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Ron was getting some items for the new school year. He first bought some textbooks with [tex]$\$[/tex]8[tex]$ more than $[/tex]\frac{1}{3}[tex]$ of his money. Then, he bought his stationery with $[/tex]\[tex]$12.20$[/tex] less than [tex]$\frac{1}{2}$[/tex] of his remaining money. Lastly, he bought some socks with [tex]$\$[/tex]2.80[tex]$ more than $[/tex]\frac{1}{2}[tex]$ of the money left. He had $[/tex]\[tex]$15.40$[/tex] with him. How much money did he have at first?

Sagot :

Let's solve this problem step-by-step.

1. Define Variables:
- Let [tex]\( x \)[/tex] be the initial amount of money Ron had.
- Ron's final amount of money is \[tex]$15.40. 2. First Purchase - Textbooks: - Ron spent \( \frac{1}{3}x + 8 \) on textbooks. - Remaining money after buying textbooks: \[ x - \left( \frac{1}{3}x + 8 \right) = \frac{2}{3}x - 8 \] 3. Second Purchase - Stationery: - Ron spent \( \frac{1}{2} \) of his remaining money minus \$[/tex]12.20 on stationery.
- Remaining money after buying textbooks is [tex]\( \frac{2}{3}x - 8 \)[/tex].
- Amount spent on stationery is [tex]\( \frac{1}{2} \left( \frac{2}{3}x - 8 \right) - 12.20 \)[/tex].
- Remaining money after buying stationery:
[tex]\[ \left( \frac{2}{3}x - 8 \right) - \left( \frac{1}{2} \left( \frac{2}{3}x - 8 \right) - 12.20 \right) \][/tex]
- Simplifying this:
[tex]\[ \frac{1}{2} \left( \frac{2}{3}x - 8 \right) + 12.20 = \frac{1}{2} \left( \frac{2}{3}x - 8 \right) + 12.20 \][/tex]

4. Third Purchase - Socks:
- Ron spent [tex]\( \frac{1}{2} \)[/tex] of the remaining money plus \[tex]$2.80 on socks. - Remaining money after buying stationery is: \[ \frac{1}{2} \left( \frac{2}{3}x - 8 \right) + 12.20 \] - Amount spent on socks is \( \frac{1}{2} \left( \frac{1}{2} \left( \frac{2}{3}x - 8 \right) + 12.20 \right) + 2.80 \). - So, the remaining money after buying socks: \[ \left( \frac{1}{2} \left( \frac{2}{3}x - 8 \right) + 12.20 \right) - \left( \frac{1}{2} \left( \frac{1}{2} \left( \frac{2}{3}x - 8 \right) + 12.20 \right) + 2.80 \right) \] - Simplifying this: \[ \frac{1}{2} \left( \frac{1}{2} \left( \frac{2}{3}x - 8 \right) + 12.20 \right) - 2.80 \] Given Ron ended with \$[/tex]15.40, we can set up the equation:
[tex]\[ \frac{1}{2} \left( \frac{1}{2} \left( \frac{2}{3}x - 8 \right) + 12.20 \right) - 2.80 = 15.40 \][/tex]

Now let's simplify:
1. Remaining Money After Buying Socks:
[tex]\[ \frac{1}{2} \left( \frac{1}{2} \left( \frac{2}{3}x - 8 \right) + 12.20 \right) - 2.80 \][/tex]

Given the result:
[tex]\[ \frac{1}{2} \left( \frac{2}{3}x - 8 \right) + 12.20 = 0.333333333333333x + 8.20 \][/tex]

Thus, equating the simplified amount with the final amount, we get:
[tex]\[ 0.166666666666667x + 1.3 = 15.40 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ 0.166666666666667x = 15.40 - 1.3 \][/tex]
[tex]\[ 0.166666666666667x = 14.10 \][/tex]
[tex]\[ x = \frac{14.10}{0.166666666666667} \][/tex]
[tex]\[ x = 84.60 \][/tex]

Ron initially had $84.60.