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) 2, 000
(b) 1,000
(c) 1, 500
(d) 1, 200
The correct answers to the above question in:
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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Read more difference in ci and si Based Quantitative Aptitude Questions and Answers
Question : 1
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 : 2
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 : 3
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
Answer »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
Question : 4
The difference between compound interest (compounded annually) and simple interest on a certain sum of money at 10% per annum for 2 years is 40. The sum is :
a) 3200
b) 4000
c) 4200
d) 3600
Answer »Answer: (b)
Let the principal be x.
Compound interest
= P$[(1 + R/100)^t - 1]$
= $x[(1 + 10/100)^2 - 1]$
= $x[(1.1)^2 - 1]$
= x (1.21 - 1) = 0.21x
SI = ${x × 2 × 10}/100 = x/5 = 0.2x$
According to the question,
0.21x - 0.2x = 40
$0.01x = 40 ⇒ x = 40/{0.01}$ = Rs.4000
Using Rule 6,
Here, C.I. - S.I. = Rs.40, R = 10%, T = 2 years, P = ?
C.I. - S.I. = P$(R/100)^2$
40 = P$(10/100)^2$ ⇒ P = Rs.4000
Question : 5
The difference between the compound interest (compounded annually) and the simple interest on a sum of 1000 at a certain rate of interest for 2 years is 10. The rate of interest per annum is :
a) 12%
b) 5%
c) 10%
d) 6%
Answer »Answer: (c)
When difference between the compound interest and simple interest on a certain sum of money for 2 years at r% rate is x, then
x = Sum$(r/100)^2$
10 = 1000$(r/100)^2$
$(r/100)^2 = 10/1000$
$r/100 = √{1/100} = 1/10$
r = $100/10$ = 10%
Using Rule 6,
Here, C.I. - S.I. = Rs.10, R = ?, T= 2 years, P = Rs.1000
C.I. - S.I. = P$(R/100)^2$
10 = 1000$(R/100)^2$
10 = 1000$ × R/100 × R/100$
$R^2$ = 100
R = $√{100}$ = 10%
Question : 6
What is the difference between compound interest on 5,000 for 1 1 2 years at 4% per annum according as the interest is compounded yearly or halfyearly?
a) 4.80
b) 2.04
c) 8.30
d) 3.06
Answer »Answer: (d)
Compound Interest (when compounded yearly)
= $5000(1 + 4/100)^(1.5) - 5000$
= $5000(26/25)^(1.5) - 5000$
= 5302.9805 - 5000 = Rs.302.9805
C.I. (When compounded halfyearly).
= $5000(1 + 2/100)^3 - 50000$
= 5306.04 - 5000 = Rs.306.04
Required difference
= Rs.(306.04 - 302.9805)
= Rs.3.059 = Rs.3.06
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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