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Integers. Word Problem

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Integers. Word Problem

by askaichin » Sun Jul 25, 2010 8:12 am
Anthony and Cindy were each given X dollars in advance for each day they were expected to perform at a community festival. Anthony eventually perfromed on all 14 days of the festival, while Cindy performed on 4 fewer days than Anthony performed. If Cindy gives Anthony y dollars of her advance payment so that they would have each recieved the same daily rate, what was Anthony paid in advance, in terms of Y?

Ans

a 2Y
b 4Y
c 5Y
d 6Y
e 10Y
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by Patrick_GMATFix » Sun Jul 25, 2010 10:55 am
Hi askaichin,

We're solving for x, the amount Anthony was paid in advance (Cindy was also paid x in advance)

Anthony worked 14 days and Cindy worked 10 days (4 fewer).

Because Cindy gave Anthony y dollars, at the end of the day Cindy has x-y and Anthony has x+y (they each started with x).

After Cindy gave some of her money, her daily rate is $/time >> (x-y)/10. Likewise, Anthony's daily rate is (x+y)/14. Since their daily rates are the same after Cindy's gift, you can set (x-y)/10 = (x+y)/14 >> x=6y

Pick D.

This question used #204 in OG12 as a template. To create drills with similar questions, set topic='Translations & Manipulations' and difficulty='700+' in the Drill Generator.

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by askaichin » Sun Jul 25, 2010 11:40 am
Patrick_GMATFix wrote:Hi askaichin,

We're solving for x, the amount Anthony was paid in advance (Cindy was also paid x in advance)

Anthony worked 14 days and Cindy worked 10 days (4 fewer).

Because Cindy gave Anthony y dollars, at the end of the day Cindy has x-y and Anthony has x+y (they each started with x).

After Cindy gave some of her money, her daily rate is $/time >> (x-y)/10. Likewise, Anthony's daily rate is (x+y)/14. Since their daily rates are the same after Cindy's gift, you can set (x-y)/10 = (x+y)/14 >> x=6y

Pick D.

This question used #204 in OG12 as a template. To create drills with similar questions, set topic='Translations & Manipulations' and difficulty='700+' in the Drill Generator.

-Patrick
I still have a doubt. Are we assuming that X is the total amount that they get.. seems so from " Because Cindy gave Anthony y dollars, at the end of the day Cindy has x-y and Anthony has x+y (they each started with x). " part.

Also Y is the total amount she gives to settle their daily rate to equal so why are we assuming x+y and x-y ( shouldnt it be 14x +y and 14 x -y? )

I am not sure I can understand the solution properly. Pls help.

Thanks
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by Patrick_GMATFix » Sun Jul 25, 2010 12:54 pm
askaichin,

You are correct. I was careless; in solving I was thinking about #204 in OG12 because this question is almost a carbon copy of that one.

In fact, this question very clearly states that C. and A. were given X dollars "for each day they were expected to perform". We have no way to know how many days they were expected to perform: all 14 days of the festival? Just some of the days? Because we don't know the expectation, it's actually impossible to mathematically solve this question.

If we assume that they were each supposed to perform on all 14 days of the festival, then again you are correct. It would mean that they each received 14x in advance. It's pretty clear that Cindy gave Anthony Y in total (there is no duration of daily gift), so as you said at the end their amounts would be (14x+y) for Anthony and (14x-y) for Cindy.

Because their daily rates are the same and Cindy only worked 10 days, their daily rates are (14x+y)/14 for Anthony and (14x-y)/10 for Cindy. Setting these equal to each other and solving for x gives us (14x+y)/14 = (14x-y)/10 >> x = 3y/7

This does not appear in the answer choices. That means that the question is poorly written. It cannot be solved mathematically unless we make a major assumption about how long they were expected to work. In addition to that, the proper mathematical solution does not match any of the answer choices.

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by rdjlar » Tue Dec 28, 2010 8:46 am
Hi,

I do get 6Y as well. My approach was as follows, please correct me if anything is wrong:

Anthony's rate: x/14
Cindy's: x/10

Set up the following equation:

x/14 +y = x/10 - y

Solving for x:

x/14 +2y = x/10

10 (x/14 +2y)= x

10x/14 +20y= x

5x/7+ 140y/7= x

5x+140y = 7x

140y= 2x

70y=x

Then substituting the X in Anthony's rate of x/14 + y:

70y/14 + y

10y/2+ 2y/2

12y/2 = 6y
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by lavinia » Tue Dec 28, 2010 2:54 pm
Could you tell us the correct answer and the source of this problem?

