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 
Monet paid in cash  = Rs. 1000. 
Balance amount  = 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. 
