Difference between revisions of "2004 AMC 10A Problems/Problem 7"

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==Video Solution==
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https://youtu.be/H2OQPsSWG4A
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Education, the Study of Everything
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==Problem==
 
==Problem==
 
A grocer stacks oranges in a pyramid-like stack whose rectangular base is <math>5</math> oranges by <math>8</math> oranges.  Each orange above the first level rests in a pocket formed by four oranges below.  The stack is completed by a single row of oranges.  How many oranges are in the stack?
 
A grocer stacks oranges in a pyramid-like stack whose rectangular base is <math>5</math> oranges by <math>8</math> oranges.  Each orange above the first level rests in a pocket formed by four oranges below.  The stack is completed by a single row of oranges.  How many oranges are in the stack?

Revision as of 16:51, 20 December 2020

Video Solution

https://youtu.be/H2OQPsSWG4A

Education, the Study of Everything


Problem

A grocer stacks oranges in a pyramid-like stack whose rectangular base is $5$ oranges by $8$ oranges. Each orange above the first level rests in a pocket formed by four oranges below. The stack is completed by a single row of oranges. How many oranges are in the stack?

$\mathrm{(A) \ } 96 \qquad \mathrm{(B) \ } 98 \qquad \mathrm{(C) \ } 100 \qquad \mathrm{(D) \ } 101 \qquad \mathrm{(E) \ } 134$

Solution

There are $5\times8=40$ oranges on the $1^{\text{st}}$ layer of the stack. The $2^{\text{nd}}$ layer that is added on top of the first will be a layer of $4\times7=28$ oranges. When the third layer is added on top of the $2^{\text{nd}}$, it will be a layer of $3\times6=18$ oranges, etc.

Therefore, there are $5\times8+4\times7+3\times6+2\times5+1\times4=40+28+18+10+4=100$ oranges in the stack. $\boxed{\mathrm{(C)}\ 100}$

See also

2004 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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