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) 4800
(b) 8000
(c) 4400
(d) 8400
The correct answers to the above question in:
Answer: (a)
Simple Approach
Difference of SI and CI for 3 years
= ${PR(300 + R)}/{100^3}$
Since, ${P × 25 × 305}/{100 × 100 × 100}$ = 36.60
P = ${36.60 × 100 × 100 × 100}/{25 × 305}$ = Rs.4800
Using Rule 6,
C.I.–S.I. = Rs.36.60, R = 5%, P =?, T = 3yrs.
C.I. - S.I.= P$(R/100)^2 × (3 + R/100)$
36.60 = P$(5/100)^2 × (3 + 5/100)$
36.60 = P × $25/{100^2} × 305/100$
P = ${36.60 × 100 × 100 × 100}/{25 × 305}$
P = $36600000/{25 × 305}$ = Rs.4800
Discuss Form
Read more difference in ci and si Based Quantitative Aptitude Questions and Answers
Question : 1
If the difference between the compound and simple interests on a certain sum of money for 3 years at 5% per annum is 15.25, then the sum is
a) 2,500
b) 2,000
c) 1,500
d) 1,000
Answer »Answer: (b)
Using Rule 6,
Difference between C.I. and S.I for 3 years
= $\text"PR"^2/(100)^2(R/100 + 3)$
15.25 = ${P × 25}/10000(5/100 + 3)$
15.25 = ${P × 305}/{400 × 100}$
P = ${15.25 × 400 × 100}/305$ = Rs.2000
Question : 2
The difference between the simple and compound interest on a certain sum of money for 2 years at 4% per annum is 4. The sum is
a) 2,000
b) 2500
c) 2,600
d) 2,400
Answer »Answer: (b)
Using Rule 6,
Sum = Difference$(100/r)^2$
= 4 × $(100/4)^2$ = Rs.2500
Question : 3
The difference between simple and compound interest (compounded annually) on a sum of money for 2 years at 10% per annum is 65. The sum is
a) 6500
b) 65650
c) 6565
d) 65065
Answer »Answer: (a)
Let the sum be x. Then,
C.I. = $x(1 + 10/100)^2 - x = {21x}/100$
S.I. = ${x × 10 × 2}/100 = x/5$
C.I. - S.I. = ${21x}/100 - x/5 = x/100$
Given that, $x/100$ = 65
x = 6500
Hence, the sum is Rs.6500.
Using Rule 6,
Here, C.I. - S.I. = Rs.65, R = 10%, T = 2 years, P = ?
C.I. - S.I. = P$(R/100)^2$
65 = P$(10/100)^2$ ⇒ P = Rs.6500
Question : 4
If the difference between the simple and compound interests on a sum of money for 2 years at 4% per annum is 80, the sum is :
a) 1000
b) 5000
c) 10000
d) 50000
Answer »Answer: (d)
Using Rule 6,
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
$x(100/r)^2$ Here x = Rs.80, r = 40%
Required sum = 80$(100/4)^2$
= 80 × 25 × 25 = Rs.50000
Question : 5
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
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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