model 4 difference in ci & si Section-Wise Topic Notes With Detailed Explanation And Example Questions
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The following question based on compound interest topic of quantitative aptitude
(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 »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 beC.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 »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 »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 beC.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 - PC.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 »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 »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 »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
compound interest Shortcuts and Techniques with Examples
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model 1 basic compound interest using formula
Defination & Shortcuts … -
model 2 at ci sum becomes ‘n’ times after ‘t’ years
Defination & Shortcuts … -
model 3 combination of si & ci
Defination & Shortcuts … -
model 4 difference in ci & si
Defination & Shortcuts … -
model 5 ci with instalments
Defination & Shortcuts … -
model 6 comparing sum in different years
Defination & Shortcuts …
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