QuestionHow many liters of $32\%$ acid solution should be added to 50 liters of $22\%$ acid to get a $28\%$ solution?

Studdy Solution

STEP 1

Assumptions1. The desired acid concentration is28% . The lab has50 liters of22% acid solution3. The lab has an unlimited amount of32% acid solution4. Mixing the two solutions will result in a solution with a concentration that is a weighted average of the two original concentrations

STEP 2

Let's denote the amount of the32% solution that needs to be added as $x$ liters. The total volume of the solution after adding $x$ liters of the32% solution to the50 liters of the22% solution is $50 + x$ liters.

STEP 3

The total amount of acid in the final solution is the sum of the acid in the22% solution and the acid in the32% solution. This can be expressed as $0.22 \times50 +0.32 \times x$

STEP 4

The concentration of the final solution is the total amount of acid divided by the total volume. We want this to be28%, or0.28. This gives us the equation $\frac{0.22 \times50 +0.32 \times x}{50 + x} =0.28$

STEP 5

To solve for $x$ , we first multiply both sides of the equation by $50 + x$ to get rid of the denominator on the left side.
$0.22 \times50 +0.32 \times x =0.28 \times (50 + x)$

STEP 6

Now, distribute the0.28 on the right side of the equation.
$0.22 \times50 +0.32 \times x =0.28 \times50 +0.28 \times x$

STEP 7

Next, simplify the equation by multiplying out the constants.
$11 +0.32x =14 +0.28x$

STEP 8

Subtract $0.28x$ from both sides of the equation.
$11 +0.04x =14$

STEP 9

Subtract11 from both sides of the equation.
$.04x =3$

STEP 10

Finally, solve for $x$ by dividing both sides of the equation by0.04.
$x = \frac{3}{0.04}$

STEP 11

Calculate the value of $x$ .
$x = \frac{3}{0.04} =75$ The teaching assistant should add75 liters of the32% acid solution to the22% acid solution to obtain a solution with the desired concentration.

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QuestionHow many liters of $32\%$ acid solution should be added to 50 liters of $22\%$ acid to get a $28\%$ solution?

Studdy Solution

STEP 1

Assumptions1. The desired acid concentration is28% . The lab has50 liters of22% acid solution3. The lab has an unlimited amount of32% acid solution4. Mixing the two solutions will result in a solution with a concentration that is a weighted average of the two original concentrations

STEP 2

Let's denote the amount of the32% solution that needs to be added as $x$ liters. The total volume of the solution after adding $x$ liters of the32% solution to the50 liters of the22% solution is $50 + x$ liters.

STEP 3

The total amount of acid in the final solution is the sum of the acid in the22% solution and the acid in the32% solution. This can be expressed as $0.22 \times50 +0.32 \times x$

STEP 4

The concentration of the final solution is the total amount of acid divided by the total volume. We want this to be28%, or0.28. This gives us the equation $\frac{0.22 \times50 +0.32 \times x}{50 + x} =0.28$

STEP 5

To solve for $x$ , we first multiply both sides of the equation by $50 + x$ to get rid of the denominator on the left side.
$0.22 \times50 +0.32 \times x =0.28 \times (50 + x)$

STEP 6

Now, distribute the0.28 on the right side of the equation.
$0.22 \times50 +0.32 \times x =0.28 \times50 +0.28 \times x$

STEP 7

Next, simplify the equation by multiplying out the constants.
$11 +0.32x =14 +0.28x$

STEP 8

Subtract $0.28x$ from both sides of the equation.
$11 +0.04x =14$

STEP 9

Subtract11 from both sides of the equation.
$.04x =3$

STEP 10

Finally, solve for $x$ by dividing both sides of the equation by0.04.
$x = \frac{3}{0.04}$

STEP 11

Calculate the value of $x$ .
$x = \frac{3}{0.04} =75$ The teaching assistant should add75 liters of the32% acid solution to the22% acid solution to obtain a solution with the desired concentration.

