40,000

c) 32,000

d) 36,000

Answer: (c)

Using Rule 6,

Time = $3/2 × 2 = 3$ half years

Rate = $10/2$ = 5% per half year

[Since, when r → $r/2$, then t → 2t]

Difference = P$(r^3/1000000 + {3r^2}/10000)$

244 = P$(125/1000000 + 75/10000)$

244 = P$(7625/1000000)$

P = ${244 × 1000000}/7625$ = Rs.32000

Question : 3

The difference between the compound and the simple interest on a sum for 2 years at 10% per annum, when the interest is compounded annually, is 28. If the interest were compounded halfyearly, the difference in the two interests will be

a) 43.29

b) 44

c) 43.41

d) 28.35

Answer: (c)

Using Rule 6,
The difference between C.I. and S.I. on a sum 'P' in 2 years at the rate of R% rate of compound interest will be
C.I - S.I. = P$(R/100)^2 = {S.I. × R}/200$
For 3 years, C.I. - S.I. = P$(R/100)^2 × (3 + R/100)$

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$[(1 + r/100)^n - 1]$

If the difference between compound interest and simple interest at the rate of r% per annum for 2 years be x, then

Principal = $x(100/r)^2$

= 28$(100/10)^2$ = Rs.2800

If the interest is compounded half yearly, then

r = $10/2$ = 5%,

Time = 4 half years

Simple interest

= ${2800 × 5 × 4}/100$ = Rs.560

Compound interest

= $2800[(1 + 5/100)^4 - 1]$

= 2800 [1.2155 - 1]

= 2800 × 0.2155 = 603.41

Difference = Rs.(603.41–560) = Rs.43.41

Question : 4

The difference between simple and compound interest compounded annually, on a certain sum of money for 2 years at 4% per annum is 1. The sum (in ) is :

a) 640

b) 650

c) 625

d) 630

Answer: (c)

Using Rule 6,

The difference between compound interest and simple interest for two years

= ${\text"Principal" × (Rate)^2}/{100 × 100}$

1 = ${\text"Principal" × (4)^2}/10000$

Principal = $10000/16$ = Rs.625

Question : 5

The difference between the compound interest and the simple interest on a certain sum at 5% per annum for 2 years is 1.50. The sum is

a) 300

b) 600

c) 400

d) 500

Answer: (b)

Using Rule 6,

Difference = ${PR}^2/(100)^2$

1.50 = ${P × 5 × 5}/(100)^2$

P = 400 × 1.5 = Rs.600

Question : 6

The difference between simple interest and compound interest of a certain sum of money at 20% per annum for 2 years is 48. Then the sum is

a) 2, 000

b) 1,000

c) 1, 500

d) 1, 200

Answer: (d)

Using Rule 6,

Difference of two years

= P$(r^2/10000)$

48 = P$(400/10000)$

48 = $P/25$

P = 48 × 25 = Rs.1200

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