A06-WordProblems-Rates.html

High School Modules > Algebra by Gregory A. Moore

Rate-Time-Distance Word Problems

An exploration of the method of setting up and solving rate/time/distance word problems.

[Directions : Execute the Code Resource section first. Although there will be no output immediately, these definitions are used later in this worksheet.]

0. Code

> restart; with(plots):

Warning, the name changecoords has been redefined

>    RateTableEmpty := proc(name1, name2)

     local    a, b, rt, M, equation;
     rt := cr ; #(a*ar + b*br)/(a+b);
     a := x;   b := c-a;

     M := array( [ [``, `Rate   *`, ` Time`,` = Distance`],
                   [ name1,     `?`,      `?`,    `?`        ],
                   [ name2,      `?`,    `?`,     `?`   ]    ]);


end proc:

>    RateTableRT := proc(name1,r1,t1,name2,r2,t2)

     local    a, b, rt, M, equation;

     M := array( [ [``, `Rate   *`, ` Time`,` = Distance`],
                   [ name1,     r1,      t1,    r1*t1        ],
                   [ name2,      r2,    t2,     r2*t2   ]    ]);

      end proc:

>    RateTableRD := proc(name1,r1,d1,name2,r2,d2)

     local    a, b, rt, M, equation;

     M := array( [ [``, `Rate   *`, ` Time`,` = Distance`],
                   [ name1,      r1,     r1/d1,    d1        ],
                   [ name2,      r2,    r2/d2,     d2   ]    ]);
      end proc:

>    RateTableTD := proc(name1,t1,d1,name2,t2,d2)

     local    a, b, rt, M, equation;

     M := array( [ [``, `Rate   *`, ` Time`,` = Distance`],
                   [ name1,     d1/t1,     t1,    d1        ],
                   [ name2,     d2/t2,     t2,    d2 ]    ]);

      end proc:

1. The Set-Up

The common thread to setting up these problems is to create a table to put the various knowns and unknowns. Usually this problems involve two trains, cars, planes, runners, or other moving objects.

> RateTableEmpty( train1, train2);

> RateTableEmpty( `1st car`, `2nd car` );

> RateTableEmpty( plane1, plane2);

> RateTableEmpty( `John`, `Jane` );

Next, we'll insert information into the various slots. If we knew the rates and times, we would have something like this :

> RateTableRT( car1, 50, 3, car2, 65, 2);

> RateTableRT( `John`, 5, 3/2, `Jane`, 4.5, 2 );

2. Same Distance - Different Rates

Problem 2.1 : Two cars drive from Mudville to Sandville and arrive at the same time. The first car goes at 52 mph, and the second car, which leaves 2 hours later, goes at 65 mph.
A. Find the time that the first car travels.
B. Find the distance between the two cities

First set up an empty chart (if doing this problem by hand).

> RateTableEmpty(car1, car2);

We can make a wild guess - which will turn out to be wrong.

> RateTableRT(car1, 52, 3, car2, 65, 2);

To solve the problem, we let t be the unknown time that car1 drives. We can then fill in the times and rates for both cars. The distances are computed as the products of these.

> RateTableRT(car1, 52, t, car2, 65, t-2);

Since the two expressions for distance refer to the same distance, they are equal. This means that we have an equation that we can solve - to get t. After getting t, we can find d.

> 52*t = 65*t - 130; t = solve( %, t);

> d = 52*10;

3. Same Distance - Different Rate and Time

Problem 3.1 : Two planes fly 1,960 miles. The second plane travels 98 mph faster but takes 1 hour less time. Find the rate and time of the first plane.

> RateTableEmpty( plane1, plane2);

If we take this job, we get :

> RateTableRT(car1, r, t, car2, r+98, t-1);

Both of the expressions for distance are equal to 1,960. This gives two equations with two unknowns.

> EQ1 := r*t = 1960;
EQ2 := (r+98)*(t-1) = 1960;

Solve each of the equations for r. Then equate these solutions, because both of them are equal to r, and solve.

> solve(EQ1,r); solve(EQ2, r);
EQ3 := % = %%;

> solve(%,t);

> t = 5;

> r = 1960/5;

4. Same Time

Problem 4.1: A bicyclist and a rollerblader travel for 3 hours. The bicyclist travels twice as fast and ends up 2 miles farther. How far did the cyclist travel?

The key thing is to put the right expressions in the right places. The rate and distance of the bicyclist are both expressed in terms of the information about the rollerblader. So we should let r = the rate of the rollerblader, and d = the distance of the rollerblader. Then we can express the rate and distance of the cyclist relative to those.

> RateTableRD( bicyclist, 2*r, d+2, rollerblader, r, d);

The two expressions for time are both equal to 3.

> 2*r/(d+2) = 3; r/d = 3;

> solve(%%, r); solve(%%,r);

We can find the d, the distance travelled by the rollerblader by equating these two expressions for r, and solving this. Then to answer the question, we need to add 2 to get the distance for the cyclist.

> d = solve( % = %%, d);

> d + 2;

5. Same Rate

Problem : Rick and Steve both run at the same rate. Steve can run 30 miles if he runs four hours longer than Rick runs 18 miles. Find their rate.

> RateTableTD( Rick, t, 18, Steve, t+4, 30);

Here are other examples.

> 18/t = 30/(t+4);

> t = solve(%, t);

> r = 18/6;

>