Problem Solving with Functions

 Daesun starts counting at 100, and he counts by fours: 100, 104, 108, . . . . Andrew
starts counting at 800, and he counts backwards by three: 800, 797, 794, . . . . They both start counting
at 1 PM, and say one number each minute. What time is it when Daesun first says a number that is
more than twice the number Andrew says?




Solution

In order to compare Daesun’s number to Andrew’s, we need an expression
for each in terms of the time. So, we define a function, D(x), for Daesun, and a function, A(x), for
Andrew:
Let D(x) be Daesun’s number x minutes after 1 PM.
Let A(x) be Andrew’s number x minutes after 1 PM.
Since Daesun starts at 100 and counts up by fours, we have
                            D(x) = 100 + 4x.
Since Andrew starts a 800 and counts down by threes, we have
                            A(x) = 800 − 3x.
We seek the first time such that
                            D(x) > 2A(x).
Our expressions for D(x) and A(x) give us
                            100 + 4x > 2(800 − 3x).
Solving this inequality gives us x > 150. The smallest such x is 151, so the first time Daesun says a
number that is more than twice Andrew’s number is 151 minutes after 1 PM, or 3:31 PM. 2