This is the second part of the Solving Investment Problems Series. In the first part, we discussed in detail the solution of a problem at two different rates of interest. In this post, we discuss another problem.
Mr. Reyes invested a part of Php70000 at 3% yearly interest and the remaining part at a 5% yearly interest. The annual interest on the 3% investment is Php100 more than the annual interest on the 5% investment. How much was invested at each rate?
If we let x be the amount invested at 3%, then, 70000 – x is the amount invested at 5%. The yearly interest is the product of the rates and the amount invested so,
(3%)(x) = yearly interest of the amount invested at 3%
(5%)(70000 – x) = yearly interest of the amount invested at 5%
Now, the annual interest at 3% is 100 more than the annual interest at 5%. This means that if we add 100 to the yearly interest at 5%, the interests will be equal. That is,
(3%)(x) = (5%)(70000 – x) + 100.
Next, we convert percent to decimal by dividing the percentage by 100. So,
(0.03)(x) = (0.05)(70000 – x) + 100.
Simplifying, we have
0.03x = 3500 – 0.05x + 100
0.03x = 3600 – 0.05x
0.03x + 0.05x = 3600
0.08x = 3600.
Tip: You can calculate better by eliminating the decimal. You can do this by multiplying both sides by 100.
Dividing both sides by 0.08, we have
x = 45000.
This means that 45000 is invested at 3% yearly interest.
Now, the remaining amount is 70000 – 45000 = 25000.
This means that 25000 is invested at 5%.
Yearly interest of 45000 at 3% interest = (45000 x 0.03) = 1350
Yearly interest of 25000 at 5% interest = (25000 x 0.05) = 1250
As we can see, the interest at 45000 at 3% interest is 100 more than the interest of 25000 at 5% interest.
Therefore, we are correct.