SOLUTION: A radiator contains 10 quarts of fluid, 30% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 40% antifreeze?
Algebra ->
Customizable Word Problem Solvers
-> Mixtures
-> SOLUTION: A radiator contains 10 quarts of fluid, 30% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 40% antifreeze?
Log On
Question 763949: A radiator contains 10 quarts of fluid, 30% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 40% antifreeze? Found 2 solutions by stanbon, josmiceli:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! A radiator contains 10 quarts of fluid, 30% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 40% antifreeze?
---------------------
Equation:
active - active + active = active
0.30*10 - 0.30x + 1.00x = 0.40*10
-------
30*10 - 30x + 100x = 40*10
70x = 10*10
x = 10/7 = 1.4286 quarts (amt. to remove and replace)
================
Cheers,
Stan H.
================
You can put this solution on YOUR website! Let = quarts of fluid to be drained and replaced
with pure antifreeze = quarts of pure antifreeze in fluid drained off quarts of pure antifreeze originally in fluid
----------------------------
( note that you start with quarts and end with quarts )
1.4286 quarts of fluid must be drained off and
replaced with pure antifreeze
check:
OK