## Arithmatical - Simple Interest

DIRECTIONS : Problems based on Simple Interest.
1. A man took a loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was A.  Rs.2000 B.  Rs.10,000 C.  Rs.15,000 D.  Rs.20,000
Solution
 Principal = Rs. (100×5400 / 12×3) ‹=› Rs. 15,000.
2. 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 A.  Rs.1050 B.  Rs.2050 C.  Rs.3000 D.  Rs.19000
Solution
 Balance amount Monet paid in cash = Rs. 1000. = Rs. (20000 - 1000) ‹=› Rs. 19000.

3. A lent Rs.5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs.2200 in all from both of them as interest. The rate of interest per annum is A.  5% B.  7% C.  8% D.  10%
Solution
 Let the rate be R% p.a. Then, (5000xRx2/100) + (3000xRx4/100) ‹=›100R+120R= 2200 ‹=›R=(2200/220) Rate ‹=›10%.
4. A money lender finds that dues to a fall in the annual rate of interest from 8% to 7x3/4%, his yearly income diminishes by Rs.61.50. His capital is A.  Rs.22,400 B.  Rs.23,800 C.  Rs.24,600 D.  Rs.26,000
Solution
 Let the capital be Rs. x. Then, = (x × 8×1/100) - (x × 31/4×1/100) = 61.50. ‹=›32x - 31x =6150×4 ‹=›x= 24600.
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