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

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

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

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: (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)

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