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Working at a constant rate, Bob can produce x/3 widgets in
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13 Jul 2020, 07:41
13 Jul 2020, 07:41
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Working at a constant rate, Bob can produce \(\frac{x}{3 }\) widgets in 8 minutes. Working at a constant rate, Jack can produce \(2x \) widgets in 40 minutes, where \(x > 0\).
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Quantity A |
Quantity B |
The number of minutes it will take Bob to produce \(5x\) widgets |
The number of minutes it will take Jack to produce \(6x \) widgets |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
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Posts: 1141
Given Kudos: 9
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Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
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WE:Engineering (Computer Software)
Working at a constant rate, Bob can produce x/3 widgets in
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17 Jul 2020, 10:26
17 Jul 2020, 10:26
1
This problem is based on Rate and Work.
Formula used will be Rate * Time = Work Done
Let Rate of Bob be B and Rate of Jack be J
Working at a constant rate, Bob can produce \(\frac{x}{3 }\) widgets in 8 minutes
=> B * 8 = \(\frac{x}{3 }\)
=> B = \(\frac{x}{{3*8} }\) = \(\frac{x}{24 }\)
Quantity A = The number of minutes it will take Bob to produce \(5x\) widgets
=> B * T = 5x
=> \(\frac{x}{24 }\) * T = 5x
=> T = \(\frac{{5x * 24}}{x }\) = 120 mins
Working at a constant rate, Jack can produce \(2x \) widgets in 40 minutes
=> J * 40 = 2x
=> J = \(\frac{2x}{40 }\) = \(\frac{x}{20 }\)
Quantity B = The number of minutes it will take Jack to produce \(6x \) widgets
=> J * Time = 6x
=> \(\frac{x}{20 }\) * Time = 6x
=> Time = \(\frac{{6x*20}}{x}\) = 120 mins
=> Clearly, Quantity A = Quantity B = 120 mins
So, Answer will be C
Hope it helps!
Watch the following video to learn How to Solve Work Rate Problems
Formula used will be Rate * Time = Work Done
Let Rate of Bob be B and Rate of Jack be J
Working at a constant rate, Bob can produce \(\frac{x}{3 }\) widgets in 8 minutes
=> B * 8 = \(\frac{x}{3 }\)
=> B = \(\frac{x}{{3*8} }\) = \(\frac{x}{24 }\)
Quantity A = The number of minutes it will take Bob to produce \(5x\) widgets
=> B * T = 5x
=> \(\frac{x}{24 }\) * T = 5x
=> T = \(\frac{{5x * 24}}{x }\) = 120 mins
Working at a constant rate, Jack can produce \(2x \) widgets in 40 minutes
=> J * 40 = 2x
=> J = \(\frac{2x}{40 }\) = \(\frac{x}{20 }\)
Quantity B = The number of minutes it will take Jack to produce \(6x \) widgets
=> J * Time = 6x
=> \(\frac{x}{20 }\) * Time = 6x
=> Time = \(\frac{{6x*20}}{x}\) = 120 mins
=> Clearly, Quantity A = Quantity B = 120 mins
So, Answer will be C
Hope it helps!
Watch the following video to learn How to Solve Work Rate Problems
gmatclubot
Working at a constant rate, Bob can produce x/3 widgets in [#permalink]
17 Jul 2020, 10:26
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