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Alems 630
thin 1
thimg 2
3) How many quarts of pure alcohol must be added to 5 quarts of a
40% alcohol solution to make a SOF alcohor solution?
A

qquad ar. 1
Ph. 754_WP_Mixture_v08-6

Alems $630$ $\newline$ thin $1$ $\newline$ thimg $2$ $\newline$ $3$ ) How many quarts of pure alcohol must be added to $5$ quarts of a $40 \%$ alcohol solution to make a SOF alcohor solution? $\newline$ A $\newline$ $\qquad$ ar. $1$ $\newline$ Ph. $754$ _WP_Mixture_v $08$ $-6$

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Convert rates and measurements: metric units

Full solution

Q. Alems

630

\newline

thin

1

\newline

thimg

2

\newline

3

) How many quarts of pure alcohol must be added to

5

quarts of a

40 \%

alcohol solution to make a SOF alcohor solution?

\newline

A

\newline

\qquad

ar.

1

\newline

Ph.

754

_WP_Mixture_v

08

-6

Calculate Alcohol Content: First, let's figure out how much alcohol is currently in the $5$ quarts of the $40\%$ solution. $\newline$ Calculation: $5$ quarts $\times 0.40 = 2$ quarts of alcohol.
Determine Pure Alcohol Needed: Now, we need to find out how many quarts of pure alcohol ( $100\%$ ) we need to add to get a $50\%$ alcohol solution. $\newline$ Let's call the number of quarts of pure alcohol we need to add $\textit{x}$ .
Calculate Total Volume: After adding ＂ $x$ ＂ quarts of pure alcohol, the total volume of the solution will be $5$ quarts + $x$ quarts. The total amount of alcohol will be $2$ quarts (from the original solution) + $x$ quarts (pure alcohol added).
Set Up Equation: We want this new solution to be $50\%$ alcohol. $\newline$ So, we set up the equation: $(2 + x) / (5 + x) = 0.50$ .
Solve for x: Now, solve for $x$ . $0.50 \times (5 + x) = 2 + x$ . $2.5 + 0.50x = 2 + x$ .
Final Solution: Subtract $2$ from both sides to isolate the terms with $x$ . $\newline$ $0.50x - x = 2.5 - 2$ . $\newline$ $-0.50x = 0.5$ .
Final Solution: Subtract $2$ from both sides to isolate the terms with $x$ . $\newline$ $0.50x - x = 2.5 - 2$ . $\newline$ $-0.50x = 0.5$ .Divide both sides by $-0.50$ to solve for $x$ . $\newline$ $x = 0.5 / -0.50$ . $\newline$ $x = -1$ .

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