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) 20 years
(b) 10 years
(c) 7 years
(d) 15 years
The correct answers to the above question in:
Answer: (d)
Let the Principal be P and rate of interest be r%.
2 P = P$(1 + r/100)^2$
2 = $(1 + r/100)^5$ ...(i)
On cubing both sides,
8 = $(1 + r/100)^15$
Time = 15 years
Using Rule 5,
Here, m = 2, t = 5 years
It becomes 8 times = $2^3$ times
in t × n = 5 × 3 = 15 years
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Read more ci becomes n times after t years Based Quantitative Aptitude Questions and Answers
Question : 1
A sum of money placed at compound interest doubles itself in 4 years. In how many years will it amount to four times itself ?
a) 16 years
b) 12 years
c) 8 years
d) 13 years
Answer »Answer: (c)
A = P$(1 + R/100)^T$
Let P be Rs.1, then A = Rs.2
2 = 1$(1 + R/100)^4$
$2^2 = (1 + R/100)^8$
Time = 8 years
Using Rule 11,
Here, $x = 2, n_1 = 4, y = 4, n_2$ = ?
Using $x^{1/n_1} = y^{1/n_2}$
$(2)^{1/4} = (4)^{1/n_2}$
$(2)^{1/4} = (2^2)^{1/n_2}$
$2^{1/4} = 2^{1/n_2}$
$1/4 = 2/n_2$
$n_2$ = 8 years
Question : 2
If a sum of money compounded annually becomes 1.44 times of itself in 2 years, then the rate of interest per annum is
a) 20%
b) 25%
c) 21%
d) 22%
Answer »Answer: (a)
A = P$(1 + R/100)^T$
1.44P = P$(1 + R/100)^2$
$(1.2)^2 = (1 + R/100)^2$
$1 + R/100$ = 1.2
R = 0.2 × 100 = 20%
Using Rule 8,
Here, n = 1.44, t = 2 years
R% = $(n^{1/6} - 1) × 100%$
= $[(1.44)^{1/2} - 1] × 100%$
= [(1.2) - 1] × 100%
= 0.2 × 100% ⇒ R% = 20%
Question : 3
A sum of money invested at compound interest doubles itself in 6 years. At the same rate of interest it will amount to eight times of itself in :
a) 10 years
b) 15 years
c) 18 years
d) 12 years
Answer »Answer: (c)
Let the sum be x. Then,
$2x = x(1 + r/100)^6$
2 = $(1 + r/100)^6$
Cubing both sides,
8 =$((1 + r/100)^6)^3$
8 = $(1 + r/100)^18$
$8x = x(1 + r/100)^18$
The sum will be 8 times in 18 years.
i.e., Time = 18 years
Using Rule 5,
Here, m = 2, t = 6 years
It will becomes 8 times of itself
= $2^3$ times of it self
in t × n years = 6 × 3 = 18 years
Question : 4
A sum of money becomes eight times in 3 years, if the rate is compounded annually. In how much time will the same amount at the same compound rate become sixteen times?
a) 5 years
b) 6 years
c) 8 years
d) 4 years
Answer »Answer: (d)
Let the principal be Rs.1.
A = P$(1 + R/100)^T$
8 = 1$(1 + R/100)^3$
$2^3 = 1(1 + R/100)^3$
2 = 1$(1 + R/100)^1$
$2^4 = (1 + R/100)^4$
Time = 4 years
Using Rule 11,
Here, $x = 8, n_1 = 3, y = 16, n_2$ = ?
Using $x^{1/n_1} = y^{1/n_2}$
$(8)^{1/3} = (16)^{1/n_2}$
$(2^3)^{1/3} = (2^4)^{1/n_2}$
$2^1 = 2^{4/n_2}$
1= $4/n_2$
$n_2$ = 4 years
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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