AMC 10 · 2014 · #6
Grade 7 rate-ratioSuppose that a cows give b gallons of milk in c days. At this rate, how many gallons of milk will d cows give in e days?
Pick an answer.
AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: $a$ cows produce $b$ gallons in $c$ days; Every cow makes milk at the same constant rate; Answer choices: (A) $\frac{bde}{ac}$, (B) $\frac{ac}{bde}$, (C) $\frac{abde}{c}$, (D) $\frac{bcde}{a}$, (E) $\frac{abc}{de}$
Unknowns: The number of gallons $d$ cows produce in $e$ days, written in terms of $a,b,c,d,e$
Understand
Restated: 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.
Givens: $a$ cows produce $b$ gallons in $c$ days; Every cow makes milk at the same constant rate; Answer choices: (A) $\frac{bde}{ac}$, (B) $\frac{ac}{bde}$, (C) $\frac{abde}{c}$, (D) $\frac{bcde}{a}$, (E) $\frac{abc}{de}$
Plan
Primary tool: #8 Analyze the Units
Secondary: #4 Introduce a Variable, #9 Solve an Easier Related Problem
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.
Execute — Answer: A
6.RP.A.2 Step 1 Find milk per cow per day
- The $b$ gallons were made by $a$ cows working for $c$ days, so the work was spread over $a\times c$ cow-days.
- Divide the gallons by that many cow-days to get the rate for a single cow in a single day.
💡 One steady rate per cow-day is the hidden constant that every version of the question shares.
7.RP.A.1 Step 2 Scale up to the new herd and time
- Now run that same rate for the new situation.
- Each of the $d$ cows makes $\frac{b}{ac}$ gallons every day, and they work for $e$ days, so multiply the rate by $d$ cows and by $e$ days.
💡 More cows and more days each multiply the output, so both new counts sit on top of the fraction.
6.EE.A.2 Step 3 Check the units and match a choice
- The units confirm the setup: $\frac{\text{gallons}}{\text{cow}\cdot\text{day}}\times\text{cows}\times\text{days}$ cancels to gallons, exactly what the question asks for.
- The expression $\frac{bde}{ac}$ matches choice (A).
💡 If the leftover unit is gallons, the formula is built correctly.
6.RP.A.2 The $b$ gallons were made by $a$ cows working for $c$ days, so the work was spre 7.RP.A.1 Now run that same rate for the new situation. Each of the $d$ cows makes $\frac{ 6.EE.A.2 The units confirm the setup: $\frac{\text{gallons}}{\text{cow}\cdot\text{day}}\t Review
Reasonableness: Sanity-check the shape of the answer: more cows ($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 $\frac{3}{8}$ gallon per cow-day, and $5$ cows over $6$ days give $\frac{3}{8}\times5\times6=\frac{90}{8}=11.25$ gallons; the formula $\frac{bde}{ac}=\frac{90}{8}$ agrees. The answer is (A).
Alternative: Use Tool #9 with concrete numbers instead of the unit rate. Let $a=2,b=3,c=4,d=5,e=6$ and read the story: "2 cows give 3 gallons in 4 days; how much do 5 cows give in 6 days?" Two cows give $\frac{3}{4}$ gallon per day, so one cow gives $\frac{3}{8}$ per day; five cows give $\frac{15}{8}$ per day; over six days that is $\frac{90}{8}=11.25$ gallons. Testing each choice with these numbers, only (A) yields $\frac{90}{8}$ — the others give $\frac{8}{90}$, $45$, $180$, and $0.8$ — so (A) is the unique match.
CCSS standards used (min grade 7)
6.RP.A.2Understand the concept of a unit rate and use rate language (Collapsing the given information into a single unit rate: gallons per cow per day.)7.RP.A.1Compute unit rates associated with ratios of fractions (Working with the fractional rate $\frac{b}{ac}$ and scaling it by two independent factors ($d$ cows and $e$ days).)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Building and reading the literal answer $\frac{bde}{ac}$ from the five letter quantities.)
⭐ 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.
⭐ 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.
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