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Bill buys two types of soda. He buys \(m\) bottles of Brand \(A\) at \(\$0.50\) each. He buys \(n\) bottles of Brand

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by M7MBA » Thu Nov 19, 2020 9:15 am
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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. \(\dfrac{50m+60n}{mn}\)

Answer: D

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Re: Bill buys two types of soda. He buys \(m\) bottles of Brand \(A\) at \(\$0.50\) each. He buys \(n\) bottles of Brand

by Scott@TargetTestPrep » Sat Nov 28, 2020 11:04 am
M7MBA wrote: ↑
Thu Nov 19, 2020 9:15 am
 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. \(\dfrac{50m+60n}{mn}\)

Answer: D

Solution:

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

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