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Find the overall balance of antifreeze: amount of antifreeze to start with + the amount of pure antifreeze to add = amount of antifreeze in the final solution

Let X be the amount of pure antifreeze to add in gallons.

(70 * 0.25) + X = 0.80 * (70+X)

17.5 + X = 56 + 0.8X

0.2X = 56 - 17.5

0.2X = 38.5

X = 38.5/0.2 = 192.5 gallons

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Q: How many gallons of antifreeze must be mixed with 70 gallons of 25 percent antifreeze to get a mixture that is 80 percent antifreeze?

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Suppose x gallons of 90% antifreeze is mixed. Then total volume of mixture = x + 70 gallons and total antifreeze in mixture = 0.9*x + 0.15*70 = 0.9x + 10.5 Concentration of mixture = (0.9x + 10.5)/(x + 70) which is 80% or 0.8 So 0.9x + 10.5 = 0.8x + 56 that is 0.1x = 45.5 or x = 455 gallons

Let X = gallons of 50% antifreeze .5X + .1(70) = .4 (X + 70) .5X + 7 = .4X + 28 .1X = 21 X = 210 GALLONS

330 gallons of 80% antifreeze mixed with 60 gallons of 15% antifreeze will provide 390 gallons of 70% antifreeze. Let the multiplier of 60 gallons be x, then considering the percentages of antifreeze in the solutions: (15 + 80x) ÷ (1 + x) = 70 ⇒ 15 + 80x = 70 + 70x ⇒ 10x = 55 ⇒ x = 5.5 Therefore mix 60 x 5.5 = 330 gallons of 80% 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.

.70x+.10(60)=.60(x+60) let x= number of gallon at 70% .7x+6=.6x+36 .1x=30 x=300

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Suppose x gallons of 90% antifreeze is mixed. Then total volume of mixture = x + 70 gallons and total antifreeze in mixture = 0.9*x + 0.15*70 = 0.9x + 10.5 Concentration of mixture = (0.9x + 10.5)/(x + 70) which is 80% or 0.8 So 0.9x + 10.5 = 0.8x + 56 that is 0.1x = 45.5 or x = 455 gallons

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

Let X = gallons of 50% antifreeze .5X + .1(70) = .4 (X + 70) .5X + 7 = .4X + 28 .1X = 21 X = 210 GALLONS

330 gallons of 80% antifreeze mixed with 60 gallons of 15% antifreeze will provide 390 gallons of 70% antifreeze. Let the multiplier of 60 gallons be x, then considering the percentages of antifreeze in the solutions: (15 + 80x) ÷ (1 + x) = 70 ⇒ 15 + 80x = 70 + 70x ⇒ 10x = 55 ⇒ x = 5.5 Therefore mix 60 x 5.5 = 330 gallons of 80% 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.

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.

You will need 3.2 gallons.

.70x+.10(60)=.60(x+60) let x= number of gallon at 70% .7x+6=.6x+36 .1x=30 x=300

You will never get it there. However much 50% you have, any amount of 20% in the mix will reduce the total below 50%. So, throw away the 20% and just use the 50% on its own.