How far is it to his office?

[Chikis]

A man drives to a car park at an average speed of 40km/hr and then walks to his office at an average speed of 6km/h. The total journey takes him 25 minutes. One day his car broke down and he has to walk three times as far, so arriving at his office 17 minutes late. How far is it to his office?

 

WHEN CAR WAS NORMAL

Speed to cark park =40km/h

Speed to office = 6km/h

total time = 5/12 h

distance to car park = xkm

distance to office = ykm

time = distance/speed

 

x/40 + y/6 = 5/12

 

3x + 20y = 50

 

WHEN CAR BROKE DOWN

He walked 3(x+y)km =

total time = 7/10 h

I think I have problem making an equation for the second case. What do I do?

[imatfaal]

You have set up variables for the distance normally walked (y) and the distance normally driven (x).

1. You know from question how to extend the distance walked - your attempt above is not what I think the question says. it is quite clear how far he walked when his car broke down.

2. Any extra distance walked is not driven (ie it is the same total distance) - so remove extra walking from normal driving

3. You know the new time for the unusual journey

 

If you work through those you should get a second equation.

[studiot]

 

He walked 3(x+y)km =

 

 

This is not quite right.

 

 

and he has to walk three times as far

 

 

as imatfaal said.

[Chikis]

2. Any extra distance walked is not driven (ie it is the same total distance) - so remove extra walking from normal driving.

So how do I get or know the extra walking? Ok, I know now, extra walking is 2y. Should I say that x could have been 2y the distance driven? Therefore x-2y = 0. That means there is no distance driven. Edited October 5, 2014 by Chikis

[studiot]

Well in part (1) you have two unknowns, x and y and two equations so you can calculate them.

I think you said he walks y km.

In part (2) your wrote down that he walks 3y km (look carefully at the words)

[Chikis]

Ok, I know now, extra walking is 2y. Should I say that x could have been 2y the distance driven? Therefore x-2y = 0. That means there is no distance driven. What do I next?

Edited October 5, 2014 by Chikis

[studiot]

 

Should I say that x could have been 2y the distance driven? Therefore x-2y = 0. That means there is no distance driven.

 

Where does this come from?

 

imatfaal has already suggested how to get your third equation (you don't need 4)

 

 

(ie it is the same total distance)

 

So in each case distance walked plus distance driven = distance walked plus distance driven = x + y

 

So if you introduce a new variable, say z for distance driven the second time

 

x + y = z + 3y

 

You now have 3 equations and 3 unknowns.

[Chikis]

I believe the extra distance walked is 2y. Why must we subtract 2y from x?

[studiot]

 

Posted Today, 09:41 PM

I believe the extra distance walked is 2y. Why must we subtract 2y from x?

 

Yes the extra distance walked is 2y.

 

I'm sorry I saw this quickly and thought you had 2 equations in x and y

I didn't realise that the second was just 120 times the first.

 

 

x/40 + y/6 = 5/12

 

3x + 20y = 50

 

 

 

So you have one equation

 

Time driving plus time walking = total travel time.

 

 

 

On the second occasion this equation still holds, but the times are different.

 

They are different because he spends more time walking than before and less time driving.

 

We know he walks three times as far so, as you said, the extra distance walked is 2y.

 

This extra distance he walks is distance he does not drive so the distance he does drive is (x-2y)

 

I will keep repeating this to you.

 

Draw a diagram it is easy to see then.

 

 

 

 

You should now be able to form your second equation in x and y.

[Chikis]

I will soon go for a mini laptop. I believe that will enable me to draw a diagram.

 

My mentality made wrote 3(x+y) as the total distance. I was thinking he did all the journey by foot as his car had broken down not knowing he had done some part of the journey with his car before it broke down.

 

Or maybe, he woke up one morning and discovered that his car was not in good condition. He then decided to trek some distance and finally boarded/hired a bus and completed the journey.

Which of the thinking is right?

Edited October 6, 2014 by Chikis

[Chikis]

Could this problem be solved in one variable?

[studiot]

You draw a diagram to help yourself, not me.

So just so long as you draw them.

It is useful, but not esential to post a diagram with the question.

 

So you could use paper. They could be not pretty, as mine are not.

 

[Chikis]

Could this problem be solved using only one variable.

[studiot]

Well one way to solve a pair of simultaneous equations is by substitution.

 

This means using the second equation to find one unknown in terms of the other and then replacing it in the first equation.

 

This will obtain a single equation in one unknown for you.

 

Edit

x and y are not variables they have particular values in your problem.

 

But they are unknown until you calculate them

 

Edited October 8, 2014 by studiot

[Chikis]

I have solve the problem.