I think the answer is (9,8)
Answer:
- 6 2/3 qt 80%
- 13 1/3 qt 20%
Step-by-step explanation:
It is often convenient to solve a mixture problem by letting a variable represent the quantity of the higher-concentration contributor to the mix.
__
We can let x represent the number of quarts of 80% solution needed. Then (20-x) is the number of quarts of 20% solution needed. The amount of salt in the final mix is ...
0.80x +0.20(20-x) = 0.40(20)
0.60x = 0.20(20) . . . . . . . . subtract 0.20(20) and simplify
x = 20/3 = 6 2/3 . . . . . . . . . divide by 0.60; quarts of 80% solution
(20 -x) = 13 1/3 . . . . . . . . . . amount of 20% solution needed
The teacher should mix 6 2/3 quarts of 80% solution with 13 1/3 quarts of 20% solution.
3+x=23t therefore the answer should be D
Find percent increase by subtracting the amounts and dividing it by the old amount, then multiply the result by 100 to make it a percent.
Increase= (277-210)/210
67/210
Then it ends up as 0.319047619...
Multiply that by 100 to make it a percent, which is..
31.9047619..
Round it off, and you get..
32%!
