+77 Bob used 1/3 of his money to buy 4 similar caps and 5 similar pairs of shorts. He used 2/3 of the remaining money to buy another 16 caps. A pair of shorts cost $30 more than a cap.
(a) What fraction of the money did he have left?
(b) How much money did Bob have at first?
+394 A)
After Bob spent \(\frac{1}{3}\)then spent 2/3 of 2/3, which is \(\frac{2}{3}*\frac{2}{3}=\frac{4}{9}\). \(1-\frac{4}{9}-\frac{1}{3}=\frac{2}{9}\). He had 2/9 left.
B)
Set the price of a cap as c, and so each pair of shorts is c+30.
Set the amount of money Bob has at the beginning as x.
We get the equation \(\frac{x}{3}=4c+5(c+30)\), after bob buys 4 caps and 5 pairs of shorts
Bob has already spend 1/3 of his money, so he only has 2/3 left, so spending 2/3 of 2/3 is \(\frac{2}{3}*\frac{2}{3}=\frac{4}{9}\).
So, \(\frac{4}{9}x=16c\).
We get the system \(\begin{cases}\frac{x}{3}=4x+5(c+30)\\\frac{4}{9}x=16c\end{cases}\).
\(\begin{cases}x=12c+15(c+30)\\x=36c\end{cases}\)
\(36c=27c+450\)
\(9c=450\)
\(c=50\)
\(x=50*36=1800\).
So, Bob had 1800 dollars at the beginning.
