Difference between revisions of "2004 AMC 10A Problems/Problem 15"
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==Problem== | ==Problem== | ||
Given that <math>-4\leq x\leq-2</math> and <math>2\leq y\leq4</math>, what is the largest possible value of <math>\frac{x+y}{x}</math>? | Given that <math>-4\leq x\leq-2</math> and <math>2\leq y\leq4</math>, what is the largest possible value of <math>\frac{x+y}{x}</math>? | ||
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Now if we if set our value of <math>x</math> to its lowest, our expression becomes <math>\frac{(-2+2)}{-2} = \frac{0}{2}</math> As we decrease our <math>x</math> value, we see that our numerator decrease from <math>0</math> and our denominator decrease from <math>-2</math> at the same rate. So decreasing our <math>x</math> value decreases the overwhelming gap between our denominator and numerator, which gives us an overall bigger number. | Now if we if set our value of <math>x</math> to its lowest, our expression becomes <math>\frac{(-2+2)}{-2} = \frac{0}{2}</math> As we decrease our <math>x</math> value, we see that our numerator decrease from <math>0</math> and our denominator decrease from <math>-2</math> at the same rate. So decreasing our <math>x</math> value decreases the overwhelming gap between our denominator and numerator, which gives us an overall bigger number. | ||
| − | So we also pick the smallest value for <math>x</math>, which is <math>-4</math>. We know have <math>\frac{(-4+2)}{-4} = \frac{-2}{-4} = | + | So we also pick the smallest value for <math>x</math>, which is <math>-4</math>. We know have <math>\frac{(-4+2)}{-4} = \frac{-2}{-4} = \frac{1}{2}</math>. |
Therefore, <math>\boxed{\text{(D)}}</math> is our greatest possible value. | Therefore, <math>\boxed{\text{(D)}}</math> is our greatest possible value. | ||
-JinhoK | -JinhoK | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/LBgCCFdCvYc | ||
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| + | Education, the Study of Everything | ||
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==See also== | ==See also== | ||
Latest revision as of 01:16, 5 July 2021
Problem
Given that 
and 
, what is the largest possible value of 
?
Solution
Rewrite 
as 
.
We also know that 
because 
and 
are of opposite sign.
Therefore, 
is maximized when 
is minimized, which occurs when 
is the largest and 
is the smallest.
This occurs at 
, so 
.
Solution 2
If the answer choice is valid, then it must satisfy . We use answer choices from greatest to least since the question asks for the greatest value.
Answer choice 
. We see that if 
then
and 
. However, 
is not in the domain of , so is incorrect.
Answer choice 
, however, we can find a value that satisfies 
which simplifies to 
, such as .
Therefore, 
is the greatest.
Solution 3
As , we know that the denominator of our given fraction is negative. So to achieve the greatest value possible, our numerator, or 
must also be as small as possible.
So we pick our smallest value for , which is 
.
Now if we if set our value of to its lowest, our expression becomes 
As we decrease our value, we see that our numerator decrease from and our denominator decrease from 
at the same rate. So decreasing our value decreases the overwhelming gap between our denominator and numerator, which gives us an overall bigger number.
So we also pick the smallest value for , which is 
. We know have 
.
Therefore, is our greatest possible value.
-JinhoK
Video Solution
Education, the Study of Everything
See also
| 2004 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
