Compound Interest- Aptitude Questions and Answers

Let Principal = P, Rate = R% per annum, Time = n years.
When time is fraction of a year, say 4³/₄ years, then amount= P(1+R/100)⁴ x (1+(3R/4)/100)

So, Amount = 2800 * (1 + 10/100) * (1 + (1*10/2) / 100)
= 2800 * (1 + 10/100) * (1 +5/100)
= 2800 * 11/10 * 21/20
= 3234

Required C.I. = (3234 - 2800) = 434

Let the sum be rs X then,
CI = X*( 1 + 5/100 )² - X = X*(441 / 400 - 1) = X*41 / 400
SI = X*5*2 / 100 = X / 10

As is given, CI - SI = 40
X* ( 41 / 400 - 1 / 10) = 40
X* (41 - 40) / 400 = 40
X = 16000

Amount = P*(1+ R/100 )ⁿ , if interest is payable annually
Amount = 7500*(1 + 4/100 )²
= 7500 * (104/100)²
= 7500 * (26/25)²
= 7500 * 26/25 * 26/25
= 8112

Let Principal = P, Rate = R% per annum, Time = n years.
Amount = P(1 + (R/4)/100 )⁴ⁿ , if interest is payable quarterly

Hence,required amount = 2000* (1 + (20/4) / 100 )^4*1/₂
= 2000*(1 + 5/100)²
= 2000*(21 / 20)²
= 2205

Let the principle is P.
As is given, compound interest= P*(1 +10/100)² - P = 420
P*(11/10)² - P = 420
P*(121/100) - P = 420
P*(121-100)/100 = 420
P = 2000

So, simple interest.= P*R*T / 100 = 2000* 2* 10 / 100 = 400

Let Principal = P, Rate = R% per annum, Time = n years.
Compound Interest, CI = P* [ 1 + (R/2)/100 ]²ⁿ - P , if compound interest is half-yearly
Simple Interest, SI = P*R*T / 100

Then,CI - SI = ( P*(1 + 5/100)⁴ - P ) - ( P* 10* 2 / 100 ) = 124.05
P*(21/20)⁴ - P - P/ 5 = 124.05
P*(194481 - 160000 - 32000 ) / 160000 = 124.05
P = 19848000 / 2481
P = 8000

Amount = P(1+R/100)ⁿ , if compound interest is payable annually
Then, 109626 = 90600*(1 + 10/100)ⁿ
109626 / 90600 = (11/10)ⁿ
(121/100) = (11/10)ⁿ
(11/10)² = (11/10)ⁿ
n = 2 years

Let Principal = P, Rate = R% per annum, Time = n years.
Compound Interest, CI=P(1+R/100)ⁿ - P
Then, 5995 = P*(1 + 20/100)² - P
5995 =P*(36/25 - 1)
5995 = P* 11/25
P = 5995*25 / 11
P = 13625

Hence Simple Interest, SI = P*R*T / 100 = 13625 * 8 * 6 / 100 = 6540

Let Principal = P, Rate = R% per annum, Time = n years.
Simple Interest , SI = P*R*T /100
Amount = P + Simple Interest =P + (P*R*T/100) = 6000
P + P*20*3 /100 = 6000
P = 6000*5/8 = 3750
Thus, Simple Interest in 3 years, SI = 6000 - 3750 = 2250

Compound Interest, CI=P*(1 + R/100)ⁿ - P
CI = 3750*(1 + 20/100)³ - 3750
=3750*[ (6/5)³ -1 ]
=3750*( 91/125 ) =2730

Hence, trader will be get more if the lent was on CI = CI - SI = 2730 - 2250 = 480

Let the sum is 100 then
S.I. = 100* 5* 2 / 100 = 10
C.I = 100 *(1 + 5/100)² - 100 = 41/4
Difference between C.I and S.I = 41/4 - 10 = 0.25

So, 0.25 : 1.50 : : 100 : P
∴ P = (1.50 * 100) / 0.25 = 600

Let the principle is P
So compound interest = P * ( 1 + 12.5/100)² - P = 170
P* (112.5/100) *(112.5/100) - P = 170
P* (12656.25 - 10000) / 10000 = 170
∴ P = 1700000 / 2656.25

Simple interest SI = P* T* R /100
= (1700000 / 2656.25)* 2* 12.5 / 100
= 160

Amount= 8000*(1 + 5/100)²
= 8000* 21/20* 21/20
= 8820

Remaining amount after 1 year = 12500 *(120/100) - 2000 = 13000
Remaining amount after 2 year = 13000*(120/100) - 2000 = 13600
Remaining amount after 3 year = 13600*(120/100) - 2000 = 14320

2P = P *(1 + R/100)⁵
2⁴ P = P *(1 + R/100)²⁰

The amount becomes 2⁴ times = 16 times and required amount = 2400*16 =38400

amount = P ( 1 + R/100 )ⁿ
As is given, 882 = 800*( 1 + 5/100 )ⁿ
882/800 = (21/20)ⁿ
441/400 = (21/20)ⁿ
(21/20)² = (21/20)ⁿ
N=2

compound interest, CI= 6400* (1 + (15 / 2) / 100 )² - 6400
= 6400* (1 + 15 / 200 )² - 6400
= 6400* (43 / 40 )² - 6400
= 6400* (1849 / 1600) - 6400
= 6400* (1849 / 1600 - 1)
= 6400* 249 / 1600 = 996

Let the sum is 100 then
S.I. = 100* 4* 2 / 100 = 8
C.I = 100 *(1 + 4/100)² - 100 = 100*( 26/25) ² - 100 = 204/25
Difference between C.I and S.I = 204/25 - 8 = 4/25 = .16

So, 0.16 : 10 : : 100 : P
∴ P = (10* 100) / 0.16 = 6250