Equation: Very Tricky Question..Please help

itsnayyer

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how many cubic centimeters of a 20% solution of a nutrient must be added to 100cc of a 2% solution of the same nutrient to make a 10% solution of the nutrient?
 
how many cubic centimeters of a 20% solution of a nutrient must be added to 100cc of a 2% solution of the same nutrient to make a 10% solution of the nutrient?

1. Let x denotes the necessary amount of the 20% solution. Re-write the percentages into rational numbers.

2. "Translate" the text of the question into an equation:

\(\displaystyle \displaystyle{\frac{x \cdot 0.2 + 100 \cdot 0.02}{x+100} = 0.1}\)

3. Solve for x.
 
Thank you very much for you help. i solved it for 'x' but didn't get the same answer as in my book which is 80cc. please help.
 
Thank you very much for you help. i solved it for 'x' but didn't get the same answer as in my book which is 80cc. please help.

I don't know how you solved the given equation ... :(

So here is how I would have done it:

\(\displaystyle \displaystyle{\frac{x \cdot 0.2 + 100 \cdot 0.02}{x+100} = 0.1}\)

a) Multiply both sides by (x + 100). Keep in mind that you have to apply the distributive law.

b) Collect like terms which should give you

\(\displaystyle 0.1 x = 8\)

c) Divide both sides by 0.1 and you'll get the result of your book.
 
Thank you very much. i made a mistake while solving the equation..now got the right answer. i would highly appreciate if you guide me the logic of " x+100" in the denominator.
 
Hello, itsnayyer!

How many cubic centimeters of a 20% solution of a nutrient
must be added to 100cc of a 2% solution of the same nutrient
to make a 10% solution of the nutrient?

We start with 100cc of a 2% solution.
. . It contains: .\(\displaystyle 0.02(100) = 2\) cc of nutrient.

We add \(\displaystyle x\) cc of a 20% solution.
. . It contains: .\(\displaystyle 0.20(x) \:=\:0.2x\) cc of nutrient.

Hence, the mixture contains: .\(\displaystyle 2 + 0.2x\) cc of nutrient. .[1]


But the mixture will be \(\displaystyle 100 + x\) cc which is 10% nutrient.
. . It contains: .\(\displaystyle 0.10(100 +x)\) cc of nutrient. .[2]


We just described the final amount of nutrient in two ways.

There is our equation!

. . \(\displaystyle 2+0.2x \;=\; 0.1(100+x)\)

. . \(\displaystyle 2+0.2x \;=\;10 + 0.1x \)

. . . . . \(\displaystyle 0.1x \;=\;8\)

. . . . . . . \(\displaystyle x \;=\;\dfrac{8}{0.1} \;=\;80\)
 
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