Practice Ci becomes n times after t years - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) 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)
(b)
(c)
(d)
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%
Q-2) 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)
(b)
(c)
(d)
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%
Q-3) A sum of money becomes eight times of itself in 3 years at compound interest. The rate of interest per annum is
(a)
(b)
(c)
(d)
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%
Q-4) A sum of money becomes 1.331 times in 3 years as compound interest. The rate of interest is
(a)
(b)
(c)
(d)
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%
Q-5) At what rate percent per annum of compound interest, will a sum of money become four times of itself in two years ?
(a)
(b)
(c)
(d)
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%
Q-6) 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)
(b)
(c)
(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
Q-7) A sum of money at compound interest amounts to thrice itself in 3 years. In how many years will it be 9 times itself ?
(a)
(b)
(c)
(d)
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
Q-8) A sum of money placed at compound interest doubles itself in 5 years. In how many years, it would amount to eight times of itself at the same rate of interest ?
(a)
(b)
(c)
(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
Q-9) 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)
(b)
(c)
(d)
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
Q-10) 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)
(b)
(c)
(d)
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