AMC 10 Step by Step

2019 AMC 10B Problem 11

P11-15~3 minPrint

Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar 11 the ratio of blue to green marbles is 9:19:1 , and the ratio of blue to green marbles in Jar 22 is 8:18:1 . There are 9595 green marbles in all. How many more blue marbles are in Jar 11 than in Jar 22 ?

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Key insight

Both jars hold the same number of marbles. Green is 1/10 of Jar 1 and 1/9 of Jar 2, so a jar holding 90 marbles is the natural unit.

Solution

The two jars are linked by having the same size, so start by finding a jar size that works for both ratios.

1. In Jar 11, blue to green is 9:19:1, so out of every 1010 marbles, 11 is green. In Jar 22, out of every 99 marbles, 11 is green.
2. The jar size must divide evenly by 1010 and by 99. Try the smallest such size, 9090 marbles per jar:
- Jar 11: 99 green, 8181 blue.
- Jar 22: 1010 green, 8080 blue.
- Green in total: 9+10=199 + 10 = 19.
3. We need 9595 green marbles in total, and 95=5⋅1995 = 5 \cdot 19. So each jar is really 55 times as big: 450450 marbles.
4. Scale up:
- Jar 11: 4545 green, 405405 blue.
- Jar 22: 5050 green, 400400 blue.
5. Jar 11 has 405−400=5405 - 400 = 5 more blue marbles.

The answer is (A) 5\boxed{\textbf{(A)}\ 5}.

Why this works

A ratio like 9:19:1 tells you what fraction of the whole each color is. When two groups share the same size, choose a trial size that fits both ratios, see what totals it gives, and then scale everything by the same factor to match the real data.

Alternative approach

Let each jar hold NN marbles. The greens are N10+N9=19N90\frac{N}{10} + \frac{N}{9} = \frac{19N}{90}. Setting this equal to 9595 gives N=450N = 450. The blues are 910⋅450=405\frac{9}{10} \cdot 450 = 405 and 89⋅450=400\frac89 \cdot 450 = 400, a difference of 55.

Concise solution

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

Let each jar hold NN marbles. In Jar 11 the greens are 110\frac{1}{10} of the marbles, and in Jar 22 they are 19\frac{1}{9}. Both fractions must be whole numbers, so NN is a multiple of 9090; write N=90kN = 90k.

  • Jar 11: 9k9k green, 81k81k blue.
  • Jar 22: 10k10k green, 80k80k blue.

Total green: 9k+10k=19k=959k + 10k = 19k = 95, so k=5k = 5 (each jar holds 450450 marbles).

Blue difference: 81k−80k=k=581k - 80k = k = 5.

The answer is (A) 5\boxed{\textbf{(A)}\ 5}.

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.