AMC 10 Step by Step

2010 AMC 10A Problem 3

P1-10~1 minPrint

Tyrone had 9797 marbles and Eric had 1111 marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?

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Key insight

Passing marbles keeps the total at 108. It ends split 2 to 1, so Tyrone keeps 72 and gave away 25.

Solution

When marbles move from one boy to the other, the total does not change. Use that total.

  1. Together they have 97+11=10897 + 11 = 108 marbles, before and after.
  2. At the end Tyrone has twice as many as Eric. Think of it as 33 equal parts: Eric has 11 part and Tyrone has 22 parts.
  3. One part is 108÷3=36108 \div 3 = 36. So Eric ends with 3636 and Tyrone ends with 7272.
  4. Tyrone started with 9797 and ended with 7272, so he gave away 97−72=2597 - 72 = 25.
  5. Check from Eric's side: 11+25=3611 + 25 = 36. It matches.

The answer is (D) 25\boxed{\textbf{(D)}\ 25}.

Why this works

When things are only passed around, the total stays the same. Knowing the total and the final ratio tells you everyone's final amount, and the amount passed is just the change. Look for a quantity that stays fixed whenever something is "given" or "moved."

Alternative approach

Let xx be the number of marbles Tyrone gives. Afterwards Tyrone has 97−x97 - x and Eric has 11+x11 + x. Tyrone has twice as many:
97−x=2(11+x)=22+2x.97 - x = 2(11 + x) = 22 + 2x.
Add xx to both sides and subtract 2222: 75=3x75 = 3x, so x=25x = 25.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Giving marbles back and forth does not change the total, which is 97+11=10897 + 11 = 108.

At the end Tyrone has twice as many as Eric, so the 108108 marbles split in the ratio 2:12 : 1: Eric has 1083=36\dfrac{108}{3} = 36 and Tyrone has 7272.

Tyrone therefore gave away 97−72=2597 - 72 = 25 marbles (equivalently, Eric gained 36−11=2536 - 11 = 25).

The answer is (D) 25\boxed{\textbf{(D)}\ 25}.

Review in the AoPS books

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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.