## Compound Interest Problems

Compound Interest Shortcuts and Tricks are used in aptitude exams to solve problems based on Compound Interest. So, it becomes very necessary to know basics of CI and know who to solve Compound Interest Problems with shortcuts. In this page Compound interest problems are presented along with detailed solutions to help the individual so, that they can easily understand it. To begin with Compound Interest Question and Answers you must know Compound Interest Formula’s which are provided below.

There are several types of interest problems but in this article we will deals with solving problems on compound Interest. When, the interest is calculated more than once in a year, then it is called as “Compound Interest”. Go through the below section of page, that is prepared by the expert team members of www.privatejobshub.in through which you can learn how to solve Compound Interest Problems. Let’s go ahead and get to know how you can solve problems on compound interest in an easy way.

### Compound Interest Formula

 Simple Interest Formula: Simple Interest = Principal * Time * Rate of interest / 100 Abbreviated as SI = PTR/100                   Sometimes, the interest is also calculated half-yearly or quarterly. When compounded semi-annually or half-yearly: Amount = P [1 + (R/2)/100]2t When compounded quarterly: Amount = P[1 + (R/4)/100]4t Present worth of Principal P due t years hence is given by: P/ [1+ R/100] t

Question 1: The simple interest on a sum of money for 3 years at 6²/ % per annum is \$ 6750. What will be the compound interest on the same sum at the same rate for the same period, compounded annually?

Solution:

Given, SI = \$ 6750, R = 203203% p.a. and T = 3 years.

Sum = 100 × SI / R × T

= \$ (100 × 6750 × ³/₂₀ × 1/3) = \$ 33750.

Now, P = \$ 33750, R = 203203% p.a. and T = 3 years.

Therefore, amount after 3 years

= \$ {33750 × (1 + (20/3 × 100)}³ [using A = P (1 + R/100)

= \$ (33750 × 16/15 × 16/15 × 16/15) = \$ 40960.

Thus, amount = \$ 40960.

Hence, compound interest = \$ (40960 - 33750) = \$ 7210.

Question 2: The difference between the compound interest, compounded annually and the simple interest on a certain sum for 2 years at 6% per annum is \$ 18. Find the sum.

Solution:

Let the sum be \$ 100. Then,

SI = \$ (100 × 6 × 2/100) = \$ 12

and compound interest = \$ {100 × (1 + 6/100)² - 100}

= \$ {(100 × 53/50 × 53/50) - 100} = \$ (2809/25 - 100) = \$ 309/25

Therefore, (CI) - (SI) = \$ (309/25 – 100) = \$ 9/25

If the difference between the CI and SI is \$ 9/25, then the sum = \$ 100.

If the difference between the CI and SI is \$ 18, then the sum = \$ (100 × 25/9 × 18)
= \$ 5000.

Hence, the required sum is \$ 5000.

Alternative method

Let the sum be \$ P.

Then, SI = \$ (P × 6/100 × 2) = \$ 3P/25

And, CI = \$ {P × (1 + 6/100)² - P}

= \$ {(P × 53/50 × 53/50) - P} = \$ (2809250028092500P - P) = \$ (309P/2500)

(CI) - (SI) = \$ (309P/2500 – 3P/25) = \$ (9P/2500)

Therefore, 9P/2500 = 18

P = 2500 × 18/9

P = 5000.

Hence, the required sum is \$ 5000.

Question 3: A man deposited \$1000 in a bank. In return he got \$1331. Bank gave interest 10% per annum. How long did he keep the money in the bank?

Solution:

Let the required time be n years. Then,

Amount = \$ {1000 × (1 + 10/100)ⁿ}

= \$ {1000 × (11/10)ⁿ}

Therefore, 1000 × (11/10)ⁿ = 1331 [since, amount = \$ 1331 (given)]

(11/10)ⁿ = 1331/1000 = 11 × 11 × 11/ 10 × 10 × 10 = (11/10)³

(11/10)ⁿ = (11/10)³

n = 3.

Thus, n = 3.

Hence, the required time is 3 years.

Question 4: Find the compound interest on \$10,000 if Ron took loan from a bank for 1 year at 8 % per annum, compounded quarterly

Solution:
Here, P = principal amount (the initial amount) = \$ 10,000

Rate of interest (r) = 8 % per annum

Number of years the amount is deposited or borrowed for (n) = 1 year

Using the compound interest when interest is compounded quarterly formula, we have that

A = P(1 + r4100r4100)4n4n
= \$ 10,000 (1 + 8410084100)4∙14∙1
= \$ 10,000 (1 + 21002100)44
= \$ 10,000 (1 + 150150)44
= \$ 10,000 × (51505150)44
= \$ 10,000 × 51505150 × 51505150 × 51505150 × 51505150
= \$ 10824.3216
= \$ 10824.32 (Approx.)

