Concept explainers
a.
To calculate: the number of liters each solution will satisfy when we use 2 liters of the 50% solution.
a.
Answer to Problem 49E
1l of 10% solution
7l of 20% solution
2l of 50% solution
Explanation of Solution
Given:
A chemist needs 10 liters of 25% acid solution. The solution is to be mixed from three solutions whose concentrations are 10%, 20%, and 50%.
Calculation:
Let
So,
According to question
Here value of z is 2
By putting value of y in above equations we get
Now multiplying equation
Subtracting equation
Put
and
So,
1l of 10% solution
7l of 20% solution
2l of 50% solution
Conclusion:
1l of 10% solution
7l of 20% solution
2l of 50% solution
b.
To calculate: the number of liters each solution will satisfy when we use as little as possible of the 50% solution.
b.
Answer to Problem 49E
0l of 10% solution
8.33l of 20% solution
1.67l of 50% solution
Explanation of Solution
Given:
A chemist needs 10 liters of 25% acid solution. The solution is to be mixed from three solutions whose concentrations are 10%, 20%, and 50%.
Calculation:
Let
So,
According to question
To use as little as possible of 50% solution means mix it with 20% solution only
So here value of x is 0
By putting value of x in above equations we get
Now multiplying equation
Multiplying equation
Subtracting equation
Put
So
0l of 10% solution
8.33l of 20% solution
1.67l of 50% solution
Conclusion:
0l of 10% solution
8.33l of 20% solution
1.67l of 50% solution
c.
To calculate: the number of liters each solution will satisfy when we use as much as possible of the 50% solution.
c.
Answer to Problem 49E
6.25l of 10% solution
0l of 20% solution
3.75l of 50% solution
Explanation of Solution
Given:
A chemist needs 10 liters of 25% acid solution. The solution is to be mixed from three solutions whose concentrations are 10%, 20%, and 50%.
Calculation:
Let x be amount of 10% solution, y be amount of 20% solution and z be amount of 50% solution.
So,
According to question
To use as much as possible of 50% solution means mix it with 10% solution only.
So here value of y is 0
By putting value of y in above equations we get
Now multiplying equation
Subtracting equation
Putting
So
6.25l of 10% solution
0 of 20% solution
3.25l of 50% solution
Conclusion:
6.25l of 10% solution
0 of 20% solution
3.25l of 50% solution
Chapter 7 Solutions
EBK PRECALCULUS W/LIMITS
- Calculus: Early TranscendentalsCalculusISBN:9781285741550Author:James StewartPublisher:Cengage LearningThomas' Calculus (14th Edition)CalculusISBN:9780134438986Author:Joel R. Hass, Christopher E. Heil, Maurice D. WeirPublisher:PEARSONCalculus: Early Transcendentals (3rd Edition)CalculusISBN:9780134763644Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric SchulzPublisher:PEARSON
- Calculus: Early TranscendentalsCalculusISBN:9781319050740Author:Jon Rogawski, Colin Adams, Robert FranzosaPublisher:W. H. FreemanCalculus: Early Transcendental FunctionsCalculusISBN:9781337552516Author:Ron Larson, Bruce H. EdwardsPublisher:Cengage Learning