Question 945776: The percentage of salt in 1L of water is 10%. If 500 mL of water is added to this mixture, what percentage of salt is there now?
Found 3 solutions by lwsshak3, ikleyn, josgarithmetic:
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! The percentage of salt in 1L of water is 10%. If 500 mL of water is added to this mixture, what percentage of salt is there now?
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1L=1000 mL
amt of salt in 1L=10%*1000=100 mL
percentage of salt in mixture=100/(1000+500)=100/1500=6 2/3%
Answer by ikleyn(53906)  (Show Source):
You can put this solution on YOUR website! .
The percentage of salt in 1L of water is 10%. If 500 mL of water is added to this mixture,
what percentage of salt is there now?
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How the solution is worded in the post by @lwsshar3, it provokes homeric laughter.
It is because the amount of salt in this problem is not measured in milliliters - it is measured in grams.
See my correct solution below, written in right terms.
In this problem, the concentration of the mixture is measured in grams per milliliter.
So, the fact that 1 liter of mixture is 10% salt MEANS that the amount of salt in the mixture
is 0.1*1000 = 100 grams.
When 500 mL of water is added to the mixture, the total volume of the mixture (of the liquid) becomes 1500 milliliters.
It contains the same 100 grams of salt - hence, the concentration of salt in the mixture is now
 =  = 0.066666..., or 0.0667 = 6.67% rounded.
ANSWER. The final concentration of salt in the mixture is about 6.67%.
Solved and presented correctly, using adequate terminology.
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What did shock me even more, is the fact that in Google Overview Artificial Intelligence,
the solution to this problem was presented in same terms and in same words as in the post by @lwsshar3,
i.e. in absolutely illiterate way. <<<---=== May 20, 2026, 12:40 AM.
Answer by josgarithmetic(39832) (Show Source):
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