SOLUTION: Please help me solve this word problem. Elise wants to make a 61% alcohol solution. She has already poured 3 gal. of a 40% alcohol solution into a beaker. How many gal. of a 70

Question 741162: Please help me solve this word problem.
Elise wants to make a 61% alcohol solution. She has already poured 3 gal. of a 40% alcohol solution into a beaker. How many gal. of a 70% alcohol solution must she add to this to create the desired mixture?
This is what I have tried.
.61x+1.2=.7x+2.1
.61x=.7x+1.1
.87x=1.1
x=1.26 gallons
Is this correct?
Found 2 solutions by stanbon, josgarithmetic:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Elise wants to make a 61% alcohol solution. She has already poured 3 gal. of a 40% alcohol solution into a beaker. How many gal. of a 70% alcohol solution must she add to this to create the desired mixture?
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Equation:
alch + alch = alch
0.40*3 + 0.70x = 0.61(3+x)
--------
40*3 + 70x = 61*3 + 61x
--------
9x = 21*3
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x = 7 gallons (amt of 70% solution needed)
==================
Cheers,
Stan H.
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
Is that beaker big enough?

Let y = gallons of 70% alcohol to add.

That is the 61% target concentration is the ratio of total amount of alcohol to the total amount of solution, expressing concentrations as percents.

Solve for y.

You could do it like, but you may still multiply both sides by 100, so you'll have the equivalent of the above highlighted equation.

Further steps,

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