A car’s radiator has 24 quarts of a 40% antifreeze solution. How many quarts should be drained and replaced with pure antifreeze if the final mixture is to be 50% antifreeze.

nadib002 wrote:A car’s radiator has 24 quarts of a 40% antifreeze solution. How many quarts should be drained and replaced with pure antifreeze if the final mixture is to be 50% antifreeze. 24 * 0.4 = 9.6 (9.6 + x)/24 = 0.5 9.6 + x = 12 x = 12 - 9.6 x = 2.4

can solve it using allegation 40% - - - 100% - - 50% - - 5:1 ratio for 40% antifreeze and 100% antifreeze since 40% antifreeze = 24 quarts or 5x=24 therefore 1x = 24/5 = 4.8 quarts of 100% antifreeze should be added

I agree with MAAJ's solution. x=2.4 Whats the OA?

nadib002 wrote:A car’s radiator has 24 quarts of a 40% antifreeze solution. How many quarts should be drained and replaced with pure antifreeze if the final mixture is to be 50% antifreeze. We could plug in the answers. Since 50% of the final mixture must be antifreeze, the correct answer will yield .5*24 = 12 quarts of antifreeze. One of the answers would say that 4 quarts of the solution should be replaced with pure antifreeze. Answer choice: 4 quarts of pure antifreeze. Antifreeze in remaining 20 quarts = .4*20 = 8. Total antifreeze = 8+4 = 12. The correct answer is 4 quarts.

srcc25anu wrote:can solve it using allegation 40% - - - 100% - - 50% - - 5:1 ratio for 40% antifreeze and 100% antifreeze since 40% antifreeze = 24 quarts or 5x=24 therefore 1x = 24/5 = 4.8 quarts of 100% antifreeze should be added Alligation is a great approach for mixture problems, but be careful: you determined the correct ratio of solution:antifreeze but not the correct amount of pure antifreeze that is needed. The total volume must remain 24 quarts. Given that the ratio of solution:antifreeze is 5:1, since 5+1 = 6, the pure antifreeze must be 1/6 of the 24 quarts. Pure antifreeze = (1/6)*24 = 4. This sort of error is easily avoided if you plug in the answers, as I suggested in my post above.

MAAJ wrote:nadib002 wrote:A car’s radiator has 24 quarts of a 40% antifreeze solution. How many quarts should be drained and replaced with pure antifreeze if the final mixture is to be 50% antifreeze. 24 * 0.4 = 9.6 (9.6 + x)/24 = 0.5 9.6 + x = 12 x = 12 - 9.6 x = 2.4 If you want to solve algebraically, here's one approach: Let x = pure antifreeze. Remaining solution = 24-x. Antifreeze in remaining solution = .4(24-x). Desired amount of antifreeze in final mixture = .5*24 = 12. Thus: x + .4(24-x) = 12. x + 9.6 - .4x = 12. 10x + 96 - 4x = 120. 6x = 24. x = 4. I find plugging in the answers much easier -- and safer. When you use algebra, you can easily make an error. When you plug in the answers, it is virtually impossible not to determine the OA, because you're proving that the answer choice is correct.

!!! missed the word quarts! my bad! GMATGuruNY wrote:nadib002 wrote:A car’s radiator has 24 quarts of a 40% antifreeze solution. How many quarts should be drained and replaced with pure antifreeze if the final mixture is to be 50% antifreeze. We could plug in the answers, one of which would say that 4 quarts of the solution should be replaced with pure antifreeze. Answer choice: 4 quarts. Antifreeze in remaining 20 quarts = .4*20 = 8. Total antifreeze = 8+4 = 12. Antifreeze/Total Solution = 12/24 = 50%. The correct answer is 4 quarts.