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Pumpkin patch

by gmatIntent » Tue Nov 22, 2011 6:17 am
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.

[spoiler]OA: C[/spoiler]

Source: 800Score.com
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by neelgandham » Tue Nov 22, 2011 6:53 am
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?

Average = 12 = (10x+ry)/(x+y) => 2x = (r-12)*y => r = (2x/y) + 12.
The question can be rephrased to, r = (2x/y) + 12, What is the value of r ?
(1) There are five more heavier pumpkins than lighter pumpkins.
Since the average is 12> 10, we can conclude that the value of r > 10. from the statement
y = x +5
r = (2x/y) + 12 = (2*(y-5)/y) + 12. Insufficient! as we don't know the value of y.
(2) The weight in pounds of each of the heavier pumpkins is 3 more than their number.
Weight of heavier pumpkin = r
Number of heavier pumpkins = y
The weight in pounds of each of the heavier pumpkins is 3 more than their number. Implies
r = 3 + y. Insufficient! as we don't know the value of y.
From 1 and 2
r = 3+y and y = x +5 => r-3 = x+5 =>
y = r-3 and x = r-8
We know 2x = (r-12)*y => 2*(r-8) = (r-3)*(r-12) => r^2-17r+52=0 => (r-13)*(r-4) = 0, Implies r =13 or 4. But we already know that r > 10, the value of r = 13. Sufficient !

Option C
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by GMATGuruNY » Tue Nov 22, 2011 7:57 am
gmatIntent 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.

[spoiler]OA: C[/spoiler]

Source: 800Score.com
Very little math is needed here.

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 + 3
No way to determine the value of r.
Insufficient.

Statements 1 and 2 together: 12 < r < 14 and r = y + 3.
Since y must be an integer, the only possible combination that satisfies both statements is y = 10 and r = 13.
SUFFICIENT.

The correct answer is C.
Last edited by GMATGuruNY on Wed Nov 23, 2011 7:44 am, edited 1 time in total.
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by gmatIntent » Wed Nov 23, 2011 2:36 am
neelgandham wrote:
(1) There are five more heavier pumpkins than lighter pumpkins.
Since the average is 12> 10, we can conclude that the value of r > 10. from the statement
y = x +5
r = (2x/y) + 12 = (2*(y-5)/y) + 12. Insufficient! as we don't know the value of y.
I don't get how you concluded that r > 10. Can you please explain. Thanks.
GMATGuruNY wrote:
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 can be determined. Since r = 13 yields a mean less than 12, and r < 14, r must be a value between 13 and 14. Sufficient.
Mitch, I couldn't understand this statement. How did you decide to choose C when no integer value satisfies and this is not a legitimate question.
Should I be much worried about not getting the answer correct for this question? :(
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by neelgandham » Wed Nov 23, 2011 2:44 am
gmatIntent wrote:
neelgandham wrote:
(1) There are five more heavier pumpkins than lighter pumpkins.
Since the average is 12> 10, we can conclude that the value of r > 10. from the statement
y = x +5
r = (2x/y) + 12 = (2*(y-5)/y) + 12. Insufficient! as we don't know the value of y.
I don't get how you concluded that r > 10. Can you please explain. Thanks.
If k is average of a and b then, a < k < b where b>a
Let me show you with an example, If x = 10, y some integer,mean of x and y = 12,
Then Arithmetic mean : (x+y)/2 = 12 => x+y=24 y = 14
y > average and average > x
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by GMATGuruNY » Wed Nov 23, 2011 7:59 am
gmatIntent wrote:
GMATGuruNY wrote:
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 can be determined. Since r = 13 yields a mean less than 12, and r < 14, r must be a value between 13 and 14. Sufficient.
Mitch, I couldn't understand this statement. How did you decide to choose C when no integer value satisfies and this is not a legitimate question.
Should I be much worried about not getting the answer correct for this question? :(
Please revisit my amended post above. The question posted here is fine. Given the two statements, all the numbers work out:
Heavier pumpkins: y = 10 and r = 13, implying a total weight of 10*13 = 130.
Lighter pumpkins: x = 5, implying a total weight of 5*10 = 50.
Average weight = (130+50)/15 = 12.

My comments about the legitimacy of the question were in reference to a different thread, in which statement 2 had apparently been transcribed incorrectly. In the other thread, the weight of the heavier pumpkins is EQUAL TO their number, an impossible scenario: https://www.beatthegmat.com/pumpkins-t39120.html).

Now that I've seen this question correctly transcribed, I very much like it. It illustrates how a DS question above weighted averages can be solved very quickly -- and with very little math -- if we understand how weighted averages work.
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by gmatIntent » Thu Nov 24, 2011 2:50 am
GMATGuruNY wrote:
My comments about the legitimacy of the question were in reference to a different thread, in which statement 2 had apparently been transcribed incorrectly. In the other thread, the weight of the heavier pumpkins is EQUAL TO their number, an impossible scenario: https://www.beatthegmat.com/pumpkins-t39120.html).

Now that I've seen this question correctly transcribed, I very much like it. It illustrates how a DS question above weighted averages can be solved very quickly -- and with very little math -- if we understand how weighted averages work.
Mitch, thanks.
Can you please about weighted average? I see lot of advanced level questions use this concept but I've no clue about it.
Thanks
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by GMATGuruNY » Thu Nov 24, 2011 4:17 am
gmatIntent wrote:
GMATGuruNY wrote:
My comments about the legitimacy of the question were in reference to a different thread, in which statement 2 had apparently been transcribed incorrectly. In the other thread, the weight of the heavier pumpkins is EQUAL TO their number, an impossible scenario: https://www.beatthegmat.com/pumpkins-t39120.html).

Now that I've seen this question correctly transcribed, I very much like it. It illustrates how a DS question above weighted averages can be solved very quickly -- and with very little math -- if we understand how weighted averages work.
Mitch, thanks.
Can you please about weighted average? I see lot of advanced level questions use this concept but I've no clue about it.
Thanks
Anand
Use the search bar in the upper right to seek out problems about "weighted averages" and "mixtures". You'll get many hits. If you also type in "GMATGuruNY", you'll get links to other weighted average problems to which I've posted solutions.

One approach you might find helpful is ALLIGATION. Check here:

https://www.beatthegmat.com/coffee-that- ... 93244.html

My second post in the thread above offers links to several problems that I've solved with alligation.
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