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pumpkins

by ketkoag » Sun Jun 21, 2009 2:11 am
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..
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by abhinav85 » Mon Jun 22, 2009 12:37 pm
Man from where u get these kind of ques???

I mean what is the source??
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by joymukhi » Mon Jun 22, 2009 4:20 pm
Solution Please , whats the source.
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by pops » Mon Jun 22, 2009 9:45 pm
My attempt at this:
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.

No. of Pumpkin weight
x 10
y r
average weight 12, hence r > 10 (to get average weight above 10)
equation: (10x+ry)/(x+y)=12 ...... (1)

statement 1. y=x+5 (insufficient to answer as above equation (1) still has two variables, namely x and r)

statement 2. r = y (insufficient to answer as above equation (1) still has two variables x and r)

but using both we can get an answer as we can replace y by x and r by x to get:
x as a fraction.. which should not be correct

(please check and correct me if i went wrong somewhere)
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by tohellandback » Mon Jun 22, 2009 10:25 pm
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
The powers of two are bloody impolite!!
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by pops » Mon Jun 22, 2009 10:33 pm
[quote="tohellandback"]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[/quote]

Gr8 !!!
I missed this... shame :(
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by spellbinder » Sat Feb 05, 2011 3:09 am
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.
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by Night reader » Sat Feb 05, 2011 5:54 am
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..
question conditions: (10x+ry)/(x+y)=12, find r-?
(1) 10x+ry=12x+12y, since we know that the average is 12, y pumpkins are heavier than x ones --> x+5=y to balance :)
However still Not Sufficient ry-2x-12y=0 <--EQ.1 and x+5=y <--EQ.2
(2) heavier pumpkin's weight] r=y [number of heavier pumpkins, Not Sufficient- as we are left with two variables either r and x OR y and x + one equation
Combined st(1&2) EQ.2/ x=y-5; EQ.2/ ry-2y+10-12y=0; st(2) r=y; Summary --> r^2-2r+10-12r=0, r^2-14r+10=0 --> approximately Sqrt(D)=12.5 and r(1)=(14-12.5)/2 which is obviously not fit as r<10 and r(2)=(14+12.5)/2=~13.25

I believe the answer is C and r=13.25 pounds
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by spellbinder » Sat Feb 05, 2011 6:03 am
Night reader 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..
question conditions: (10x+ry)/(x+y)=12, find r-?
(1) 10x+ry=12x+12y, since we know that the average is 12, y pumpkins are heavier than x ones --> x+5=y to balance :)
However still Not Sufficient ry-2x-12y=0 <--EQ.1 and x+5=y <--EQ.2
(2) heavier pumpkin's weight] r=y [number of heavier pumpkins, Not Sufficient- as we are left with two variables either r and x OR y and x + one equation
Combined st(1&2) EQ.2/ x=y-5; EQ.2/ ry-2y+10-12y=0; st(2) r=y; Summary --> r^2-2r+10-12r=0, r^2-14r+10=0 --> approximately Sqrt(D)=12.5 and r(1)=(14-12.5)/2 which is obviously not fit as r<10 and r(2)=(14+12.5)/2=~13.25

I believe the answer is C and r=13.25 pounds

thats a messy calculation, what i did was just applied the root formula real quick, and just by looking at the values ull be able to tell that one of the roots is negative, no calculations required. hence both were sufficient. do you get questions like these on the gmat??
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by Night reader » Sat Feb 05, 2011 6:21 am
@spellbinder - what you call messy is actually showing that r and y are in linear function without any additional parameters. While y which is the number of pumpkins :) can be only non-negative integer, r can be non-negative non-integer. If you substitute r with y in the equation then you get answer E. I doubt that you did a second job figuring out three constraints - non-negative r, r-y linear relationship with the r prevalent over y (you need quadratic equation ready to substitute (variable+a)(variable+b)=0 and see there non-integer roots ...), one root (as quadratic equation has two roots) is less than the required condition r>10

anyway, the answer is C and I was not as smart as you ARE.
spellbinder wrote:
Night reader 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..
question conditions: (10x+ry)/(x+y)=12, find r-?
(1) 10x+ry=12x+12y, since we know that the average is 12, y pumpkins are heavier than x ones --> x+5=y to balance :)
However still Not Sufficient ry-2x-12y=0 <--EQ.1 and x+5=y <--EQ.2
(2) heavier pumpkin's weight] r=y [number of heavier pumpkins, Not Sufficient- as we are left with two variables either r and x OR y and x + one equation
Combined st(1&2) EQ.2/ x=y-5; EQ.2/ ry-2y+10-12y=0; st(2) r=y; Summary --> r^2-2r+10-12r=0, r^2-14r+10=0 --> approximately Sqrt(D)=12.5 and r(1)=(14-12.5)/2 which is obviously not fit as r<10 and r(2)=(14+12.5)/2=~13.25

I believe the answer is C and r=13.25 pounds

thats a messy calculation, what i did was just applied the root formula real quick, and just by looking at the values ull be able to tell that one of the roots is negative, no calculations required. hence both were sufficient. do you get questions like these on the gmat??
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by cyrwr1 » Sat Feb 05, 2011 6:04 pm
I spent some time on this Question:

With 1), It is insufficient as :

when x=1, r=(10+ 14/6) and y=6
and when x=2, r = (10+18/7) and y =7

this is infinite from what i can get.

2) this tells you r=y and nothing about x as x can be 0 and both r and y =12.... i don't see this reasonable.

Insuff.

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.


Any thoughts and insights?
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by gmatmachoman » Sun Feb 06, 2011 2:28 am
Night reader wrote:
ketkoag wrote:
I believe the answer is C and r=13.25 pounds
AGreed C is the answer as 13.25 fits the liner equation for variable r.If that is the case then y will also be 13.25..

But how come the number of pumpkins can be a decimal?? Isn't supposed to be a Integer going by general logic??
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by GMATGuruNY » Sun Feb 06, 2011 4:31 am
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.
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by vaflaly » Mon Feb 07, 2011 9:28 am
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 cyrwr1 » Mon Feb 07, 2011 10:35 am
@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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