Lesson Entertainment problems on mixtures

Entertainment problems on mixtures

Problem 1

There are 3/4 of the bottle volume of the pure alcohol and 1/4 of the bottle volume of water in the 1-st bottle.


There are 4/5 of the bottle volume of the pure alcohol and 1/5 of the bottle volume of water in the 2-nd bottle.



So, the total volume of the pure alcohol is   +  =  =  of the bottle volume.


    The total volume of water is   +  =  =  of the bottle volume.



The ratio alcohol to water volumes in the mixture is thus  .    ANSWER

Problem 2

(a)  It is intuitively clear that if you want to use as little as possible of the 50% solution,

     you should not use 10% solution at all, and should use the 20% solution, mixing it with the 50% solution.


    In this case, let x liters be the amount of the 50% solution to use 
              and let (8-x) liters be the amount of the 20% to add.


    Then you have this equation for the pure acid volume

        0.5x + 0.2*(8-x) = 0.25*8


    From the equation,  x =  =  = 1 .


    ANSWER.  1  liters of the 50% acid solution and the rest, 6  liters of the 20% acid solution.



(b)  It is intuitively clear that if you want to use as much as possible of the 50% solution,

     you should not use 20% solution at all, and should use the 10% solution, mixing it with the 50% solution.


    In this case, let x liters be the amount of the 50% solution to use 
              and let (8-x) liters be the amount of the 10% to add.


    Then you have this equation for the pure acid volume

        0.5x + 0.1*(8-x) = 0.25*8


    From the equation,  x =  =  = 3.


    ANSWER.  3 liters of the 50% acid solution and the rest, 5 liters of the 10% acid solution.

Problem 3

The mass percentage formulas for participating materials are

    Sornog = (x   g,  y   t,  0   m)

    Yotril = (0.75g,  0.13t,  0.12m)

    Trophy = (0.72g,  0.20t,  0.08m)


         Here letters g, t, and m designate gold, Tustrite and Motrium, as elementary components;
         x and y symbolize the quantities. 


We combine "a" grams of Sornog and "b" grams of Yotril and get a+b grams of Trophy.


Looking for the Motrium component, we see that 

    0*a + 0.12b = 0.08(a+b),

or, simplifying

    0.12b = 0.08a + 0.08b  --->  0.12b - 0.08b = 0.08a  --->  0.04b = 0.08a  --->  b = 2a.

It tells us that "a" grams of the Sornog and 2a grams of the Youtril were melted to get a+2a = 3a grams of the Trophy.


Having this, we can write the mass equation for the gold components

    ax + (2a)*0.75 = (3a)*0.72.


From it, we express "x"

    x =  = 3*0.72-2*0.75 = 0.66.


It tells us that the percentage of gold in the used Sornog is 66%.    ANSWER

Problem 4

The amount of the pure antifreeze in 15 liters of the 40% mixture 
is 0.4*15 = 6 liters.


The amount of the pure antifreeze in 15 liters of the 60% mixture 
is 0.6*15 = 9 liters.


So, the original mixture contained 6 liters of the pure antifreeze;
the final mixture contains 9 liters of the pure antifreeze.


Now, Let V be the volume of the mixture to replace.


When we drain V liters of the 40% mixture, 0.4*V liters of the pure antifreeze go out.
We replace it with V liters of the pure antifreeze.


So, V should be 3 liters more than 0.4V, which means

    V - 0.4V = 3,  or  0.6V = 3,  hence  V =  = 5 liters.


At this point, the solution is complete, and we get the 


ANSWER.  5 liters of the of the original 40% mixture should be drained 

         and replaced by 5 liters of pure antifreeze.