Question

it is necessary to have a 70% antifreeze solution in the radiator of a certain car. the radiator now has 50 liters of 60% solution. how many liters of this should be drained and replaced with 100% antifreeze to get the desired strength? round to the nearest tenth. a. 20 l b. 12.5 l c. 25 l d. 16.7 l

Explanation:

Step1: Define variables

Let \( x \) be the liters of the 60% solution to be drained and replaced with 100% antifreeze.

Step2: Analyze the amount of antifreeze

• Initially, the radiator has 50 liters of 60% solution, so the amount of antifreeze is \( 0.6 \times 50 \) liters.
• When we drain \( x \) liters of the 60% solution, the amount of antifreeze drained is \( 0.6x \) liters, and the amount of solution left is \( 50 - x \) liters, with antifreeze \( 0.6(50 - x) \) liters.
• Then we add \( x \) liters of 100% (which is 1 in decimal) antifreeze, so the new amount of antifreeze is \( 0.6(50 - x) + 1 \times x \) liters.
• The desired solution is 50 liters of 70% antifreeze, so the amount of antifreeze should be \( 0.7 \times 50 \) liters.

Step3: Set up the equation

\[
0.6(50 - x) + x = 0.7 \times 50
\]

Step4: Solve the equation

First, expand the left - hand side:
\[
0.6\times50-0.6x + x=0.7\times50
\]
\[
30 + 0.4x=35
\]
Subtract 30 from both sides:
\[
0.4x=35 - 30
\]
\[
0.4x = 5
\]
Divide both sides by 0.4:
\[
x=\frac{5}{0.4}=12.5
\]

Answer:

B. 12.5 L