Lesson Typical word problems on mixtures from the archive

Typical word problems on mixtures from the archive

Problem 1

Let  x  be the required amount (volume) of a  40% acid solution, in liters.

Then the equation for the pure acid volume is 

0.4x + 0.25*8 = 0.3*(x + 8).

Simplify and solve it:

0.4x + 2 = 0.3x + 2.4,

0.4x - 0.3x = 2.4 - 2,

0.1x = 0.4,

x =  = 4.

Answer.  4 liters of the 40% acid solution is needed.

Problem 2

    x +    y = 90,       (1)    (volume balance, in liters)
0.25x + 0.7y = 0.4*90    (2)    (pure acid volume balance, in liters)

From (1), express y = 90 - x and substitute it into equation (2). You will get

0.25x + 0.7*(90 - x) = 36.

Simplify and solve for x:

0.25x + 63 - 0.7x = 36,

-0.45x = 36 - 63,

-0.45x = -27,

x =  = 60.

Answer.  60 liters of the 25% acid solution and 90-60 = 30 liters of the 70% acid solution.

Problem 3

    x +     y = 310,        (1)    (milk volume balance, in liters)
0.07x + 0.02y = 0.06*310    (2)    (butterfat balance)

From (1), express y = 310 - x and substitute it into equation (2). You will get

0.07x + 0.02*(310 - x) = 18.6.

Simplify and solve for x:

0.07x + 6.2 - 0.02x = 18.6,

0.05x = 18.6 - 6.2,

0.05x = 12.4,

x =  = 248.

Answer.  248 liters of the 0.06% butterfat and 310-248 = 62 liters of the 0.02% butterfat.

Problem 4

Let x = # teaspoons of the 10% HR;
    y = # teaspoons of the 60% HR.

Then the system of equations is 

   x +     y = 4,        (1)     (total volume, in teaspoons)
0.1x + 0.60y = 0.55*4    (2)     (the pure HR contents)

 
Multiply the second eqn by 100 to get

   x +     y = 4,        (1')    
 10x +   60y = 220       (2')


Express x = 4 - y from (1), and then substitute into (2'). You will get

10*(4-y) + 60y = 220.

40 - 10y + 60y = 220,  ====>  50y = 220 - 40 = 180  ====>  y =  =  .