Question

A

Answer 1

50 liters for 50%  antifreeze solution and 150 liters for 90% antifreeze solution.

Further explanation

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.

$\boxed{ \ (50\% \times V_1) + (90\% \times V_2) = 80\% \times 200 \ }$

See, the formula is similar to how we want to calculate the combined average.

$\boxed{ \ (0.5 \times V_1) + (0.9 \times V_2) = 0.8 \times 200 \ }$

$\boxed{ \ 0.5V_1 + 0.9V_2 = 160 \ }$

Both sides are multiplied by two.

$\boxed{ \ V_1 + 1.8V_2 = 320 \ }$ ... (Equation-1)

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Part-2

We know that the total volume of the two solutions after mixing is 200 liters.

$\boxed{ \ V_1 + V_2 = 200 \ }$ ... (Equation-2)

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Let us solve the two equations by elimination.

$\boxed{ \ V_1 + 1.8V_2 = 320 \ }$

$\boxed{ \ V_1 + V_2 = 200 \ }$

____________ ( - )

$\boxed{ \ 0.8V_2 = 120 \ }$

Both sides are multiplied by 0.8.

Thus, we get $\boxed{ \ V_2 = 150 \ liters \ }$

Substitution V₂ to select one equation, we choose Equation-2.

$\boxed{ \ V_1 + 150 = 200 \ }$

Both sides are subtracted by 150.

$\boxed{ \ V_1 = 200 - 150 \ }$

Thus, we get $\boxed{ \ V_1 = 50 \ liters \ }$

Conclusion:

• 50% antifreeze solution = 50 liters  
• 90% antifreeze solution = 150 liters

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Keywords: 50%  antifreeze solution, to be mixed, with, 90%, to get,  200  liters, 80%, how many, solution, will be used, total volume

Answer 2

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