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Difference between revisions of "2002 AMC 8 Problems/Problem 11"
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==Problem==
  
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== Problem 11 ==
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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?
  
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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?
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<asy>
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path p=origin--(1,0)--(1,1)--(0,1)--cycle;
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draw(p);
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draw(shift(3,0)*p);
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draw(shift(3,1)*p);
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draw(shift(4,0)*p);
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draw(shift(4,1)*p);
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draw(shift(7,0)*p);
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draw(shift(7,1)*p);
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draw(shift(7,2)*p);
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draw(shift(8,0)*p);
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draw(shift(8,1)*p);
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draw(shift(8,2)*p);
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draw(shift(9,0)*p);
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draw(shift(9,1)*p);
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draw(shift(9,2)*p);</asy>
  
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<math>\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15</math>
  
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<math> \text{(A)}\ 11\qquad\text{(B)}\ 12\qquad\text{(C)}\ 13\qquad\text{(D)}\ 14\qquad\text{(E)}\ 15 </math>
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==Solution==
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The first square has a sidelength of <math>1</math>, the second square <math>2</math>, and so on. The seventh square has <math>7</math> and is made of <math>7^2=49</math> unit tiles. The sixth square has <math>6</math> and is made of <math>6^2=36</math> unit tiles. The seventh square has <math>49-36=\boxed{\text{(C)}\ 13}</math> more tiles than the sixth square.
  
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[[Solution]]
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==See Also==
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{{AMC8 box|year=2002|num-b=10|num-a=12}}

Revision as of 18:38, 23 December 2012

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

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.

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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