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A pumpkin patch contains x pumpkins that weigh 10 pounds each and y
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24 Jul 2015, 00:52
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A pumpkin patch contains x pumpkins that weigh 10 pounds each and y
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24 Jul 2015, 02:14
Hello,
I don't think I have found it. But I want to be notified by the answer, so I will only post my reasoning. I will go for E.
The stem gives us this information: x(10) + y(r) = 12 (x+y) > just diretly moved x+y from the denominator so it is easier to read. On the side, this also tells us that x+y are multiples of 12, even though I wasn't able to use this anywhere..
[1] Based on the first stem, we know that the heavier pumpkins are 5 more than the lighter ones, h = l+5. However, we don't know if r>10 or r<10. It cannot be 10 because then there wouldn't be any heavier pumkins. So, NS.
[2] Says that the weight of the heavier pumkins is 3 more than their number. We still don't know if r is more or less than 10 and we also don't know x and y (so the number of each type of pumkins). So, NS.
[1,2] Tell us thatt the weight of the heavier pumkins is l+5+3=l+8.
What I did then is replace x+y with l+h, because it is basically the same. So, this gives: x+y = l+h = (l+5)(l+8)+l(r).
However, we still don't know if x or y are the heavier ones. So, in the end we still end up with r, y, x and l to solve for. So, NS.
So, I think we need to know something more to solve it.



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Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y
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24 Jul 2015, 03:52
Bunuel wrote: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pumpkins that weigh r pounds each. If the average (arithmetic mean) weight of the pumpkins is 12 pounds, what is the value of r?
(1) There are five more heavier pumpkins than lighter pumpkins. (2) The weight in pounds of each of the heavier pumpkins is 3 more than their number.
Kudos for a correct solution. Solution  Average is greater than 10, it leads to r > 10. Average > (10x+ry)/(x+y)=12 > r=(2x/y)+12? Stmt1  Given that, y=x+5. r = (2x/y)+12 = (2*(y5)/y)+12. We do not know the value of Y. Not Sufficient. Stmt2  Given that, r = 3+y. We do not know the value of Y. Not Sufficient. 1+2  r = 3+y and y = x+5 > r3 = x+5 > y = r3 and x = r8 Substitute x and y in r = (2x/y)+12 > 2*(r8) = (r3)*(r12) > r^217r+52=0 > (r13)*(r4) = 0 > r =13 or 4. The value of r = 13. Sufficient. ANS C.
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Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y
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24 Jul 2015, 04:12
pacifist85 wrote: Hello,
I don't think I have found it. But I want to be notified by the answer, so I will only post my reasoning. I will go for E.
The stem gives us this information: x(10) + y(r) = 12 (x+y) > just diretly moved x+y from the denominator so it is easier to read. On the side, this also tells us that x+y are multiples of 12, even though I wasn't able to use this anywhere..
[1] Based on the first stem, we know that the heavier pumpkins are 5 more than the lighter ones, h = l+5. However, we don't know if r>10 or r<10. It cannot be 10 because then there wouldn't be any heavier pumkins. So, NS.
[2] Says that the weight of the heavier pumkins is 3 more than their number. We still don't know if r is more or less than 10 and we also don't know x and y (so the number of each type of pumkins). So, NS.
[1,2] Tell us thatt the weight of the heavier pumkins is l+5+3=l+8.
What I did then is replace x+y with l+h, because it is basically the same. So, this gives: x+y = l+h = (l+5)(l+8)+l(r).
However, we still don't know if x or y are the heavier ones. So, in the end we still end up with r, y, x and l to solve for. So, NS.
So, I think we need to know something more to solve it. I am critiquing your logic for statement 1. If you are given that h=l+5, you can translate it to say y=x+5 as the average of x pumpkins is 10 and the overall is 12. So x can not be the heavier ones. The heavier ones have to be 'y'. On its own, this statement is still not sufficient as : x(10) + y(r) = 12 (x+y) and y=x+5 > ry = 14y+10. Still not sufficient. Again, for your statement 2, heavier = y , lighter = x Thus, per statement 2, r=y+3. Thus on its own, not sufficient. Combining, you will be able to find out the value of r. Hope this helps.



