Problem Solving

There are 15 children going to a movie. Tickets cost 5 dollars each, popcorn costs $3 per serving, and candy costs $2.50 per serving. If each child gets one snack, popcorn or candy, and $21 are spent for popcorn, what is the average (arithmetic mean) cost of the trip per child to the nearest cent?

A. $2.73
B. $3.86
C. $5.96
D. $7.70
E. $7.73

The OA is E.

$21 for popcorn = 21/3 = 7 servings of popcorn
i.e. remaining 8 children got one candy each = 8*2.5 = $20

Ticket cost = 15*5 = $75
Average cost = total cost / total children = (75+21+20)/15 = $7.73.

Has anyone another strategic approach to solve this PS question? Regards!

Total number of children = 15
Spending on popcorn = $21
Cost of popcorn = $3 per serve.
$$Total\ serving\ =\ \frac{21}{3}=\ 7$$
Each child had either popcorn or candy
Therefore, Children had candy = 15-7 = 8
Total spending on candy = 2.5 * 8 = $20
Total spending on snacks = 20 + 21 = $41
$$Average\ spending\ on\ snacks\ =\ \frac{41}{15}=\ \text{2.733}$$
Average spending on Ticket + Snacks = $(5 + 2.733)
= $7.733
Final answer is option E.