Rate problems: distance and time, Work, mixture, and Cost

196

Rate Problems: Distance and Time, Work,

Mixture, and Cost

Word Problem Setup

200. Some problems require translation of words into algebraic expressions or equations.

For example: 8 more than 7 times a number is 22. Find the number. Let n 5 the number. We

have

7n 1 8 5 22

7n 5 14

n 5 2

Another example: There are 3 times as many boys as girls in a class. What is the ratio of boys to the total number of students? Let n 5 number of girls. Then

3n 5 number of boys 4n 5 total number of students

n_ _ tuo_mt_ab_l_es_rt _uo _df_eb_no_ty_ss_5 _43_nn_ 5 _43_ 201. Rate problems concern a special type of relationship that is very common: rate 3 input 5 output. This results from the definition of rate as the ratio between output and input. In these problems, input may represent any type of "investment," but the most frequent quantities used as inputs are time, work, and money. Output is usually distance traveled, work done, or money spent.

Note that the word per, as used in rates, signifies a ratio. Thus a rate of 25 miles per hour signifies the ratio between an output of 25 miles and an input of 1 hour.

Frequently, the word per will be represented by the fraction sign, thus _2 1_5_hm_o_i_ule_rs_ . Example: Peter can walk a mile in 10 minutes. He can travel a mile on his bicycle in 2 minutes. How far away is his uncle's house if Peter can walk there and bicycle back in 1 hour exactly?

To solve a rate problem such as the one above, follow these steps:

STEP 1. Determine the names of the quantities that represent input, output, and rate in the problem you are doing. In the example, Peter's input is time, and his output is distance. His rate will be distance per unit of time, which is commonly called speed.

STEP 2. Write down the fundamental relationship in terms of the quantities mentioned, making each the heading of a column. In the example, set up the table like this:

speed 3 time 5 distance

STEP 3. Directly below the name of each quantity, write the unit of measurement in terms of the answer you want. Your choice of unit should be the most convenient one, but remember, once you have chosen a unit, you must convert all quantities to that unit.

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We must select a unit of time. Since a minute was the unit used in the problem, it is the most logical choice. Similarly, we will choose a mile for our unit of distance. Speed (which is the ratio of distance to time) will therefore be expressed in miles per minute, usually abbreviated as mi/min. Thus, our chart now looks like this:

speed 3 time 5 distance

mi/min minutes miles

STEP 4. The problem will mention various situations in which some quantity of input is used to get a certain quantity of output. Represent each of these situations on a different line of the table, leaving blanks for unknown quantities.

In the sample problem, four situations are mentioned: Peter can walk a mile in 10 minutes; he can bicycle a mile in 2 minutes; he walks to his uncle's house; and he bicycles home. On the diagram, with the appropriate boxes filled, the problem will look like this:

speed 3 time 5 distance

1. walking 2. bicycling 3. walking 4. bicycling

mi/min

minutes 10 2

miles 1 1

STEP 5. From the chart and from the relationship at the top of the chart, quantities for filling some of the empty spaces may become obvious. Fill in these values directly.

In the example, on the first line of the chart, we see that the walking speed 3 10 equals 1. Thus, the walking speed is 0.1 mi/min (mi/min 3 10 5 1 mi; mi/min 5 _1_01__mm__iin_ 5 0.1). Similarly, on the second line we see that the bicycle speed equals 0.5 mi/min. Furthermore, his walking speed shown on line 3 will be 0.1, the same speed as on line 1; and his bicycling speed shown on line 4 will equal the speed (0.5) shown on line 2. Adding this information to our table, we get:

speed 3 time 5 distance

1. walking 2. bicycling 3. walking 4. bicycling

mi/min 0.1 0.5 0.1 0.5

minutes 10 2

miles 1 1

STEP 6. Next, fill in the blanks with algebraic expressions to represent the quantities indicated, being careful to take advantage of simple relationships stated in the problem or appearing in the chart.

Continuing the example, we represent the time spent traveling shown on line 3 by x. According to the fundamental relationship, the distance traveled on this trip must be (0.1)x. Similarly, if y represents the time shown on line 4, the distance traveled is (0.5)y. Thus our chart now looks like this:

198 ? Gruber's Complete SAT Guide 2015

1. walking 2. bicycling 3. walking 4. bicycling

speed 3 time 5 distance

mi/min 0.1 0.5 0.1 0.5

minutes 10 2 x y

miles 1 1

(0.1)x (0.5)y

STEP 7. Now, from the statement of the problem, you should be able to set up enough equations to solve for all the unknowns. In the example, there are two facts that we have not used yet. First, since Peter is going to his uncle's house and back, it is assumed that the distances covered on the two trips are equal. Thus we get the equation (0.1)x 5 (0.5)y. We are told that the total time to and from his uncle's house is one hour. Since we are using minutes as our unit of time, we convert the one hour to 60 minutes. Thus we get the equation: x 1 y 5 60. Solving these two equations (0.lx 5 0.5y and x 1 y 5 60) algebraically, we find that x 5 50 and y 5 10. (See Section 407 for the solution of simultaneous equations.)

STEP 8. Now that you have all the information necessary, you can calculate the answer required. In the sample problem, we are required to determine the distance to the uncle's house, which is (0.1)x or (0.5)y. Using x 5 50 or y 5 10 gives us the distance as 5 miles.

