pumpkins

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by Night reader » Mon Feb 07, 2011 10:45 am
ONE vote for Mitch
@vaflaly, this solution is wrong :( I am sorry

firstly, you've missed that (14-r) may equal not only 13,4 but also 12 and 9

secondly, in your equation y=10/(14-r) although y is an integer [AGREE], r can be non-integer [YOU agree?]
thus, r=13.5 which is r>12 gives us -----> y=10(14-13.5) OR y=20 and x=15
(15x+ry)/(x+y)=12 OR (15*10 + 13.5*20)/(15+20)=12

It's not A, since we found at least two values for r and y
vaflaly wrote:Correct answer is A

10x+ry / x+y =12 ----> 10x+ry=12x+12y----> 2x+(12-r)y=0 (1)

1-----> y=x+5---> x=y-5
replace in (1), 2y-10 + (12-r)y=0 -----> 14y-ry-10=0 ----- y=10/(14-r)
since y is integer, 14-r = 1,5,10
r =13,11,4

But r must be > 12, so r=13

2-------> r=y alone is INSUF

Answer is A

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by Night reader » Mon Feb 07, 2011 10:58 am
Let me advance Mitch on this, as one doesn't require too much expertise - my level ;)
The problem says "x pumpkins 10 pounds each..." Now can it be <x=0> pumpkins 10 pounds each? I guess not, otherwise the problem's statement would be meaningless
cyrwr1 wrote:@Mitch,

There is no where in the problem that tells you x cannot = 0.

If x=0, y=5, and r=12.

This is what I did when combining the statements :

With both I did: 10(x)+(x+5)^2 = 12(2x+5)

x^2 + 10x +25 + 10x = 24x +60

simplifies to :

x^2 - 4x -35 = 0..... x is an integer so hence, my choice would E.


COuld you tell me if my approach is right? Thanks.

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by prachich1987 » Wed Feb 09, 2011 3:45 am
How can x be 8 & y=r=13
The average won't be 12 then.It contradicts with the question stem???
IMO this a wrong question.

GMATGuruNY wrote:
ketkoag wrote:there are x pumpkins of 10 pounds each, and y pumpkins of r pounds each. The average weight of a pumpkin is 12 pounds. What is the value or r?
1)there are 5 heavier pumpkins more than lighter pumpkins.
2)the weight in the pounds of the heavier pumpkins is equal to their numbers.
Please lemme know the value of r as well. i think mistake in calculations or something, i am getting 2 values.. please lemme know if u are getting a unique value..
If we understand how weighted averages work, we can approach this question without performing almost any math.

If r = 14, then x = y: since 10 and 14 are equidistant from the mean of 12, we will need the same number of lighter pumpkins as heavier pumpkins:
Image

If r > 14, then x > y: since the weight of the heavier pumpkins will be further from the mean of 12, we will need fewer heavier pumpkins and more lighter pumpkins.
Image

If r < 14, then y > x: since the weight of the heavier pumpkins will be closer to the mean of 12, we will need more heavier pumpkins and fewer lighter pumpkins.
Image

Statement 1: y = x+5
Since y > x, r < 14.
Thus, 12 < r <14.
Insufficient.

Statement 2: r = y
No way to determine the value of r.
Insufficient.

Statements 1 and 2 together: 12 < r < 14 and r = y.
Since y must be an integer, r = y = 13.
Since y = x+5, x = 8.
But these values don't work. If we have 8 pumpkins that are 10 pounds each and 13 pumpkins that are 13 pounds each, the average = (8*10 + 13*13)/21 = 249/21 ≈ 11.85.

Since no integer value for r = y satisfies both statements, this is not a legitimate GMAT question. Still, we can see that the value of r could be determined. Since r = 13 yields a mean less than 12, and r < 14, r must be a value between 13 and 14. Sufficient.

The correct answer is C.

I think that the approach above is the most GMAT friendly. Since on the GMAT the number of pumpkins (and thus the value of r) would be an integer, there would no need to set up messy quadratic equations to solve for r.
Thanks!
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by tohellandback » Wed Feb 09, 2011 3:57 am
spellbinder wrote:
tohellandback wrote:IMO A
1) sufficient
given y=x+5

(10x+(x+5)r)/(x+x+5)=12
10x+(x+5)r=24x+60
14x+60=rx+5r

x=5r-60/14-r
now x must be a number(non negative and r>12)
only possible value of r in the equation is 13 which gives us
x=5
y=10
r=13

2) NOT sufficient
10x+y^2=12x+12y which has two variables
na man u got it wrong.
nowhere in the question does it say that weight of the pumpkin cant be a fraction.
the answer is c
you need both of get rid of one of the variables, and hence solve for r.
never said anything about the weight of the pumpkin. X must be a number. So A.
The powers of two are bloody impolite!!

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by GMATGuruNY » Wed Feb 09, 2011 4:35 am
tohellandback wrote:
spellbinder wrote:
tohellandback wrote:IMO A
never said anything about the weight of the pumpkin. X must be a number. So A.
Incontrovertible proof that statement 1 is insufficient:

Plug in x = 1.
Then y = 1+5 = 6.
Total number of pumpkins = 1+6 = 7.
Total weight of all 7 pumpkins = 7*12 = 84.
Weight of 1 10-pound pumpkin = 1*10 = 10.
Weight of 6 r-pound pumpkins = 84-10 = 74.
r = 74/6 = 37/3.

Plug in x = 10.
Then y = 10+5 = 15.
Total number of pumpkins = 10+15 = 25.
Total weight of all 25 pumpkins = 25*12 = 300.
Weight of 10 10-pound pumpkins = 10*10 = 100.
Weight of 15 r-pound pumpkins = 300-100 = 200.
r = 200/15 = 40/3.

Since r=37/3 in the first case and r=40/3 in the second case, insufficient.

As I noted in my original response, statement 1 tells us only that 12<r<14.
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by cyrwr1 » Wed Feb 09, 2011 10:55 am
So is the answer to this question C or E?

A) is incorrect as the statement is true when

x=0,y=5 with r =12
and
x=5,y=10 with r=13

My explanation for the combination :
There is no where in the problem that tells you x cannot = 0.

If x=0, y=5, and r=12.

This is what I did when combining the statements :

With both I did: 10(x)+(x+5)^2 = 12(2x+5)

x^2 + 10x +25 + 10x = 24x +60

simplifies to :

x^2 - 4x -35 = 0..... x is an integer so hence, my choice would E.


COuld you tell me if my approach is right? Thanks.