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Difference between revisions of "2022 MMATHS Problems/Problem 1"

(Created page with "=Problem= Suppose that <math>a+b = 20, b+c = 22,</math> and <math>a+c = 2022</math>. Compute <math>\frac {a-b}{c-a}</math>.")
 
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Suppose that <math>a+b = 20, b+c = 22,</math> and <math>a+c = 2022</math>. Compute <math>\frac {a-b}{c-a}</math>.
 
Suppose that <math>a+b = 20, b+c = 22,</math> and <math>a+c = 2022</math>. Compute <math>\frac {a-b}{c-a}</math>.
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==Solution 1==
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We solve everything in terms of <math>a</math>. <math>b = 20 - a</math> and <math>c = 2022 - a</math>. Therefore, <math>20-a + 2022-a = 22</math>. Solving for <math>a</math>, we get that <math>a</math> = 1010. Since <math>c = 2022-a, c = 1012</math>. Since <math>b = 20-a, b = -990</math>. Computing <math>\frac {1010-(-990)}{1012-1010}</math> gives us the answer of <math>\boxed {1000}</math>.
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~Arcticturn

Revision as of 19:54, 18 December 2022

Problem

Suppose that $a+b = 20, b+c = 22,$ and $a+c = 2022$. Compute $\frac {a-b}{c-a}$.

Solution 1

We solve everything in terms of $a$. $b = 20 - a$ and $c = 2022 - a$. Therefore, $20-a + 2022-a = 22$. Solving for $a$, we get that $a$ = 1010. Since $c = 2022-a, c = 1012$. Since $b = 20-a, b = -990$. Computing $\frac {1010-(-990)}{1012-1010}$ gives us the answer of $\boxed {1000}$.

~Arcticturn

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