SOLUTION: Please help me to solve this problem.
A certain paint mixture weighing 305 kg contains 22% solids suspended in water. How many kilograms of water must be allowed to evaporate to
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A certain paint mixture weighing 305 kg contains 22% solids suspended in water. How many kilograms of water must be allowed to evaporate to
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Question 1054224: Please help me to solve this problem.
A certain paint mixture weighing 305 kg contains 22% solids suspended in water. How many kilograms of water must be allowed to evaporate to raise the concentration of solids to 27%?
Treat the percents given in this exercise as exact numbers, and work to three significant digits.
Quantity of water =
You can put this solution on YOUR website! 305*0.22=67.1 kg of solids
That will still be present and will become 27% of x
67.1/x=0.27/1
cross-multiply
67.1=0.27x
divide by 0.27 and get 248.51
249 kg
From 305 kg, this requires 56.5 kg water to evaporate
You can put this solution on YOUR website! .
Please help me to solve this problem.
A certain paint mixture weighing 305 kg contains 22% solids suspended in water. How many kilograms of water must be allowed
to evaporate to raise the concentration of solids to 27%?
Treat the percents given in this exercise as exact numbers, and work to three significant digits.
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Let "w" be the mass of water evaporated.
Then the concentration equation is
= 0.27, or
0.22*305 = 0.27*(305-w).
Simplify and solve for "w":
67.1 = 82.35 - 0.27w, or
0.27w = 82.35 - 67.1 ---> 0.27w = 15.25 ---> w = = 56.48.
Answer. 56.481 kg of water must evaporate.