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QuestionHow many liters of 32%32\%32% acid solution should be added to 50 liters of 22%22\%22% acid to get a 28%28\%28% solution?

Studdy Solution

STEP 1

Assumptions1. The desired acid concentration is28% . The lab has50 liters of22% acid solution3. The lab has an unlimited amount of32% acid solution4. Mixing the two solutions will result in a solution with a concentration that is a weighted average of the two original concentrations

STEP 2

Let's denote the amount of the32% solution that needs to be added as xxx liters. The total volume of the solution after adding xxx liters of the32% solution to the50 liters of the22% solution is 50+x50 + x50+x liters.

STEP 3

The total amount of acid in the final solution is the sum of the acid in the22% solution and the acid in the32% solution. This can be expressed as0.22×50+0.32×x0.22 \times50 +0.32 \times x0.22×50+0.32×x

STEP 4

The concentration of the final solution is the total amount of acid divided by the total volume. We want this to be28%, or0.28. This gives us the equation0.22×50+0.32×x50+x=0.28\frac{0.22 \times50 +0.32 \times x}{50 + x} =0.2850+x0.22×50+0.32×x​=0.28

STEP 5

To solve for xxx, we first multiply both sides of the equation by 50+x50 + x50+x to get rid of the denominator on the left side.0.22×50+0.32×x=0.28×(50+x)0.22 \times50 +0.32 \times x =0.28 \times (50 + x)0.22×50+0.32×x=0.28×(50+x)

STEP 6

Now, distribute the0.28 on the right side of the equation.0.22×50+0.32×x=0.28×50+0.28×x0.22 \times50 +0.32 \times x =0.28 \times50 +0.28 \times x0.22×50+0.32×x=0.28×50+0.28×x

STEP 7

Next, simplify the equation by multiplying out the constants.11+0.32x=14+0.28x11 +0.32x =14 +0.28x11+0.32x=14+0.28x

STEP 8

Subtract 0.28x0.28x0.28x from both sides of the equation.11+0.04x=1411 +0.04x =1411+0.04x=14

STEP 9

Subtract11 from both sides of the equation..04x=3.04x =3.04x=3

STEP 10

Finally, solve for xxx by dividing both sides of the equation by0.04.x=30.04x = \frac{3}{0.04}x=0.043​

STEP 11

Calculate the value of xxx.x=30.04=75x = \frac{3}{0.04} =75x=0.043​=75The teaching assistant should add75 liters of the32% acid solution to the22% acid solution to obtain a solution with the desired concentration.

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QuestionHow many liters of $32\%$ acid solution should be added to 50 liters of $22\%$ acid to get a $28\%$ solution?

Let's denote the amount of the32% solution that needs to be added as $x$ liters. The total volume of the solution after adding $x$ liters of the32% solution to the50 liters of the22% solution is $50 + x$ liters.

The total amount of acid in the final solution is the sum of the acid in the22% solution and the acid in the32% solution. This can be expressed as $0.22 \times50 +0.32 \times x$

The concentration of the final solution is the total amount of acid divided by the total volume. We want this to be28%, or0.28. This gives us the equation $\frac{0.22 \times50 +0.32 \times x}{50 + x} =0.28$

To solve for $x$ , we first multiply both sides of the equation by $50 + x$ to get rid of the denominator on the left side.
$0.22 \times50 +0.32 \times x =0.28 \times (50 + x)$

Now, distribute the0.28 on the right side of the equation.
$0.22 \times50 +0.32 \times x =0.28 \times50 +0.28 \times x$

Next, simplify the equation by multiplying out the constants.
$11 +0.32x =14 +0.28x$

Subtract $0.28x$ from both sides of the equation.
$11 +0.04x =14$

Subtract11 from both sides of the equation.
$.04x =3$

Finally, solve for $x$ by dividing both sides of the equation by0.04.
$x = \frac{3}{0.04}$

Calculate the value of $x$ .
$x = \frac{3}{0.04} =75$ The teaching assistant should add75 liters of the32% acid solution to the22% acid solution to obtain a solution with the desired concentration.