A bag contains 10 white gold balls and 6 striped gold balls. A golfer wants to add 112 golf balls to the bag. He wants the ratio of whit to striped golf balls to remain the same. How many of each should he add?

Answer 1

Let's say the golfer adds $w$ white golf balls and $s$ striped golf balls to the bag.

The current ratio of white golf balls to striped golf balls is $10/6 = 5/3$.

The ratio after adding $w$ white golf balls and $s$ striped golf balls should still be $5/3$.

So, we can set up the following equation based on the problem statement:

\[
\frac{10+w}{6+s} = \frac{5}{3}
\]

Let's cross multiply and simplify:

\[
3(10+w) = 5(6+s)
\]
\[
30 + 3w = 30 + 5s
\]
\[
3w = 5s
\]

Since we want to add a total of 112 golf balls, we also know that $w + s = 112$.

We have two equations with two variables:

\begin{align*}
3w &= 5s \\
w + s &= 112
\end{align*}

Multiplying the second equation by 3, we can substitute the value of $112 - s$ for $w$:

\begin{align*}
3w &= 5s \\
3(112-s) + s &= 336 \\
336 - 3s + s &= 336 \\
-2s &= 0 \\
\end{align*}

This gives us $s = 0$, which is not a valid solution for this problem. It means there are already enough white golf balls in the bag to maintain the ratio.

Therefore, the golfer should add 0 white golf balls and 112 striped golf balls.