Difference between revisions of "2007 AMC 8 Problems/Problem 13"

Revision as of 13:57, 16 January 2023

Problem

Sets $A$ and $B$ , shown in the Venn diagram, have the same number of elements. Their union has $2007$ elements and their intersection has $1001$ elements. Find the number of elements in $A$ .

$[asy] defaultpen(linewidth(0.7)); draw(Circle(origin, 5)); draw(Circle((5,0), 5)); label("$A$", (0,5), N); label("$B$", (5,5), N); label("$1001$", (2.5, -0.5), N);[/asy]$

$\mathrm{(A)}\ 503 \qquad \mathrm{(B)}\ 1006 \qquad \mathrm{(C)}\ 1504 \qquad \mathrm{(D)}\ 1507 \qquad \mathrm{(E)}\ 1510$

Solution

Let $x$ be the number of elements in $A$ and $B$ which is equal.

Then we could form equation $2x-1001 = 2007$

$2x = 3008$

$x = 1504$ .

The answer is $\boxed{\textbf{(C)}\ 1504}$

Solution 2

Let $x$ be the number of elements in $A$ not including the intersection. $2007-1001=1006$ total elements excluding the intersection. Since we know that $A=B$ , we can find that $x=\frac{1006}2=503$ . Now we need to add the intersection. $503+1001=\boxed{\textbf{(C)} 1504}$ .

Video Solution by WhyMath

https://youtu.be/3LtGb3KjhoU

~savannahsolver

Video Solution

https://www.youtube.com/watch?v=6F9x1XBOAeo

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 29: / Line 29: @@
 ==Solution 2==
-Let <math>x</math> be the number of elements in <math>A</math> not including the intersection. <math>2007-1001=1006</math> total elements excluding the intersection. Since we know that <math>A=B</math>, we can find that <math>x=\frac{1006}2=503</math>. Now we need to add the intersection. <math>503+1001=1504</math>, so the answer is <math>\boxed{\textbf{(C}) 1504}</math>.
+Let <math>x</math> be the number of elements in <math>A</math> not including the intersection. <math>2007-1001=1006</math> total elements excluding the intersection. Since we know that <math>A=B</math>, we can find that <math>x=\frac{1006}2=503</math>. Now we need to add the intersection. <math>503+1001=\boxed{\textbf{(C)} 1504}</math>.
 ==Video Solution by WhyMath==

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