ICSE Class 8 Maths Chapter 11 Interest Simple and Compound

Chapter 11: Interest

Simple And Compound

11.1 Review

The interest (I) depends on: Principal (P), Rate or rate percent (R), and The time (T).

The formula for calculating interest is \[I = \frac{P \times R \times T}{100}\]

Since, Amount = Principal + Interest

\[\Rightarrow A = P + I\]

\[\Rightarrow A = P + \frac{P \times R \times T}{100} \text{ i.e. } A = P \left( 1 + \frac{RT}{100} \right)\]

Teacher's Note

When you borrow money from a bank for education or a home, the bank charges interest. Understanding how much extra you will pay helps you make smart financial decisions about loans.

Example 1

Find the simple interest on 1,300 from December 23, 2002 to May 18, 2003 at \[7\frac{1}{2}\]% per annum.

Solution

Given: P = 1,300 and R = \[\frac{15}{2}\]%

Also, T = 146 days = \[\frac{146}{365}\] years = \[\frac{2}{5}\] years

To calculate time (T):

December: 8 days (31 - 23)

January: 31 days

February: 28 days

March: 31 days

April: 30 days

May: 18 days

Total = 146 days

\[\therefore S.I. = \frac{1,300 \times 15 \times 2}{100 \times 2 \times 5} = 39\] (Answer)

Note: The day on which the money is borrowed is not included in the time.

Note: The day on which the money is paid back to the money lender is included in the time.

11.2 To Find The Principal (P), The Rate Per Cent (R) and The Time (T)

The formula for interest, \[I = \frac{P \times R \times T}{100}\] can be re-written as:

(i) \[P = \frac{100 \times I}{R \times T}\] i.e. Principal = \[\frac{100 \times \text{Interest}}{\text{Rate} \times \text{Time}}\]

(ii) \[R = \frac{100 \times I}{P \times T}\] i.e. Rate = \[\frac{100 \times \text{Interest}}{\text{Principal} \times \text{Time}}\]

(iii) \[T = \frac{100 \times I}{P \times R}\] i.e. Time = \[\frac{100 \times \text{Interest}}{\text{Principal} \times \text{Rate}}\]

Test Yourself

1. If interest (I) = 318.50, time (T) = \[3\frac{1}{2}\] years and rate (R) = 2.6%, Principal (P) = _________________

2. If P = 6,000, I = 2,100 and T = 5 years. Rate (R) = __________________% = __________%

3. If P = 650, I = 312 and R = 6%, Time (T) = ___________________ = __________

Teacher's Note

Being able to calculate the principal, rate, or time helps you work backwards from interest amounts, useful when comparing different loan or investment offers.

Exercise 11 (A)

1. Find the interest and the amount on:

(i) 750 in 3 years 4 months at 10% per annum.

(ii) 5,000 at 8% per year from 23rd December 2011 to 29th July 2012.

(iii) 2,600 in 2 years 3 months at 1% per month.

(iv) 4,000 in \[1\frac{3}{4}\] years at 2 paise per rupee per month.

2. Rohit borrowed 24,000 at 7.5 per cent per year. How much money will he pay at the end of 4 years to clear his debt?

3. The interest on a certain sum of money is 1,480 in 2 years and at 10 per cent per year. Find the sum of money.

4. On what principal will the simple interest be 7,008 in 6 years 3 months at 5% per year?

5. Find the principal which will amount to 4,000 in 4 years at 6.25% per annum.

6. (i) At what rate per cent per annum will 830 produce an interest of 126 in 4 years?

(ii) At what rate per cent per year will a sum double itself in \[6\frac{1}{4}\] years?

7. (i) In how many years will 950 produce 399 as simple interest at 7%?

(ii) Find the time in which 1,200 will amount to 1,536 at 3.5% per year.

8. The simple interest on a certain sum of money is \[\frac{3}{8}\] of the sum in \[6\frac{1}{4}\] years. Find the rate per cent charged.

9. What sum of money borrowed on 24th May will amount to 10,210.20 on 17th October of the same year at 5 per cent per annum simple interest.

10. In what time will the interest on a certain sum of money at 6% be \[\frac{5}{8}\] of itself?

11. Ashok lent out 7,000 at 6% and 9,500 at 5%. Find his total income from the interest in 3 years.

12. Raj borrows 8,000; out of which 4,500 at 5% and remaining at 6%. Find the total interest paid by him in 4 years.

13. Mohan lends 4,800 to John for \[4\frac{1}{2}\] years and 2,500 to Shyam for 6 years and receives a total sum of 2,196 as interest. Find the rate per cent per annum, it being the same in both the cases.

14. John lent 2,550 to Mohan at 7.5 per cent per annum. If Mohan discharges the debt after 8 months by giving an old black and white television and 1,422.50. Find the price of the television.

Example 2

Find the rate of interest per year, if the interest charged for 8 months be 0.06 times of the money borrowed.

Solution

Let the money borrowed be 100 i.e. P = 100

Given: Interest (I) charged = 0.06 × 100 = 6 and T = \[\frac{8}{12}\] years = \[\frac{2}{3}\] years

\[\therefore \text{Rate} = \frac{I \times 100}{P \times T} = \frac{6 \times 100}{100 \times \frac{2}{3}} \% = 9\%\] (Answer)

Example 3

A sum of money lent out at 9 per cent for 5 years produces twice as much interest as 4,800 in \[4\frac{1}{2}\] years at 10 per cent. Find the sum.

Solution

Let the required sum be x. According to the given statement:

\[\frac{x \times 9 \times 5}{100} = 2 \times \frac{4,800 \times 10 \times 9}{100 \times 2}\]

On solving, we get: x = 9,600

\[\therefore\] The required sum = 9,600 (Answer)

Example 4

A certain sum amounts to 9,440 in 3 years and to 10,400 in 5 years. Find the sum and the rate per cent.

Solution

Amount in 3 years = 9,440 => P + I of 3 years = 9,440 .......(I)

Amount in 5 years = 10,400 => P + I of 5 years = 10,400 .......(II)

\[\therefore\] Eq. II - Eq. I => Interest of 2 years = 10,400 - 9,440 = 960

\[\Rightarrow \text{Interest of 1 year} = \frac{960}{2} = 480\]

And, Interest of 3 years = 480 × 3 = 1,440

From equation I, we get:

P + 1,440 = 9,440 => P = 8,000

Taking P = 8,000, I = 480 and T = 1 year

We get, \[\text{rate} = \frac{I \times 100}{P \times T} \% = \frac{480 \times 100}{8000 \times 1} \% = 6\%\]

\[\therefore\] The sum = 8,000 and rate per cent = 6% (Answer)

Teacher's Note

Problems like these teach you to think backwards and solve for unknowns, skills that help when comparing the actual cost of loans or investments with different terms and rates.

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