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Hi! I really need help in solving this qn.

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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?

Feb 21, 2024

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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.

Feb 21, 2024
edited by hairyberry  Feb 21, 2024