COIN PROBLEM INVOLVING LINEAR EQUATIONS IN ONE VARIABLE

Mary has P10.50 in twenty-five centavo coins and ten centavo coins. She has 7 more ten centavo coins than twenty-five centavo coins. How many coins of each kind does she have?

Solution:

Let x = number of twenty-five centavo coins
x + 4 = number of ten centavo coins
.25x + .10(7 + x) = 10.50            - equation
100 [.25x + .10(7 + x)] = 10.50(100)
25x + 10(7 + x) = 1050
25x + 10x = 1050 - 70
35x = 980
35       35
x = 28                  - number of twenty-five centavo coins
x + 7 = 35            - number of ten centavo coins

Hence, there are 28 twenty-five centavo coins and 35 ten centavo coins amounting to P10.50.

Check:
.25(28) + .10(35) = 10.50
7 + 3.5 = 10.50
10.50 = 10.50