Difference between revisions of "2001 AMC 8 Problems/Problem 2"
m (Added question.) |
Talkinaway (talk | contribs) |
||
| Line 5: | Line 5: | ||
<math>\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 12</math> | <math>\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 12</math> | ||
| − | ==Solution== | + | ==Solution 1== |
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>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | Use the answers to attempt to "reverse engineer" an appropriate pair of numbers. Looking at option <math>A</math>, guess that one of the numbers is <math>3</math>. If the sum of two numbers is <math>11</math> and one is <math>3</math>, then other must be <math>11 - 3 = 8</math>. The product of those numbers is <math>3\cdot 8 = 24</math>, which is the second condition of the problem, so our number are <math>3</math> and <math>8</math>. | ||
| + | |||
| + | However, <math>3</math> is the smaller of the two numbers, so the answer is <math> 8</math> or <math>\boxed{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 23:25, 30 July 2011
Contents
Problem
I'm thinking of two whole numbers. Their product is 24 and their sum is 11. What is the larger number?
Solution 1
Let the numbers be 
and 
. Then we have 
and 
. Solving for in the first equation yields 
, and substituting this into the second equation gives 
. Simplifying this gives 
, or 
. This factors as 
, so 
or 
, and the corresponding values are 
and 
. These are essentially the same answer: one number is 
and one number is 
, so the largest number is 
.
Solution 2
Use the answers to attempt to "reverse engineer" an appropriate pair of numbers. Looking at option 
, guess that one of the numbers is . If the sum of two numbers is 
and one is , then other must be 
. The product of those numbers is 
, which is the second condition of the problem, so our number are and .
However, is the smaller of the two numbers, so the answer is or 
.
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 | ||
