How to Solve Investment Problems Part 3

By | August 31, 2015

This is the third part of the Solving Investment Problems Series. In this part, we discuss solve invest problem which is very similar to the second part. We discuss an investment at two different interests.

Problem

A government employee invested a part of Php60000 in bonds at 6% yearly interest and the remaining part in stocks at a 5% yearly interest. The annual interest in stocks is Php850 less than the annual interest in bonds. How much was invested at each rate?

Solution and Explanation

Let x = amount invested on bonds (6% yearly interest)
60,000 – x = amount invested on stocks (5% yearly interest).

The interest in bonds is 6% per year times the amount invested or

(0.06)(x)

when the percentage is converted to decimals.

In addition, the interest in stocks is 5% per year times the amount invested or

(0.05)(60000 – x)

when the percentage is converted to decimals.

Now, the interest in stocks is 850 less than the interest in bonds which means that if we subtract 850 from the interest in bonds they will be equal. That is

interest in bonds – 850 = interest in stocks.

Substituting the expressions of each, we have

(0.06)(x) – 850 = (0.05)(60000 – x).

Simplifying, we have

0.06x – 850 = 3000 – 0.05x.

To eliminate the decimal numbers, we multiply everything by 100. The equation becomes

6x – 85000 = 300000 – 5x

We simplify by adding 85000 and 5x to both sides. The equation becomes.

11x = 385000.

Dividing both sides by 11, we have
x = 35000.

So, 35000 was invested in bonds and 60000-35000 = 25000 was invested in stocks.

Check:

35000 × 0.06 = 2100
25000 × 0.05 = 1250

Indeed, the amount invested in bonds is 2100 – 1200 = 850 less than the interest in stocks.

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