# Problem of the Week Problem E and Solution Red and Blue Chips

## Problem

Jane and Fred each have their own collection of red and blue bingo chips. The ratio of the number of Jane’s chips to the number of Fred’s chips is $$3:2$$. When they combine their chips, the ratio of the number of red chips to the number of blue chips is $$7:3$$. For Jane’s chips, the ratio of the number of red chips to the number of blue chips is $$4:1$$.

What is the ratio of the number of red chips to the number of blue chips for Fred’s bingo chips?

## Solution

Solution 1

Suppose that Jane and Fred have a total of $$100$$ bingo chips. (We may assume any convenient total number of chips.)

Since the ratio of the number of Jane’s chips to the number of Fred’s chips is $$3:2$$, then Jane has $$\frac{3}{5}$$ of the $$100$$ chips, or $$60$$ chips. Fred has the remaining $$40$$ chips.

When they combine their chips, the ratio of the number of red chips to the number of blue chips of $$7:3$$. Therefore, $$\frac{7}{10}$$ of the $$100$$ chips, or $$70$$ chips, are red and the remaining $$30$$ chips are blue.

For Jane’s chips, the ratio of the number of red chips to the number of blue chips is $$4:1$$, so $$\frac{4}{5}$$ of her $$60$$ chips, or $$48$$ chips, are red and the remaining $$12$$ chips are blue.

Since there are $$70$$ red chips in total, then Fred has $$70 - 48 = 22$$ red chips.
Since there are $$30$$ blue chips in total, then Fred has $$30 - 12 =18$$ blue chips.

Therefore, the ratio of the number of red chips to the number of blue chips for Fred’s chips is $$22 : 18 = 11 : 9$$.

Solution 2

Suppose that Jane and Fred have a total of $$x$$ chips.

Since the ratio of the number of Jane’s chips to the number of Fred’s chips is $$3:2$$, then Jane has $$\frac{3}{5}$$ of the chips, or $$\frac{3}{5}x$$ chips. Fred has the remaining $$\frac{2}{5}x$$ chips.

When they combine their chips, the ratio of the number of red chips to the number of blue chips is $$7:3$$. Therefore, $$\frac{7}{10}x$$ chips are red and the remaining $$\frac{3}{10}x$$ chips are blue.

For Jane’s chips, the ratio of the number of red chips to the number of blue chips is $$4:1$$, so $$\frac{4}{5}$$ of her $$\frac{3}{5}x$$ chips, or $$\frac{4}{5} \left( \frac{3}{5}x \right)= \frac{12}{25}x$$, are red and the remaining $$\frac{3}{5}x -\frac{12}{25}x = \frac{3}{25}x$$ chips are blue.

Since there are $$\frac{7}{10}x$$ red chips in total, then Fred has $$\frac{7}{10}x - \frac{12}{25}x = \frac{11}{50}x$$ red chips.

Since there are $$\frac{3}{10}x$$ blue chips in total, then Fred has $$\frac{3}{10}x - \frac{3}{25}x = \frac{9}{50}x$$ blue chips.

Therefore, the ratio of the number of red chips to the number of blue chips for Fred’s chips is $$\frac{11}{50}x : \frac{9}{50}x = 11 : 9$$.