Question

Luke has blue and red balls.Every day, he wins 2 blue balls and loses 3 red ones.After 5 days, he has the same amount of blue and red balls.After 9 days, he has twice as many blue balls as red balls.How many red balls did he have at the beginning?

Answer 1

Given that;

Luke has blue and red balls.

And, every day, he wins 2 blue balls and loses 3 red ones.

Let a represent the blue balls and b represent the read balls

After 5 days, he has the same amount of blue and red balls, i.e.

For 5 days, the number of blue balls won and red balls lost by Luke will be

`\begin{gathered} For\text{ the blue balls won:} \\ 2*5=10 \\ For\text{ the red balls lost:} \\ 3*5=15 \end{gathered}`

After 5 days, he has the same amount of blue and red balls,

The number of blue and red balls will be

`\begin{gathered} a+10=b-15 \\ a-b=-10-15 \\ a-b=-25...(1) \end{gathered}`

After 9 days, he has twice as many blue balls as red balls.

For 9 days, the number of blue balls won and red balls lost by Luke will be

`\begin{gathered} For\text{ the blue balls won:} \\ 2*9=18 \\ For\text{ the red balls lost:} \\ 3*9=27 \end{gathered}`

After 9 days, he has twice as many blue balls as red balls.

The number of blue and red balls will be

`\begin{gathered} a+18=2(b-27) \\ a+18=2b-54 \\ a-2b=-18-54 \\ a-2b=-72...(2) \end{gathered}`

Solving the equation simltaneously, to find b (the number of red balls)

Appplying the elimonation method

`\begin{gathered} a-b=-25 \\ - \\ a-2b=-72 \\ -b+2b=-25+72 \\ b=72-25 \\ b=47 \end{gathered}`

Hence, the number of red balls, b, at the beginning is 47