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How many gallons of a 50% antifreeze solution must be mixed with 60 gallons of 30% antifreeze to get a mixture that is 42% antifreeze?

 Aug 26, 2020
 #1
avatar+1039 
+1

We can form a "proportion" with information. First, let's find out a general "formula":

 

a%         b%           c%

----   +    ----   =    ---------

 x            y             x+y


Multiplying the percentages with the corresponding gallon, we see:    ax + by = c(x+y)

 

From here, we can plug in numbers:

 

0.5x + 0.3(60) = (0.42)(x + 60)

 

=> 0.5x + 18 = 0.42x + 25.2

 

=> 0.5x = 0.42x + 7.2

 

=> 0.08x = 7.2

 

=> x = 90

 

Therefore, we must add 90 gallons.

 

Edit: Thanks to guest's answer, I corrected my numbers (but my answer remains the same...)

 

:)

 Aug 26, 2020
edited by ilorty  Aug 26, 2020
 #2
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Let the number of gallon of 50% antifrreeze =G

 

0.50G + 0.30*60 =0.42 *[G + 60]

0.50G +  18 =0.42G + 25.2

0.50G - 0.42G =25.2 - 18

0.08G = 7.2

G = 7.2 / 0.08

G = 90 - gallons of 50% antifreeze must be mixed with 60 gallons of 30% antifreeze to make [90 + 60]=150 gallons of 0.42% of antifreeze.

 Aug 26, 2020
 #3
avatar+1039 
+1

Remember, it's not asking for how many gallons of 42%, but 50%!

 

:)

ilorty  Aug 26, 2020

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