Bill buys two types of soda. He buys m bottles of Brand A at $0.50 each. He buys n bottles of Brand B at $0.60 each...

[BTGmoderatorLU]

Source: Princeton Review

Bill buys two types of soda. He buys m bottles of Brand A at $0.50 each. He buys n bottles of Brand B at $0.60 each. What is Bill’s average cost in cents for a bottle of soda, in terms of m and n?

A. \(\dfrac{0.5m+0.6n}{m+n}\)

B. \(\dfrac{m+n}{110}\)

C. \(\dfrac{1.1}{m+n}\)

D. \(\dfrac{50m+60n}{m+n}\)

E. \(\dfrac{50m+60n}{mn}\)

The OA is D

[deloitte247]

What is Bill’s average cost in cents for a bottle of soda, in terms of m and n?

He buys m bottles of brand A at $0.50
$0.50 = 0.5 * 100 = 50 cents
Total amount spent on m bottles of brand A = 50m
He buys m bottles of brand B at $0.60
$0.60 = 0.6 * 100 = 60 cents
Total amount spent on n bottles of B = 60n
$$Average\ \cos t=\frac{Total\ \cos t\ of\ all\ bottles\ of\ soda}{No.\ of\ bottles\ of\ soda}$$ .$$=\frac{50m+60n}{m+n}$$

This validates option D as our choice of answer.

[Scott@TargetTestPrep]

Source: Princeton Review

Bill buys two types of soda. He buys m bottles of Brand A at $0.50 each. He buys n bottles of Brand B at $0.60 each. What is Bill’s average cost in cents for a bottle of soda, in terms of m and n?

A. \(\dfrac{0.5m+0.6n}{m+n}\)

B. \(\dfrac{m+n}{110}\)

C. \(\dfrac{1.1}{m+n}\)

D. \(\dfrac{50m+60n}{m+n}\)

E. \(\dfrac{50m+60n}{mn}\)

The OA is D

Note that $0.50 is 50 cents and $0.60 is 60 cents. Thus, the total cost of the (m + n) bottles of soda, in cents, is 50m + 60n. Thus, the average cost is:

(50m + 60n) / (m + n)

Answer: D