Quick Answer: 4 Liters of 80% Acid should be used.
Comment
<em>One</em>: 50% of the final volume must be acid.
<em>Two</em>: We don't know the volume of 80% acid needed.
<em>Three</em>: We don't know the volume of the final mixture.
What we do know.
One: We know how much of the 30% acid we have. that's really all we do know.
Volume of the 30% acid = 6 L
Volume of the acid = (30/100 * 6) = 180/100 = 1.8 liters of the 6 liters is acid.
What we can set up.
Let the volume of the 80% acid = x Liters.
Let the acid content = (80/100) * x and this will be in liters
Let the acid content = 0.8*x
Let the total volume = 6 + x (of both acid and water).
Let the acid content of this volume = 50% of the total volume
So the acid content of the result = 50% (x + 6) = 50/100(x + 6) = 0.5(x + 6)
That's the right side of the equation.
The left side is what we know about the acid itself
We know that the 30% content will contribute 6* 30/100 = 1.8 L of acid.
The 80% acid will contribute 0.8*x
Total acid content = 0.8x + 1.8 This is the left hand side.
Equation
0.8x + 1.8 = 0.5(x + 6)
Solve
0.8x + 1.8 = 0.5x +3 Subtract 0.5x from both sides.
0.8x - 0.5x + 1.8 = 3 Combine like terms.
0.3x + 1.8 = 3 Subtract 1.8 from both sides
0.3x = 1.2 Divide by 0.3
x = 4 <<<< answer 4 liters of 80% acid should be used.
Check
Amount of acid contributed by each acid
1.8 L of the 30% acid
3.2 L of the 80% acid
Total volume of acid in the mixture = 5 L
The total volume of 80% + 30% acid = 6 + 4 = 10 L
50% of that amount is 5 Liters. We have solved the problem correctly
Note
You may wonder where I got that 3.2 L from for the 80% acid.
You have 4L total. 80% of it is acid. 80/100 * 4 = 3.2 Liters are acid.
X^2-8x+21
=x^2-2(4)(x) +16+5
=(x-4)^2+5
here a=1
h=4
k=5
area = x^2 - 4x + 3
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If the length of rectangular school hall is 4 times the width and the perimeter be 55m, then the length is 22m and the width of the school hall be 5.5m.
Given that the length of rectangular school hall is 4 times the width.
We are required to find the length and breadth of the rectangle which has the length be 4 times the width.
Perimeter of rectangle is the sum of all the length and breadth of that rectangles.
Perimeter of rectangle is given as 55m
Perimeter=2(L+B)
let the width of rectangle be x.
Length of that rectangle be 4x.
According to question,
2(x+4x)=55
2*5x=55
10x=55
x=55/10
x=5.5 m
Width be 5.5 m.
Length of rectangle be 4*5.5=22m.
Hence if the length of rectangular school hall is 4 times the width then the length is 22m and the width of the school hall be 5.5m.
Learn more about perimeter at brainly.com/question/19819849
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Answer:
24
Step-by-step explanation: