Answer:
The number of liters of 50% antifreeze solution is 50 and the number of liters of 90% antifreeze solution is 150
Step-by-step explanation:
Let
x ----> number of liters of 50% antifreeze solution
y ----> number of liters of 90% antifreeze solution
we know that
50%=50/100=0.50
90%=90/100=0.90
80%=80/100=0.80
x+y=200
x=200-y ------> equation A
0.50x+0.90y=0.8(200) -----> equation B
substitute equation A in equation B and solve for y
0.50(200-y)+0.90y=160
100-0.50y+0.90y=160
0.40y=160-100
0.40y=60
y=150 liters
Find the value of x
x=200-y
x=200-150=50 liters
therefore
The number of liters of 50% antifreeze solution is 50
The number of liters of 90% antifreeze solution is 150
50 liters for 50% antifreeze solution and 150 liters for 90% antifreeze solution.
Given:
A 50% antifreeze solution is to be mixed with a 90% antifreeze solution to get 200 liters of a 80% solution.
The Problem:
How many liters of the 50% solution and how many liters of the 90% solution will be used?
The Process:
Let V₁ as the volume of 50% antifreeze solution, and V₂ as the volume of 90% antifreeze solution.
After mixing, 200 liters of 80% solution is formed.
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Part-1
Let us arrange Equation-1 from mixing the two solutions.
[tex]\boxed{ \ (50\% \times V_1) + (90\% \times V_2) = 80\% \times 200 \ }[/tex]
See, the formula is similar to how we want to calculate the combined average.
[tex]\boxed{ \ (0.5 \times V_1) + (0.9 \times V_2) = 0.8 \times 200 \ }[/tex]
[tex]\boxed{ \ 0.5V_1 + 0.9V_2 = 160 \ }[/tex]
Both sides are multiplied by two.
[tex]\boxed{ \ V_1 + 1.8V_2 = 320 \ }[/tex] ... (Equation-1)
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Part-2
We know that the total volume of the two solutions after mixing is 200 liters.
[tex]\boxed{ \ V_1 + V_2 = 200 \ }[/tex] ... (Equation-2)
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Let us solve the two equations by elimination.
[tex]\boxed{ \ V_1 + 1.8V_2 = 320 \ }[/tex]
[tex]\boxed{ \ V_1 + V_2 = 200 \ }[/tex]
____________ ( - )
[tex]\boxed{ \ 0.8V_2 = 120 \ }[/tex]
Both sides are multiplied by 0.8.
Thus, we get [tex]\boxed{ \ V_2 = 150 \ liters \ }[/tex]
Substitution V₂ to select one equation, we choose Equation-2.
[tex]\boxed{ \ V_1 + 150 = 200 \ }[/tex]
Both sides are subtracted by 150.
[tex]\boxed{ \ V_1 = 200 - 150 \ }[/tex]
Thus, we get [tex]\boxed{ \ V_1 = 50 \ liters \ }[/tex]
Conclusion:
Keywords: 50% antifreeze solution, to be mixed, with, 90%, to get, 200 liters, 80%, how many, solution, will be used, total volume
