# How to Solve Investment Problems Part 2

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.

**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?

**Solution**

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%.

**Check:**

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.