Thanks!
askaichin wrote:
Patrick_GMATFix wrote:Hi askaichin,

We're solving for x, the amount Anthony was paid in advance (Cindy was also paid x in advance)

Anthony worked 14 days and Cindy worked 10 days (4 fewer).

Because Cindy gave Anthony y dollars, at the end of the day Cindy has x-y and Anthony has x+y (they each started with x).

After Cindy gave some of her money, her daily rate is $/time >> (x-y)/10. Likewise, Anthony's daily rate is (x+y)/14. Since their daily rates are the same after Cindy's gift, you can set (x-y)/10 = (x+y)/14 >> x=6y

Pick D.

This question used #204 in OG12 as a template. To create drills with similar questions, set topic='Translations & Manipulations' and difficulty='700+' in the Drill Generator.

-Patrick
I still have a doubt. Are we assuming that X is the total amount that they get.. seems so from " Because Cindy gave Anthony y dollars, at the end of the day Cindy has x-y and Anthony has x+y (they each started with x). " part.


Also Y is the total amount she gives to settle their daily rate to equal so why are we assuming x+y and x-y ( shouldnt it be 14x +y and 14 x -y? )

I am not sure I can understand the solution properly. Pls help.

Thanks
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by sumayahlaura » Wed Dec 29, 2010 9:06 am
I would like to know the correct answer for this problem. when I plug in numbers I get answer D as well

Here is how:
A works 14 days
C works 10 days

total days worked 24

Because A and C where EACH GIVEN x dollars, i plugged in a total of $240 dollars given, or $120 given in advanced to both A and C. This breaks down into a daily rate $10 a day for the total 24 days worked (maybe it should have been 28 days worked, given that C was supposed to work the total 14 days)

At a daily rate of $10 a day, C has been overpaid by $20 or y. $120 (original payment)/$20 (y) = 6
So 6y = $120
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by shovan85 » Wed Dec 29, 2010 9:43 am
sumayahrose wrote:I would like to know the correct answer for this problem. when I plug in numbers I get answer D as well

Here is how:
A works 14 days
C works 10 days

total days worked 24

Because A and C where EACH GIVEN x dollars, i plugged in a total of $240 dollars given, or $120 given in advanced to both A and C. This breaks down into a daily rate $10 a day for the total 24 days worked (maybe it should have been 28 days worked, given that C was supposed to work the total 14 days)

At a daily rate of $10 a day, C has been overpaid by $20 or y. $120 (original payment)/$20 (y) = 6
So 6y = $120
You are correct. The correct answer is D. Great way of solving by plugging the values. However, we can get the answer with the help of x and y itself. But which ever is faster for you ... you can follow :)

Anthony worked for 14 days.
Cindy worked for 10 days.

Cindy gives y from her advance of x to Anthony.

Thus Anthony has x+y and Cindy has x-y as their payments respectively.

Cindy pays y to Anthony in order to get the SAME rate.

Thus,

Daily Rate of Anthony's performance = Daily Rate of Cindy's performance

=> (x+y)/14 = (x-y)/10
=> 10x + 10y = 14x - 14y (After Cross multiplication)
=> -4 x = -24 y
=> x = 6y
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by thebigkats » Sun Jan 02, 2011 11:57 am
Hi:

I am not sure if I am able to follow Patrick's explanation. I also come at the same conclusion as other - 6Y

I agree that the stem is unclear, it uses the word expected no of days but doesn;t talk about what the expected was. So we have to assume maximum days (14 days) and in that case, it becomes very simple:

(14X + Y) ==> A's amount
(14X - Y) ==> C's amount

Given that their daily rate becomes - (14X + Y) / 14 and (14X - Y) / 10
and these are same meaning that (14X + Y) / 14 = (14X - Y) / 10
==> 14X (14 - 10) = (14 + 10) Y
==> 56X = 24Y
==> X = 3Y/7

Now Q is what was A piad in advance (14X), so substituting above: 14 * (3Y/7) = 6Y
Hence D
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by GMATGuruNY » Sun Jan 02, 2011 1:39 pm
askaichin wrote:Anthony and Cindy were each given X dollars in advance for each day they were expected to perform at a community festival. Anthony eventually perfromed on all 14 days of the festival, while Cindy performed on 4 fewer days than Anthony performed. If Cindy gives Anthony y dollars of her advance payment so that they would have each recieved the same daily rate, what was Anthony paid in advance, in terms of Y?

