An alloy containing 30%gold is mixed with a 60% gold alloy to get 300 kilograms of alloy that is 50%gold . How many kilograms of each alloy were used?

Question

 how much of each was used?

Answer 1

Let a=mass of 30% alloy

Let b= mass of 60% alloy

 

Since the total mass after mixing is 300 kg, a + b = 300 kg

 

The new mixture is a 50% alloy, so         0.3a + 0.6b = 0.5(300)

 

We have 2 equations with 2 unknowns, which can be solved by substitution or elimination

 

Substitution: b = 300 - a

                  

                  0.3a + 0.6(300 - a) = 150

 

                  0.3a + 180 - 0.6a = 150

 

Combine like terms and subtract 180 from each side,

 

                 - 0.3a = -30

 

                        a = 100 kg

 

b = 300 - a,       b = 200 kg

 

Elimination: Multiply the first equation by -0.3 to get:   -0.3a - 0.3b = -90

                                         The second equation is:     0.3a + 0.6b = 150

 

                             Add the two equations together:       0 + 0.3b = 60

 

                                                                                  b = 200 kg

                    

                     Using the first equation a + 200 = 300, so a = 100 kg

Answer 2

Hey Johnathan -- since a 1:1 "half-and-half" mix takes us "halfway" between

the 30-60 at 45% ... a 2:1 mix takes us "two-thirds" thru the 30-60 spread ...

Apply the double portion at the "heavier" end ==> 200kg 60% : 100kg 30% :)

Answer 3

Jonathan:

 

Let a=the number of kilograms of the first alloy

Let b=the number of kilograms of the second allow

 

From the statement of the problem, we can write two equations:

 

1) Since we know that the total number of kilograms in the combined mixture is equal to 300: a+b=300

2) The combination of the two alloys can be expressed as:  .30a+.60b=(.50)(300)

 

From equation 1 above, b=300-a.  Substitution this for b in equation 2, we get: .3a+.6(300-a)=150

 

Simplifying and solving, we get:

.3a+180-.6a=150

            -.3a= 150-180

            -.3a=-30

                a=100 kilograms of the first alloy used in the mixture

 

Since we know that a+b=300:

                b=300-a

                b=300-100

                b=200 kilograms of the second alloy used in the mixture

 

We can know check our results by substituting the values we computed for a and b back into the second equation to verify that both sides match:

.3a+.6b=150

(.3)(100)+(.6)(200) = 150

    30      +     120   = 150

                     150   = 150 (check!!)

  

Let me know if this helps.

 

George T

Answer 4

Let x = mass of 30% alloy

So 300-x = mass of 60% alloy

 

Equation for total amount of gold

Gold in 30% + Gold in 60% = Gold in 50%

 

.3x + .6(300-x) = .5(300)

.3x + 180 - .6x = 150

-.3x + 180 = 150

-.3x = -30

x = -30/-.3 = 30/.3 = 300/3 = 100

 

100 kg of 30%

300-x = 300-100 = 200 kg of 60%