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# Mixed Numbers in Applications

## Change mixed numbers to improper fractions, Like signs +, Unlike signs

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Mixed Numbers in Applications

### Mixed Numbers in Applications

#### Real-World Application: Chemistry

In the chemistry lab there is a bottle with two liters of a 15% solution of hydrogen peroxide (H2O2)\begin{align*}(H_2 O_2)\end{align*}. John removes one-fifth of what is in the bottle, and puts it in a beaker. He measures the amount of H2O2\begin{align*}H_2 O_2\end{align*} and adds twice that amount of water to the beaker. Calculate the following measurements.

a) The amount of H2O2\begin{align*}H_2 O_2\end{align*} left in the bottle.

To determine the amount of H2O2\begin{align*}H_2 O_2\end{align*} left in the bottle, we first determine the amount that was removed. That amount was 15\begin{align*}\frac{1}{5}\end{align*} of the amount in the bottle (2 liters). 15\begin{align*}\frac{1}{5}\end{align*} of 2 is 25\begin{align*}\frac{2}{5}\end{align*}.

The amount remaining is 225\begin{align*}2 - \frac{2}{5}\end{align*}, or 10525=85\begin{align*}\frac{10}{5} - \frac{2}{5} = \frac{8}{5}\end{align*} liter (or 1.6 liters).

There are 1.6 liters left in the bottle.

b) The amount of diluted H2O2\begin{align*}H_2 O_2\end{align*} in the beaker.

We determined that the amount of the 15% H2O2\begin{align*}H_2 O_2\end{align*} solution removed was 25\begin{align*}\frac{2}{5}\end{align*} liter. The amount of water added was twice this amount, or 45\begin{align*}\frac{4}{5}\end{align*} liter. So the total amount of solution in the beaker is now 25+45=65\begin{align*}\frac{2}{5} + \frac{4}{5} = \frac{6}{5}\end{align*} liter, or 1.2 liters.

There are 1.2 liters of diluted H2O2\begin{align*}H_2 O_2\end{align*} in the beaker.

c) The concentration of the H2O2\begin{align*}H_2 O_2\end{align*} in the beaker.

The new concentration of H2O2\begin{align*}H_2 O_2\end{align*} can be calculated.

John started with 25\begin{align*}\frac{2}{5}\end{align*} liter of 15% H2O2\begin{align*}H_2 O_2\end{align*} solution, so the amount of pure H2O2\begin{align*}H_2 O_2\end{align*} is 15% of 25\begin{align*}\frac{2}{5}\end{align*} liters, or 0.15×0.40=0.06\begin{align*}0.15 \times 0.40 = 0.06\end{align*} liters.

After he adds the water, there is 1.2 liters of solution in the beaker, so the concentration of H2O2\begin{align*}H_2 O_2\end{align*} is 0.061.2=120\begin{align*}\frac{0.06}{1.2} = \frac{1}{20}\end{align*} or 0.05. To convert to a percent we multiply this number by 100, so the beaker’s contents are 5% H2O2\begin{align*}H_2 O_2\end{align*}.

#### Real-World Application: Splitting Candy

Anne has a bar of chocolate and she offers Bill a piece. Bill quickly breaks off 14\begin{align*}\frac{1}{4}\end{align*} of the bar and eats it. Another friend, Cindy, takes 13\begin{align*}\frac{1}{3}\end{align*} of what was left. Anne splits the remaining candy bar into two equal pieces which she shares with a third friend, Dora. How much of the candy bar does each person get?

First, let’s look at this problem visually.

Anne starts with a full candy bar.

Bill breaks off 14\begin{align*}\frac{1}{4}\end{align*} of the bar.

Cindy takes 13\begin{align*}\frac{1}{3}\end{align*} of what was left.

Dora gets half of the remaining candy bar.

We can see that the candy bar ends up being split four ways, with each person getting an equal amount.

Each person gets exactly 14\begin{align*}\frac{1}{4}\end{align*} of the candy bar.

We can also examine this problem using rational numbers. We keep a running total of what fraction of the bar remains. Remember, when we read a fraction followed by of in the problem, it means we multiply by that fraction.

We start with 1 bar. Then Bill takes 14\begin{align*}\frac{1}{4}\end{align*} of it, so there is 114=34\begin{align*}1 - \frac{1}{4} = \frac{3}{4}\end{align*} of a bar left.

