Word Problem

I

imala

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Six young people, Anna and Annika, Brooke and Breanna, Chloe and Claire, must make the 10 km trip back home with only their 4 seater car to carry them. Each of the six can drive at a steady 45 km an hour carrying three of the others but the remaining two must travel on foot. Fortunately, Anna and Annika can each run at 15 km an hour, Brooke and Breanna can both jog at a brisk 9 km an hour, and Chloe and Claire will happily walk at 5 km an hour. By cleverly combining their resources, they have worked out how they can reach their destination in the minimum possible time. How many minutes is that?


I started answering this question and now an stuck.

They start off with 4 people aboard, while Breanne and Brooke jog at 9 km an hour. The car then drops off Anna and Annika who run at 15 km an hour and returns to pick them up en route, Now their is four aboard again. This way, all four arrive at home at the same time .

Any ideas on how to figure out how many minutes would be the fastest?
 
I never got it. I didn't see my answer or my reply up. So I reposted it.
 


You never got it ?

Your first post had no strategy. Denis posted one.

Now, you reposted the exercise with the strategy that Denis posted.

Anyways, where is your work ? What times or distances have you calculated, so far ?

 
A LOT of posts have vanished from this section: from March 10th...
 
I had all the work I did in the last message I sent to you yesterday. I calculated the distance and time it would take using a triangle diagram.
 


I worked it two ways, and they both came out the same: 24 minutes.

I hope that's what Denis got; I can't remember.

So, I broke the total time into three parts.

#1 The car takes the Bs to drop them off along the way. This takes x hours.

#2 The car turns around and heads back to meet the As (who started running when the car left). This takes y hours.

#3 The car turns around and finishes the trip. This takes z hours.

During x hours, the car goes 45x km and the As go 15x km.

During y hours, the car goes 45y km and the As go 15y km.

During z hours, the car goes 45z km.

If you draw line segments to represent these distances, and align them over 10 km, you can see how different combinations of these segments equal sums of others. This process generates two equations.

15x + 15y + 45y = 45x, and this simplifies to x = 2y

15x + 15y + 45z = 10, and this simplifies to 3x + 3y + 9z = 2

The As run during y + z hours, and the distance they run is 10 - 45x km. This gives us a third equation.

9(y + z) = 10 - 45x

Solve this system of three equations for x, y, and z. (Those are the fractional hours.) Add them together, to get the total time. Finally, convert the sum from hours to minutes.

 
Mark, agree with 24 minutes.

Note: (the C's always in car)

1: Car leaves @ 45 (carrying B's); A's leave @ 15 (at same time)

2: After 10 2/3 minutes, car drops off B's: B's continue @ 9 ; car returns @ 45

3: After another 5 1/3 minutes, car meets A's and A's jump in: car turns around and goes @ 45

4: After another 8 minutes, car and B's arrive at 10km point at SAME TIME

So total time = 10 2/3 + 5 1/3 + 8 = 24 minutes

EDIT: I earlier goofed (changed speeds to kpm, changed one wrong) :cry:
 


Well, Denis, I must have goofed up because I followed the same stages.

I have x + y + z = 2/5 hours.

I better check my work (later) !

 
Subhotosh Khan said:
Corner ..... Dennis ..... Corner .....
Click to expand...
YES...I was already there :?
How come you were gone: you still had 10 minutes left ?
 
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