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

Latest revision as of 13:33, 8 July 2024

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

Video Solution by AliceWang

https://youtu.be/ThBO09fGBgM

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 17: / Line 17: @@
 == Solution ==
-Let <math>x</math> be the number of elements in <math>A</math> and <math>B</math>.
+Let <math>x</math> be the number of elements in <math>A</math> and <math>B</math> which is equal.
-Since the union is the sum of all elements in <math>A</math> and <math>B</math>,
-and <math>A</math> and <math>B</math> have the same number of elements then,
+Then we could form equation
 <math>2x-1001 = 2007</math>
@@ Line 31: / Line 28: @@
 The answer is <math>\boxed{\textbf{(C)}\ 1504}</math>
-== Solution 2 ==
+==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=\boxed{\textbf{(C)} 1504}</math>.
-First find the number of elements in <math>A</math> without including the intersection. There are 2007 elements in total, so there are <math>1006</math> elements in <math>A</math> and <math>B</math> excluding the intersection (<math>2007-1001</math>). There are <math>503</math> elements in set A after dividing <math>1006</math> by <math>2</math>. Add the intersection (<math>1001</math>) to get <math>\boxed{\textbf{(C)}\ 1504}</math>
-- spoamath321
 ==Video Solution by WhyMath==
@@ Line 41: / Line 35: @@
 ~savannahsolver
+==Video Solution==
+https://www.youtube.com/watch?v=6F9x1XBOAeo
+==Video Solution by AliceWang==
+https://youtu.be/ThBO09fGBgM
 ==See Also==
 {{AMC8 box|year=2007|num-b=12|num-a=14}}
 {{MAA Notice}}

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