Question 105·200 Super-Hard SAT Math Questions·Problem Solving and Data Analysis

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A jeweler mixes Alloy A (35% gold by mass) and Alloy B (80% gold by mass) to create 400 grams of an alloy. She then adds 50 grams of pure gold (100% gold) and 50 grams of pure silver (0% gold). The final mixture has a gold content of 56% by mass.

Which choice is the mass, in grams, of Alloy B used in the original 400-gram alloy?

A mixture equation tracks the amount of one ingredient, rather than adding percentages together. The cue is that two alloys contribute different shares of gold, and later additions change the total mass. Find the final gold mass, remove the gold added later, then use a one-variable Desmos regression for the original alloys. The trap is using the original mass as the base for the final percent.

Hints

Hint 1
In a mixture, a percentage is taken of a particular total. Both additions change the final mass, even though only pure gold adds gold. What total mass does the final $56\%$ use?
Hint 2
A gold balance counts grams of gold, not grams of every metal. Pure silver contributes zero gold, while the pure gold contributes $50$ grams. How much gold was present before either addition?
Hint 3
Let $x$ be grams of Alloy B. Alloy A then weighs $400-x$ grams. Multiply each alloy’s mass by its gold fraction, then add the results to account for the gold in the original alloy.

Step-by-step

Track the gold before and after the additions
Step 1
Find the final mass
Type $400+50+50$ in Desmos. It prints $500$ , so the final mass is $500$ grams. Pure silver adds no gold, but it still adds mass.
Step 2
Find the gold in the final mixture
Type $56\%\operatorname{of}500$ . Desmos prints $280$ , so the final mixture contains $280$ grams of gold. The $56\%$ applies to all $500$ grams, not only the original $400$ .
Step 3
Find the gold in the original alloy
The pure gold adds $50$ grams of gold; the pure silver adds none. Subtract the gold added later:
$280-50=230.$
So the original $400$ grams contained $230$ grams of gold.
Step 4
Write an equation for the original gold
Let $x$ mean grams of Alloy B. Because the original alloys total $400$ grams, Alloy A weighs $400-x$ grams. A gold fraction is the share of an alloy that is gold: $35\%=0.35$ and $80\%=0.80$ . Multiply each fraction by its alloy’s mass, then add grams of gold, not the percentages:
$0.35(400-x)+0.80x=230.$
Each side now measures gold in the original alloy.
Step 5
Solve for the mass of Alloy B
Type $0.35(400-x_1)+0.80x_1\sim230$ in Desmos. The $\sim$ tells Desmos to find the value of $x_1$ that makes both sides match. Under PARAMETERS, it shows $x_1=200$ . Since $x_1$ is the mass of Alloy B, the jeweler used $200$ grams of Alloy B. Choice C.

Question 105·200 Super-Hard SAT Math Questions·Problem Solving and Data Analysis

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