2017 AMC 10B Problem 5
Camilla had twice as many blueberry jelly beans as cherry jelly beans. After eating pieces of each kind, she now has three times as many blueberry jelly beans as cherry jelly beans. How many blueberry jelly beans did she originally have?
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Key insight
Call the cherry count c, so blueberry is 2c. After eating 10 of each, 2c - 10 = 3(c - 10), so c = 20 and blueberry is 40.
Solution
There are fewer cherry jelly beans, so give that number a name and write everything else in terms of it.
1. Let be the number of cherry jelly beans at the start. Blueberry is twice as many: .
2. After eating of each, she has cherry and blueberry.
3. Now blueberry is three times cherry:
4. Add to both sides and subtract : .
5. The question asks for blueberry at the start: .
6. Check: at the start and (twice as many). After eating, and (three times as many).
The answer is .
Why this works
Two "times as many" facts about the same two amounts, one before and one after a change, give one equation with one unknown. Name the smaller amount, write the larger one using the first fact, then use the second fact on the changed amounts. Before answering, re-read which amount the question wants.
Alternative approach
Test the choices as the blueberry count. With blueberry there were cherry; after eating of each there are and , exactly three times. Checking the others: blueberry and cherry leaves and , not three times; the smaller choices fail the same way.
Concise solution
A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.
Let be the original number of cherry jelly beans, so there were blueberry ones. After eating of each, the counts are and , and the new ratio is :
So , and the original blueberry count is . Check: blueberry and cherry remain, a ratio.
The answer is .
Review in the AoPS books
- Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
- Introduction to Algebra, Chapter 7: Proportion (§7.4)
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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.