Cai collected most and Mark collected least
Solution:
Let Jen collected be "x"
Given that,
Mark collected 2/3 as much as Jen
[tex]\text{Mark collected } = \frac{2}{3} \times \text{Jen collected}\\\\\text{Mark collected } = \frac{2}{3} \times x\\\\\text{Mark collected } = \frac{2x}{3} = 0.67x[/tex]
Also given that,
Cai collected 2 1/2 times as much as mark
Therefore,
[tex]\text{Cai collected } = 2\frac{1}{2} \times \text{ mark collected }\\\\\text{Cai collected } = 2\frac{1}{2} \times \frac{2x}{3}\\\\\text{Cai collected } = \frac{5}{2} \times \frac{2x}{3}\\\\\text{Cai collected } = \frac{5x}{3} = 1.67x[/tex]
Who collected the most? Who collected the least?
Jen collected = x
Mark collected = 0.67x
Cai collected = 1.67x
Thus,
1.67x > x > 0.67x
Thus, Cai collected most and Mark collected least