Cindy takes 13\begin{align*}\frac{1}{3}\end{align*} of what’s left, or 1334=14\begin{align*}\frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4}\end{align*} of a whole bar. That leaves 3414=24\begin{align*}\frac{3}{4} - \frac{1}{4} = \frac{2}{4}\end{align*}, or 12\begin{align*}\frac{1}{2}\end{align*} of a bar.

That half bar gets split between Anne and Dora, so they each get half of a half bar: 1212=14\begin{align*}\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}\end{align*}.

So each person gets exactly 14\begin{align*}\frac{1}{4}\end{align*} of the candy bar.

Extension: If each person’s share is 3 oz, how much did the original candy bar weigh?

#### Real-World Application: Newton's Law

Newton’s second law (F=ma)\begin{align*}(F = ma)\end{align*} relates the force applied to a body in Newtons (F)\begin{align*}(F)\end{align*}, the mass of the body in kilograms (m)\begin{align*}(m)\end{align*} and the acceleration in meters per second squared (a)\begin{align*}(a)\end{align*}. Calculate the resulting acceleration if a Force of 713\begin{align*}7\frac{1}{3}\end{align*} Newtons is applied to a mass of 15 kg\begin{align*}\frac{1}{5} \ kg\end{align*}.

First, we rearrange our equation to isolate the acceleration, a\begin{align*}a\end{align*}. If F=ma\begin{align*}F = ma\end{align*}, dividing both sides by m\begin{align*}m\end{align*} gives us a=Fm\begin{align*}a = \frac{F}{m}\end{align*}. Then we substitute in the known values for F\begin{align*}F\end{align*} and m\begin{align*}m\end{align*}:

\begin{align*}a = \frac{7 \frac{1}{3}}{\frac{1}{5}} = \frac{22}{3} \div \frac{1}{5} = \frac{22}{3} \times \frac{5}{1} = \frac{110}{3}\end{align*}

The resultant acceleration is \begin{align*}36 \frac{2}{3} \ m/s^2\end{align*}.

### Example

Andrew is driving down the freeway. He passes mile marker 27 at exactly mid-day. At 12:35 he passes mile marker 69. At what speed, in miles per hour, is Andrew traveling?

To find the speed, we need the distance traveled and the time taken. If we want our speed to come out in miles per hour, we’ll need distance in miles and time in hours.

The distance is \begin{align*}69 - 27 \end{align*} or 42 miles. The time is 35 minutes, or \begin{align*}\frac{35}{60}\end{align*} hours, which reduces to \begin{align*}\frac{7}{12}\end{align*}. Now we can plug in the values for distance and time into our equation for speed.

\begin{align*}\text{Speed} = \frac{42}{\frac{7}{12}} = 42 \div \frac{7}{12} = \frac{42}{1} \times \frac{12}{7} = \frac{6 \cdot 7 \cdot 12}{1 \cdot 7} = \frac{6 \cdot 12}{1} = 72\end{align*}

Andrew is driving at 72 miles per hour.

### Review

For 1-8, perform the operations of multiplication and division.

1. \begin{align*}(3 - x)\end{align*}
2. \begin{align*}\frac{1}{13} \times \frac{1}{11}\end{align*}
3. \begin{align*}\frac{7}{27} \times \frac{9}{14}\end{align*}
4. \begin{align*}\left ( \frac{3}{5} \right )^2\end{align*}
5. \begin{align*}\frac{1}{2} \div \frac{x}{4y}\end{align*}
6. \begin{align*}\left ( - \frac{1}{3} \right ) \div \left( - \frac{3}{5} \right )\end{align*}
7. \begin{align*}\frac{7}{2} \div \frac{7}{4}\end{align*}
8. \begin{align*}11 \div \frac{-x}{4}\end{align*}

For 9-11, solve the real-world problems using multiplication and division.

1. The label on a can of paint says that it will cover 50 square feet per pint. If I buy a \begin{align*}\frac{1}{8}\end{align*} pint sample, it will cover a square two feet long by three feet high. Is the coverage I get more, less or the same as that stated on the label?
2. The world’s largest trench digger, "Bagger 288", moves at \begin{align*}\frac{3}{8}\end{align*} mph. How long will it take to dig a trench \begin{align*}\frac{2}{3}\end{align*} mile long?
3. A \begin{align*}\frac{2}{7}\end{align*} Newton force applied to a body of unknown mass produces an acceleration of \begin{align*}\frac{3}{10} \ m/s^2\end{align*}. Calculate the mass of the body.

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