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Working together at their respective constant rates, robot A
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06 Jul 2018, 16:20
06 Jul 2018, 16:20
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Question Stats:
72% (02:20) correct
27% (02:43) wrong based on 130 sessions
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Working together at their respective constant rates, robot A and robot B polish 88 pounds of gemstones in 6 minutes. If robot A’s rate of polishing is \(\frac{3}{5}\)that of robot B, how many minutes would it take robot A alone to polish 165 pounds of gemstones?
(A) 15.75
(B) 18
(C) 18.75
(D) 27.5
(E) 30
(A) 15.75
(B) 18
(C) 18.75
(D) 27.5
(E) 30
Re: Working together at their respective constant rates, robot A
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07 Jul 2018, 19:10
07 Jul 2018, 19:10
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Let us assume,
Machine B can do x work in 1 min
or, Machine B can do \(6x\) work in \(6 min\)
It follows that Machine A is only 3/5 efficient as machine A hence,
Machine A can do \(\frac{18}{5} x work in 6 min.\)
From question,
\(\frac{18}{5} x + 6x = 88\)
because working together the two machines can polish \(88 pounds\) of gemstone in \(6 min.\)
solving for x we get \(x = \frac{440}{48}\)
This is the pound of gemstone machine B can polish in 1 min.
The amount of gemstone that machine A can polish in 1 min is \(\frac{440}{48} * \frac{3}{5}\)
Hence, machine A can polish \(\frac{1320}{240}\) pound of gemstone in 1 min.
or, Machine A can polish 1 pound of gemstone in \(\frac{240}{1320}\) min
or, Machine A can polish 165 pound of gemstone in \(\frac{240}{1320} * 165\) min = 30 min
Machine B can do x work in 1 min
or, Machine B can do \(6x\) work in \(6 min\)
It follows that Machine A is only 3/5 efficient as machine A hence,
Machine A can do \(\frac{18}{5} x work in 6 min.\)
From question,
\(\frac{18}{5} x + 6x = 88\)
because working together the two machines can polish \(88 pounds\) of gemstone in \(6 min.\)
solving for x we get \(x = \frac{440}{48}\)
This is the pound of gemstone machine B can polish in 1 min.
The amount of gemstone that machine A can polish in 1 min is \(\frac{440}{48} * \frac{3}{5}\)
Hence, machine A can polish \(\frac{1320}{240}\) pound of gemstone in 1 min.
or, Machine A can polish 1 pound of gemstone in \(\frac{240}{1320}\) min
or, Machine A can polish 165 pound of gemstone in \(\frac{240}{1320} * 165\) min = 30 min
Re: Working together at their respective constant rates, robot A
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10 Jul 2018, 06:12
10 Jul 2018, 06:12
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Expert Reply
Explanation
When rate problems involve multiple situations, it can help to set up an initial “skeleton” W= RT chart for the solution. That way, you can determine what data is needed and fill in that data as you find it. Since the question asks how long robot A will take alone, the chart will look like this:
image1.jpg [ 26.04 KiB | Viewed 12053 times ]
Work is known and the question asks for time, so robot A’s rate is needed. Call the rates a and b. Now set up another chart representing what you know about the two robots working together.
image1.jpg [ 50.99 KiB | Viewed 12084 times ]
Now, 6(a + b) = 88 and, from the question stem, robot A’s rate is \(\frac{3}{5}\) of B’s rate. This can be written as \(a = \frac{3}{5}b\). To solve for a, substitute for b:
\(a = \frac{3}{5} b\)
\(\frac{5}{3}a = b\)
\(6(a+ \frac{5}{3}a) = 88\)
\(6 \frac{8}{3} a = 88\)
\(a = \frac{11}{2}\)
So A’s rate is pounds per minute. Now just plug into the original chart:
image1.jpg [ 28.05 KiB | Viewed 12083 times ]
The time robot A takes to polish 165 pounds of gems is \(165/\frac{11}{2} =\frac{330}{165} = 30\) minutes.
When rate problems involve multiple situations, it can help to set up an initial “skeleton” W= RT chart for the solution. That way, you can determine what data is needed and fill in that data as you find it. Since the question asks how long robot A will take alone, the chart will look like this:
Attachment:
image1.jpg [ 26.04 KiB | Viewed 12053 times ]
Work is known and the question asks for time, so robot A’s rate is needed. Call the rates a and b. Now set up another chart representing what you know about the two robots working together.
Attachment:
image1.jpg [ 50.99 KiB | Viewed 12084 times ]
Now, 6(a + b) = 88 and, from the question stem, robot A’s rate is \(\frac{3}{5}\) of B’s rate. This can be written as \(a = \frac{3}{5}b\). To solve for a, substitute for b:
\(a = \frac{3}{5} b\)
\(\frac{5}{3}a = b\)
\(6(a+ \frac{5}{3}a) = 88\)
\(6 \frac{8}{3} a = 88\)
\(a = \frac{11}{2}\)
So A’s rate is pounds per minute. Now just plug into the original chart:
Attachment:
image1.jpg [ 28.05 KiB | Viewed 12083 times ]
The time robot A takes to polish 165 pounds of gems is \(165/\frac{11}{2} =\frac{330}{165} = 30\) minutes.