Therefore, compound interest \$ (10824.32 - \$ 10,000) = \$ 824.32

Question 5: A certain sum amounts to \$ 72900 in 2 years at 8% per annum compound interest, compounded annually. Find the sum.

Solution:

Let the sum be \$ 100. Then,

amount = \$ {100 × (1 + 8/100)²}

= \$ (100 × 27/25 × 27/25) = \$ (2916/25)

If the amount is \$ 2916/25 then the sum = \$ 100.

If the amount is \$ 72900 then the sum = \$ (100 × 25/2916 × 72900) = \$ 62500.

Hence, the required sum is \$ 62500.

Alternative method

Let the sum be \$ P. Then,

amount = \$ {P × (1 + 8/100)²}

= \$ {P × 27/25 × 27/25} = \$ (729P/625)

Therefore, 729P/625 = 72900

P = (72900 × 625)/729

P = 62500.
Hence, the required sum is \$ 62500.

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Question 6: Find the amount and the compound interest on \$ 7,500 in 2 years and at 6% compounded yearly.

Solution:

Here,

Principal (P) = \$ 7,500

Number of years (n) = 2

Rate of interest compounded yearly (r) = 6%

A = P(1 + r100r100)nn

= \$ 7,500(1 + 61006100)22

= \$ 7,500 × (106100106100)22

= \$ 7,500 × 11236100001123610000

= \$ 8,427

Therefore, the required amount = \$ 8,427 and

Compound interest = Amount - Principal

= \$ 8,427 - \$ 7,500

= \$ 927

Question 7: Find the rate of compound interest for \$ 10,000 amounts to \$ 11,000 in two years.

Solution:

Let the rate of compound interest be r% per annum.

Principal (P) = \$ 10,000

Amount (A) = \$ 11,000

Number of years (n) = 2

Therefore,

A = P(1 + r100r100)nn

10000(1 + r100r100)22 = 11664

(1 + r100r100)22 = 11664100001166410000

(1 + r100r100)22 = 729625729625

(1 + r100r100)22 = (27252725)

1 + r100r100 = 27252725

r100r100 = 27252725 - 1

r100r100 = 225225

25r = 200

r = 8

Therefore, the required rate of compound interest is 8 % per annum.

Question 8: A sum of money becomes \$ 2,704 in 2 years at a compound interest rate 4% per annum. Find

(i) The sum of money at the beginning

(ii) The interest generated.

Solution:

Let the sum of money at the beginning = \$ P

Here,

Amount (A) = \$ 2,704

Rate of interest compounded yearly (r) = 4

Number of years (n) = 2

(i) A = P(1 + r100r100)nn

2,704 = P(1 + 41004100)22

2,704 = P(1 + 125125)22

2,704 = P(26252625)22

2,704 = P × 676625676625

P = 2,704 × 625676625676

P = 2,500

Therefore, the sum of money at the beginning was \$ 2,500

(ii) The interest generated = Amount – Principal

= \$2,704 - \$2,500

= \$ 204

Question 9: Find the amount and the compound interest on \$ 4,000 is 11212 years at 10 % per annum compounded half-yearly.

Solution:

Here, the interest is compounded half-yearly. So,

Principal (P) = \$ 4,000

Number of years (n) = 11212 × 2 = 3232 × 2 = 3

Rate of interest compounded half-yearly (r) = 102102% = 5%

Now, A = P (1 + r100r100)ⁿ

A = \$ 4,000(1 + 51005100³

A = \$ 4,000(1 + 120120)³

A = \$ 4,000 × (21202120)³

A = \$ 4,000 × 9261800092618000

A = \$ 4,630.50 and

Compound interest = Amount - Principal

= \$ 4,630.50 - \$ 4,000

= \$ 630.50

Therefore, the amount is \$ 4,630.50 and the compound interest is \$ 630.50

Question 10: Find the amount and the compound interest on \$ 8,000 at 10 % per annum for 11212 years if the interest is compounded half-yearly.

Solution:

Here, the interest is compounded half-yearly. So,

Principal (P) = \$ 8,000

Number of years (n) = 11212 × 2 = 3232 × 2 = 3

Rate of interest compounded half-yearly (r) = 102102% = 5%

Now, A = P (1 + r100r100)ⁿ

A = \$ 8,000(1 + 51005100)³

A = \$ 8,000(1 + 120120)³

A = \$ 8,000 × (21202120)³

A = \$ 8,000 × 9261800092618000
A = \$ 9,261 and

Compound interest = Amount - Principal

= \$ 9,261 - \$ 8,000

= \$ 1,261

Therefore, the amount is \$ 9,261 and the compound interest is \$ 1,261

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Final Note: