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

Question

nowhereman · Answer

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 · Answer

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

anonymous · Answer

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
$$50v_1 + 40v_2 = 46(80) $$

nowhereman · Answer

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

anonymous · Answer

I keep getting 32 gallons is that right?

anonymous · Answer

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 · Answer

The correct solution should be 48

anonymous · Answer

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

nowhereman · Answer

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 · Answer

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

anonymous · Answer

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

radar · Answer

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