SOLUTION: Question:
A mixture of 20% acid and 50% acid must be mixed together to obtain 100mL of 44% acid solution. How much of each acid will be required?
Answer:
Let x represent amoun
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Question 695427: Question:
A mixture of 20% acid and 50% acid must be mixed together to obtain 100mL of 44% acid solution. How much of each acid will be required?
Answer:
Let x represent amount (in mL) of 20% acid required
Let y represent amount (in mL) of 50% acid required
100 = x + y
44 = 0.2x + 0.5y
I got the right answer
x = 20
y = 80
I just don't understand why 44 isn't changed into 0.44 and the other numbers are. Would you mind explaining?
Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!
Question:
A mixture of 20% acid and 50% acid must be mixed together to obtain 100mL of 44% acid solution. How much of each acid will be required?
Answer:
Let x represent amount (in mL) of 20% acid required
Let y represent amount (in mL) of 50% acid required
100 = x + y
44 = 0.2x + 0.5y
I got the right answer
x = 20
y = 80
I just don't understand why 44 isn't changed into 0.44 and the other numbers are. Would you mind explaining?
x + y = 100 ----- eq (i)
0.2x + 0.5y = .44(100) ------ 0.2x + 0.5y = 44
Do you now see how 44 was derived?
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