Practice Test on Compound Interest | Maths Compound Interest Questions with Solutions

Given that the compound interest on $ 25000 at 4 % per annum for 2 years.
Principal amount = $ 25000
Rate of Interest = 4 %
Time = 2 years
The formula to find the compound interest is Compound Interest = P(1 + \(\frac { r }{ 100 } \))ⁿ – P where P is the Principal, r is the rate of interest, and n is the time.
Substitute the given details in the above formula.
Compound Interest = $ 25000(1 + \(\frac { 4 }{ 100 } \))² – $ 25000)
CI = $ 25000{(1 + \(\frac { 1 }{ 25 } \))² – $1)}
CI = $ 25000{(\(\frac { 26 }{ 25 } \))² – 1}
Compound Interest = $ 25000 {\(\frac { 676 }{ 625 } \) – 1}
CI = $ 25000{\(\frac { 51 }{ 625 } \)} = $2040

Therefore, the Compound Interest is (d) $ 2040.

Let the rate percent be r.
Given that at r rate percent per annum will a sum of Rs 28224 amount to Rs 5625 in 2 years.
Sum = Rs 28224
Amount = Rs 5625
Time = 2 years
The formula to find the rate percent is Amount A = P{(1 + \(\frac { r }{ 100 } \))ⁿ}where P is the Principal, r is the rate of interest, A is the amount, and n is the time.
Substitute the given details in the above formula.
Rs 28224 = Rs 5625{(1 + \(\frac { r }{ 100 } \))²}
Rs 28224/Rs 5625 = {(1 + \(\frac { r }{ 100 } \))²}
(168/75)² = (1 + \(\frac { r }{ 100 } \))²
168/75 = 1 + \(\frac { r }{ 100 } \)
168/75 – 1 = \(\frac { r }{ 100 } \)
93/75 = \(\frac { r }{ 100 } \)
1.24 = \(\frac { r }{ 100 } \)
r = 1.24 * 100 = 124%
At 124 rate percent per annum will a sum of Rs 28224 amount to Rs 5625 in 2 years, compounded annually.

Therefore, the answer is (c) 124 %.

Let the period be n years.
Given that Michael deposits $ 10000 in a bank at 8% per annum compound interest for a certain time is $ 1664.
Then, amount after n years = $(10000 + 1664) = $11664
Principal = $10000
Rate of Interest = 8%
Compound Interest = $ 1664.
The formula to find the time is the amount after n years = P{(1 + \(\frac { r }{ 100 } \))ⁿ} where P is the Principal, r is the rate of interest, and n is the time.
Substitute the given details in the above formula.
$ 11664 = $ 10000{(1 + \(\frac { 8 }{ 100 } \))ⁿ}
$ 11664/$ 10000 = {(1 + \(\frac { 8 }{ 100 } \))ⁿ}
(11664/10000) = (\(\frac { 108 }{ 100 } \))ⁿ
(108/100)² = (\(\frac { 108 }{ 100 } \))ⁿ
n = 2 years.

Therefore, the time is (b) 2 years.

Given that the principal that amounts to $ 9826 in 3 years at \(\frac { 50 }{ 8 } \) % per annum compound interest.
Then, amount A = $ 9826
Time n = 3 years
Rate of Interest = \(\frac { 50 }{ 8 } \) %
The formula to find the compound interest is A = P{(1 + \(\frac { r }{ 100 } \))ⁿ} where P is the Principal, r is the rate of interest, A is the amount, and n is the time.
Substitute the given details in the above formula.
9826 = P{(1 + \(\frac { 50 }{ 800 } \))³}
9826 = P{(1 + \(\frac { 1 }{ 16 } \))³}
9826 = P{ (\(\frac { 17 }{ 16 } \))³}
9826= P(\(\frac { 4913 }{ 4096 } \))ⁿ
P = 2 * 4096 = 8192.

Therefore, the answer is (c) 8192.

Given that the simple interest on a sum of money at 10% per annum for 2 years is $ 600.
Then, Simple Interest (S.I.) = $ 600
Time n = 2 years
Rate of Interest = 10 %
The formula to find the compound interest on the same sum for the same period at the same rate will be is A = P{(1 + \(\frac { r }{ 100 } \))ⁿ} where P is the Principal, r is the rate of interest, A is the amount, and n is the time.
Principal = \(\frac { 100 * 600}{ 2 * 10 } \) = 3000
Substitute the given details in the above formula.
Amount = 3000{(1 + \(\frac { 10 }{ 100 } \))²}
Amount = 3000{(\(\frac { 110 }{ 100 } \))²}
A = 3000{(\(\frac { 11 }{ 10 } \))²}
A = 3000{ (\(\frac { 121 }{ 100 } \))}
Amount = 3630.
C.I. = 3630 – 3000 = $630

Therefore, the answer is (a) $630.

