Aptitude Compound Interest Test Yourself

Explanation:

Simple Interest =$\dfrac{PRT}{100}$ = $\dfrac{900 \times 10 \times 1}{100} $= Rs.90

Amount after 1 year on Rs.900 at 10% per annum when interest is reckoned half-yearly

=P $\left(1 + \dfrac{\left(R/2\right)}{100}\right)^2T$ = 900$\left(1 + \dfrac{(10/2)}{100}\right)^{2 \times 1}$ = 900$\left(1 + \dfrac{5}{100}\right)^{2}$ = 900$\left(\dfrac{105}{100}\right)^{2}$

$= \dfrac{900\times 105 \times 105}{100 \times 100}$ = Rs.992.25

Compound Interest = 992.25 - 900 = 92.25

Required difference between simple interest and compound interest = 92.25 - 90 = Rs.2.25

Explanation:

Amount of Rs. 100 for 1 year when compounded half-yearly= Rs $\left[100\times\left(1+\dfrac{3}{100}\right)^2\right]$= Rs. 106.09

$\therefore$ Effective rate = (106.09 - 100)% = 6.09%

Explanation:

= Rs. $\left[1600\times\left(1+\dfrac{5}{2\times100}\right)^2+1600\times\left(1+\dfrac{5}{2\times100}\right)\right]$

= Rs. $\left[1600+\dfrac{41}{40}\times\dfrac{41}{40}+1600\times\dfrac{41}{40}\right]$

= Rs. $\left[1600+\dfrac{41}{40}\left(\dfrac{41}{40}+1\right)\right]$

= Rs. $\left[\dfrac{1600\times41\times81}{40\times40}\right]$

= Rs. 3321

$\therefore$ C.I. = Rs. (3321 - 3200) = Rs. 121

Explanation:

Amount after 3 years = P $\left(1 + \dfrac{R}{100}\right)^T$ = $40000\left(1 + \dfrac{11}{100}\right)^3$ =$ 40000\left(\dfrac{111}{100}\right)^3$

= $\dfrac{40000 \times 111 \times 111 \times 111}{100 \times 100 \times 100}$ = $\dfrac{4\times 111 \times 111 \times 111}{ 100} $= 54705.24

Compound Interest = 54705.24 - 40000 = Rs. 14705.24

Explanation:

Let the rate be R% p.a.

Then, 1200 $ \times\left(1 +\dfrac{R}{100} \right)^2$= 1348.32

$\Rightarrow \left(1 +\dfrac{R}{100} \right)^2$=$ \dfrac{134832}{120000} $=$ \dfrac{11236}{10000} $

$\therefore \left(1 +\dfrac{R}{100} \right)^2$=$ \left(\dfrac{106}{100} \right)^2$

$\Rightarrow$ 1 +$ \dfrac{R}{100} $=$ \dfrac{106}{100} $

$\Rightarrow$ R = 6%

Explanation:

Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.

$\therefore$ R =$ \left(\dfrac{100 \times 60}{100 \times 6} \right) $= 10% p.a.

Now, P = Rs. 12000. T = 3 years and R = 10% p.a.

$\therefore$ C.I.	= Rs.$ \left[12000 \times\left\{\left(1 +\dfrac{10}{100} \right)^3- 1\right\}\right] $
	= Rs.$ \left(12000 \times\dfrac{331}{1000} \right) $
	= 3972.

Explanation:

Simple Interest = Rs. 80

Rate of interest =$\dfrac{100 \times SI}{PT} = \dfrac{100 \times 80}{100 \times 8} $= 10per annum

Now lets find out the compound interest of Rs. 14,000 after 3 years at 10%

P = Rs.14000

T = 3 years

R = 10%

Amount after 3 years = P$\left(1 + \dfrac{R}{100}\right)^T$ = 14000$\left(1 + \dfrac{10}{100}\right)^3$

= $14000\left(\dfrac{110}{100}\right)^3 = 14000\left(\dfrac{11}{10}\right)^3 = 14 \times 11^3 = 18634$

Compound Interest = Rs.18634 - Rs.14000 = Rs.4634

Explanation:

C.I. when interest is compounded half-yearly	= Rs.$ \left[5000 \times\left(1 +\dfrac{4}{100} \right)\times\left(1 +\dfrac{\dfrac{1}{2} \times 4}{100} \right)\right] $
	= Rs.$ \left(5000 \times\dfrac{26}{25} \times\dfrac{51}{50} \right) $
	= Rs. 5304.

C.I. when interest is compounded half-yearly	= Rs.$ \left[5000 \times\left(1 +\dfrac{2}{100} \right)^3\right] $
	= Rs.$ \left(5000 \times\dfrac{51}{50} \times\dfrac{51}{50} \times\dfrac{51}{50} \right) $
	= Rs. 5306.04

$\therefore$ Difference = Rs. (5306.04 - 5304) = Rs. 2.04

Explanation:

Let Principal = 2500, r=6%=0.06616=0.03, n=8×2=16

We know that
A=P(1+r)n
=2500(1+.03)16
=4011.73
The compound interest =4011.73−2500=1511.73

Explanation:
Use the formula:
FV = PV × (1 + r)n
Substitute PV = $2,500, r = 2% = 0.02 and n = 15
FV = $2,500 × (1 + 0.02)15 = $2,500 × (1.02)15
= $2,500 × 1.34586...
= $3364.6708...
So the value after 6 years = $3,364.67