## Saturday, March 9, 2013

### Money - Investment Problem, 3

Category: Algebra

"Published in Newark, California, USA"

Noel invested ₱ 25,000 in 2 business ventures one earning 6% and the other 8% at the end of the year. If his proceeds amounted to ₱ 1,900.00, how much did he invest in each?

Solution:

The given word problem is about money and investment problem. Let's analyze the given word problem as follows

Let x = amount of money invested in business A
y = amount of money invested in business B
₱ 25,000.00 = total amount of money invested
6% = annual interest in business A
8% = annual interest in business B
0.06x = annual earning in business A
0.08y = annual earning in business B
₱ 1,900.00 = annual total earning

From the word statement, "Noel invested ₱ 25,000 in 2 business ventures....", then the working equation will be

x + y = 25,000

From the word statement, "If his proceeds amounted to ₱ 1,900.00.....", then the working equation will be

0.06x + 0.08y = 1,900

We can solve for the value of x and y using the two equations, two unknowns. Consider the first equation

+ y = 25,000

or           y = 25,000 - x

Substitute the value of y to the second equation, we have

0.06x + 0.08y = 1,900

0.06x + 0.08(25,000 - x) = 1,900

0.06x + 2,000 - 0.08x = 1,900

- 0.02x = 1,900 - 2,000

- 0.02x = - 100

x = 5,000

Substitute the value of x to the first equation, we have

y = 25,000 - x

y = 25,000 - 5,000

y = 20,000

Therefore, Noel invested ₱ 5,000.00 in business A and ₱ 20,000.00 in business B.

Note: The monetary sign, ₱ means Philippine Pesos and the word problem was in 1964.