MHB Mixture to create a 50% acid solution?

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SUMMARY

The discussion focuses on calculating the volume of a 10% acid solution required to mix with 55 milliliters of an 80% acid solution to achieve a 50% acid solution. The equation set up is confirmed as correct: 0.10x + 55(0.80) = 0.50(x + 55). The solution for x is determined to be 41.25 milliliters, which is mathematically accurate despite concerns regarding the decimal result compared to the whole number input.

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mathdad
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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?
 
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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?
 

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