a distributor has two gasohol blends: one contains 7%alcohol and the other 16%alcohol. how many gallons of each(first then second) must be mixed to make 100 gallons of gasohol containing 14%alcohol?
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Hi,
x + y = 100
7x + 16y = 14(100)
-7(x + y = 100)
7x + 16y = 1400
-7x - 7y = -700
7x + 16y = 1400
--------------------------
9y = 700
y = 700/9 or 77 7/9 gallons of 16% alcohol and x = 22 2/9 gallons of 7% alcohol <==ANSWER
I hope that helps!! :-)
x + y = 100
7x + 16y = 14(100)
-7(x + y = 100)
7x + 16y = 1400
-7x - 7y = -700
7x + 16y = 1400
--------------------------
9y = 700
y = 700/9 or 77 7/9 gallons of 16% alcohol and x = 22 2/9 gallons of 7% alcohol <==ANSWER
I hope that helps!! :-)
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Let the first gasahol blend have x gallons.
Let the second gasahol blend have y gallons.
Given:
x + y = 100
so x = 100 - y and y = 100 - x
x gallons of 7% alcohol + y gallons of 16% alcohol = 100 gallons of 14% alcohol
0.07x + 0.16y = 0.14 * 100
0.07x + 0.16 (100 - x) = 14
0.07x + 16 - 0.16x = 14
2 = 0.09x
22 2/9 = x
y = 100 - x
= 100 - 22 2/9
y = 77 7/9
so there is 22 2/9 gallons of 7% alcohol and 77 7/9 gallons of 16% alcohol
Let the second gasahol blend have y gallons.
Given:
x + y = 100
so x = 100 - y and y = 100 - x
x gallons of 7% alcohol + y gallons of 16% alcohol = 100 gallons of 14% alcohol
0.07x + 0.16y = 0.14 * 100
0.07x + 0.16 (100 - x) = 14
0.07x + 16 - 0.16x = 14
2 = 0.09x
22 2/9 = x
y = 100 - x
= 100 - 22 2/9
y = 77 7/9
so there is 22 2/9 gallons of 7% alcohol and 77 7/9 gallons of 16% alcohol