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# A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pu

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Math Expert
Joined: 02 Sep 2009
Posts: 50585
A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pu  [#permalink]

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29 Jul 2015, 02:24
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Difficulty:

95% (hard)

Question Stats:

38% (02:08) correct 62% (02:30) wrong based on 93 sessions

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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 six more heavier pumpkins than lighter pumpkins.
(2) There are four times fewer lighter pumpkins than heavier ones.

Kudos for a correct solution.

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Joined: 14 Mar 2014
Posts: 147
GMAT 1: 710 Q50 V34
A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pu  [#permalink]

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29 Jul 2015, 04:58
1
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 six more heavier pumpkins than lighter pumpkins.
(2) There are four times fewer lighter pumpkins than heavier ones.

Kudos for a correct solution.

IMO: B

Question Stem:
$$\frac{(10x + ry)}{(x + y)}$$ = 12
10x + ry = 12x + 12y
Since value of r is asked r = $$\frac{(2x + 12y)}{y}$$ --(i)

Since avg lies between min and max values
10<12<r
Thus there are y heavier pumpkins

Statement 1: There are six more heavier pumpkins than lighter pumpkins.

y = x+6
r = $$\frac{2x+ 12x+72}{x+6}$$
x is unknown
Hence not suff

Statement 2: There are four times fewer lighter pumpkins than heavier ones.

y = heavier pumpkins
then x = y/4
Sub in the eq--(i)
we get r = 12.5
Hence suff
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Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pu  [#permalink]

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29 Jul 2015, 05:24
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 six more heavier pumpkins than lighter pumpkins.
(2) There are four times fewer lighter pumpkins than heavier ones.

Kudos for a correct solution.

Given : Average Weight of all Pumpkins = (10x + ry)/(x+y) = 12
i.e. (10x + ry) = 12(x+y)
i.e. (r-12)y = 2x
i.e. r = 2(x/y) + 12
So we need to calculate the value of (x/y) to calculate the value of r

Also Since Average i.e. 12 is greater than 10 therefore r must be Greater than 12 because the average liest between the bigger and smaller number

Question : Find r i.e. find (x/y) ?

Statement 1: There are six more heavier pumpkins than lighter pumpkins.
y = x+6
but (x/y) can't be calculated. Hence,
NOT SUFFICIENT

Statement 2: There are four times fewer lighter pumpkins than heavier ones.
i.e. 4x = y
i.e. (x/y) = (1/4)
i.e. r = 2(1/4)+12 = 12.5 Hence,
SUFFICIENT

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Posts: 50585
Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pu  [#permalink]

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17 Aug 2015, 08:30
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 six more heavier pumpkins than lighter pumpkins.
(2) There are four times fewer lighter pumpkins than heavier ones.

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. If the y pumpkins weighed less than 12, then the average weight of all the pumpkins would be less than 12 as well, because all the pumpkins would simply weigh less than 12.

The equation for the average weight of the pumpkins is (10x + ry) / (x + y) = 12. Solve for r.
10x + ry = 12x + 12y
r = (2x + 12y) / y
r = 12 + 2x/y

Statement (1) tells us y = x + 6 . Plug it in the formula for r:
r = 12 + 2x/(x + 6)
Different values of x yield different values of r. For example the two following situations are possible: x = 2, y = 8, r = 12.5 and x = 6, y = 12, r = 13. Thus statement (1) is not sufficient.

Statement (2) tells us x = y/4 . Rewrite it as y = 4x and plug into the formula for r.
r = 12 + 2x/4x
r = 12.5
The value of r is definite, so statement (2) is sufficient.

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Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pu  [#permalink]

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10 Nov 2018, 22:26
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 six more heavier pumpkins than lighter pumpkins.
(2) There are four times fewer lighter pumpkins than heavier ones.

Kudos for a correct solution.

From the question-
10x + ry = 12(x+y).....Equation 1
The equation has 4 variables. So 3 equations are needed for solving.

From statement 1: y = x+ 6....Equation 2
Cannot be solved for r.

From statement 2: 4y = x....Equation 3
Cannot be solved for r.

Combining:
We have 3 equations and 3 variables. Can be solved.

Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pu &nbs [#permalink] 10 Nov 2018, 22:26
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