Quantitative Aptitude - SPLessons

Simple Interest

Chapter 22

SPLessons 5 Steps, 3 Clicks
5 Steps - 3 Clicks

Simple Interest

shape Introduction

Simple Interest deals with the following terms: principal, interest, rate of interest, time, amount. The term Intrtest is actually one of the most fundamental business terms, and without it, the financial trading of the world would come to standstill. Interest is defined as the “time value of money”.


shape Methods

1. Simple interest(S.I): If the interest on a sum borrowed for a certain period is calculated uniformly, then it is called simple interest.


2. Principal(P): The money borrowed or lent out for a certain period is called principal.


3. Interest(I): Paying of extra money for using other’s money is called interest.


4. Rate(R): A rate of interest is the rate at which the interest is calculated on the principal amount and it is a special ratio in which the two terms are in different units.


5. Time(T): The specific period for which the money is borrowed is called as time/ time period.


6. Amount(A): The money to be assured or matured or paid finally is called amount which is equal to the principal+simple interest.

Example 1
Find the simple intrest on Rs. 68,000 at 16\(\frac{2}{3}\)% pen annum for 9 months.


Solution:

    P = Rs. 68000, R = \(\frac{50}{3}\)% p.a and T = \(\frac{9}{12}\) years = \(\frac{3}{4}\) years.

    ∴ S.I = (\(\frac{P \times R \times T}{100}\)) = Rs. (68000 x \(\frac{50}{3}\) x \(\frac{3}{4}\) x \(\frac{1}{100}\)) = Rs. 8500.


Example 2
Find the simple intrest on Rs. 3000 at 6\(\frac{1}{4}\)% per annum for the period from 4th Feb, 2005 to 18th April, 2005.


Solution:

    Time = (24 + 31 + 18) days = 73 days = \(\frac{73}{365}\) year = \(\frac{1}{5}\) year.

    P = Rs. 3000 and R = 6\(\frac{1}{4}\)% p.a. = \(\frac{25}{4}\)% p.a.

    ∴ S.I = Rs. (3000 x \(\frac{25}{4}\) x \(\frac{1}{5}\) x \(\frac{1}{100}\)) = Rs. 37.50.

    Remark: The day on which money is deposited is not counted while the day on which money is withdrawn is counted.


Example 3
If Rs. 64 amounts to Rs. 83.20 in 2 years, what will Rs. 86 amount to in 4 yewars at the same rate percent per annum?


Solution:

    P = Rs. 64, S.I. = Rs. (83.20 – 64) = Rs. 19.20, T = 2 years.

    So, rate = (\(\frac{100 \times 19.20}{64 \times 2}\))% = 15%.

    Now, P = Rs. 86,. R = 15%, T = 4 years.

    ∴ S.I. = Rs. (\(\frac{86 \times 15 \times 4}{100}\)) = Rs. 51.60


Example 1
A man took loan from a bank at the rate of 12% p.a. simple intrest. After 3 years he had to pay Rs.5400 intrest only for the period. The principal amount borrowed by him was:


Solution:

    Principal = Rs. (\(\frac{100 \times 5400}{12 \times 3}\)) = Rs. 15000.


Example 2
A sum fetched a total simple intrest of Rs. 4016.25 at the rate of 9 p.c.p.a in 5 years. What is the sum?


Solution:

    Principal = Rs. (\(\frac{100 \times 4016.25}{9 \times 5}\)) = Rs. (\(\frac{401625}{45}\)) = Rs. 8925.


Example 1
A certain sum of money amounts to Rs. 1008 in 2 uears and to Rs. 1164 in 3\(\frac{1}{2}\) years. Find the sum and the rate of intrest.


Solution:

    S.I for 1\(\frac{1}{2}\) years = Rs. (1164 – 1008) = Rs. 156.

    S.I for 2 years = Rs. (156 x \(\frac{2}{3}\) x 2) = Rs. 208.

    ∴ Principle = Rs. (1008 – 208) = Rs. 800.

    Now, P = 800, T = 2 and S.I. = 208.

    Rate = (\(\frac{100 \times 208}{800 \times 2}\)) % = 13%.


Example 2
At what rate percent per annum will a sum of money double in 15 years?


Solution:

    Let principal = P. Then, S.I. = P and T = 16 yrs.

    ∴ Rate = (\(\frac{100 \times P}{P \times 16}\))% = 6\(\frac{1}{4}\)% p.a.


Example 1
In how many years, Rs. 150 will produce the same intrest @ 8% as Rs. 800 produce in 3 years @ 4\(\frac{1}{2}\)%?


Solution:

    P = Rs. 800, R = 4\(\frac{1}{2}\)% = \(\frac{9}{2}\)%, T = 3 years. Then,

    S.I. = Rs. (800 x \(\frac{9}{2}\) x \(\frac{3}{100}\)) = Rs. 108.

    Now, P = Rs. 150, S.I. = Rs. 108, R = 8%.

    ∴ Time = (\(\frac{100 \times 108}{150 \times 8}\)) years = 9 years.


Example 2
In how many years will a sum of money double itself at 12% per annum?


