MHB Mixture to create a 50% acid solution?

  • Thread starter Thread starter mathdad
  • Start date Start date
  • Tags Tags
    Mixture
AI Thread Summary
To create a 50% acid solution, the equation set up is correct: 0.10x + 55(0.80) = 0.50(x + 55). After solving for x, the result is 41.25 milliliters of a 10% acid solution needed. This decimal answer raises questions about consistency with the whole number of 55 milliliters provided. The calculation is mathematically sound, but the decimal result prompts a discussion about practical implications in solution mixing.
mathdad
Messages
1,280
Reaction score
0
How many milliliters of a 10% acid solution must be added to 55 milliliters of a 80% acid solution to create a 50% acid solution?

This is my equation set up:

0.10x + 55(0.80) = 0.50(x + 55)

Is this correct?
 
Mathematics news on Phys.org
Yes, you have the amount of acid (in mL) on both sides of the equation. (Yes)
 
MarkFL said:
Yes, you have the amount of acid (in mL) on both sides of the equation. (Yes)

All I have to do is solve for x, right?

- - - Updated - - -

I got x = 41.25.Is this right?
 
RTCNTC said:
All I have to do is solve for x, right?

- - - Updated - - -

I got x = 41.25.Is this right?

Mathematically, yes.
 
Joppy said:
Mathematically, yes.

Ok. So, the answer is 41.25 milliliters of a 10% acid solution. Does this even make sense? Why is this answer for milliliters a decimal number while the given milliliter is 55, a whole number? See my point?
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top