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

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.

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:


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.


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.


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.

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.

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.