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Difference between revisions of "2002 AMC 8 Problems/Problem 11"

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==Solution 2==
 
==Solution 2==
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The edge of each square is one tile length longer than the edge of the previous square, which means that each square has <math>2*</math> edge length <math>- 1</math> more tiles than the previous square, because each square is just the previous square with two added edges of 1 more tile length than the previous with one overlapping tile (the top right). Then the seventh square has <math>2*7-1=13</math> more tiles than the sixth square, which is <math>\boxed{\text{(C)}\ 13}</math>.
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The edge of each square is one tile length longer than the edge of the previous square, which means that each square has <math>2*</math> edge length <math>- 1</math> more tiles than the previous square, because each square is just an one edge added on the top and on the right to the previous square, with one overlapping tile.. Then the seventh square has <math>2*7-1=13</math> more tiles than the sixth square, which is <math>\boxed{\text{(C)}\ 13}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2002|num-b=10|num-a=12}}
 
{{AMC8 box|year=2002|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:05, 28 July 2013

Problem

A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth?

[asy] path p=origin--(1,0)--(1,1)--(0,1)--cycle; draw(p); draw(shift(3,0)*p); draw(shift(3,1)*p); draw(shift(4,0)*p); draw(shift(4,1)*p); draw(shift(7,0)*p); draw(shift(7,1)*p); draw(shift(7,2)*p); draw(shift(8,0)*p); draw(shift(8,1)*p); draw(shift(8,2)*p); draw(shift(9,0)*p); draw(shift(9,1)*p); draw(shift(9,2)*p);[/asy]

$\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$

Solution

Solution 1

The first square has a sidelength of $1$, the second square $2$, and so on. The seventh square has $7$ and is made of $7^2=49$ unit tiles. The sixth square has $6$ and is made of $6^2=36$ unit tiles. The seventh square has $49-36=\boxed{\text{(C)}\ 13}$ more tiles than the sixth square.

Solution 2

The edge of each square is one tile length longer than the edge of the previous square, which means that each square has $2*$ edge length $- 1$ more tiles than the previous square, because each square is just an one edge added on the top and on the right to the previous square, with one overlapping tile.. Then the seventh square has $2*7-1=13$ more tiles than the sixth square, which is $\boxed{\text{(C)}\ 13}$.

See Also

2002 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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 AJHSME/AMC 8 Problems and Solutions

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