model 2 at ci sum becomes ‘n’ times after ‘t’ years Practice Questions Answers Test with Solutions & More Shortcuts

Question : 11 [SSC CGL Tier-I 2010]

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: (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 : 12 [SSC MTS 2013]

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: (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 : 13 [SSC CGL Tier-II 2012]

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) 3 years

b) 9 years

c) 6 years

d) 27 years

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

Question : 14 [SSC CHSL 2013]

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: (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 : 15 [SSC Investigator 2010]

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: (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%

IMPORTANT quantitative aptitude EXERCISES

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