AMC 10 · 2014 · #6

Grade 7 rate-ratio

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

rateratio-proportion dimensional-analysisconvert-to-algebraeasier-related-problem ↑ Prerequisites: ratio-proportion

📏 Short solution 💡 1 insight

Problem

A group of a cows together produce b gallons of milk over c days. Assuming every cow keeps making milk at the same steady rate, write a formula for how many gallons d cows produce over e days.

Pick an answer.

(A)

$\frac{bde}{ac}$

(B)

$\frac{ac}{bde}$

(C)

$\frac{abde}{c}$

(D)

$\frac{bcde}{a}$

(E)

$\frac{abc}{de}$

AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Analyze the Units

The phrase "at this rate" is the signal for Tool #8: track the units. Milk is measured in gallons, but the thing that stays constant is gallons per cow per day. If we boil the given information down to that one unit rate, the rest is just multiplying it back up by the new number of cows and the new number of days. Tool #4 helps keep the five letters straight — treat each as a placeholder and build the expression piece by piece. Tool #9 is the safety net: if the letters feel slippery, replace them with easy numbers, solve the concrete version, and confirm the formula reproduces that number.

1STEP 1

Find milk per cow per day

The b gallons came from a cows working c days, that is a×c cow-days, so dividing gives b/ac gallons per cow per day.

rate = (b gallons)/(a cows × c days) = b/ac gallons per cow per day

2STEP 2

Scale up to the new herd and time

Run that same rate forward: d cows each making b/ac gallons a day for e days yields bde/ac gallons.

b/ac × d × e = bde/ac gallons

3STEP 3

Check the units and match a choice

Units confirm it: gallons/(cow·day)×cows×days leaves plain gallons, so bde/ac is choice (A).

gal/(cow·day)·cow·day=gal → bde/ac → (A)

Answer

bde/ac

Sanity-check the shape of the answer: more cows (d) or more days (e) should give more milk, and both appear in the numerator, so the amount grows when they grow — good. The starting cows a and days c appear in the denominator, which is right because if the original a cows had needed more days c to make the same b gallons, each cow would be slower, so the output should shrink. Plugging in a=2,b=3,c=4,d=5,e=6: the rate is 3/8 gallon per cow-day, and 5 cows over 6 days give 3/8×5×6=90/8=11.25 gallons; the formula bde/ac=90/8 agrees. The answer is (A).

💡Key takeaway

For any "at this rate" problem, shrink everything down to one unit (one cow, one day), find that single rate, then multiply it back up by the new amounts.

Find milk per cow per day
Scale up to the new herd and time
Check the units and match a choice

Find milk per cow per day

Scale up to the new herd and time

Check the units and match a choice

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