salt on potato chips

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realolman

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This is a question that seems to me should be easy, but for the life of me I have not been able to figure it out.

Potato chips contain 0.6 percent salt by weight naturally.

I want to add salt until the total weight is 2.6 percent salt.

I would like an equation that will tell me how many grams of salt I need to add to a weight of potato chips to bring the total to 2.6 percent salt.

thank you
 
realolman said:
This is a question that seems to me should be easy, but for the life of me I have not been able to figure it out.

Potato chips contain 0.6 percent salt by weight naturally.

I want to add salt until the total weight is 2.6 percent salt.

I would like an equation that will tell me how many grams of salt I need to add to a weight of potato chips to bring the total to 2.6 percent salt.

thank you
Click to expand...


Let's start by NAMING things.

Let w = weight of the natural potato chips in grams

Let x = grams of salt added

You're told that the percent of salt occurring naturally in the potato chips is 0.6%. So, if you have w grams of potato chips, the salt contained in them is 0.6%, or 0.006w grams.

Let x = number of grams of pure salt added

This is 100% salt.

Now...if you take "w" grams of potato chips and add "x" grams of salt, you'll have a mixture containing w + x grams, which you'd like to contain 2.6% salt.

salt occurring naturally + salt added = salt in final mixture

0.006w + 1.00x = 0.026(w + x)

You can solve this for x in terms of w...and you'll get an equation that tells you how much salt in grams (x) you need to add to a given number of grams of potato chips (w) to achieve the desired percentage of salt.
 
Hello, realolman!

Mrspi is absolutely correct!
We can treat this like a "mixture problem".

Potato chips contain 0.6 % salt by weight naturally.
I want to add salt until the total weight is 2.6 % salt.
How much salt do I need to add to a weight of potato chips to bring the total to 2.6 % salt?
Click to expand...

Let \(\displaystyle W\) = original weight of the package (potato chips and salt combined).
Let's say it's measured in pounds.

We begin with \(\displaystyle W\) pounds of stuff which is 0.6% salt.
. . It contains: .\(\displaystyle 0.006W\) pounds of salt.

We add \(\displaystyle x\) pounds of salt.

The final mix has: .\(\displaystyle W + x\) pounds of stuff
. . of which \(\displaystyle 0.006W + x\) pounds is salt.

And this mix should be 2.6% salt.

\(\displaystyle \text{So we have: }\;\frac{0.006W + x}{W + x} \:=\:0.026\)

. . \(\displaystyle 0.006W + x \;=\;0.026W + 0.026x \quad\Rightarrow\quad x-0.026x \:=\:0.026W - 0.006W\)

. . \(\displaystyle 0.975x \:=\:0.02W \quad\Rightarrow\quad x \:=\:\frac{0.02}{0.974}W \quad\Rightarrow\quad x \:=\:(0.020533881)W\)


If the original package weighs \(\displaystyle W\) pounds, the amount of salt we will add is about 2% of \(\displaystyle W\).

 
thank you

I had tried several times to approach it by naming things and what not, but for some reason I just couldn't get my head around it.

It IS easy now that I see how you've shown me to do it.
:D
thank you again
 
I don't know what's the matter with me . I can't seem to get my head wrapped around this.

If you had a known quantity of the aforementioned potato chips that were aleardy salted to the 2.6% specification,
How could you find out how much of that weight was added salt?

I have tried for hours and I can't get it. It seems to me that you should be able to substitute the total weight for the w + x in the previous equations that you have shown me, but I always end up with two variables ... the original weight that I dont know, and the amount of salt added... that I don't know..

please help
 
what I want to know is the equation to determine how much salt has been added to a specific quantity of potato chips AFTER the salt has been added.

In the above posts, they showed me how I can determine how much salt I need to add to a quantity of chips that already contain a certain amount of natural salt in order to achive a desired percentage of salt... and that works fine.

I thought I could manipulate that equation around to tell me how much salt has been added to a known quantity of potato chips, but I can't figure it out.

