2. 30,000.
Problems on Simple Interest:
1. A sum of 500 is lent at the beginning of a year at a certain rate of interest. After 9 months, a sum of 400 more is lent but at rate thrice the former. At the end of the year, 50 is earned as interest from both the loans. What was the original rate of interest?
Solution:
(500*R*1)/100 + (400*3R*3)/100*4 = 50
R+9R = 50
R = 5%.
2. The difference between the simple interest received from 2 different sources on 1000 for 2 years is 10. The difference between their rates of interest is:
Solution:
(1000*R1*2)/100 –(1000*R2*2)/100 = 10
R1-R2 = 0.5%
Comments
Post a Comment