This page contains word problems.
Note: these are not simple, but are within the range of your mathematical ability.

Word Problems

Cycling around a track

Bernice is cycling around a track at 15 km/h. Betty starts at the same time, but only goes 12 km/h. How many minutes after they start will Bernice pass Betty if the track is 1/2 km long? They are moving in the same direction. Remember that distance=rate of speed * time (d=r*t) **

Cycling around a track

d = rt
follows naturally from the definition of:
r = d/t.

If Bernice goes 15 km/h, how many miles will she travel in an hour?
How many times around the 1/2 km track is that?
If Betty goes 12 km/h, how many km will she travel in an hour?
How many times around the 1/2 km track is that?
Once students understand that, they should appreciate that Bernice "laps" (i.e. catches up to and passes, or "meets") Betty 6 times every hour, or once every 10 minutes.

Hot Air Balloon

A hot air balloon covered 2100 km in 7 days. If it covered 50 km more each day than the day before, how many km did it cover each day?

Hot Air Balloon

As discussed in class, here is one possible way to get an answer:


day 1: x
day2: x + 50
day 3: x + 100
day 4: x + 150
and so on, until you get to day 7, which is x+300,
then 7+1050 = 2100-1050 = 1050.
Then, 1050 divided by 7 is 150
so the answer is:
day 1 = 150 km
day 2 = 200 km
day 3 = 250 km
day 4 = 300 km
day 5 = 350 km
day 6 = 400 km
day 7 = 450 km

A Plane Flying with a Tailwind

A jet is flying 3, 862 km from Hawaii to San Francisco. In still air, it flies at 966 km/h. There is a 64 km/hr tailwind. How many hours after takeoff would it be faster to just go to San Francisco than to return to Hawaii in the case of an emergency?

A Plane Flying with a Tailwind

Flying to San Francisco, the plane's speed is 1030 km/hr.
If it were flying back to Hawaii, its speed would be 902 km/hr.

Let t be the number of hours of flight.
Then it has flown a distance of 1030t km and the distance yet to go is 3, 862 - 1030t.
The time left to fly is then (3, 862-1030t)/1030.

However, if it were to return to Hawaii, it would have to fly 1030t km at 902 km/hr which would then take 1030t/902 hrs.
If we equate these times we have (3, 862-1030t)/1030 = 1030t/902

If you solve this equation for t you get t = 1.75 hr.

Jack and the Beanstalk

Jack climbed up the beanstalk at a uniform rate. At 2 P.M. he was one-sixth the way up and at 4 P.M. he was three fourths the way up. What fractional part of the entire beanstalk had he climbed by 3 P.M. At what time did he start climbing? When will he get to the top? How long was his trip? And how tall was the beanstalk anyway?

Jack and the Beanstalk

At 2 pm Jack is 4/24 (or 1/6) of the way up and at 4 pm, two hours later, he is 18/24 (or 3/4) of the way up...

So Jack goes from 4/24 of the way up to 18/24 of the way up in two hours. That's 14/24 every two hours... reduce that to 7/12 every two hours.

Now let's divide that rate by two so we know how much of the stalk he climbs every hour. (7/12) / 2 = 7/24

So Jack climbs 7/24 of the stalk each hour. We know where he was at 2:00 and how fast he climbs, so now we can figure out where he was at 3:00.. 4/24 (2:00 position) + 7/24 (his hourly climbing rate) = 11/24 of the way up at 3:00

Now, when did he start climbing? At 2:00 he was 4/24 of the way up. At 1:00, he would have been -3/24 of the way up (if the stalk continued underground).

So we know that he started sometime between 1:00 and 2:00.

Note that it takes Jack ONE HOUR to go from -3/24 to 4/ 24. The difference between 4 and -3 is seven so let's divide the hour into seven equal segments of 8.57 minutes

(THIS IS NOT 8 MINUTES, 57 SECONDS!! IT IS 8 MINUTES, 34.2 SECONDS)

After three of these 8.57 minute segments, Jack would be at ground zero.

That's 25.71 minutes. Let's convert that to minutes and seconds: 25 minutes, 42.6 seconds.

Here's what all that means: 25 minutes and 42.6 seconds after 1:00, Jack leaves the ground.

So Jack started climbing at 1:25:42.6 Now, when will he get to the top? Well, at 4:00, he was 18/24 of the way up. At 5:00, he would be at 25/24 (I get this using the predetermined climbing rate of 7/24 per hour).

Using a little common sense, we can see that he reaches the top a little before 5:00. Let's figure out exactly when. 6/7 of the way between 4:00 and 5:00, Jack hits the top. 6/7 of 60 minutes is 51.43 minutes.

Convert to minutes and seconds: 51 minutes, 25.8 seconds. So Jack reaches the top at 4:51:25.8 How long was his trip? It lasted from 1:25:42 to 4:51:25. That's 3 hours, 25 minutes, 43 seconds.

 

This page was updated February 4, 2007 10:18 PM