AMC 10 Step by Step

2012 AMC 10A Problem 5

P1-10~2 minPrint

Last year 100100 adult cats, half of whom were female, were brought into the Smallville Animal Shelter. Half of the adult female cats were accompanied by a litter of kittens. The average number of kittens per litter was 44 . What was the total number of cats and kittens received by the shelter last year?

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Key insight

Follow the halves one at a time: 50 females, 25 litters, 25 times 4 = 100 kittens, then add the 100 adult cats.

Solution

Work through the story one step at a time and write down each number.

  1. Adult cats: 100100. Half are female, so there are 5050 females.
  2. Half of those 5050 females came with a litter, so there were 2525 litters.
  3. The litters average 44 kittens each. Some litters may be bigger and some smaller, but an average of 44 over 2525 litters always means 25×4=10025 \times 4 = 100 kittens in total.
  4. The question asks for all the animals, adults and kittens: 100+100=200100 + 100 = 200.

The answer is (B) 200\boxed{\textbf{(B)}\ 200}.

Why this works

An average times the number of groups gives the total, even when the groups are different sizes. In a word problem with several steps, writing each number down as you go keeps you from stopping one step early. Before choosing an answer, reread what the question actually asks for; here it is the total of adults and kittens.

Concise solution

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

Follow the chain of halves.

Adult females: half of 100100, so 5050.

Litters: half of the 5050 females brought one, so 2525 litters.

Kittens: an average of 44 per litter over 2525 litters means 25⋅4=10025 \cdot 4 = 100 kittens in all (the average times the number of litters is the total, whatever the individual litter sizes).

Total animals received: 100100 adults +100+ 100 kittens =200= 200.

The answer is (B) 200\boxed{\textbf{(B)}\ 200}.

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.