SOLUTION: You have 10 cups of a drink that contains 97% water. How many cups of a drink containing 43% water needs to be added in order to have a final drink that is 61% water?

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: You have 10 cups of a drink that contains 97% water. How many cups of a drink containing 43% water needs to be added in order to have a final drink that is 61% water?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1121791: You have 10 cups of a drink that contains 97% water. How many cups of a drink containing 43% water needs to be added in order to have a final drink that is 61% water?
Found 2 solutions by josmiceli, greenestamps:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+10%2A.97+=+9.7+ cups of water initially
Let +x+ = number of cups of 43% water mixture needed
+.43x+ = number of cups of water in the 43% mixture
————————
+%28+9.7+%2B+.43x+%29%2F+%28+10+%2B+x+%29+=+.61+
+9.7+%2B+.43x+=+.61%2A%28+10+%2B+x+%29+
+9.7+%2B+.43x+=+6.1+%2B+.61x+
+.18x+=+3.6+
+x+=+20+
———————
Check answer
+%28+9.7+%2B+8.6+%29%2F%28+10+%2B+20+%29+=+.61+
+18.3%2F30+=+.61+
+.61+=+.61+
OK

Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


The formal algebraic solution by the other tutor is fine. But for me, the way to start the problem that makes it most clear what is being done is

0.97%2810%29%2B0.43%28x%29+=+0.61%2810%2Bx%29

But here is a method for solving this kind of problem that is much faster and easier, if you understand it.

(1) The percentage of the final mixture (61%) is twice as close to 43% as it is to 97%: 61-43 = 18; 96-61 = 36.

(2) Therefore the mixture needs to have twice as much of the 43% water as it has of the 97% water.

(3) Since you have 10 cups of the 97% water ingredient, you need 20 cups of the 43% water ingredient.