AMC 10 · 2010 · #8
Easy mode Grade 4One ticket to a play costs a whole number of dollars, x. Every ticket costs the same. A group of 9th graders spends 48$ in all on tickets. A group of 10th graders spends64inall.Howmanydifferentvaluescouldx$ be?
Pick an answer.
AMC 10 2010 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 ticket costs a whole number of dollars, x. One group spent $48 in total on tickets and another group spent $64 in total. Count how many different whole-number ticket prices x could make both totals possible.
Givens: The ticket price x is a whole number of dollars.; The 9th graders spent $48 in total.; The 10th graders spent $64 in total.
Unknowns: How many different values of x are possible.
Understand
Restated: A ticket costs a whole number of dollars, x. One group spent $48 in total on tickets and another group spent $64 in total. Count how many different whole-number ticket prices x could make both totals possible.
Givens: The ticket price x is a whole number of dollars.; The 9th graders spent $48 in total.; The 10th graders spent $64 in total.
Plan
Primary tool: #2 Make a Systematic List
Secondary: #4 Introduce a Variable, #8 Analyze the Units
The question asks how many prices work, so the plan is to nail down the rule every valid price must obey, then list all numbers that obey it and count them. A price times a whole number of tickets equals a total, so the price must divide each total exactly. Listing the numbers that divide both totals turns the puzzle into a short, checkable list.
Execute — Answer: E
4.OA.B.4 Step 1 Price divides $48
- Let t be the number of tickets the 9th graders bought.
- Their spending is the price times the count of tickets, so 48 = t times x.
- Because t is a whole number, x has to divide 48 exactly.
💡 If the money splits into whole tickets, the price must go into the total evenly.
4.OA.B.4 Step 2 Price divides $64
- The same reasoning applies to the 10th graders.
- If they bought s tickets, then 64 = s times x with s a whole number, so x must also divide 64 exactly.
💡 The one price has to fit both totals, so it must divide each of them.
4.OA.B.4 Step 3 List common factors
- So x must be a factor of 48 and a factor of 64 at the same time.
- Write out the factors of each, then keep the ones that appear in both lists.
💡 A price that works must live in both factor lists at once.
4.OA.B.4 Step 4 Count the values
- The numbers that divide both totals are 1, 2, 4, 8, and 16.
- That is 5 possible ticket prices, so the answer is (E).
💡 Counting the shared factors directly answers how many prices are possible.
4.OA.B.4 Let t be the number of tickets the 9th graders bought. Their spending is the pri 4.OA.B.4 The same reasoning applies to the 10th graders. If they bought s tickets, then 6 4.OA.B.4 So x must be a factor of 48 and a factor of 64 at the same time. Write out the f 4.OA.B.4 The numbers that divide both totals are 1, 2, 4, 8, and 16. That is 5 possible t Review
Reasonableness: Each listed price really works: for x = 16, the 9th graders buy 3 tickets ($48) and the 10th graders buy 4 tickets ($64), both whole numbers. A price like 5 fails because 48 is not a multiple of 5, so not every small number counts, only the 5 shared factors do. Five matches choice (E).
Alternative: The numbers that divide both 48 and 64 are exactly the divisors of their greatest common factor. The greatest common factor of 48 and 64 is 16, and 16 has the divisors 1, 2, 4, 8, 16, which is again 5 values.
CCSS standards used (min grade 4)
4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Recognizing the price must be a factor of each total, then listing the factors of 48 and 64 to find the ones they share.)
⭐ A ticket price has to divide every group's total exactly, so just count the numbers that go evenly into both totals.
⭐ A ticket price has to divide every group's total exactly, so just count the numbers that go evenly into both totals.
More like this
Same archetype — closest grade level first.