## anonymous 5 years ago 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

1. 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

2. anonymous

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

3. 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 $50v_1 + 40v_2 = 46(80)$

4. nowhereman

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

5. anonymous

I keep getting 32 gallons is that right?

6. 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.

7. nowhereman

The correct solution should be 48

8. anonymous

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

9. 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 $x = 80\cdot 60\% = 48$

10. anonymous

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

11. anonymous

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