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

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

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%

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%

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

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%

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%