question on math destroyer test13:33

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Menthol

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How many grams of salt must be added to m grams of a n% salt solution to make it a p% solution.
anw is M(p-n)/(100-p)
any help woulbe very appreciated!
 
Designate "x" as your amount of grams that you need to add.

If you have n% salt solution and m grams total, you have m*(n%) grams of salt, = m*(n/100) g of salt.

Then if you add "x" amount to m*(n/100) g of salt and then divide by the total amount, you should get a p/100 concentration.

So...

[x+m*(n/100)]/(x+m) = p/100

The reason why you divide the total amount of salt by x+m and not m is because you're adding x grams to the total amount m, so the total amount has to change.

So then if you work it out:

[x+m*(n/100)] = (p*(m+x))/100

Isolate the x variable by itself:

x = (m*p + x*p)/100 - m*(n/100)

x - (x*p)/100 = (m*p)/100 - m*(n/100)

x(1 - p/100) = (m*p - m*n)/100

x(100 - p) = m*p - m*n <--- Multiply both sides by 100

x = (m*p - m*n)/(100-p) = m(p-n)/(100-p)


Tough problem, but good practice. Let me know if you're confused about any of the steps above.
 
Designate "x" as your amount of grams that you need to add.

If you have n% salt solution and m grams total, you have m*(n%) grams of salt, = m*(n/100) g of salt.

Then if you add "x" amount to m*(n/100) g of salt and then divide by the total amount, you should get a p/100 concentration.

So...

[x+m*(n/100)]/(x+m) = p/100

The reason why you divide the total amount of salt by x+m and not m is because you're adding x grams to the total amount m, so the total amount has to change.

So then if you work it out:

[x+m*(n/100)] = (p*(m+x))/100

Isolate the x variable by itself:

x = (m*p + x*p)/100 - m*(n/100)

x - (x*p)/100 = (m*p)/100 - m*(n/100)

x(1 - p/100) = (m*p - m*n)/100

x(100 - p) = m*p - m*n <--- Multiply both sides by 100

x = (m*p - m*n)/(100-p) = m(p-n)/(100-p)


Tough problem, but good practice. Let me know if you're confused about any of the steps above.
omg thank you so much.. you really are a dat destroyer..👍
 
No problem. PM me if you have other questions, going to bed for the night.
 
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