Tyrone had 97 marbles and Eric had 11 marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?
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Toolkit + CCSS Solution
Understand
Restated: Tyrone starts with 97 marbles and Eric with 11. Tyrone hands some of his marbles to Eric until Tyrone has exactly twice as many marbles as Eric. Find how many marbles Tyrone handed over.
Givens: Tyrone begins with 97 marbles.; Eric begins with 11 marbles.; After the gift, Tyrone has twice as many marbles as Eric.
Unknowns: The number of marbles Tyrone gave to Eric.
Understand
Restated: Tyrone starts with 97 marbles and Eric with 11. Tyrone hands some of his marbles to Eric until Tyrone has exactly twice as many marbles as Eric. Find how many marbles Tyrone handed over.
Givens: Tyrone begins with 97 marbles.; Eric begins with 11 marbles.; After the gift, Tyrone has twice as many marbles as Eric.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #16 Change Focus / Count the Complement, #11 Work Backwards
Instead of tracking the messy transfer, focus on the ending totals. The marbles are just shared between two boys, so the pile stays the same size the whole time. Naming Eric's final count as a variable turns 'twice as many' into one short equation, and once the ending amounts are known, a single subtraction recovers how many marbles actually moved.
Execute — Answer: D
#16 Change Focus / Count the Complement 4.NBT.B.4Step 1
Find the unchanging total
Handing marbles from one boy to the other never creates or destroys any, so the two piles always add up to the same number.
Add the starting amounts: $97+11=108$.
There are 108 marbles at every moment.
$$97+11=108$$
💡 Moving marbles around never changes how many there are in total.
#4 Introduce a Variable 6.EE.B.6Step 2
Name the ending amounts
Let $E$ stand for the number of marbles Eric ends with.
Tyrone ends with twice as many, so Tyrone has $2E$.
Together they still hold all 108 marbles, which gives the equation $2E+E=108$.
$$2E+E=108$$
💡 'Twice as many' just means one share for Eric and two equal shares for Tyrone.
#4 Introduce a Variable 6.EE.B.7Step 3
Solve for the ending amounts
Combine the like terms: $2E+E=3E$, so $3E=108$.
Divide both sides by 3 to get $E=36$.
Eric ends with 36 marbles and Tyrone ends with $2\cdot36=72$.
$$3E=108\Rightarrow E=36,\quad 2E=72$$
💡 Three equal shares make 108, so each share is a third of 108.
#11 Work Backwards 4.NBT.B.4Step 4
Answer the real question
The question asks how many marbles were given, not the final counts.
Eric climbed from 11 marbles to 36, so he received $36-11=25$ marbles.
Checking Tyrone: he dropped from 97 to 72, also a change of 25.
So Tyrone gave 25 marbles, which is choice (D).
$$36-11=25$$
💡 To find what was handed over, compare each boy's before and after.
[1]
#16 4.NBT.B.4Handing marbles from one boy to the other never creates or destroys any, so the
[2]
#4 6.EE.B.6Let $E$ stand for the number of marbles Eric ends with. Tyrone ends with twice a
[3]
#4 6.EE.B.7Combine the like terms: $2E+E=3E$, so $3E=108$. Divide both sides by 3 to get $E
[4]
#11 4.NBT.B.4The question asks how many marbles were given, not the final counts. Eric climbe
Review
Reasonableness: The final counts 72 and 36 fit the rule exactly, since $72=2\cdot36$, and they still add to 108. The transfer of 25 is under half of Tyrone's original 97, which makes sense: he only needs to give enough to fall to twice Eric's amount, not to hand over most of his marbles. Among the choices, only (D) makes the ending split work.
Alternative: Track the transfer directly. Let $x$ be the marbles given and write $97-x=2(11+x)$. Expanding gives $97-x=22+2x$, so $75=3x$ and $x=25$, matching (D).
CCSS standards used (min grade 6)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Adding 97 and 11 to get the total 108, and subtracting 11 from 36 to find the marbles given.)
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Letting $E$ be Eric's final count and writing Tyrone's count as $2E$.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Solving $3E=108$ to find $E=36$.)
⭐ When things are only shared back and forth, the total stays fixed, so split that total into equal shares and read off the answer.
⭐ When things are only shared back and forth, the total stays fixed, so split that total into equal shares and read off the answer.