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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:

72% (01:07) correct 27% (01:22) wrong based on 85 sessions
Working at constant rates, machine R completely presses X records in 0.5 hours, and machine S completely presses X records in 0.75 hours (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
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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] New post 14 Aug 2019, 13:01
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I've attached a file showing how I did it.
Attachments

daum_equation_1565817301052.png
daum_equation_1565817301052.png [ 32.04 KiB | Viewed 1425 times ]


Last edited by arc601 on 06 May 2020, 08:56, edited 2 times in total.
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Re: Working at constant rates [#permalink] New post 14 Aug 2019, 13:09
Expert's post
pclawong wrote:
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!


No. I'm just applying some number sense in my solution.

Cheers,
Brent
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Re: Working at constant rates [#permalink] New post 15 Aug 2019, 13:43
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For A, x records 30 minute
2x records 60 minute
6x records 180 minute
18x records 3 hour

For S, X records 45 minute
x/9 records 5 minute
48*(x/9) or 16x/3 records 48*5=240 minute=4 hour
16x records 4 hour

So A is greater.
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Re: Working at constant rates [#permalink] New post 03 May 2020, 06:09
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We could also imagine that x=10

leading us to 10....0,5 hr for R-->60 in 3 hr
and 10......0,75 hr for S---> 50 something in 4 hr
Hence A is bigger
Re: Working at constant rates   [#permalink] 03 May 2020, 06:09
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