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Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y
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24 Jul 2015, 04:15
My answer is C. Stem: X pumpkins weight 10 pound each> [Total weight]= 10x Y pumpking weight r pound each> [Total weight] = Yr. total weight of all pumpkins= 10x+Yr. Average weight of all pumpins = (10x+Yr) / X+Y. Given avg= 12. 10x+yr = 12x+ 12 Y. =>Yr= 2x+ 12Y. equation 1. [This is a 3 variable equation, hence we need 3 equations to solve] Option 1: There are five more heavier pumpkins than lighter pumpkins From the stem we know that Average of X pumpkins will be 10. From the statment, as the average gone up, the average of r pumpkins must be higher. So statement 1 can be written as y= x+5 Equation 2. and Statement 2 can be written as r=3+Y  Equation 3. so either statment is not sufficient, but both are required. Also by solving Eq2 and Eq3, we can establish a relationship between X and r and that become ours equ 4. Hope its correct. Bunuel wrote: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pumpkins that weigh r pounds each. If the average (arithmetic mean) weight of the pumpkins is 12 pounds, what is the value of r?
(1) There are five more heavier pumpkins than lighter pumpkins. (2) The weight in pounds of each of the heavier pumpkins is 3 more than their number.
Kudos for a correct solution.
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Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y
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24 Jul 2015, 10:29
Bunuel wrote: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pumpkins that weigh r pounds each. If the average (arithmetic mean) weight of the pumpkins is 12 pounds, what is the value of r?
(1) There are five more heavier pumpkins than lighter pumpkins. (2) The weight in pounds of each of the heavier pumpkins is 3 more than their number.
Kudos for a correct solution. given: x have 10 pounds each and y have r pounds each and avg. wt is 12 pounds, hence r>10 12 = (10x+ry) / (x+y), rearranging we get r=12+2x/y or r12=2x/y stmt1 : y = x+5...clearly insufficient, r12=2x/(x+5), x is unknown stmt2 : r = 3+y...insufficient, as y is unknown stmt1+stmt2 : from 1 we get x=y5 & from 2 we get y=r3 hence, x=r35=r8 now we have x&y both in terms of r, hence r=12+2(r8)/(r3) this gives, (r13)(r4)=0 as r>10 (given) ...r=13 Ans. C
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Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y
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26 Jul 2015, 12:51
Bunuel wrote: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pumpkins that weigh r pounds each. If the average (arithmetic mean) weight of the pumpkins is 12 pounds, what is the value of r?
(1) There are five more heavier pumpkins than lighter pumpkins. (2) The weight in pounds of each of the heavier pumpkins is 3 more than their number.
Kudos for a correct solution. 800score Official Solution:The original statement says the average weight of a pumpkin is 12 pounds. Since x pumpkins weigh 10 pounds each, the y pumpkins must be heavier. Writing an equation for the average weight of the pumpkins is (10x + ry) / (x + y) = 12. This equation has three unknowns. Solve for r to get: r = 12 + (2x/y) Statement (1) defines the relationship between x and y as y = x + 5 or y = x – 5. Plug in the formula for r to get r = 12 + (2(y – 5)/y) r = 14 – 10/y Plug in couple values of y (greater than 5) to see that they yield different results for r. Statement (1) is not sufficient. Statement (2) gives the relationship between y and r as r = y + 3. This isn't enough information as you can plug it into the original equation and see that too many variables remain. Plug in couple values for y to see that we get different possible valuee of r and x. Using the given with the equations from (1) and (2), the system is (10x + ry) / (x + y) = 12 y = x + 5 r = y + 3 Rewrite r = y + 3 as r – 3 = y. Substitute r – 3 = y into y = x + 5 and get r – 3 = x + 5 so r = x + 8. This gives an equation for y in terms of x, and an equation for r in terms of x. Substitute these equations into the original equation. (10x + ry) / (x + y) = 12. Substitute y = x + 5 and r = x + 8. (10x + (x + 8)(x + 5)) / (x + (x + 5)) = 12 10x + (x² + 13x + 40) = 12(2x + 5) x² + 23x + 40 = 24x + 60 x² – x – 20 = 0 (x – 5)(x + 4) = 0 x = 5 or x = 4 = 0 So there are 5 of the smaller pumpkins. Using y = x + 5, y = 5 + 5 = 10, so there are 10 of the larger pumpkins. And using r = y + 3, r = 13, the larger pumpkins weigh 13 pounds.
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Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y
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04 Aug 2015, 16:12
Bunuel wrote: Bunuel wrote: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pumpkins that weigh r pounds each. If the average (arithmetic mean) weight of the pumpkins is 12 pounds, what is the value of r?
(1) There are five more heavier pumpkins than lighter pumpkins. (2) The weight in pounds of each of the heavier pumpkins is 3 more than their number.
