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Re: Gmatprep. Find another way to solve this problem! [#permalink]
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if bottom layer is 3 by 5
next layer will be 1 less in each dimension, same with the next layer => 1 by 3 = 3 oranges.
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Re: Gmatprep. Find another way to solve this problem! [#permalink]
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sondenso wrote:
A grocer stacked oranges in a pile. The bottom layer was rectangular with 3 rows of 5 oranges each. In the second layer from the bottom, each orange rested on 4 oranges from the bottom layer, and in the third layer, each orange rested on 4 oranges from the second layer. Which of the following is the maximum number of oranges that could have been in the third layer?

(A) 5
(B) 4
(C) 3
(D) 2
(E) 1

Can you solve this prob by making reasoning only, not by drawing it on the paper sheet? Thanks!


5 * 3 - bottom
2 * 4 - second
1 * 3 - top

Thus 3.

I didnt understand why the question uses word "maximum" though.
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Re: from Gmat Prep Question.... [#permalink]
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As mentioned in the question the first layer of the pile looks like this (each symbol o represents one organge):

o o o o o

o o o o o

o o o o o

Now since each orange of the second layer has to be supported by 4 oranges of the first layer there will be two rows of 4 oranges each (shown in blue font):

o o o o o
o o o o
o o o o o
o o o o
o o o o o

Repeat the same step for the third layer and you'll see that there are 3 oranges in the thrird layer (shown in red font):

o o o o
o o o
o o o o

Hence option c is correct.
Edit:
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Re: from Gmat Prep Question.... [#permalink]
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gmatJP wrote:
Please help solve this... I dont understand how it comes to 3...

Grocer stacked oranges in a pile. Bottom layer was rectangular with 3 rows of 5 oranges each. In 2nd layer , from bottom , each orange rested on 4 oranges from bottom layer and in the 3rd layer each orange rested on 4 oranges from 2nd layer. Which of the following is max.no. of oranges that could have been in 3rd layer.

a. 5
b. 4
c. 3
d. 2
e. 1

thanks...


My approach:

I believe if you understand that the every single orange above the bottom most layer can rest between the two oranges in the bottom most layer then it is easy to solve this question. Again the 3rd layer requires that the every single orange be rested on top of 4 oranges which is in the immediate bottom (2nd) layer.

stanford2012 has provided the correct approach but the diagrammatic representation is not correctly aligned and hence it is difficult to understand.
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Re: A grocer stacked oranges in a pile. The bottom layer was [#permalink]
What is this question testing its from GMAT prep...
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Re: A grocer stacked oranges in a pile. The bottom layer was [#permalink]
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fozzzy wrote:
What is this question testing its from GMAT prep...


I'd say logic and 3D visualization...
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Re: A grocer stacked oranges in a pile. The bottom layer was [#permalink]
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Bottom Layer = 5 x 3 = 15
IInd Layer = (5-1) x (3-1) = 8
IIIrd Layer = (4-1) x (2-1) = 3

Answer = 3 = C
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Re: A grocer stacked oranges in a pile. The bottom layer was [#permalink]
somalwar wrote:
Bottom Layer = 5 x 3 = 15
IInd Layer = (5-1) x (3-1) = 8
IIIrd Layer = (4-1) x (2-1) = 3

Answer = 3 = C


Why do you subtract 1 for each dimension? Would you kindly explain the logic behind this?
Also when they say between 4 oranges does it mean that an orange can be serving to hold to others oranges that might be adjacent? Or is it that one orange is only allowed to 'touch' one other orange from the second layer?

Cheers
J :)
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A grocer stacked oranges in a pile. The bottom layer was [#permalink]
Bunuel could you please refer to some similar (visualization) questions?Thanks.
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Re: A grocer stacked oranges in a pile. The bottom layer was [#permalink]
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sadikabid27 wrote:
Bunuel could you please refer to some similar (visualization) questions?Thanks.


Check GEOMETRY: 3-D Geometry Questions.

Hope it helps.
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Re: A grocer stacked oranges in a pile. The bottom layer was [#permalink]
sondenso wrote:
A grocer stacked oranges in a pile. The bottom layer was rectangular with 3 rows of 5 oranges each. In the second layer from the bottom, each orange rested on 4 oranges from the bottom layer, and in the third layer, each orange rested on 4 oranges from the second layer. Which of the following is the maximum number of oranges that could have been in the third layer?

(A) 5
(B) 4
(C) 3
(D) 2
(E) 1


Bunuel

I got the answer, but I do not understand in what sense did we achieve five oranges in each row ? there's five only at the bottom one. Also, what's with the word maximum ?
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Re: A grocer stacked oranges in a pile. The bottom layer was [#permalink]
KarishmaB is there another way of solving this question other than visualizing? Given the numbers, one could draw and understand BUT what if the numbers were huge, let's say 100 (starting from the bottom layer)?
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Re: A grocer stacked oranges in a pile. The bottom layer was [#permalink]
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Hoozan wrote:
KarishmaB is there another way of solving this question other than visualizing? Given the numbers, one could draw and understand BUT what if the numbers were huge, let's say 100 (starting from the bottom layer)?


You don't have to visualise the whole thing - only a part of it to get the pattern. Obviously, once you get the pattern, it can be used for any number.

Think of 4 balls placed together. We can place 1 ball on top of them in the middle. If we add another 2 balls such that we have 2 rows of 3 balls each, we can place another ball on top. So a 3*2 first layer has 2*1 second layer. Put another 3 balls at the back such that the first layer is 3*3. Now you can place 2 more balls on the top.
So a 3*3 first layer has a 2*2 second layer.
So the dimensions of the layer on the top is 1 less for both row and column.

So say 25 * 30 first layer will have 24*29 second layer and 23 * 28 third layer etc.
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Re: A grocer stacked oranges in a pile. The bottom layer was [#permalink]
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