model 2 at ci sum becomes ‘n’ times after ‘t’ years Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (a)

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%

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%

Answer: (a)

A = P$(1 + R/100)^T$

24000 = 12000$(1 + R/100)^5$

2 = $(1 + R/100)^5$

$2^4 = (1 + R/100)^20$

i.e. The sum amounts to Rs.192000 after 20 years.

Here, x = 2, $n_1 = 5, y = ?, n_2$ = 20

$x^{1/n_1} = y^{1/n}$

$2^{1/5} = y^{1/20}$

$y = (2^{1/5})^20$

$y = 2^4 ⇒ y$ = 16 times

Sum = 16 × 12000 = Rs.1,92,000

Answer: (d)

A = P$(1 + R/100)^T$

Let P. Rs., A = Rs.2

2 = 1$(1 + R/100)^3$

On squaring both sides.

4 = 1$(1 + R/100)^6$

Time = 6 years

Using Rule 11,

Here, $x = 2, n_1 = 3, y = 4, n_2$ = ?

$x^{1/n_1} = y^{1/n_2}$

$2^{1/3} = 4^{1/n_2}$

$2^{1/3} = (2^2)^{1/n_2}$

$2^{1/3} = 2^{2/n_2}$

$1/3 = 2/n_2$

$n_2$ = 6 Years

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%

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

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