Solution:

    Let Sum = \(x\). Then, S.I = \(x\).

    ∴ Time = (\(\frac{100 \times S.I.}{P \times R}\)) = (\(\frac{100 \times x}{x \times 12}\)) years = 8\(\frac{1}{3}\) years = 8 years 4 months.


Example 1
A sum at simple intrest at 13\(\frac{1}{2}\)% per annum amounts to Rs. 2502.50 after 4 years. Find the sum.


Solution:

    Let sum be Rs. \(x\). Then, S.I. = Rs. (\(x\) x \(\frac{27}{2}\) x 4 x \(\frac{1}{100}\)) = Rs. \(\frac{27x}{50}\).

    ∴ Amount = Rs. (\(x\) + \(\frac{27x}{50}\)) = Rs. \(\frac{77x}{50}\).


Example 2
A sum invested at 5% simple intrest per annum grows to Rs. 504 in 4 years. The same amount at 10% simple intrest per annum in 2\(\frac{1}{2}\) years will grow to:


Solution:

    Let the sum be Rs. \(x\). Then, S.I. = Rs. (504 – \(x\)).

    ∴ (\(\frac{x \times 5 \times 4}{100}\)) = 504 – \(x\) ⇒ 20\(x\) = 50400 – 100\(x\) ⇒ 120\(x\) = 50400 ⇒ \(x\) = 420.

    Now, P = Rs. 420, R = 10%, T = \(\frac{5}{2}\) years.

    S.I. = Rs. (\(\frac{420 \times 10}{100}\) x \(\frac{5}{2}\)) = Rs. 105.

    ∴ Amount = Rs. (\(\frac{420 \times 10}{100}\) x \(\frac{5}{2}\)) = Rs. 105.

    Amount = Rs. (420 + 105) = Rs. 525.

shape Formulae

1. Simple interest(S.I) = \(\frac{principal(P) \times rate(R) \times time(T)}{100}\)

2. Principal(P) = \(\frac{100 \times Simple interest}{Rate \times time}\)

3. Rate = \(\frac{100 \times Simple interest}{principal \times time}\)

4. Time = \(\frac{100 \times Simple interest}{principal \times rate}\)

5. Amount = Principal+ simple interest (or) Amount = P(1 + \(\frac{R \times T}{100}
\))

shape Samples

1. Find the principal on a certain sum of money at 5% per annum for 2 \(\frac{2}{5}\) years if the amount being Rs. 1120?


Solution:

    Given,

    Amount = Rs. 1120

    Rate = 5 %

    Time = \(\frac{12}{5}\) years

    Now, Amount = P(1 + \( \frac{R * T}{100} \))

    ⇒ 1120 = P(1 + \( \frac{5 * \frac{12}{5}}{100} \))

    ⇒ P = 1000

    Therefore, principal = Rs. 1000


2. Find the simple interest on Rs. 625 at 6\(\frac{1}{2}\) % per annum for 2\(\frac{1}{2}\) years?


Solution:

    Given,

    Principal= Rs. 625

    Rate = \(\frac{13}{2}\)

    Time = \(\frac{5}{2}\)

    Now, Simple interest(S.I.) = \(\frac{P * R * T }{100}\)

    S.I. = \(\frac{625 * 13 * 5}{100}\)

    S.I. = Rs. 101.56


3. A shopkeeper borrowed Rs. 25000 from the two money-lenders. For one loan paid 12 % per annum and for other 14 % per annum. The total interest paid for one year was Rs. 3260. How much did shopkeeper borrow at each rate?


Solution:

    Suppose money borrowed at 12 % = Rs.\( x \)

    Then, the money borrowed at 14 % = Rs. (25000 – \( x \))

    Therefore,

    \(\frac{x * 21 * 1}{100}\) = \(\frac{(25000 – x) * 14 * 1}{100}\) = 3260

    ⇒ 12\( x\) + 350000 – 14 \(x\) = 326000

    ⇒ 2\(x\) = 24000

    ⇒ \(x\) = 12000

    Money borrowed at 12 % = Rs. 12000

    Money borrowed at 14 % = Rs. 13000


4. What installment will discharge a debt of Rs. 1092 due in 3 years at 12 % S.I.?


Solution:

    Let each installment be \(x\),

    Then,

    (\( x + \frac{x * 12 * 1}{100}\)) + (\( x + \frac{x * 12 * 2}{100}\)) + \(x\) = 1092

    ⇒ \(\frac{28x}{25}\) + \(\frac{31x}{25}\) + \(x\)

    ⇒ \(x\) = 325

    Therefore, each installment = Rs. 325


5. A sum was put at simple interest at a certain rate for 2 years. Had it been put at 3 % higher rate, it would have fetched Rs. 300 more. Find the sum?


Solution:

    Let,

    sum = \(x\), original rate = R%

    Then,

    \(\frac{x * (R + 3) * 2}{100}\)

    ⇒ \(\frac{x * R * 2}{100}\)

    ⇒ 2 R\(\)x\(\) + 6\(\)x\(\) – 2R\(\)x\(\) = 300 x 100

    ⇒ \(\)x\(\) = 5000

    Therefore, sum = Rs. 5000