How many gallons of pure alcohol should be mixed with 20 gal of a 15?

Let x gallon of pure alcohol added to 15 % mixture to make it upto 25 %

15 % alcohol of 20 gallon can be shown as

(15 /100) * 20 + x = (25 /100) (x +20)

3 + x = 0.25 x + 5

0.75 x = 2

X = 2.66 gallon

Hence, 2.66 gallon of pure alcohol need to added to 20 gallon 15 % alcohol to obtain 25 % solution.

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explanation:

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied