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
x + 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.
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