QuestionTo make a $5 \%$ ethylene glycol solution from $20$ oz of $2 \%$ solution, how much of the $12 \%$ solution is needed?

Studdy Solution

STEP 1

Assumptions1. The initial solution contains% ethylene glycol and its volume is20 fluid ounces. . The solution to be added contains12% ethylene glycol.
3. The final solution should contain5% ethylene glycol.
4. The percentage of ethylene glycol in the final solution is the weighted average of the percentages in the initial and added solutions.

STEP 2

Let's denote the volume of the12% solution to be added as $x$ fluid ounces. The total volume of the final solution will be $20 + x$ fluid ounces.

STEP 3

The total amount of ethylene glycol in the final solution will be the sum of the amounts in the initial solution and the added solution. This can be expressed as $0.02 \cdot20 +0.12 \cdot x =0.05 \cdot (20 + x)$

STEP 4

We can simplify this equation by multiplying each term by100 to get rid of the decimals.
$2 \cdot20 +12 \cdot x = \cdot (20 + x)$

STEP 5

Expand the right side of the equation.
$40 +12x =100 +5x$

STEP 6

Subtract $5x$ from both sides of the equation to isolate the terms with $x$ on one side.
$12x -5x =100 -40$

STEP 7

implify the equation.
$7x =60$

STEP 8

olve for $x$ by dividing both sides of the equation by7.
$x =60 /7$

STEP 9

Calculate the value of $x$ .
$x =8.57$

STEP 10

Round the value of $x$ to the nearest tenth.
$x \approx8.6$ So, the students should add approximately8.6 fluid ounces of the12% ethylene glycol solution to obtain a5% ethylene glycol solution.

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QuestionTo make a $5 \%$ ethylene glycol solution from $20$ oz of $2 \%$ solution, how much of the $12 \%$ solution is needed?

Studdy Solution

STEP 1

STEP 2

Let's denote the volume of the12% solution to be added as $x$ fluid ounces. The total volume of the final solution will be $20 + x$ fluid ounces.

STEP 3

The total amount of ethylene glycol in the final solution will be the sum of the amounts in the initial solution and the added solution. This can be expressed as $0.02 \cdot20 +0.12 \cdot x =0.05 \cdot (20 + x)$

STEP 4

We can simplify this equation by multiplying each term by100 to get rid of the decimals.
$2 \cdot20 +12 \cdot x = \cdot (20 + x)$

STEP 5

Expand the right side of the equation.
$40 +12x =100 +5x$

STEP 6

Subtract $5x$ from both sides of the equation to isolate the terms with $x$ on one side.
$12x -5x =100 -40$

STEP 7

implify the equation.
$7x =60$

STEP 8

olve for $x$ by dividing both sides of the equation by7.
$x =60 /7$

STEP 9

Calculate the value of $x$ .
$x =8.57$

STEP 10

Round the value of $x$ to the nearest tenth.
$x \approx8.6$ So, the students should add approximately8.6 fluid ounces of the12% ethylene glycol solution to obtain a5% ethylene glycol solution.

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QuestionTo make a 5%5 \%5% ethylene glycol solution from 202020 oz of 2%2 \%2% solution, how much of the 12%12 \%12% solution is needed?

Studdy Solution

STEP 1

STEP 2

Let's denote the volume of the12% solution to be added as xxx fluid ounces. The total volume of the final solution will be 20+x20 + x20+x fluid ounces.

STEP 3

The total amount of ethylene glycol in the final solution will be the sum of the amounts in the initial solution and the added solution. This can be expressed as0.02⋅20+0.12⋅x=0.05⋅(20+x)0.02 \cdot20 +0.12 \cdot x =0.05 \cdot (20 + x)0.02⋅20+0.12⋅x=0.05⋅(20+x)

STEP 4

We can simplify this equation by multiplying each term by100 to get rid of the decimals.2⋅20+12⋅x=⋅(20+x)2 \cdot20 +12 \cdot x = \cdot (20 + x)2⋅20+12⋅x=⋅(20+x)

STEP 5

Expand the right side of the equation.40+12x=100+5x40 +12x =100 +5x40+12x=100+5x

STEP 6

Subtract 5x5x5x from both sides of the equation to isolate the terms with xxx on one side.12x−5x=100−4012x -5x =100 -4012x−5x=100−40

STEP 7

implify the equation.7x=607x =607x=60

STEP 8

olve for xxx by dividing both sides of the equation by7.x=60/7x =60 /7x=60/7

STEP 9

Calculate the value of xxx.x=8.57x =8.57x=8.57

STEP 10

Round the value of xxx to the nearest tenth.x≈8.6x \approx8.6x≈8.6So, the students should add approximately8.6 fluid ounces of the12% ethylene glycol solution to obtain a5% ethylene glycol solution.

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QuestionTo make a $5 \%$ ethylene glycol solution from $20$ oz of $2 \%$ solution, how much of the $12 \%$ solution is needed?

Let's denote the volume of the12% solution to be added as $x$ fluid ounces. The total volume of the final solution will be $20 + x$ fluid ounces.

The total amount of ethylene glycol in the final solution will be the sum of the amounts in the initial solution and the added solution. This can be expressed as $0.02 \cdot20 +0.12 \cdot x =0.05 \cdot (20 + x)$

We can simplify this equation by multiplying each term by100 to get rid of the decimals.
$2 \cdot20 +12 \cdot x = \cdot (20 + x)$

Expand the right side of the equation.
$40 +12x =100 +5x$

Subtract $5x$ from both sides of the equation to isolate the terms with $x$ on one side.
$12x -5x =100 -40$

implify the equation.
$7x =60$

olve for $x$ by dividing both sides of the equation by7.
$x =60 /7$

Calculate the value of $x$ .
$x =8.57$

Round the value of $x$ to the nearest tenth.
$x \approx8.6$ So, the students should add approximately8.6 fluid ounces of the12% ethylene glycol solution to obtain a5% ethylene glycol solution.