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ps

Expert replies
by baller12q » Thu Apr 02, 2009 7:46 am
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 the maximum number of oranges that could have been in the third layer.

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

OA C
how do i get the correct anwser
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Source: — Problem Solving |

Re: ps

by Stuart@KaplanGMAT » Thu Apr 02, 2009 10:53 am
baller12q 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 the maximum number of oranges that could have been in the third layer.

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

OA C
how do i get the correct anwser
I'm guessing the quickest way is to draw it out. It took me about 20 seconds to draw the diagram and find the correct answer.

A good general rule to follow is that when you've got a word problem that translates into a picture, draw it out. Many test takers find that visual representations make questions much easier to handle.
Image

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by kaf » Thu Apr 02, 2009 11:28 am
Stuart could you please post the diagram you used for your solution so some of us can understand the solution

thanks
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by mike22629 » Thu Apr 02, 2009 12:11 pm
F = First row oranges
S = Second row oranges
T = third row oranges


F F F F F
S S S S
F F F F F
S S S S
F F F F F

Since the second row is

S S S S
T T T
S S S S

There are maximum of 3 T
Hope this helps
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by mike22629 » Thu Apr 02, 2009 12:12 pm
oops that didnt post the way i expected it would

sorry
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by vittalgmat » Thu Apr 02, 2009 12:34 pm
Could not find any way other than to draw a pic

First layer has 3 rows of 5 oranges each. Draw _ and a space for each orange in row 1.
Each orange in Second layer touches 4 oranges in 1st layer.
so max of 8 oranges. Draw squares connecting 4 _. This becomes ur 2nd row.

For third row, draw another set of squares which connect 4 squares from the 2nd layer. If u want u can choose to draw a rhombus or X to denote the connection.
Count the squares/X in 3d row. It will be 3.

Let me know if u can t picture this.

thanks
-V
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by Stuart@KaplanGMAT » Thu Apr 02, 2009 2:00 pm
Here's my diagram - fear my art skills!
Attachments
Scan001.PDF
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Image

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