How To Solve Quickly Simple Interest Questions

July 14, 2023

How to Solve Simple Interest Questions Quickly in Aptitude

As we know that Simple Interest is the most important topic in mathematics. But it is important to know that the topic consists multiple level of Questions. Let us review thoroughly How To Solve Simple Interest Questions Quickly.

What is considered Simple Interest? Simple interest is when interest is calculated only on the initial money put in (the principal). In addition, the interest is not continually added to this total.

Definition of Simple Interest

If the interest on a sum borrowed for certain period is calculated uniformly, then it is called simple interest.
It is simply obtained by multiplying principal amount with rate and with given time interval.
Formula of Simple Interest = $\mathbf{\frac{P * R * T}{100}}$

Example – What is simple interest of Rs 5000/- for 5 years at 5% interest per annum.

Solution – SI= $\mathbf{\frac{P * R * T}{100}}$

= $\frac{5000 *5 *5}{100}$

=Rs 1250.

How to Solve for Interest Rate?

The above equation can be used to solve for any of the variables: interest, principal, rate, or time. To solve the equation the known information needs to be plugged in and then solve for the unknown variable. The equation can be rearranged so each variable is set to equal the known variables:

Solve for interest earned (initial equation): $I = P\times R \times T$
Solve for principal: $P = \frac{I}{R \times T}$
Solve for interest rate: $R = \frac{I}{P \times T}$
Solve for time: $T = \frac{I}{P \times R}$

Type 1: How to Solve Problems On Simple Interest (SI)

Question 1. Find the simple interest on Rs. 60,000 at 13/5 % p.a. for a period of 9 months?

Options:

A. 1170

B. 1710

C. 11700

D. 1017

Solution: We know that, SI = $\mathbf{\frac{P * R * T}{100}}$

P = 60000

R = $\frac{13}{5}$ %

T = 9 months = $\frac{ 3}{4}$ th year

SI = $\frac{60000 * 13 * 3}{5 * 4 * 100}$

SI = $\frac{117000}{100}$

SI = 1170

Correct option: A

Question 2. What will be the ratio of simple interest earned on a certain amount at the same rate of interest for 6 years and that for 24 years?

Options:

A. 1:2

B. 3:5

C. 1:3

D. 1:4

Solution: Required Ratio = Simple Interest for 6 years/Simple Interest for 24 years

= $\frac{T_{1}}{T_{2}}$

= $\frac{6}{24}$

= $\frac{1}{4}$

=1:4

Correct option: D

Question 3.Find the simple interest on Rs.500 for 10 months at 5 paisa per month?

Options:

A. Rs. 2500

B. Rs. 250

C. Rs. 25

D. Rs. 25.5

Solution: SI = $\mathbf{\frac{P * R * T}{100}}$

SI = $\frac{500 * 5 * 10}{100}$

= Rs. 250

Correct option: B

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Type 2: Problems On Simple Interest (SI) When Rates are different for different years

Question 1. Nisha borrowed some money at the rate of 5% p.a. for the first two years. She again borrowed at the rate of 10% p.a. for the next three years. Later at the rate of 15% p.a. for the rest of the years. Total interest paid by her was Rs. 15000 at the end of 9 years. Calculate the amount of money she borrowed?

Options:

A. Rs. 15005
B. Rs. 20000
C. Rs. 15000
D. Rs. 10000

Solution: According to the question,

r₁ = 5% , T₁ = 2 years

r₂ = 10% , T₂ = 3 years

r₃ = 15% , T₃ = 4 years

(since, beyond 5 years rate is 15%)

Simple interest = 15000

Therefore, P = $\frac{15000 * 100}{5*2 +10*3 +15*4 }$

= $\frac{1500000}{10 + 30 + 60 }$

= $\frac{1500000}{100}$

= Rs. 15000

Correct option: C

Question 2. Rahul invests some amount of money in three different schemes for 5 years, 10 years and 15 years at 10%, 12% and 15% Simple Interest respectively. At the completion of each scheme, he gets the same interest. Find out the ratio of his investment?

Options:

A. 3: 9: 15
B. 10: 24: 45
C. 10: 24: 40
D. 9: 24: 45

Solution: If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained at simple interest on each part where interest rates are 10 %, 12 %,15% respectively and time periods are 5 years , 10 years , 15 years respectively.

Let the three amounts be Rs. x, Rs. y and Rs. z,
Then , According to question

$\frac{x × 10 × 5}{100} = \frac{y× 12 × 10}{100} = \frac{z × 15 × 15}{100}$
50 x = 120 y = 225 z = k(say)

10 x = 24 y = 45 z = k

\frac{k}{10}

\frac{k}{24}

\frac{k}{45}

Hence, the ratio of his investment will be

\frac{k}{10}

\frac{k}{24}

\frac{k}{45}

10 : 24 : 45

Correct option: B

Question 3. Ram tool a loan for Rs.10000 from bank for a period of 3 years. The bank charged the interest rates as 5% for first year, 7% for second year and 9% for the third year. Find the amount he has to pay back to the bank after three years?

Options:

A. Rs. 18500
B. Rs. 145000
C. Rs. 15100
D. Rs 12100.

Solution: Simple Interest = 5% + 7% + 9% = 21%.

Amount = Principal + Rate of Interest.

Principal is 100% of the amount.

Therefore, Amount = 100% + 21%.= 121%

According to the question, total amount to be paid

A = $\frac{121 * 10000}{100}$

A = 12100

Therefore, the amount that Ram has to pay back to the bank after three years Rs 12100.

Correct option: D

Type 3: How to Solve Problems On Simple Interest (SI) to find the Rate of Interest, Time period and Principal

Question 1. A sum of money becomes six times in 30 years. Calculate the rate of interest.

Options:

A. 16.66%

B. 4.5%

C. 15.45%

D. 15%

Solution: We know that, if sum of money becomes x times in n years at some rate of interest, then rate of interest is calculated as,

R = 100 ( $\frac{x-1}{n}$ ) %

R = 100 ( $\frac{6-1}{30}$ )

R = 100 ( $\frac{5}{30}$ )

R = $\frac{500}{30}$

R = 16.66%

Correct option: A

Question 2. How much time will it take for an amount of Rs. 630 to yield Rs. 72 as interest at 5.4 % p.a. of simple interest?

Options:

A. 2 years and 10 months

B. 1 years and 11 months

C. 2 years and 11 months

D. 1 years and 11 months

Solution: We know that, Time = $\frac{100 * SI }{R * P }$

T = $\frac{100 * 72}{630 * 5.4}$

T = 7200/3402 = $\frac{7200}{3402}$

T = 2 years and 11 months

Correct option: C

Question 3. Mr. Tata invested Rs. 13,900 in two different schemes I and II. The rate of interest for both the schemes were 14% and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme II?

Options:

A. Rs. 2400

B. Rs. 6200

C. Rs. 6400

D. Rs. 4600

Solution: Let the amount invested in schemes I = x

Therefore, in schemes II = (13900 – x)

SI (scheme I) = $\frac{P * R * T}{100}$

Then For scheme I,

SI = $\frac{x * 14 * 2}{100}$

Then For scheme II,

SI (scheme II) = $\frac{(13900-x) * 11 * 2}{100}$

SI (scheme I) + SI (scheme II) = 3508

$\frac{( x * 14 * 2)}{100}+\frac{((13900-x) * 11 * 2)}{100}$ = 3508

\frac{28x}{100}+\frac{(13900 – x) * 22}{100}

= 3508

6x = 45000

x = 45000/6

x = 7500

Hence, the sum invested in Scheme II = 13900 – 7500 = Rs. 6400

Correct option: C

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