How many liters of a 12% alcohol solution must be added to 6 liters of
[#permalink]
20 Jul 2021, 15:10
Hey guys!
We are told that two solutions will be mixed to produce a final solution. The unknowns are the total volume of the second solution and the final total volume of the combination of the two solutions.
Use variables for unknowns
Let:
x = volume of the second solution
y = final volume of the combination of the two solutions
6 + x = y
Next, set up a second equation using what we know about the alcohol percentage of each solution
This is going to end up as a two equation problem where you will use the trick of subtracting one equation from another to leave just one variable
The second equation is:
.15y = 6(.24) + .12x
We know that the first solution had 6(.24) liters of alcohol, the second one was 12% alcohol, and the resulting combination was 15% alcohol
When trying to subtract one equation from another, place them above each other to see how you can manipulate them to better subtract
y = 6 + x
.15y = 6(.24) + .12x
Nasty right?
This is the time to look for a time saver, do you notice anything about either equation that can help break it down or simplify it?
15, 12, and 24 are all divisible by 3
Divide both sides of the second equation by .03:
5y = 6(8) + 4x
Now place the two equations next to each other again
5y = 48 + 4x
y = 6 + x
Multiply both sides of the second equation by 4:
5y = 48 + 4x
4y = 24 + 4x
Now subtract the bottom equation from the top one:
y = 24
If y = 24, x = y - 6 = 18
18 liters of the 12% alcohol solution must be added to the 6 liters to make the 15% alcohol solution
The answer is (C)