anonymous
  • anonymous
How much 50% antifreeze solution should be mixed with a 40% antifreeze solution to give 80 gallons of a 46% antifreeze solution? pls anyone can help
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
nowhereman
  • nowhereman
Say you'll take x of 50% antifreeze. Then you need of course take 80 - x of 40% antifreeze and in the end you want \[x\cdot 50\% + (80-x)\cdot 40\% = 46 \% \] so just solve that for x
anonymous
  • anonymous
my problem is i dont know the formua to find the x
anonymous
  • anonymous
If v is the volume of a solution and c is the concentration of antifreeze in that solution, then the amount of antifreeze in a solution is given by vc. Since the total amount of antifreeze is conserved.. \[ v_1c_1 + v_2c_2 = v_3c_3 \] Since the total volume is the sum of the individual volumes.. \[v_1 + v_2 = v_3 = 80\] Therefore \[50v_1 + 40v_2 = 46(80) \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

nowhereman
  • nowhereman
oh yeah, I forgot the 80 on the right side, sorry But then it's really kindergarten math.
anonymous
  • anonymous
I keep getting 32 gallons is that right?
anonymous
  • anonymous
Then just take the fact that: \[v_1 + v_2 = 80 \rightarrow v_1 = 80-v_2\] to substitute out v1 in the other equation and find v2, then plug back in here to find v1.
nowhereman
  • nowhereman
The correct solution should be 48
anonymous
  • anonymous
how did you get the 48? i keep getting 32 or 42 gallons did I do wrong in my solution?
nowhereman
  • nowhereman
Unfortunately I can't read your calculations from here. From the above equations, it should be straight-forward, but I can show it to you: \[x\cdot 50\% + (80-x)\cdot 40\% = 80\cdot 46\%\] than you get \[x\cdot 10\% = 80\cdot 6\%\] and thus \[x = 80\cdot 60\% = 48\]
anonymous
  • anonymous
ok thats help me thank you this is my first time to encounter this problem.thank you
anonymous
  • anonymous
You still need to plug back in to find the other volume as well.
radar
  • radar
I like that approach nowhereman, it gave the result that was requested

Looking for something else?

Not the answer you are looking for? Search for more explanations.