SOLUTION: An industrialist has 450 litres of a chemical which is 70% pure. he mixes it with a chemical of the same type but 90% pure so as to obtain a mixture which is 75% . pure. find the

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: An industrialist has 450 litres of a chemical which is 70% pure. he mixes it with a chemical of the same type but 90% pure so as to obtain a mixture which is 75% . pure. find the       Log On


   

Question 1101892: An industrialist has 450 litres of a chemical which is 70% pure. he mixes it with a chemical of the same type but 90% pure so as to obtain a mixture which is 75% . pure. find the amount of the 90% pure chemical used.
Found 2 solutions by ikleyn, richwmiller:
Answer by ikleyn(52869) About Me  (Show Source):
You can put this solution on YOUR website!
.
an industrialist has 450 litres of a chemical which is 70% pure. he mixes it with a chemical of the same type but 90% pure
so as to obtain a mixture which is 75% . pure. find the amount of the 90% pure chemical used.
~~~~~~~~~~~~~~~~~~ 

Let x be the volume of the 90% solution, in liters.


450 liters of the 70% solution contain 0.7*450 liters of the pure.

x   liters of the 90% solution contain 0.9*x liters   of the pure.

Final (450+x) liters solution contain  0.75*(450+x)  liters of the pure.


PURE    + PURE   = PURE           ========================>

0.7*450 + 0.9*x  =  0.75*(450+x) 


You just got your balance equation for "the pure".

Now solve it:


0.7*450 + 0.9x = 0.75*450 + 0.75x  ====>

0.9x - 0.75x = 0.75*450 - 0.7*450,

0.15x = 0.05*450  ====>  x = %280.05%2A450%29%2F0.15 = 150.


Answer.  150 liters of the 90% solution should be added.


Check.  There are  0.7*450 + 0.9*150 = 450 liters of the pure in the input components.   

        There are  0.75*(450+150) = 450 liters of the pure in the FINAL solution.   ! Correct !


There is entire bunch of introductory lessons covering various types of mixture problems
    - Mixture problems
    - More Mixture problems
    - Solving typical word problems on mixtures for solutions
    - Word problems on mixtures for antifreeze solutions
    - Word problems on mixtures for alloys
    - Typical word problems on mixtures from the archive
in this site.

You will find there ALL TYPICAL mixture problems with different methods of solutions,
explained at the different levels of detalization, from very detailed to very short.

Read them and become an expert in solution mixture word problems.

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook in the section "Word problems" under the topic "Mixture problems".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.


Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
An industrialist has 450 litres of a chemical which is 70% pure. He mixes it with a chemical of the same type but 90% pure so as to obtain a mixture which is 75% . pure. Find the amount of the 90% pure chemical used.

The industrialist has no pure chemical as far as we know.
He has on hand a mixture which is 70% and a mix that is 90% and wants to blend them to get a mix that is 75%
Pure means 100%
The 70% mix is a blend of 30%*450 = 125 L something else and 70 %*450 =315 L pure
Otherwise, the math presented is correct.
We have on hand 450 L of 70% solution.
Using the method of alligation. 

90         5 5/15*450=150 L of 90% solution
      75            600 L of 75% solution
70         15        450 L of 70% solution
           20

check
0.9 * 150 + 0.7 * 450 = 0.75*(450+150)
135.0 + 315.0 = 0.75*(600)
450.0 = 450.0
ok