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- 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

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- anonymous

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- 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

my problem is i dont know the formua to find the x

- 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
\[50v_1 + 40v_2 = 46(80) \]

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- nowhereman

oh yeah, I forgot the 80 on the right side, sorry
But then it's really kindergarten math.

- anonymous

I keep getting 32 gallons is that right?

- 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

The correct solution should be 48

- anonymous

how did you get the 48? i keep getting 32 or 42 gallons did I do wrong in my solution?

- 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

ok thats help me thank you this is my first time to encounter this problem.thank you

- anonymous

You still need to plug back in to find the other volume as well.

- radar

I like that approach nowhereman, it gave the result that was requested

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