Ans

a 2Y
b 4Y
c 5Y
d 6Y
e 10Y
We can plug in for the daily rate. Let rate = $10 per day.
Pay for Anthony's 14 days of work = 14*10 = $140.
Pay for Cindy's 10 days of work = 10*10 = $100.
Combined pay = 140+100 = $240.
Since in advance each was paid X dollars, X = .5*240 = 120.
If Cindy was paid $120 in advance, to Anthony she must give Y = 120-100 = $20.
The problem asks for the amount that Anthony was paid in advance. With our numbers, Anthony was paid 120. This is our target.

Now we plug Y=20 into all the answer choices to see which yields our target of 120.

Only answer choice D works:
6Y = 6*20 = 120.

The correct answer is D.
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by shrikantpatel » Wed Jun 01, 2011 6:34 am
My approach using equation,

Lets Anthony and Cindy were paid for Z days in advance, So each was paid in total of $(X*Z)


The final pay for Anthony for 14 days can be written in equation format as XZ + Y = 14X (Equation 1)
The final pay for Cindy for 10 day can be written in equation format as XZ - Y = 10X (Equation 2)

Subtracting Equation 1 from Equation 2 we get 2Y=4X or X=Y/2

Now subsituting this value in equation 1, XZ = 14X - Y = 7*(Y/2) - Y = 7Y - Y = 6Y.

So final advance that Anthony was paid is XZ = 6Y, So D.
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by mishelk » Mon Jul 18, 2011 10:20 pm
GMATGuruNY wrote:
askaichin wrote:Anthony and Cindy were each given X dollars in advance for each day they were expected to perform at a community festival. Anthony eventually perfromed on all 14 days of the festival, while Cindy performed on 4 fewer days than Anthony performed. If Cindy gives Anthony y dollars of her advance payment so that they would have each recieved the same daily rate, what was Anthony paid in advance, in terms of Y?

Ans

a 2Y
b 4Y
c 5Y
d 6Y
e 10Y
We can plug in for the daily rate. Let rate = $10 per day.
Pay for Anthony's 14 days of work = 14*10 = $140.
Pay for Cindy's 10 days of work = 10*10 = $100.
Combined pay = 140+100 = $240.
Since in advance each was paid X dollars, X = .5*240 = 120.
If Cindy was paid $120 in advance, to Anthony she must give Y = 120-100 = $20.
The problem asks for the amount that Anthony was paid in advance. With our numbers, Anthony was paid 120. This is our target.

Now we plug Y=20 into all the answer choices to see which yields our target of 120.

Only answer choice D works:
6Y = 6*20 = 120.

The correct answer is D.
How can u know from the question that both were expected to work on 12 days of the 14 days event and not on all the 14 days or that Cindy received 2 days of extra payment not 4 days? We are also having to assume that Anthony covered for Cindy to match our answer with that of the answer choices. Wouldn't it take longer time to workout all the assumptions?
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by shingik » Thu Sep 08, 2011 12:58 pm
This question seems pretty straightforward. But then again I am a novice. I do not think how many days they were expected to work in the begining matters. What matters is whether they were given an equal amount at the begining and whether they worked for the same daily amount or not in the end. The answer changes with the number of days that they each worked. The answer is the dependent variable. The BEST answer has to be X = 6Y. If this were a data sufficiency question then maybe the answer could then be E.
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by chris558 » Tue Aug 28, 2012 9:44 am
RATE * TIME = DISTANCE
RATE * TIME = TOTAL EARNED

A: Rate * 14 = (X+Y)
C: Rate * 10 = (X-Y)

(X+Y)/14 = (X-Y)/10
10(X+Y)=14(X-Y)
5(X+Y)=7(x-Y)
5X+5Y=7X-7Y
12Y=2X
X=6Y

Answer is D
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by rajeshsinghgmat » Sat Mar 09, 2013 5:41 pm
(D) 6Y

(x+y)/14=(x-y)/10

10x+10y=14x-14y

24y=4x

6y=x

x=6y
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