What if you had a jar of pennies? You take out a handful that is equal to \begin{align*}\frac{1}{10}\end{align*} the total. Your friend Zoey then takes a handful that is equal to \begin{align*}\frac{2}{9}\end{align*} the amount that remains. What fraction of the total does Zoey take? After completing this Concept, you'll be able to solve real-world problems like this one.
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CK-12 Foundation: 0207S Real-World Multiplication and Division (H264)
Guidance
Using the skills learned in the last two concepts, you now ready to solve real-world problems.
Example A
In the chemistry lab there is a bottle with two liters of a 15% solution of hydrogen peroxide \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 \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 \begin{align*}H_2 O_2\end{align*} left in the bottle.
b) The amount of diluted \begin{align*}H_2 O_2\end{align*} in the beaker.
c) The concentration of the \begin{align*}H_2 O_2\end{align*} in the beaker.
Solution
a) To determine the amount of \begin{align*}H_2 O_2\end{align*} left in the bottle, we first determine the amount that was removed. That amount was \begin{align*}\frac{1}{5}\end{align*} of the amount in the bottle (2 liters). \begin{align*}\frac{1}{5}\end{align*} of 2 is \begin{align*}\frac{2}{5}\end{align*}.
The amount remaining is \begin{align*}2 - \frac{2}{5}\end{align*}, or \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) We determined that the amount of the 15% \begin{align*}H_2 O_2\end{align*} solution removed was \begin{align*}\frac{2}{5}\end{align*} liter. The amount of water added was twice this amount, or \begin{align*}\frac{4}{5}\end{align*} liter. So the total amount of solution in the beaker is now \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 \begin{align*}H_2 O_2\end{align*} in the beaker.
c) The new concentration of \begin{align*}H_2 O_2\end{align*} can be calculated.
John started with \begin{align*}\frac{2}{5}\end{align*} liter of 15% \begin{align*}H_2 O_2\end{align*} solution, so the amount of pure \begin{align*}H_2 O_2\end{align*} is 15% of \begin{align*}\frac{2}{5}\end{align*} liters, or \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 \begin{align*}H_2 O_2\end{align*} is \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% \begin{align*}H_2 O_2\end{align*}.
Example B
Anne has a bar of chocolate and she offers Bill a piece. Bill quickly breaks off \begin{align*}\frac{1}{4}\end{align*} of the bar and eats it. Another friend, Cindy, takes \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 \begin{align*}\frac{1}{4}\end{align*} of the bar.
Cindy takes \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.
Solution
Each person gets exactly \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 \begin{align*}\frac{1}{4}\end{align*} of it, so there is \begin{align*}1 - \frac{1}{4} = \frac{3}{4}\end{align*} of a bar left.
Cindy takes \begin{align*}\frac{1}{3}\end{align*} of what’s left, or \begin{align*}\frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4}\end{align*} of a whole bar. That leaves \begin{align*}\frac{3}{4} - \frac{1}{4} = \frac{2}{4}\end{align*}, or \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: \begin{align*}\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}\end{align*}.
So each person gets exactly \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?
Example C
Newton’s second law \begin{align*}(F = ma)\end{align*} relates the force applied to a body in Newtons \begin{align*}(F)\end{align*}, the mass of the body in kilograms \begin{align*}(m)\end{align*} and the acceleration in meters per second squared \begin{align*}(a)\end{align*}. Calculate the resulting acceleration if a Force of \begin{align*}7\frac{1}{3}\end{align*} Newtons is applied to a mass of \begin{align*}\frac{1}{5} \ kg\end{align*}.
Solution
First, we rearrange our equation to isolate the acceleration, \begin{align*}a\end{align*}. If \begin{align*}F = ma\end{align*}, dividing both sides by \begin{align*}m\end{align*} gives us \begin{align*}a = \frac{F}{m}\end{align*}. Then we substitute in the known values for \begin{align*}F\end{align*} and \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*}.
Watch this video for help with the Examples above.
CK-12 Foundation: Solving Real-World Problems Using Multiplication and Division
Vocabulary
When multiplying an expression by negative one, remember to multiply the entire expression by negative one.
To multiply fractions, multiply the numerators and multiply the denominators: \begin{align*}\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}\end{align*}
The multiplicative properties are:
- Commutative Property: The product of two numbers is the same whichever order the items to be multiplied are written. Example: \begin{align*}2 \cdot 3 = 3 \cdot 2 \end{align*}
- Associative Property: When three or more numbers are multiplied, the sum is the same regardless of how they are grouped. Example: \begin{align*} 2 \cdot (3 \cdot 4) = (2 \cdot 3) \cdot 4\end{align*}
- Multiplicative Identity Property: The product of any number and one is the original number. Example: \begin{align*}2 \cdot 1 = 2\end{align*}
- Distributive Property: The product of an expression and a sum is equal to the sum of the products of the expression and each term in the sum. For expressions \begin{align*}a, b, \end{align*} and \begin{align*}c\end{align*}, \begin{align*}a(b+c)=ab+ac\end{align*}. Example: \begin{align*}4(2 + 3) = 4(2) + 4(3)\end{align*}
The multiplicative inverse of a number is the number which produces 1 when multiplied by the original number. The multiplicative inverse of \begin{align*}x\end{align*} is the reciprocal \begin{align*}\frac{1}{x}\end{align*}. To find the multiplicative inverse of a fraction, simply invert the fraction: \begin{align*}\frac{a}{b}\end{align*} inverts to \begin{align*}\frac{b}{a}\end{align*}.
To divide fractions, invert the divisor and multiply: \begin{align*}\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}\end{align*}.
An object moving at a certain speed will cover a fixed distance in a set time. The quantities speed, distance and time are related through the equation \begin{align*}\text{Speed} = \frac{\text{Distance}}{\text{Time}}\end{align*}.
Guided Practice
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?
Solution
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.
Practice
For 1-8, perform the operations of multiplication and division.
- \begin{align*}(3 - x)\end{align*}
- \begin{align*}\frac{1}{13} \times \frac{1}{11}\end{align*}
- \begin{align*}\frac{7}{27} \times \frac{9}{14}\end{align*}
- \begin{align*}\left ( \frac{3}{5} \right )^2\end{align*}
- \begin{align*}\frac{1}{2} \div \frac{x}{4y}\end{align*}
- \begin{align*}\left ( - \frac{1}{3} \right ) \div \left( - \frac{3}{5} \right )\end{align*}
- \begin{align*}\frac{7}{2} \div \frac{7}{4}\end{align*}
- \begin{align*}11 \div \frac{-x}{4}\end{align*}
For 9-11, solve the real-world problems using multiplication and division.
- 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?
- 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?
- 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.