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
(a) 1,70,000
(b) 1,00,000
(c) 1,20,000
(d) 1,10,000
The correct answers to the above question in:
Answer: (c)
Using Rule 6,
Simple Approach
Sum = (CI - SI)$(100/r)^2$
= 768 × $(100/8)^2$ = Rs.1,20,000
Discuss Form
Read more difference in ci and si Based Quantitative Aptitude Questions and Answers
Question : 1
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 »Answer: (b)
Using Rule 6,
Difference = ${PR}^2/10000$
25 = ${10000 × R^2}/10000$ ⇒ R = 5%
Question : 2
The difference between simple and compound interests on a sum of money at 4% per annum for 2 years is 8. The sum is
a) 5,000
b) 400
c) 4,000
d) 800
Answer »Answer: (a)
Using Rule 6,
Let the sum be x.
When difference between the compound interest and simple interest on a certain sum of money for 2 years at r% rate is x, then the sum is given by:
Sum = Difference × $(100/\text"Rate")^2$
= Rs.8 × $(100/4)^2$
= Rs.8 × 25 × 25 = Rs.5000
Question : 3
If the difference between the compound interest, compounded every six months, and the simple interest on a certain sum of money at the rate of 12% per annum for one year is 36, the sum is :
a) 9,000
b) 10,000
c) 15,000
d) 12,000
Answer »Answer: (b)
Easy Trick
As the interest was compounded half-yearly,
we changed r to $r/2$ and t to 2t.
T = 1 year & R 6%
Sum = ${36 × 100 × 100}/{6 × 6}$ = Rs.10000
Question : 4
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 »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 : 5
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
Answer »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
Question : 6
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 »Answer: (b)
Using Rule 6,
Difference = ${PR^2}/10000$
25 = ${P × 5 × 5}/10000$
P = Rs.10000
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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