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09 Sep 2024, 13:19
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50% (01:48) correct
50% (01:13) wrong based on 4 sessions
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Machine A, working alone at its constant rate, produces x pounds of peanut butter in 12 minutes. Machine B, working alone at its constant rate, produces x pounds of peanut butter in 18 minutes. How many minutes will it take machines A and B, working simultaneously at their respective constant rates, to produce x pounds of peanut butter?
?=minutes
?=minutes
09 Dec 2024, 13:11
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Translating the given information into an algebraic expression, you see that machine A produces x/12 pounds of peanut butter in 1 minute, and machine B produces x/18 pounds of peanut butter in 1 minute. Terefore, working simultaneously, machine A and machine B produce x/12 +x/18 pounds of peanut butter in 1 minute.
Letting t be the number of minutes it takes machines A and B, working simultaneously, to produce x pounds of peanut butter, you can set up the following equation.
\((\frac{x}{12}+\frac{x}{18}) t=x\)
Solving for t, you get
\((\frac{1}{12}+\frac{1}{18})=t=x\)
\(\frac{5}{36}t=1\)
\(t=\frac{36}{5}=7.2\)
The correct answer is 7.2.
Translating the given information into an algebraic expression, you see that machine A produces x/12 pounds of peanut butter in 1 minute, and machine B produces x/18 pounds of peanut butter in 1 minute. Terefore, working simultaneously, machine A and machine B produce x/12 +x/18 pounds of peanut butter in 1 minute.
Letting t be the number of minutes it takes machines A and B, working simultaneously, to produce x pounds of peanut butter, you can set up the following equation.
\((\frac{x}{12}+\frac{x}{18}) t=x\)
Solving for t, you get
\((\frac{1}{12}+\frac{1}{18})=t=x\)
\(\frac{5}{36}t=1\)
\(t=\frac{36}{5}=7.2\)
The correct answer is 7.2.