Now that we have shown the fundamental steps in solving a rate problem, we shall discuss various types of rate problems.

Distance and Time

202.In distance and time problems the fundamental relationship that we use is speed 3 time 5 distance. Speed is the rate, time is the input, and distance is the output. The example in Section 201 was this type of problem.

Example: In a sports-car race, Danica gives Pablo a head start of 10 miles. Danica's car goes 80 miles per hour and Pablo's car goes 60 miles per hour. How long should it take Danica to catch up to Pablo if they both leave their starting marks at the same time?

STEP 1. Here the fundamental quantities are speed, time, and distance.

STEP 2. The fundamental relationship is speed 3 time 5 distance. Write this at the top of the chart.

STEP 3. The unit for distance in this problem will be a mile. The unit for speed will be miles per hour. Since the speed is in miles per hour, our time will be in hours. Now our chart looks like this:

speed 3 time 5 distance

mi/hr

hours

miles

STEP 4. The problem offers us certain information that we can add to the chart. First we must make two horizontal rows, one for Pablo and one for Danica. We know that Pablo's speed is 60 miles per hour and that Danica's speed is 80 miles per hour.

STEP 5. In this case, none of the information in the chart can be used to calculate other information in the chart.

STEP 6. Now we must use algebraic expressions to represent the unknowns. We know that both Pablo and Danica travel for the same amount of time, but we do not know for how much

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time, so we will place an x in the space for each driver's time. Now from the relationship of speed 3 time 5 distance, we can calculate Pablo's distance as 60x and Danica's distance as 80x. Now the chart looks like this:

speed 3 time 5 distance

Pablo Danica

mi/hr 60 80

hours x x

miles 60x 80x

STEP 7. From the statement of the problem we know that Danica gave Pablo a 10-mile head start. In other words, Danica's distance is 10 more miles than Pablo's distance. This can be stated algebraically as 60x 1 10 5 80x. That is, Pablo's distance 1 10 miles 5 Danica's distance. Solving for x gives us x 5 _12_.

STEP 8. The question asks how much time is required for Danica to catch up to Pablo. If we look at the chart, we see that this time is x, and x has already been calculated as _21_, so the answer is _21_ hour.

Work

203.In work problems the input is time and the output is the amount of work done. The rate is the work per unit of time.

Example: Jack can chop down 20 trees in 1 hour, whereas it takes Ted 1_21_hours to chop down 18 trees. If the two of them work together, how long will it take them to chop down 48 trees?

Solution: By the end of Step 5 your chart should look like this:

rate 3 time 5 work

1. Jack 2. Ted 3. Jack 4. Ted

trees/hr 20 12 20 12

hours 1 112?

trees 20 18

In Step 6, we represent the time that it takes Jack by x in line 3. Since we have the relationship that rate 3 time 5 work, we see that in line 3 the work is 20x. Since the two boys work together (therefore, for the same amount of time), the time in line 4 must be x, and the work must be 12x. Now, in Step 7, we see that the total work is 48 trees. From lines 3 and 4, then, 20x 1 12x

5 48. Solving for x gives us x 5 1_21_. We are asked to find the number of hours needed by the boys to chop down the 48 trees together, and we see that this time is x, or 1_21_ hours.

200 ? Gruber's Complete SAT Guide 2015

Mixture

204.In mixture problems you are given a percent or a fractional composition of a substance, and you are asked questions about the weights and compositions of the substance. The basic relationship here is that the percentage of a certain substance in a mixture 3 the amount of the mixture 5 the amount of substance.

Note that it is often better to change percentages to decimals because it makes it easier to avoid errors.

Example: A chemist has two quarts of 25% acid solution and one quart of 40% acid solution. If he mixes these, what will be the concentration of the mixture?

Solution: Let x 5 concentration of the mixture. At the end of Step 6, our table will look like this:

concentration 3 amount of sol 5 amount of acid

25% solution 40% solution mixture

_qq_t_t(_(a_sc_o_idl_)_) 0.25 0.40 x

qts (sol)

2 1 3

qts (acid)

0.50 0.40

3x

We now have one additional bit of information: The amount of acid in the mixture must be equal to the total amount of acid in each of the two parts, so 3x 5 0.50 1 0.40. Therefore x is equal to 0.30, which is the same as a 30% concentration of the acid in the mixture.

Cost

205.In cost problems the rate is the price per item, the input is the number of items, and the output is the value of the items considered. When you are dealing with dollars and cents, you must be very careful to use the decimal point correctly.

Example: Jim has $3.00 in nickels and dimes in his pocket. If he has twice as many nickels as he has dimes, how many coins does he have altogether?

Solution: After Step 6, our chart should look like this (where c is the number of dimes Jim has):

rate 3 number 5 value

nickels dimes

cents/coin 5 10

coins 2c c

cents 10c 10c

Now we recall the additional bit of information that the total value of the nickels and dimes is $3.00, or 300 cents. Thus, 5(2c) 1 10c 5 300; 20c 5 300; so c 5 15, the number of dimes. Jim has twice as many nickels, so 2c 5 30.

The total number of coins is c 1 2c 5 3c 5 45.

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