Question 596·Hard·Algebra

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A jar contained some identical marbles. Evan removed one-third of the marbles plus 4 more from the jar. Later, Maya removed half of the marbles that remained, leaving 20 marbles in the jar. How many marbles were originally in the jar?

Your answer

A sequence of removals is a one-variable equation story: choose a letter for the starting count, then track what's left. Work backward to undo the last removal, and equate that result to what the first person left. A Desmos regression can solve the finished equation. Each fraction applies to the amount present at that moment, not automatically to the starting amount.

Hints

Hint 1
A half means two equal parts. Maya took one part and left 20 as the other. How many marbles were there immediately before she took any?
Hint 2
Let $x$ be the original count. Evan goes first, so the fraction he takes is $\frac{x}{3}$ . How do his extra 4 marbles change the amount left for Maya?
Hint 3
Keep everything Evan took together in parentheses. A one-variable equation can set the amount Evan left equal to the count you found by undoing Maya's removal.

Step-by-step

Approach 1: Model what Evan left, then solve
Step 1
Undo Maya's removal
Maya took half of the marbles she found, so the 20 left is the other half. Double it to find what she found:
$2(20)=40.$
That is the count after Evan, not the original count.
Step 2
Describe what Evan left
Let $x$ be the original number of marbles. Evan went first, so he took $\frac{x}{3}+4$ . Subtract his entire removal from $x$ to get the count Maya found:
$x-\left(\frac{x}{3}+4\right).$
The parentheses keep the extra 4 among the marbles taken, rather than added back.
Step 3
Match the two counts
The expression for what Evan left must equal the 40 marbles Maya found. So write the one-variable equation:
$x-\left(\frac{x}{3}+4\right)=40.$
Don't take half of the original $x$ for Maya; she took half of what remained after Evan.
Step 4
Solve for the original count
Type $x_{1}-\left(\frac{x_{1}}{3}+4\right)\sim40$ in Desmos. Use $x_{1}$ instead of $x$ so Desmos solves for the count rather than graphing it; $\sim$ tells it to match the two sides. Under PARAMETERS, Desmos reports $x_{1}=66$ . Since $x_{1}$ stands for the starting count, the jar originally held 66 marbles. Grid in 66.

Approach 2: Undo Evan's removals in reverse order
Step 1
Put back the extra marbles
Start with the 40 marbles Maya found. Undo Evan's extra 4 first:
$40+4=44.$
This is what would have remained if Evan had taken only one-third of the original count.
Step 2
Undo the one-third removal
Taking one-third leaves two-thirds, so 44 is $\frac{2}{3}$ of the original count. Multiply by its reciprocal, $\frac{3}{2}$ , which reverses multiplication by $\frac{2}{3}$ :
$44\times\frac{3}{2}=66.$
The jar originally contained 66 marbles. Grid in 66.

Question 596·Hard·Algebra

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