model 2 at ci sum becomes ‘n’ times after ‘t’ years Practice Questions Answers Test with Solutions & More Shortcuts
compound interest PRACTICE TEST [6 - EXERCISES]
model 1 basic compound interest using formula
model 2 at ci sum becomes ‘n’ times after ‘t’ years
model 3 combination of si & ci
model 4 difference in ci & si
model 5 ci with instalments
model 6 comparing sum in different years
Question : 6 [SSC CGL Prelim 1999]
If the amount is 2.25 times of the sum after 2 years at compound interest (compound annually), the rate of interest per annum is :
a) 50%
b) 25%
c) 45%
d) 30%
Answer »Answer: (a)
Suppose P = Rs.100
and amount A = Rs.225
A = P$(1 + r/100)^t$
or 225 = $100(1 + r/100)^2$
or $225/100 = [1 + r/100]^2$
or 1 + $r/100 = 15/10$
or ${100 + r}/100 = 15/10$
or 100 + r = 150
or r = 50%
Using Rule 8,If a sum becomes 'n' times of itself in 't' years on compound interest, then R% = $[n^{1/t} - 1] × 100%$
Here, n = 2.25, t = 2 years
R% = $(n{1/t} - 1) × 100%$
R% = $[(2.25)^{1/2} - 1]$ × 100%
= [1.5 - 1] × 100%
= 0.5 × 100% = 50%
Question : 7 [SSC CGL Tier-I 2011]
A sum of Rs.12,000, deposited at compound interest becomes double after 5 years. How much will it be after 20 years ?
a) Rs.1,92,000
b) Rs.1,44,000
c) Rs.1,50,000
d) Rs.1,20,000
Answer »Answer: (a)
A = P$(1 + R/100)^T$
24000 = 12000$(1 + R/100)^5$
2 = $(1 + R/100)^5$
$2^4 = (1 + R/100)^20$
i.e. The sum amounts to Rs.192000 after 20 years.
Using Rule 11,A certain sum at C.I. becomes x times in $n_1$ year and y times in $n_2$ years then $x^{1/n_1} = y^{1/n_2}$
Here, x = 2, $n_1 = 5, y = ?, n_2$ = 20
$x^{1/n_1} = y^{1/n}$
$2^{1/5} = y^{1/20}$
$y = (2^{1/5})^20$
$y = 2^4 ⇒ y$ = 16 times
Sum = 16 × 12000 = Rs.1,92,000
Question : 8 [SSC CPO S.I.2010]
A sum of money becomes double in 3 years at compound interest compounded annually. At the same rate, in how many years will it become four times of itself ?
a) 7.5 years
b) 4 years
c) 6.4 years
d) 6 years
Answer »Answer: (d)
A = P$(1 + R/100)^T$
Let P. Rs., A = Rs.2
2 = 1$(1 + R/100)^3$
On squaring both sides.
4 = 1$(1 + R/100)^6$
Time = 6 years
Using Rule 11,
Here, $x = 2, n_1 = 3, y = 4, n_2$ = ?
$x^{1/n_1} = y^{1/n_2}$
$2^{1/3} = 4^{1/n_2}$
$2^{1/3} = (2^2)^{1/n_2}$
$2^{1/3} = 2^{2/n_2}$
$1/3 = 2/n_2$
$n_2$ = 6 Years
Question : 9 [SSC CGL Prelim 2002]
A sum borrowed under compound interest doubles itself in 10 years. When will it become fourfold of itself at the same rate of interest ?
a) 40 years
b) 15 years
c) 24 years
d) 20 years
Answer »Answer: (d)
Let the sum be x which becomes 2x in 10 years.
Hence, 4x in 20 years
Method 2 :
Unitary Method can also be used.
Using Rule 5,
Here, m = 2, t = 10
Time taken to become 4 times = $2^2$ times
= t × n = 10 × 2 = 20 years
Question : 10 [SSC CGL Prelim 2004]
A sum of money becomes eight times of itself in 3 years at compound interest. The rate of interest per annum is
a) 10%
b) 100%
c) 20%
d) 80%
Answer »Answer: (b)
Let the principal be x and the rate of compound interest be r% per annum. Then,
$8x = x(1 + r/100)^3$
8 = $(1 + r/100)^3$
$2^3 = (1 + r/100)^3$
2 = $1 + r/100 ⇒ r/100 = 1 ⇒ r = 100%$
Using Rule 8,If a sum becomes 'n' times of itself in 't' years on compound interest, then R% = $[n^{1/t} - 1] × 100%$
Here, n = 8, t = 3 years.
R% = $[n^{1/t} - 1] × 100%$
= $[8^{1/3} - 1] × 100%$
= $[(2^3)^{1/3} - 1] × 100%$ = 100%
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model 2 at ci sum becomes ‘n’ times after ‘t’ years Shortcuts »
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Click to Start..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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