model 2 at ci sum becomes ‘n’ times after ‘t’ years 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) 3 years
(b) 9 years
(c) 6 years
(d) 27 years
The correct answers to the above question in:
Answer: (c)
A = P$(1 + R/100)^T$
Let P = Rs.1, then A = Rs.3
3 = 1$(1 + R/100)^3$
On squaring both sides,
9 = 1$(1 + R/100)^6$
Time = 6 years
Using Rule 11,
Here, $x = 3, n_1 = 3, y = 9, n_2$ = ?
Using, $x^{1/n_1} = y^{1/n_2}$
$(3)^{1/3} = (9)^{1/n_2}$
$3^{1/3} = (3^2)^{1/n_2}$
$3^{1/3} = 3^{2/n_2}$
$1/3 = 2/n_2 ⇒ n_2$ = 6 years
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Read more ci becomes n times after t years Based Quantitative Aptitude Questions and Answers
Question : 1
If the amount is 3$3/8$ times the sum after 3 years at compound interest compounded annually, then the rate of interest per annum is
a) 33$1/3$%
b) 25%
c) 16$2/3$%
d) 50%
Answer »Answer: (d)
A = P$(1 + R/100)^T$
$27/8x = x(1 + R/100)^3$
$(3/2)^3 = (1 + R/100)^3$
$1 + R/100 = 3/2$
$R/100 = 3/2 - 1 = 1/2$
R = $1/2$ × 100 ⇒ R = 50%
Using Rule 8,
n= $27/8$, t = 3 years
R% = $(n^{1/t} - 1) × 100%$
= $((27/8)^{1/3} - 1) × 100%$
= $[(3/2) - 1]$ × 100% ⇒ R% = 50%
Question : 2
At what rate percent per annum of compound interest, will a sum of money become four times of itself in two years ?
a) 20%
b) 100%
c) 50%
d) 75%
Answer »Answer: (b)
A = P$(1 + R/100)^T$
4 = $(1 + R/100)^2$
1 + $R/100$ = 2
$R/100$ = 1 ⇒ R = 100%
Using Rule 8,
Here, n = 4, t = 2 years
R% = $(n^{1/t} - 1)$ × 100%
= $((4)^{1/2} - 1)$ × 100% = 100%
Question : 3
A sum of money doubles itself in 4 years at compound interest. It will amount to 8 times itself at the same rate of interest in :
a) 24 years
b) 18 years
c) 16 years
d) 12 years
Answer »Answer: (d)
A sum of Rs.x becomes Rs.2x in 4 years.
Similarly, Rs.2x will become 2 × 2x
= Rs.4x in next 4 years and Rs.4x will become
2 × 4x = Rs.8x in yet another 4 years.
So, the total time = 4 + 4 + 4 = 12 years
Using Rule 5,A certain sum becomes 'm' times of itself in 't' years on compound interest then the time it will take to become mn times of itself is t × n years.
Here, m = 2, t = 4
Time taken to become
$2^3$ = n × t years
= 3 × 4 = 12 years
Note : If a sum of money becomes n times in t years, it will become $t^1 = n^x$ times at the same rate of interest in $t^1$ years given by,$t^1$ = xt
Question : 4
A sum of money becomes 1.331 times in 3 years as compound interest. The rate of interest is
a) 50%
b) 8%
c) 10%
d) 7.5%
Answer »Answer: (c)
If principal = Rs.1000, amount = Rs.1331
A = P$(1 + R/100)^T$
$1331/1000 = (1 + R/100)^3$
$(11/10)^3 = (1 + R/100)^3$
$1 + R/100 = 11/10$
$R/100 = 1/10$
R = $1/10 × 100$ = 10%
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 = 1.331, t = 3 years
R% = $(n^{1/t} - 1) × 100%$
= $[(1.331)^{1/3} - 1] × 100%$
= [1.1 - 1] × 100%
= 0.1 × 100% ⇒ R% = 10%
Question : 5
A sum of money at compound interest doubles itself in 15 years. It will become eight times of itself in
a) 60 years
b) 45 years
c) 54 years
d) 48 years
Answer »Answer: (b)
A = P$(1 + R/100)^T$
2 = 1$(1 + R/100)^15$
Cubing on both sides, we have
8 = 1$(1 + R/100)^45$
Required time = 45 years
Using Rule 5,
Here, m = 2, t = 15 years
It becomes 8 times = $2^3$ times
in t × n years= 15 × 3 = 45 years
Question : 6
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%
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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