Mixture Problems

Q:  How many mL of a 3% acid solution should be mixed with a 7% acid solution to create 120 mL of a 5% acid solution?


There is a generic formula we can use for mixture problems:

(quantity of solution A)*(% of A) + (quantity of solution B)*(% of B) = (final quantity)*(% of final)

So, for this problem, we know the final quantity is 120 mL.

Let’s use “x” to represent the amount of solution A that we need.  Since we need 120 total, we know that the amount of solution B can be represented by “120 – x” (the left-over).

So, plugging the values and variables into the equation:

(quantity of solution A)*(% of A) + (quantity of solution B)*(% of B) = (final quantity)*(% of final)

(x)(.03)+(120-x)(.07) = (120)*(.05)

Now we need to simplify and do the math to solve for x:

.03x + 8.4 – .07x = 6 [I distributed through the .07 on the left side and multiplied the numbers on the right side of the equation]

Now, combine the x’s on the left side:

-.04x + 8.4 = 6[now solve for x]

-.04x = -2.4

x = 60 mL

So, we should use 60 mL of solution A.

And, we should use 120 – x of solution B.  Since x = 60, we need 120 – 60 of solution B.  This is also 60 mL.

So, we should use 60 mL of solution A (the 3% solution) and 60 mL of solution B (the 7% solution) to get 120 mL of a 5% solution.

Note:  You could also solve this problem by observation:  Since 5% is exactly in between 3% and 7%, we know we need an equal amount of each solution for a perfect balance.  Since there is 120 mL total needed, we would need half to be 3% and half to be 7%

This type of logic only works when the numbers are “nice” like in this problem.  The work, shown above, can be used no matter what the numbers are.


Q:  A scientist has 256 g of goo. After 195 minutes, her sample has decayed to 16 g. What’s the half-life of the goo in minutes? (Assume exponential decay)


First we need to set up the problem with our decay formula.  Depending on your school / teacher / book, the variables in the formulas may be different letters, but they all mean the same thing:

P = A*e^(rt)

P is the ending amount, A is the initial amount, r is the growth rate/decay, and t is time.

So, the set-up with our info:

16 = 256*e^(r*195)

Now we need to solve for r:

16 = 256*e^(r*195)

1/16 = e^(r*195)

Take the natural log of both sides:

ln(1/16) = ln(e^(r*195))

Exponents can come down as multipliers (log rules):

ln(1/16) = (195r)*ln(e)

And, ln(e) = 1, so:

ln(1/16) = (195r)

ln(1/16) / 195 = r

Plug into a calculator to get:

r = -.01422

Now, we have r.  We now need to find the half-life.  So, if we started with 256 g, we want to know how long (t) it takes to get to 128 grams.  Plug the info in to our equation:

128 = 256*e^(-.01422*t)

and solve for t:

1/2 = e^(-.01422*t)

ln(1/2) = ln(e^(-.01422*t))

ln(1/2) = -.01422*t

ln(1/2)/-.01422 = t

So, t = 48.74 minutes

The half-life of the goo is approximately 48.74 minutes.

Algebra Word Problem Example

Q:  The total cost of a holiday package to a warm climate in January can be modeled using the formula C=x+0.12x+0.15x+38. C is the total cost in dollars. x is the list price of the package in dollars. 0.12 is the tax rate, 12%. 0.15 is the service charge rate,15%. 38 is the airport tax in dollars.  Find the list price when the total cost of the holiday package is $3338.73.

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