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

Answer: (d)

Let the sum be x which becomes 2x in 10 years.

Hence, 4x in 20 years

Method 2 :

Unitary Method can also be used.

Using Rule 5,

Here, m = 2, t = 10

Time taken to become 4 times = $2^2$ times

= t × n = 10 × 2 = 20 years

Answer: (b)

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 = $1 + r/100 ⇒ r/100 = 1 ⇒ r = 100%$

Here, n = 8, t = 3 years.

R% = $[n^{1/t} - 1] × 100%$

= $[8^{1/3} - 1] × 100%$

= $[(2^3)^{1/3} - 1] × 100%$ = 100%

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: (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: (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: (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