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

Ad: Over 600 Algebra Word Problems at edhelper.com


   



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) About Me  (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.
================

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +x+ = quarts of fluid to be drained and replaced
with pure antifreeze
+.3x+ = quarts of pure antifreeze in fluid drained off
+.3%2A10+=+3+ quarts of pure antifreeze originally in fluid
----------------------------
+%28+3+-+.3x++%2B+x+%29+%2F+10+=+.4+
( note that you start with +10+ quarts and end with +10+ quarts )
+%28+3+%2B+.7x+%29+%2F+10+=+.4+
+3+%2B+.7x+=+4+
+.7x+=+1+
+x+=+1.4286+
1.4286 quarts of fluid must be drained off and
replaced with pure antifreeze
check:
+%28+3+-+.3x++%2B+x+%29+%2F+10+=+.4+
+%28+3+-+.3%2A1.4286++%2B+1.4286+%29+%2F+10+=+.4+
++3+-+.4286+%2B+1.4286+=+4+
+4+=+4+
OK