Depth · one concept from first principles · step 2 of 4
There are three containers (A), (B), and (C) holding water. The amount of water in (B) is of (A) more than the amount in (A), and the amount of water in (C) is of the amount in (B). The amount of water in (C) is how many times the amount in (A)? Find the answer.
Pick a friendly amount for A so both fractions divide evenly; 20 cups works.
Trying a concrete easy number turns an abstract 'how many times' question into plain arithmetic a fifth grader can do.
5.NF.B.6Solve An Easier Related ProblemPicking a convenient amount for A is safe, because the answer asked for is a comparison and not an amount.
Every relation in the problem says one container holds some fraction of another, so making A larger or smaller stretches B and C by the very same factor.
A comparison between two amounts does not move when both are stretched alike, so the answer cannot depend on which starting amount was chosen.
Scaling both sides of a comparison by the same number is multiplying it by that number over itself, which is one.
B is A plus one-fifth of A. One-fifth of 20 is 4, so B is 20 + 4 = 24 cups.
A fraction of a quantity is the whole times the fraction, so one-fifth of 20 is just 20 divided into 5 equal parts.
5.NF.B.4Identify SubproblemsC is three-fourths of B. Three-fourths of 24 is 24 divided by 4 (which is 6) times 3, giving 18 cups.
Three-fourths of an amount means split it into 4 equal parts and take 3 of them.
5.NF.B.4Identify SubproblemsC is 18 cups and A is 20, so C is 18/20 = 9/10 of A.
Because the answer is a ratio, the specific easy number I chose cancels out and the result 9/10 holds for any starting amount.
5.NF.B.6Solve An Easier Related ProblemTake a fraction of an amount by multiplying, do it step by step, and the comparison 9/10 stays the same no matter how much A starts with.
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The same shape shows up later on the AMC, as Compound Area by Decomposition / Difference. Why the shapes repeat →