Hello, Kyle!
This is a tricky one ... different from most "work" problems.
One painter works 1 1/2 times as fast as another.
Find their individual rates for painting a room if together it takes them 4 hour.
Let "A" be the faster painter, "B" be the slower one.
A's painting speed is \(\displaystyle 1\frac{1}{2}\) times faster than B's.
. . This means that A can paint the room in
two-thirds of the time that B takes.
.*
Let \(\displaystyle x\) = time it takes B to paint the room alone.
. . In one hour, B paints
.\(\displaystyle \frac{1}{x}\) of the room.
. . In 4 hours, B paints
.\(\displaystyle 4\,\times\,\frac{1}{x}\:=\:\frac{4}{x}\) of the room.
Let \(\displaystyle \frac{2}{3}x\) = time it takes A to paint the room alone.
. . In one hour, A paints
.\(\displaystyle \frac{1}{\frac{2}{3}x}\,=\;\frac{3}{2x}\) of the room.
. . In 4 hours, A paints
.\(\displaystyle 4\,\times\,\frac{3}{2x}\:=\:\frac{6}{x}\) of the room.
So, in 4 hours, they paint: \(\displaystyle \frac{4}{x}\,+\,\frac{6}{x}\,=\,\frac{10}{x}\) of the room . . . or "one whole room".
There is our equation:
. \(\displaystyle \frac{10}{x}\:=\:1\)
Therefore: B takes 10 hours to paint the room alone
. . . . . .and A takes \(\displaystyle \frac{20}{3}\) hours to paint the room alone.
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* . I'll illustrate this reasoning.
Suppose B must drive 120 miles . . . and his speed is 40 mph.
. . B's trip will require 3 hours.
Suppose A' must drive 120 miles . . . and his speed is \(\displaystyle 1\frac{1}{2}\) times B's speed.
. . A's speed is: \(\displaystyle \frac{3}{2}\,\times\,40\:=\:60\) mph.
. . A's trip will require 2 hours.
See? . . . It takes A only <u>two-third</u> of B's time.
