MAT 130 College Algebra Milestone 3 2026 (pdf) | College Algebra | SNHU

Sophia College Algebra
MILESTONE 3 RETAKE GUIDE
Southern New Hampshire University
Actual Questions & Verified Answers with Rationales

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, SCORE
UNIT 3 — MILESTONE 3 17/18



17/18  that's 94% RETAKE



 17 questions were answered correctly.
 1 question was answered incorrectly.



1 
●

On her way home from the school board meeting, Kelly fills up her car. She likes the idea of using
gasoline with ethanol, but thinks her car can only handle 50% ethanol. At the gas station, she can
use regular gas with 15% ethanol or E78 fuel with 78% ethanol.


How many gallons of each type of fuel should Kelly use if she wants to fill up her car with 9
gallons of fuel containing 50% ethanol?



5 gallons of regular gas with 15% ethanol;
 4 gallons of E78 fuel with 78% ethanol



3 gallons of regular gas with 15% ethanol;
 6 gallons of E78 fuel with 78% ethanol



6 gallons of regular gas with 15% ethanol;

,  3 gallons of E78 fuel with 78% ethanol



4 gallons of regular gas with 15% ethanol;
  5 gallons of E78 fuel with 78% ethanol



RATIONALE


We will use the variables x and y to represent the types of fuel: x
represents the gallons of 15% ethanol gas, and y represents the gallons
of 78% gas. The first equation is the total amount of gas Kelly will use to
fill up her car.

She can use the two types of fuel, and together, she puts 9 gallons of gas
in her car, so x + y = 9. The second equation will represent the amount of
ethanol from the two fuels.

The coefficient to x is 0.15 because that is the 15% ethanol fuel. The
coefficient to y is 0.78 because that is the 78% ethanol fuel. Finally, she
wants to have 9 gallons of 50% ethanol, which can be expressed as 0.50
times 9. We now need to solve this system of equations.

Since we have two equations that represent this equation, one way to
solve is to use substitution to rewrite one variable in terms of the other,
and solve one variable at a time. Let's take a look at the second equation.

In equation x + y = 9, subtracting x from both sides gives us y = 9 – x. We
can use this in the other equation to write y as 9 – x.

This is the other equation in the system, but y has been replaced with an
equivalent expression of 9 – x. Now this is a single-variable equation and
we can solve for x. First, distribute 0.78 into (9 – x).

0.78 times 9 is 7.02 and 0.78 times x is 0.78x. Next, evaluate the
multiplication on the right side.

0.50 times 9 is 4.5. Now, combine like terms on the left side.