Question 67·200 Super-Hard SAT Math Questions·Algebra
A metalworker creates a 180-gram alloy by mixing some alloy that is 30% aluminum with a second alloy that is 60% aluminum. He then melts this 180-gram alloy with 20 grams of a third alloy that is 80% aluminum to create a 200-gram alloy that is 50% aluminum.
Which choice is the mass, in grams, of the 60% aluminum alloy used to make the 180-gram alloy?
For multi-step mixture problems, track the amount of the key ingredient (here, grams of aluminum), not just total mass. Assign a variable to one unknown mass, express all other masses in terms of it, add up ingredient amounts from each part, and set that equal to the ingredient amount implied by the final percent and final total mass. This keeps the work to one clean linear equation.
Hints
Choose a variable for one part of the first mixture
Let represent the grams of the 60% alloy used in the 180-gram mixture. Then express the grams of the 30% alloy using and 180.
Write an equation using aluminum amounts (not total mass)
Compute the grams of aluminum contributed by each component: 60% of , 30% of , and 80% of 20.
Use the final mixture to get the target aluminum amount
The final mixture is 200 grams at 50% aluminum. Set the sum of the aluminum from all parts equal to 50% of 200, then solve for .
Desmos Guide
Create an expression for total aluminum in the final mixture
Let be the grams of 60% alloy in the first mixture. Enter the total aluminum (in grams) as
A(x)=0.6x+0.3(180-x)+0.8(20)
Graph the target aluminum amount
Since the final alloy is 200 grams at 50% aluminum, enter the constant target:
T=0.5(200)
Find where the amounts match
Graph y=A(x) and y=T. Click their intersection and read the -coordinate (this is the grams of 60% alloy).
Step-by-step Explanation
Define the variable and write aluminum amounts
Let be the mass (grams) of the 60% aluminum alloy in the 180-gram mixture.
Then the mass of the 30% alloy is .
Aluminum (in grams) in the 180-gram mixture is
The 20 grams of 80% alloy contributes
grams of aluminum.
Set up the equation using the final percent
The final mixture is 200 grams and 50% aluminum, so it contains
grams of aluminum.
Set “aluminum from the parts” equal to “aluminum in the final mixture”:
Solve the linear equation
Simplify:
So the mass of the 60% aluminum alloy is 100 grams.