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Working at constant rates

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Working at constant rates [#permalink] New post 27 Dec 2016, 08:21
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Question Stats:

84% (01:00) correct 16% (01:21) wrong based on 25 sessions
Working at constant rates, machine R completely presses X records in 0.5 hour, and machine S completely presses X records in 0.75 hour (x > 0 )

Quantity A
Quantity B
The number of records completely pressed by R in 3 hours
The number of records completely pressed by S in 4 hours


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.
[Reveal] Spoiler: OA

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Re: Working at constant rates [#permalink] New post 27 Dec 2016, 10:52
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Carcass wrote:
Working at constant rates, machine R completely presses X records in 0.5 hour, and machine S completely presses X records in 0.75 hour (x > 0 )

Quantity A
Quantity B
The number of records completely pressed by R in 3 hours
The number of records completely pressed by S in 4 hours



Machine R presses X records in 0.5 hour
So, Machine R presses 2X records in 1 hour
So, Machine R presses 4X records in 2 hours
So, Machine R presses 6X records in 3 hours

Machine S presses X records in 0.75 hours
So, Machine S presses 2X records in 1.5 hours
So, Machine S presses 4X records in 3 hours
So, Machine S presses 6X records in 4.5 hours
Given this, we can see that, in 4 hours, Machine S presses fewer than 6X records

We get:
Quantity A: 6X
Quantity B: fewer than 6X

Answer: A
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Re: Working at constant rates [#permalink] New post 13 Jan 2018, 23:57
GreenlightTestPrep wrote:
Carcass wrote:
Working at constant rates, machine R completely presses X records in 0.5 hour, and machine S completely presses X records in 0.75 hour (x > 0 )

Quantity A
Quantity B
The number of records completely pressed by R in 3 hours
The number of records completely pressed by S in 4 hours



Machine R presses X records in 0.5 hour
So, Machine R presses 2X records in 1 hour
So, Machine R presses 4X records in 2 hours
So, Machine R presses 6X records in 3 hours

Machine S presses X records in 0.75 hours
So, Machine S presses 2X records in 1.5 hours
So, Machine S presses 4X records in 3 hours
So, Machine S presses 6X records in 4.5 hours
Given this, we can see that, in 4 hours, Machine S presses fewer than 6X records

We get:
Quantity A: 6X
Quantity B: fewer than 6X

Answer: A


Dear friend,
is there a math formula in this question?
thanks!
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Re: Working at constant rates [#permalink] New post 29 Jun 2018, 22:48
pclawong wrote:

Dear friend,
is there a math formula in this question?
thanks!



You can use :

Work = No. of Individual * Rate * TIme
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Re: Working at constant rates   [#permalink] 29 Jun 2018, 22:48
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