one math word problem!

[Danny289]

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solve this problem:
A motorman should have taken a distance from town A to town B for exact time. Two hours after he left, he noticed that he covered 80 km and if he keeps that speed he will arrive in B with 15 min delay. So he increased the speed with 10km/h and arrived in town B 36 minutes earlier. Find:
à) The distance between the two towns;
b) The exact time that the motorman should have taken the distance from A to B

 

[Contach]

If I understood the crooked english correctly:

a) Distance =~ 80km + 500/3km =~ 80 + 165 = 245 km
b) I don't know what it means to "take a distance"

 

[Danny289]

If I understood the crooked english correctly:

a) Distance =~ 80km + 500/3km =~ 80 + 165 = 245 km
b) I don't know what it means to "take a distance"

from A to B take exact time, your answer for a is not correct but it is closed.

 

[PooyaH]

Danny,

Is the distance 204 Km? I'm not quite sure though!

 

[Contach]

from A to B take exact time, your answer for a is not correct but it is closed.

Do you mean:
Exactly how long did it take to get from A to B?

 

[Danny289]

Danny,

Is the distance 204 Km? I'm not quite sure though!

No

Do you mean:
Exactly how long did it take to get from A to B?

this is why everybody complaining about math section. i didn't make up this question and IMO the wording for this problem is perfect. just try understand the problem, it is funny even after understanding the problem I had a problem for solving my equations(stupid mistakes)

 

[Contach]

Here is how I attemped this: (maybe some1 can help me)

a) 15minutes + 37 = 51minutes .. 51 minutes is the disparity between going at 40km/h and 50km/h for the last segment.

51 =~5/6 of an hour.

So then I produce a rate equation for the remaining distance at each speed:
And we know the distances are equal
D = R x T and T1 = T and T2 = T - 5/6
D = 40(T)
D = 50(T - 5/6)
set these equal to one another. And you get T = 25/6 is about 4 hours and a bit.. sounds reasonable...

plug T = 25/6 H into the first Rate equation.
D = 500/3 km
Add this to the first segment.. Total Distance = 500/3 km + 80 km =~245 km

b) first segment was 2 hours.
the second segment was T - 5/6 = 25/6 - 5/6 = 20/6 = 3 1/3 H
It took a total of 5 Hours and 20 minutes..

 

[Danny289]

Here is how I attemped this: (maybe some1 can help me)

a) 15minutes + 37 = 51minutes .. 51 minutes is the disparity between going at 40km/h and 50km/h for the last segment.

51 =~5/6 of an hour.

So then I produce a rate equation for the remaining distance at each speed:
And we know the distances are equal
D = R x T and T1 = T and T2 = T - 5/6
D = 40(T)
D = 50(T - 5/6)
set these equal to one another. And you get T = 25/6 is about 4 hours and a bit.. sounds reasonable...

plug T = 25/6 H into the first Rate equation.
D = 500/3 km
Add this to the first segment.. Total Distance = 500/3 km + 80 km =~245 km

b) first segment was 2 hours.
the second segment was T - 5/6 = 25/6 - 5/6 = 20/6 = 3 1/3 H
It took a total of 5 Hours and 20 minutes..

noch!!, if we see something like this in DAT. it is better skip that or go with our gut.

 

[td4azklz]

Here is how I attemped this: (maybe some1 can help me)

a) 15minutes + 37 = 51minutes .. 51 minutes is the disparity between going at 40km/h and 50km/h for the last segment.

51 =~5/6 of an hour.

So then I produce a rate equation for the remaining distance at each speed:
And we know the distances are equal
D = R x T and T1 = T and T2 = T - 5/6
D = 40(T)
D = 50(T - 5/6)
set these equal to one another. And you get T = 25/6 is about 4 hours and a bit.. sounds reasonable...

plug T = 25/6 H into the first Rate equation.
D = 500/3 km
Add this to the first segment.. Total Distance = 500/3 km + 80 km =~245 km

b) first segment was 2 hours.
the second segment was T - 5/6 = 25/6 - 5/6 = 20/6 = 3 1/3 H
It took a total of 5 Hours and 20 minutes..

Where are you getting your 500/3 from? I'm sorry, I'm not really following you.

 

[Danny289]

this the answer:
We mark the distance from A to B with x km. Because the motorman took 80km for 2 hours his speed is V = 80/2 = 40 km/h. With that speed he would have taken the whole distance for x/40 h, delaying with 15min, i.e. the exact time is x/40 – 15/60 h. The rest of the distance (x - 80) km. he took with V = 40 + 10 = 50 km/h
So the time he took the distance from A to B, is 2 +(x - 80)/50 h. and it is with 36 min. earlier than expected. Therefore the expected time is 2 + (x -80)/50 + 36/60 When we equalize the expressions for the expected time, we get the equation :
x/40 &#8211; 15/60 = 2 + (x -80)/50 + 36/60 <=> (x - 10)/40 = (100 + x - 80 + 30)/50 <=> (x - 10)/4 = (x +50)/5 <=> 5x - 50 = 4x + 200 <=> x = 250
So the searched distance is 250 km. The exact time we will find by substituting x with 250 in of the sides of the first equation, for example;
x/40 &#8211; 15/60 = 250/40 &#8211; 1/4 = 25/4 &#8211; 1/4 = 24/4 = 6 hours

 

[doc3232]

this the answer:
We mark the distance from A to B with x km. Because the motorman took 80km for 2 hours his speed is V = 80/2 = 40 km/h. With that speed he would have taken the whole distance for x/40 h, delaying with 15min, i.e. the exact time is x/40 – 15/60 h. The rest of the distance (x - 80) km. he took with V = 40 + 10 = 50 km/h
So the time he took the distance from A to B, is 2 +(x - 80)/50 h. and it is with 36 min. earlier than expected. Therefore the expected time is 2 + (x -80)/50 + 36/60 When we equalize the expressions for the expected time, we get the equation :
x/40 – 15/60 = 2 + (x -80)/50 + 36/60 <=> (x - 10)/40 = (100 + x - 80 + 30)/50 <=> (x - 10)/4 = (x +50)/5 <=> 5x - 50 = 4x + 200 <=> x = 250
So the searched distance is 250 km. The exact time we will find by substituting x with 250 in of the sides of the first equation, for example;
x/40 – 15/60 = 250/40 – 1/4 = 25/4 – 1/4 = 24/4 = 6 hours

I agree, this is a very tough question and the wording is fine...the logic that I can't make for it is depressing.
Nice post Danny

This is how I did it (took me about 4 mins, definitely would skip on the test)
a=time the second part of his trip would take
50(a-15)=40(a+36)
this gave a to be
[So worked on it another 3 or 4 mins]

then a=204
add 36 mins
you get 360 then add the 120 he already drove and you get 6 hours...finally
This is a dif way to do it.

 

[DDSguyLA]

these are the questions im going to put all "C"s on

😀

 

[Danny289]

these are the questions im going to put all "C"s on

😀

"C" is my choise you put "B" plz! 🙂

 

[Contach]

Where are you getting your 500/3 from? I'm sorry, I'm not really following you.

D=40*T
plug in T= 25/6 into the above.. and you get D = 500/3