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Post by happyface375 on Apr 21, 2004 23:10:43 GMT -5
Bob has a 20% antifreeze mix in this 13 gal. radiator. How much must he drain out and replace with an 80% antifreeze mix to obtain a 50% mix in his radiator?
can anyone answer this one?
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mo0
New Member
thx to pashy for my siggy :P
Posts: 52
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Post by mo0 on Apr 22, 2004 0:35:34 GMT -5
He should remove half  because if the final concentration is 50%, there should be 6.5 gallons of pure antifreeze. removing half: 6.5gal @ 20% = 1.3gal of antifreeze remaining in the radiator adding 6.5gal @ 80% = 5.2gal of antifreeze (added to radiator) and 1.3 + 5.2 = 6.5gallons, or 50% concentration  Just call me mo0 einstein   *cough cough* we use metric units nowdays *waits to be insulted* |
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Post by happyface375 on Apr 22, 2004 16:34:32 GMT -5
thank for the help einstein ;D i mean Mo0  Later |
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