Rate Problems
Sample Problem 1. Car A travels at a speed of 25 mph. If car B starts 5 miles behind car A, at what speed does car B travel if it catches up with car A in 30 minutes?
For this problem, we can use the equation d=rt, where d is the distance, r is the rate, and t is the time. If this equation doesn't appear to be obviously true immediately, think about how the units will cancel out. We will want to introduce some variables to represent what we already know and what we want to know. In general, when doing word problems, it is often a good idea to start by defining variables. Let \(d_a\), \(r_a\), and \(t_a\) be the distance car A travels, the rate at which car A travels, and the amount of time car A travels. We will define \(d_b\), \(r_b\), and \(t_b\) similarly. Next, we will look at the information we are given. We know that \(r_a=25 mph \). We are also given that when \(t= 1/2 hours\), \(d_b=d_a+5\). Thus, we know that \(d_b=\frac{1}{2}\cdot 25=12.5 miles\). By writing the equation \(d_b=12.5+5=17.5=\frac{r_b}{2}\), we can see that \(r_b=35 mph\). Another way to look at this problem, that is essentially the same, but a little quicker, is to see that car B travels 5 more miles than car A in half an hour, so car B has a speed \(10 mph\) greater than car A's speed. Thus, car B has a speed of \(35 mph\), as we saw before. One important thing to note, and to be careful of, is that we are using mph as speed, so \(d=rt\) is only true if all of the units are the same, meaning we must note that we used \(30 minutes = \frac{1}{2} hours \neq 30 hours\).
Sample Problem 2. Fred runs at a speed of 7 mph. Ned runs at a speed of 6 mph. If Fred and Ned both start at the same place on a 169 mile track, how long until Fred and Ned meet for the first time once they start running.
We will, again, use d=rt. For Fred and Ned to meet, they will have had to travel the full distance of the track when the distances each ran is combined. In one hour, we can see they travel a total of \(7+6=13 miles\) closer to each other. Thus, we see that 169=13t, so they will meet each other for the first time in exactly 13 hours.
Sample Problem 3. If Fred biked up a hill at \(r_1\) mph and biked down the same hill at \(r_2\) mph, find Fred's average speed across the whole ride in terms of \(r_1\) and \(r_2\).
When we first see this problem, we may think that the answer is just \(\frac{r_1+r_2}{2}\). However, we can intuitively see that his is wrong because Fred is biking for longer when he bikes at a slower rate. For this problem, we already have two variables defined for us and, as we have seen previously, it is a good idea to assign variables to everything. Thus, we will call the distance from the bottom to the top of the hill \(d\) miles. We can then easily see that Fred traveled \(2d\) miles in total. We want to find Fred's average speed, which is imply total distance divided by total time. We have the distance, so now we just want to find the time. For the first part of his ride, Fred traveled \(d\) miles at a rate of \(r_1\) mph. Thus, for the first part he took \(\frac{d}{r_1}\) hours. Similarly, for going down the hill, it took Fred \(\frac{d}{r_2}\) hours. Thus, Fred's total time was \( \frac{d}{r_1} \)+\(\frac{d}{r_2}\) = \(\frac{(r_1+r_2)d}{r_1\cdot r_2}\) hours. Finally, we can easily find Fred's average rate by taking the distance divided by the time to get his rate to be \[\frac{2d}{(\frac{d(r_1+r_2)}{r_1\cdot r_2})} = \frac{2\cdot r_1\cdot r_2}{r_1 + r_2}.\]. This quantity is called the harmonic mean of \(r_1\) and \(r_2\).
Work Problems
Sample Problem 4. If 7 workers can paint 7 houses in 7 hour, how long would it take for 3 workers to paint 3 houses? Assume each worker works at the same rate and and each house needs the same amount painted.
Similar to our previous two problems, we will use the equation w=rt where w is the amount of work, r is the rate, and t is the time. In this problem, we let the units for w be houses painted, units for r be in houses per hour for one worker, and t be in hours. Thus, \(7=7r\cdot7\), so each worker can paint \(\frac{1}{7}\) of a house per hour. Thus, 3 workers can a paint \(\frac{3}{7}\) of a house in an hour. Thus, to paint 3 house, it takes 3 workers \(\frac{3}{\frac{3}{7}}=7\) hours. Before moving on, it is important to note that it took 7 not 3 hours slightly unintuitively. We can notice that this task took the same amount of time as it did for 7 workers to paint 7 houses, so we wonder if there is a nicer explanation for this. Indeed, there is. Using the equation w=rt, and solving for each variable, it can be easily seen that when we change the number of workers from a to b, the amount of time gets multiplied by \(\frac{a}{b}\). Similarly, when we change the number of houses needing painting from c to d, we multiply the time by \(\frac{d}{c}\). Thus, in the situation of the problem, we are multiplying 7 hours by \(\frac{3}{7}\cdot\frac{7}{3}=1\). Make sure you understand what happens when you change the number of workers and the amount of work, it can be very easily generalized to other problems and will often pop up.
Ratio Problems
Sample Problem 5. In a town of 98 people, \(\frac{1}{2}\) of the men and \(\frac{2}{3}\) of the women are married. Assuming each marriage consists of one man and one woman and consists of only people in the town, how many men are in the town.
Again, we will start by defining variables. Let w be the number of women and m be the number of men in the town. Thus, \(\frac{m}{2}=\frac{2w}{3}\implies m=\frac{4w}{3}\), or equivalently, \(\frac{4}{7}\) of the total population. Thus, there are \(\frac{4\cdot 98}{7}=56\) men. Just like in many AMC 10 level word problems, once we define variables the rest of the steps are fairly intuitive and obvious.
Excercises
- Albert is running a lemonade stand. He got 2 more cups than than he had ounces of lemonade. The cooler he got could unfortunately only hold one less than half of the number of cups he got. Finally, the cooler he had could hold one more than one fourth of the number of ounces of lemonade he. How many ounces of lemonade did he have?
- Mary is running exactly one time around a track. She ran the first half of the track in only 3 minutes. She was then tired and decreased her speed by 1 mph for the rest of distance to have a total time of 7 minutes. How long is the track?
Solution 1
We will simply turn the words in the problems into equations and then solve. We will let the number of cups he has be \(c\), the number of ounces of lemonade he had be \(x\), and the number of cups his cooler be \(s\). From our first given we have that \(c=x+2\). Second, we have that \(s=\frac{c}{2}-1\). Third, we have that \(s=\frac{x}{4}+1\). By substituting the first equation into the second for \(c\) and multiplying by 2, we get \(2s=x+2-2=x\implies s=\frac{x}{2}.\) Thus, using the last equation, we get \(\frac{x}{2}\) = \(\frac{x}{4}+1\) \(\implies\) \(2x=x+4\) \(\implies\) \(x=4.\) Therefore, our answer is 4.
Solution 2
We will again start by defining variables. Let Mary's rate for the first half of the track be \(r\) mph and the distance of the total track be \(d\) miles. We have the the first half of the track takes 3 minutes or \(\frac{1}{20}\), so \(\frac{d}{2r}=\frac{1}{20}\). Thus, \(r=10d\). Next, we know she decreased her rate by 1 mph and traveled the second half of the track in 4 minutes or \(\frac{1}{15}\) hours. Thus, we have \(\frac{d}{2(r-1)}=\frac{1}{15}\) and \(r=\frac{15d+2}{2}\). Equation \our two expressions for \(r\), we get \(10d=\frac{15d+2}{2}\implies 20d=15d+2\). Therefore, we finally achieve \(d=\frac{2}{5}\) miles.