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) 650
(b) 600
(c) 560
(d) 625
The correct answers to the above question in:
Answer: (d)
Using Rule 6,
When difference between the CI and SI on a certain sum of money for 2 years at r % rate is x, then
Sum = x × $(100/r)^2$
= 1 × $(100/4)^2$ = Rs.625
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Read more difference in ci and si Based Quantitative Aptitude Questions and Answers
Question : 1
A sum of 6,000 is deposited for 3 years at 5% per annum compound interest (compounded annually). The difference of interests for 3 and 2 years will be
a) 375.00
b) 75.00
c) 330.75
d) 30.75
Answer »Answer: (c)
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$[(1 + r/100)^n - 1]$
C.I. after 3 years
= 6000$[(1 + 5/100)^3 - 1]$
= 6000$({9261 - 8000}/8000)$
= 6000 × $1261/8000$ = Rs.945.75
CI after 2 years
= $6000[(1 + 5/100)^2 - 1]$
= 6000$({441 - 400}/400)$
= 6000 × $41/400$ = Rs.615
Required difference
= Rs.(945.75 - 615) = Rs.330.75
Question : 2
The difference between compound interest and simple interest on 2500 for 2 years at 4% per annum is
a) 4
b) 40
c) 14
d) 45
Answer »Answer: (a)
S.I. = Rs.${2500 × 2 × 4}/100$ = Rs.200
C.I. = Rs.2500$[(1 + 4/100)^2 - 1]$
= Rs.2500$[(26/25)^2 - 1]$
= Rs.${(676 - 625)}/625$ × 2500
= Rs.$51/625 × 2500$ = Rs.204
The required difference
= C.I. - S.I. = Rs.(204 - 200) = Rs.4
Using Rule 6,
Here, C.I. - S.I.= ?, P = Rs.2500, R = 4%, T = 2
C.I. - S.I.= P$(R/100)^2$
= 2500$(4/100)^2$
= 2500 × $1/25 × 1/25$
C.I.–S.I. = Rs.4
Question : 3
The difference between the simple and compound interest on a certain sum of money at 5% rate of interest per annum for 2 years is 15. Then the sum is :
a) 7,000
b) 6,500
c) 6,000
d) 5,500
Answer »Answer: (c)
Let the sum Rs.x. Then,
C.I. = $x(1 + 5/100)^2 - x$
= ${441x}/400 - x = {441x - 400x}/400$
= $41/400$x
Now, S.I. = ${x × 5 × 2}/100 = x/10$
(C.I.) - (S.I.)= ${41x}/400 - x/10$
= ${41x - 40x}/400 = x/400$
$x/400 = 15$
x = 15 × 400 = 6000
Hence, the sum is Rs.6000
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$(R/100)^2 × (3 + R/100)$
C.I. - S.I. = Rs.15, R = 5%, T = 2 years, P =?
C.I. - S.I. = P$(R/100)^2$
15 = P$(5/100)^2$
P = 15 × 400 = Rs.6000
Question : 4
On a certain sum of money lent out at 16% p.a. the difference between the compound interest for 1 year, payable half yearly, and the simple interest for 1 year is 56. The sum is
a) 5780
b) 1080
c) 8750
d) 7805
Answer »Answer: (c)
Using Rule 6,
Rate of interest = 8% per halfyear
Time = 2 half years
Difference of interests = ${PR^2}/100$
56 = $P × (8)^2/(100)^2$
P = ${56 × 10000}/64$ = 8750
Question : 5
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 : 6
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
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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