2009 CEMC Gauss (Grade 8) Problems/Problem 10: Difference between revisions
Created page with "==Problem== If two numbers differ by <math>2</math> and their sum is <math>20</math>, the larger number is <math> \text{ (A) }\ 11 \qquad\text{ (B) }\ 10 \qquad\text{ (C) }\..." |
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==Solution 3 (answer choices)== | ==Solution 3 (answer choices)== | ||
We can notice that the number must be greater than <math>10</math> because <math>\frac{20}{2} = 10</math>, as the number wouldn't be the larger number otherwise. | We can notice that the number must be greater than <math>10</math> because <math>\frac{20}{2} = 10</math>, as the number wouldn't be the larger number otherwise. This leaves only answer choices A and D. | ||
Trying answer choice A, we have: | Trying answer choice A, we have: | ||
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~anabel.disher | ~anabel.disher | ||
{{CEMC box|year=2009|competition=Gauss (Grade 8)|num-b=9|num-a=11}} | |||
Latest revision as of 20:15, 19 October 2025
Problem
If two numbers differ by 
and their sum is 
, the larger number is
Solution 1
Let 
be the larger number. Since the numbers differ by , the smaller number must be 
. Thus, we can express the sum in terms of :
~anabel.disher
Solution 2
Let be the smaller number. We can use similar logic to solution 1 to conclude that 
is the larger number, and we can find that 
.
The larger number is therefore 
.
~anabel.disher
Solution 3 (answer choices)
We can notice that the number must be greater than 
because 
, as the number wouldn't be the larger number otherwise. This leaves only answer choices A and D.
Trying answer choice A, we have:
for the other number
These differ by , so the answer is 
~anabel.disher
| 2009 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
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| CEMC Gauss (Grade 8) | ||