In other words in the above equation I want to know how much salt has been added ( or what the original weight (W) was )if I know the finished amount, the finished percentage of salt and the percentage of salt contained in the natural chips before salting.

If you could help me I'd certainly appreciate it.
 
you need to use same equations. Analyse such problem in following manner:
Initially amount of chips and salt separately. Assign a variable for each of them.
After some process of addition how quantities change and alter the variable accordingly and take ratio of initial to final situation and solve the variables.
 
OIC.

How about picking a new name for some weight of chips that contains a total of 2.6% salt.

Let T = the weight (in ounces) of some chips to which the required salt has already been added

Then we have the following: T = W + x.

This means that W (in terms of T) equals T - x.

Let's substitute this expression for W into the relationship derived above.

\(\displaystyle \frac{0.006W + x}{W + x} \;=\; 0.026\)

\(\displaystyle \frac{0.006(T - x) + x}{(T - x) + x} \;=\; 0.026\)

Solve this equation for x, and you'll get an expression for the ounces of added salt in terms of the total weight T.
 
Re:

mmm4444bot said:
OIC.

How about picking a new name for some weight of chips that contains a total of 2.6% salt.

Let T = the weight (in ounces) of some chips to which the required salt has already been added

Then we have the following: T = W + x.

This means that W (in terms of T) equals T - x.

Let's substitute this expression for W into the relationship derived above.

\(\displaystyle \frac{0.006W + x}{W + x} \;=\; 0.026\)

\(\displaystyle \frac{0.006(T - x) + x}{(T - x) + x} \;=\; 0.026\)

Solve this equation for x, and you'll get an expression for the ounces of added salt in terms of the total weight T.
Click to expand...

That certainly seems like what I was trying to do.... I will try to figure it out right now.... thank you very much
 
I'm sorry ... I am afraid I don't know how to solve it..... :oops: would you be so kind as to show me an example....

suppose T= 100 grams and the finished salt was 2.6 percent and the naturally occurring salt was 0.6 percent in the quantity before x was added.

I really am trying... I just don't know how to do it

would the answer be x=2.006?


thank you
 
\(\displaystyle \frac{0.006(T - x) + x}{(T - x) + x} \;=\; 0.026\)

We want to solve for x. We see that x appears inside of some parthentheses, so we need to get rid of the parentheses.

In the numerator, (T - x) is multiplied by the number 0.006, so we have to use the Distributive Property, to get rid of those parentheses.

But, in the denominator, (T - x) is not multiplied by anything, so we can simply remove the parentheses.

\(\displaystyle \frac{(0.006)(T) - (0.006)(x) + x}{T - x + x} \;=\; 0.026\)

\(\displaystyle \frac{0.006T - 0.006x + x}{T} \;=\; 0.026\)

\(\displaystyle \frac{0.006T + 0.994x}{T} \;=\; 0.026\)

Now we get rid of the ratio, on the lefthand side; multiply both sides of the equation by T.

\(\displaystyle 0.006T + 0.994x \;=\; 0.026T\)

Next, Isolate the term containing x; subtract 0.006T from both sides.

\(\displaystyle 0.994x \;=\; 0.02T\)

Solve for x by dividing both sides by 0.994 .

\(\displaystyle x \;=\; 0.02012T\)

The amount of salt added is 2.012% of the total weight.

So, with your example of 100 ounces of chips (to which salt has already been added), we have the following.

T = 100

0.02012T = 2.012 (this is x)

2.012 ounces of salt were added, to bring the weight up to 100 ounces.

T - x = 100 - 2.012 = 97.998

The chips weighed 97.998 ounces before the extra salt was added (this is W).

0.006(97.998) = 0.58793

The chips contain 0.58793 ounces of naturally-occurring salt.

0.58793 + 2.012 = 2.5999

The total amount of salt in the 100 ounces of chips is 2.6 ounces (2.6% of 100).
 
Hey, thank you so much. I really appreciate it. :D
That was very kind of you. thank you.
 
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