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Difference between revisions of "2001 AMC 8 Problems/Problem 2"

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==Problem==
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I'm thinking of two whole numbers. Their product is 24 and their sum is 11. What is the larger number?
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<math>\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 12</math>
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==Solution==
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Let the numbers be <math> x </math> and <math> y </math>. Then we have <math> x+y=11 </math> and <math> xy=24 </math>. Solving for <math> x </math> in the first equation yields <math> x=11-y </math>, and substituting this into the second equation gives <math> (11-y)(y)=24 </math>. Simplifying this gives <math> -y^2+11y=24 </math>, or <math> y^2-11y+24=0 </math>. This factors as <math> (y-3)(y-8)=0 </math>, so <math> y=3 </math> or <math> y=8 </math>, and the corresponding <math> x </math> values are <math> x=8 </math> and <math> x=3 </math>. These are essentially the same answer: one number is <math> 3 </math> and one number is <math> 8 </math>, so the largest number is <math> 8, \boxed{\text{D}} </math>.
 
Let the numbers be <math> x </math> and <math> y </math>. Then we have <math> x+y=11 </math> and <math> xy=24 </math>. Solving for <math> x </math> in the first equation yields <math> x=11-y </math>, and substituting this into the second equation gives <math> (11-y)(y)=24 </math>. Simplifying this gives <math> -y^2+11y=24 </math>, or <math> y^2-11y+24=0 </math>. This factors as <math> (y-3)(y-8)=0 </math>, so <math> y=3 </math> or <math> y=8 </math>, and the corresponding <math> x </math> values are <math> x=8 </math> and <math> x=3 </math>. These are essentially the same answer: one number is <math> 3 </math> and one number is <math> 8 </math>, so the largest number is <math> 8, \boxed{\text{D}} </math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2001|num-b=1|num-a=3}}
 
{{AMC8 box|year=2001|num-b=1|num-a=3}}

Revision as of 17:18, 19 May 2011

Problem

I'm thinking of two whole numbers. Their product is 24 and their sum is 11. What is the larger number?

$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 12$

Solution

Let the numbers be $x$ and $y$. Then we have $x+y=11$ and $xy=24$. Solving for $x$ in the first equation yields $x=11-y$, and substituting this into the second equation gives $(11-y)(y)=24$. Simplifying this gives $-y^2+11y=24$, or $y^2-11y+24=0$. This factors as $(y-3)(y-8)=0$, so $y=3$ or $y=8$, and the corresponding $x$ values are $x=8$ and $x=3$. These are essentially the same answer: one number is $3$ and one number is $8$, so the largest number is $8, \boxed{\text{D}}$.

See Also

2001 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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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