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How many gallons of a 9 percent salt solution must be mixed with 25 gallons of a 31 percent solution to obtain a 20 percent solution?
25 gallons
What is 15 percent of ten gallons?
15% of 10 gallons = 10*15/100 = 1.5 gallons.
How much is tweny five percent of 10 gallons?
25% of 10 gallons = 2.5 gallons
How many gallons of 50 percent antifreeze solution must be mixed with 70 gallons of 25 percent antifreeze to get a mixture that is 40 percent antifreeze?
105 gallons. Let the multiple of 70 gallons to add be x, then using the percentages of antifreeze in the solutions: (25 + 50x) ÷ (1 + x) = 40 ⇒ 25 + 50x = 40 + 40x ⇒ 10x = 15 ⇒ x = 1.5 Therefore add 70 x 1.5 = 105 gallons.
What is 10 percent of one gallon?
1.6 cups is ten percent of a gallon
What is 25 percent of 10 gallons?
25 percent of 10 gallons is 2.5 gallons.
How many gallons of a 9 percent salt solution must be mixed with 25 gallons of a 31 percent solution to obtain a 20 percent solution?
25 gallons
How many gallons is 25 percent of 30 gallons?
7.5 gallons
What is 15 percent of ten gallons?
15% of 10 gallons = 10*15/100 = 1.5 gallons.
How much is tweny five percent of 10 gallons?
25% of 10 gallons = 2.5 gallons
What is ten percent out of 250?
25
How many gallons of a 90 percent antifreeze solution must be mixed with 80 gallons of 25 percent antifreeze to get a mixture that is 80 percent antifreeze?
614
What is 25 percent of a 30 gallon tank?
7.5 gallons.
What is 25 percent in gallons?
25% = 1/4 1/4 gallon = 1 quart
How many gallons of a 80 percent antifreeze solution must be mixed with 90 gallons of 25 percent antifreeze to get a mixture that is 70 percent antifreeze?
Let X = gallons of 80% antifreeze Then 0.80X + .25 (90) = .70( 90 + X) .80X + 22.5 = 63 + .70X .1X = 41.5 X = 415 gallons
How many gallons of a 50 percent antifreeze solution must be mixed with 90 gallons of 25 percent antifreeze to get a mixture of 40 percent antifreeze?
135 gallons. Let the multiple of 90 gallons to add be x, then using the percentages of antifreeze in the solutions: (25 + 50x) ÷ (1 + x) = 40 ⇒ 25 + 50x = 40 + 40x ⇒ 10x = 15 ⇒ x = 1.5 Therefore add 90 x 1.5 = 135 gallons.
How many gallons of 50 percent antifreeze solution must be mixed with 70 gallons of 25 percent antifreeze to get a mixture that is 40 percent antifreeze?
105 gallons. Let the multiple of 70 gallons to add be x, then using the percentages of antifreeze in the solutions: (25 + 50x) ÷ (1 + x) = 40 ⇒ 25 + 50x = 40 + 40x ⇒ 10x = 15 ⇒ x = 1.5 Therefore add 70 x 1.5 = 105 gallons.