Kudos for a correct solution. 800score Official Solution:The original statement says the average weight of a pumpkin is 12 pounds. Since x pumpkins weigh 10 pounds each, the y pumpkins must be heavier. Writing an equation for the average weight of the pumpkins is (10x + ry) / (x + y) = 12. This equation has three unknowns. Solve for r to get: r = 12 + (2x/y) Statement (1) defines the relationship between x and y as y = x + 5 or y = x – 5. Plug in the formula for r to get r = 12 + (2(y – 5)/y) r = 14 – 10/y Plug in couple values of y (greater than 5) to see that they yield different results for r. Statement (1) is not sufficient. Statement (2) gives the relationship between y and r as r = y + 3. This isn't enough information as you can plug it into the original equation and see that too many variables remain. Plug in couple values for y to see that we get different possible valuee of r and x. Using the given with the equations from (1) and (2), the system is (10x + ry) / (x + y) = 12 y = x + 5 r = y + 3 Rewrite r = y + 3 as r – 3 = y. Substitute r – 3 = y into y = x + 5 and get r – 3 = x + 5 so r = x + 8. This gives an equation for y in terms of x, and an equation for r in terms of x. Substitute these equations into the original equation. (10x + ry) / (x + y) = 12. Substitute y = x + 5 and r = x + 8. (10x + (x + 8)(x + 5)) / (x + (x + 5)) = 12 10x + (x² + 13x + 40) = 12(2x + 5) x² + 23x + 40 = 24x + 60 x² – x – 20 = 0 (x – 5)(x + 4) = 0 x = 5 or x = 4 = 0 So there are 5 of the smaller pumpkins. Using y = x + 5, y = 5 + 5 = 10, so there are 10 of the larger pumpkins. And using r = y + 3, r = 13, the larger pumpkins weigh 13 pounds. For 2, why did not you consider 10=x+3?



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Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y
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02 Sep 2015, 09:06
The original statement says the average weight of a pumpkin is 12 pounds. Since x pumpkins weigh 10 pounds each, the y pumpkins must be heavier. I didn't understand this part . Please help to understand . that if one is 10 pound and other is y pound and avg is 12 . How can i conclude that  y >10 taking to into account the avg 12.
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A pumpkin patch contains x pumpkins that weigh 10 pounds each and y
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03 Sep 2015, 04:50
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution. A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pumpkins that weigh r pounds each. If the average (arithmetic mean) weight of the pumpkins is 12 pounds, what is the value of r? (1) There are five more heavier pumpkins than lighter pumpkins. (2) The weight in pounds of each of the heavier pumpkins is 3 more than their number. In the original condition, this question is typical 2by2 question in GMAT test. (10x+ry/x+y)=12, and thus there are 3 variables (x,y,r) and 1 equation (10x+ry/x+y)=12. We need to match the number of equations and varibles and thus we need 2 more. Since there is 1 each in 1) and 2), we need both of them and thus C is likely the answer, and it turns out that C actually is the answer. Using both 1) & 2) together, let's start from the assumption that 10>r. x=y+5, 10=x+3, then x=7, y=2, (10*7+2r/9)=12, 70+2r=12*9=108, 2r=38, r=19. Here we assumed that pumpkin1 is more heavier, and we have contradiction. Assuming 10<r, y=x+5, r=y+3, 10x+ry=12(x+y)=12x+12y, ry=2x+12y, r=y+3. substituting x=y5 gives us (y+3)y=2(y5)+12y, y^2+3y=2y10+12y, y^211y+10=0, (y10)(y1)=0, y=10, y=1. since y=1 gives us x=4, this is not our answer, and having y=10 gives us x=5, r=13 which is the answer. However, solving the question directly such as the above is inappropriate for 2 minute deadline, and it's more efficient to solve through variable approach method. The answer is C.
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Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y
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17 Mar 2016, 14:38
didn't understand the combine statement 1 and statement 2 part



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Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y
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17 Mar 2016, 15:06
rhine29388 wrote: didn't understand the combine statement 1 and statement 2 part Quote: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pumpkins that weigh r pounds each. If the average (arithmetic mean) weight of the pumpkins is 12 pounds, what is the value of r?
(1) There are five more heavier pumpkins than lighter pumpkins. (2) The weight in pounds of each of the heavier pumpkins is 3 more than their number. Hi, since you have asked about combined solution, i'll just speak on that..
INFO from Q there are x pumpkins of 10 pounds each and y pumpkins that weigh r pounds each.. avg weight =12 so 10x+ry=12(x+y)
Info from statements 1) y=x+5 2) r=y+3= x+5+3=x+8
Combined we have all values in terms of x.. lets substitute them in 10x+ry=12(x+y).. so \(10x + (x+8)(x+5) = 12 (x+x+5)\) => \(10x + x^2 +13x + 40 = 24x + 60\).. \(x^2  x 20 =0\).. \(x^2  5x + 4x 20=0\).. \(x(x5)+4(x5)=0\).. \((x5)(x+4)=0\). so\(x=5,4\).. x cannot be ive so x= 5 and \(r = x + 8 = 13\)Hope it helps
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Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y
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