Cai collected most and Mark collected least
Solution:
Let Jen collected be "x"
Given that,
Mark collected 2/3 as much as Jen
![\text{Mark collected } = (2)/(3) * \text{Jen collected}\\\\\text{Mark collected } = (2)/(3) * x\\\\\text{Mark collected } = (2x)/(3) = 0.67x](https://img.qammunity.org/2021/formulas/mathematics/middle-school/miktvd6uje56dkblj3ik483wmwgtrz21po.png)
Also given that,
Cai collected 2 1/2 times as much as mark
Therefore,
![\text{Cai collected } = 2(1)/(2) * \text{ mark collected }\\\\\text{Cai collected } = 2(1)/(2) * (2x)/(3)\\\\\text{Cai collected } = (5)/(2) * (2x)/(3)\\\\\text{Cai collected } = (5x)/(3) = 1.67x](https://img.qammunity.org/2021/formulas/mathematics/middle-school/2apabpg0ufarw2i9vmii1mjf10p9r5h464.png)
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