If the interest on a sum borrowed for certain period is reckoned uniformly, then it is called simple interest. Let the principal = P, Rate = R% per annum (p.a) and Time = T years. Then ,

1. S.I. = (P x R x T / 100)

2. P = (100 x S.I. / R x T)

3. R = (100 x S.I / P x T)

4. T = (100 x S.I. / P x R)

The money borrowed or lent out for a certain period is called the principal or the sum

2.Interest :Extra money paid for using other’s money is called interest.

Example 1:

What is the present worth of Rs. 132 due in 2 years at 5% simple interest per annum?

Solution:

Let the present worth be Rs.x

Then,S.I.= Rs.(132 - x)

=> (x× 5 × 2/100)

=> 132 - x

=> 10x = 13200 - 100x

=> 110x = 13200

x= 120.

Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. 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 B?

Solution:

Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x)

Then, [(x * 14* 2)/100]+[((13900-x)*11*2)/100] = 3508

28x - 22x = 350800 - (13900 x 22)

6x = 45000

x = 7500.

So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400

A sum of Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest?

Solution:

Let the original rate be R%. Then, new rate = (2R)%

Note: Here, original rate is for 1 year(s); the new rate is for only 4 months i.e. 1/3 year(s)

=> [(725*R*1)/100]+[(362.50*2R*1)/100*3] = 33.50

=> (2175 + 725) R = 33.50 x 100 x 3

=> (2175 + 725) R = 10050

=> (2900)R = 10050

=> R = 10050/2900 =3.46

Original rate = 3.46%

The price of a T.V set worth Rs.20,000 is to be paid in 20 instalments of Rs. 1000 each. If the rate of interest be 6% per annum, and the first instalments be paid at the time of purchase, then the value of the last instalments covering the interest as well will be:

Solution:

Monet paid in cash = Rs. 1000

Balance amount = Rs. (20000 - 1000)

=> Rs. 19000

- 1. Problems on Train
- 2. Time and Work
- 3. Percentage
- 4. Time and Distance
- 5. Problems on ages
- 6. Profit and Loss
- 7. Average
- 8. Permutation and Combination
- 9. Problems on H.C.F and L.C.M
- 10. Square Root and Cube Root
- 11. Chain Rule
- 12. Alligation or Mixture
- 13. Stocks and Shares
- 14. Banker's Discount
- 15. Simple Interest
- 16. Compound Interest
- 17. Partnership
- 18. Height and Distance
- 19.Calendar

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