Given that a certain sum of money gives $ 1020 as compound interest 7¹/₂ % per annum for 2 years.
Then, compound interest (C.I.) = $ 1020
Time n = 2 years
Rate of Interest = 7¹/₂ %
The formula to find the principal is CI = P{(1 + \(\frac { r }{ 100 } \))ⁿ} where P is the Principal, r is the rate of interest, A is the amount, and n is the time.
Substitute the given details in the above formula.
Compound interest = P{(1 + \(\frac { r }{ 100 } \))ⁿ} – P
1020 = P{(1 + \(\frac { 15 }{ 200 } \))²} – P
1020 = P{(\(\frac { 215 }{ 200 } \))²} – P
1020 = P{(\(\frac { 43 }{ 40 } \))²} – P
1020 = P{\(\frac { 1849 }{ 1600 } \)} – P
1020 = \(\frac { 249P }{ 1600 } \)
P = 6554.2168
Simple Interest = (Principal * rate * time)/100 = 983.1325

Therefore, the Simple Interest is about (d) $980.

Given that on a sum of $ 30000 for 2 years, if the difference between compound interest and simple interest is $ 192.
Then, compound interest (C.I.) = P{(1 + \(\frac { r }{ 100 } \))ⁿ – 1}
Simple Interest = Prt/100 = (30000 * r * 2)/100 = 600r
C.I. = 30000{(1 + \(\frac { r }{ 100 } \))² – 1} = 30000{1 + \(\frac { r² }{ 1000 } \) + \(\frac { 2r }{ 100 } \) – 1} = 30r² + 600r
C.I. – S.I. = 192
30r² + 600r – 600r = 192
30r² = 192
r = 2.529

Therefore, the rate of interest percent per annum is about (b) 2.5%.

Given that Alex borrowed Rs. 3,500 from John at 6 % per annum compound interest. After 2 years Alex gave Rs. 348 and an old dinner set to clear his account.
Sum = Rs. 3,500
Rate = 6%
Time = 2 years
Sum after 2 years = P{(1 + \(\frac { r }{ 100 } \))ⁿ} = Rs. 3,500 {(1 + \(\frac { 6 }{ 100 } \))²} = Rs. 3,500 * \(\frac { 106 }{ 100 } \) * \(\frac { 106 }{ 100 } \) = 3932.6
After 2 years Alex gave Rs. 348 and an old dinner set to clear his account.
Cost of the dinner set = Rs. 3932.6 – Rs. 348 = 3584.6

Therefore, the Cost of the dinner set is about (b) Rs. 3584.6.

Given that Principal = Rs. 3,000
r1 = 4%, r2 = 6%, and r3 = 9%.
Time n = 3 years
Amount = P{(1 + \(\frac { r1 }{ 100 } \))* (1 + \(\frac { r2 }{ 100 } \)) * (1 + \(\frac { r3 }{ 100 } \))}
Substitute the given details in the above formula.
Amount = 3000{(1 + \(\frac { 4 }{ 100 } \))* (1 + \(\frac { 6 }{ 100 } \)) * (1 + \(\frac { 9 }{ 100 } \))} = 3604.848
Compound Interest C.I. = Amount – Principal
Compound Interest C.I. = Rs. 3604.848 – Rs. 3000 = Rs. 604.848

Therefore, the answer is (a) Rs. 604.848

Given that Amount = Rs. 1468
Rate of interest (r%) = 4% p.a. or 2% half yearly
Time n = 2 ½ years or 5 half years.
Amount = P{(1 + \(\frac { r }{ 100 } \))ⁿ}
Substitute the given details in the above formula.
Rs. 1468 = P{(1 + \(\frac { 2 }{ 100 } \))⁵}
Rs. 1468 = P{(\(\frac { 51 }{ 50 } \))⁵}
Rs. 1468 = P{(\(\frac { 51 }{ 50 } \))⁵}
P = Rs. 1329.612
Hence, the principal amount is Rs. 1329.612

Therefore, the answer is (b) Rs. 1329.612