AMC 10 · 2014 · #2

Grade 6 algebra
Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Direct Ratio Scale-Up
linear-equations-one-varratio-proportiondecimal-arithmetic convert-to-algebraidentify-subproblems ↑ Prerequisites: linear-equations-one-varratio-proportion
📏 Short solution 💡 2 insights
Problem
One price is half the other and a known mix has a known total. Find another mix's total.

Pick an answer.

(A)
$\$35$
(B)
$\$38.50$
(C)
$\$40$
(D)
$\$42$
(E)
$\$42.50$
How to solve
Strategy Convert to Algebra

Two prices are floating around, but the sentence "children get in for half price" locks one to the other. That is the hinge of the whole problem: it turns two unknowns into one, and one unknown is exactly what a single given total can pin down. So name the adult price (Tool #4), translate the $24.50 purchase into one equation (Tool #13), and solve it. Then re-count the second purchase in the same currency of "adult tickets" (Tool #8). Afterwards, recount both purchases in child tickets instead (Tool #15) — a route with no equation at all — to check the result by a different structure, and use the choice list (Tool #3) as a final guard.

1STEP 1

Name one price, get both

One name gives both prices.

a = adult price, child price = a/2
2STEP 2

Turn the purchase into an equation

The known mix becomes one equation.

5a + 4·a/2 = 5a + 2a = 7a = 24.50
3STEP 3

Solve for the adult price

Solving gives the larger price directly.

a = 24.50/7 = 3.50, a/2 = 1.75
4STEP 4

Price the second purchase

Pricing the second mix follows immediately.

8(3.50) + 6(1.75) = 28.00 + 10.50 = 38.50
5STEP 5

Cross-check by counting in child tickets

Counting in small units confirms it, choice (B).

(22 child tickets)/(14 child tickets) = 11/7, 24.50 × 11/7 = 38.50 → (B)
Answer
$38.50
First check the prices against the given purchase: 5(3.50) + 4(1.75) = 17.50 + 7.00 = 24.50, as required, and 3.50/3.50 /1.75 are sane ticket prices that land on whole cents. Now the point the whole argument turns on: the phrase "children get in for half price" is load-bearing, not decoration. Suppose it were dropped and only 5a + 4c = 24.50 were known. Then c = (24.50 - 5a)/4, and the target total becomes 8a + 6c = 8a + (6(24.50 - 5a))/4 = 36.75 + a/2, which still depends on a — for instance a = 4.00 would force c = 1.125 and a total of 38.75,not38.75, not38.50. So one equation alone does not determine the answer; the reason it is determined here is that half price collapses two unknowns into one, making 7a = 24.50 solvable outright. The check that the answer is not merely necessary but correct is the second route in the last step, which recomputes $38.50 without ever using the first route's value of a.
💡Key takeaway

When one price is a fixed multiple of another, measure the whole purchase in one kind of ticket — then the mixed order becomes a single multiplication.

  • Name one price, get both
  • Turn the purchase into an equation
  • Solve for the adult price
  • Price the second purchase
  • Cross-check by counting in child tickets

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