[deepboy]

Originally Posted by catdog

Assuming distance to be 60 KM .....it takes 1 hr without stoppage ....if he takes stoppage he takes
60/48 hrs i.e (60/4*60 minutes .......75 minutes ................this implies he stopped for 15 minutes ...............


75 minutes 15 minutes stops

so in 60 minutes

60/75*15

12 minutes /hr ..........

I feel this can be interpreted just by logic and not involving quants at all.Here is my stand:

When it travels non-stop it covers 60 kms in 1 hr.
Since its speed is not changed i can easily say that it will take 48 mins of travelling
to cover 48 kms.
There it stopped for 12 mins in 1 hr.
Hence stoppage time is 12 mins/hr...............

Cheers!!!!!!!!!!!!!!



"Today is a good day to be random"

[deepboy]

Originally Posted by ganeshiyer

i guess for these type of problems u can directly use the following formula

time taken to overtake= [total distance]/[relative speed]
i.e t=900/40 =22.5

Just like to clarify one thing.............
I guess it shud be put as
rel distance/rel speed.
And ya 22.5 sec shul be the answer.

[sriram5874]

Originally Posted by Vedvyas

Yet another Simple method:

Assume that the distance is 60 * 48 km (No need to calculate the value)

When average speed = 60 kmph => time = 48 hrs. (Without stoppages)
When average speed = 48 kmph => time = 60 hrs. (With stoppages)

Hence stoppage time for same distance = 60 - 48 = 12 hrs.

Total time with stoppage = 60 hrs.

Hence average stoppage time = 12/60 hrs = 1/5 hrs = 12 mins.

No calculations needed at all.

Cheers,

Ved

hai vedvyas ..its totallly correct but explaining a littlemoree cud suffice...
let x be total distance...
x/60 is time without stoppages..
x/48 is time with stoppages..
x/60-x/48=x/240 is the timefor staoppages...
thus (x/240)/(x/4=1/5 which is the ans...
this shows the stoppage timeavg..does notdepend on distance travelled...
ok bye sriram...the questiona posted here are cool...

[rainbowsparks]

:infinity: A train travelling at 10 m/sec frm A to B at 7.00 am meets a train leaving B at 7:20 am & coming to A at a speed 1/3rd times faster than the first train.if the dist frm A to B is 68 km then at what dist frm A will the two trains meet. ans is 36km .but how do u solve it?

[Arsenal]

Originally Posted by rainbowsparks

:infinity: A train travelling at 10 m/sec frm A to B at 7.00 am meets a train leaving B at 7:20 am & coming to A at a speed 1/3rd times faster than the first train.if the dist frm A to B is 68 km then at what dist frm A will the two trains meet. ans is 36km .but how do u solve it?

first calculate how much train A will cover till 7.20 that is in 20 min =20/60 *10*18/5=12km

now from 7.20 train A will cover 3/7 of the distance 3/7*56=24km

therefore toatll distance from A =12+24=36km

[Wanderlust]

Originally Posted by rainbowsparks

:infinity: A train travelling at 10 m/sec frm A to B at 7.00 am meets a train leaving B at 7:20 am & coming to A at a speed 1/3rd times faster than the first train.if the dist frm A to B is 68 km then at what dist frm A will the two trains meet. ans is 36km .but how do u solve it?

A:train from A ,B:Train from B
first lets convert the units...for A: 10m/sec is 600m/min(.6km/min)
B : 800m/sec <1/3 times faster>(.8 km/min)

AT 7:20 : A has travelled 600x20 =12000m=12km.Now distance between the 2 trains is
68-12=56km
this distance can be covered by the two together in 56/.8+.6 = 40 mins

in 40 mins A will travel another .6 x40 = 24 kms.Hence total distance from point A is 12+24=36 as required.

Hope That helps!

[rainbowsparks]

hey ppl thanks a lot fr replying.here r a few more bouncers.
i hav got answers fr all the following questions but I just don't understand the
explanation given in my book .plz help me out.plz explain me each & every step.
make it as simple as possible.

1.A car which was driven in fog passed a man walking at 3 km/hr in the same direction.
he could see the car for 4 min & upto a distance of 100 m.what was the speed of the car.
(Ans 4&1/2 km/hr)
2.Two trains A & B start frm stations X & y towards each other.B leaves station Y half an
hour aftr train A leaves station X.two hrs aftr train A has strtd,the dist between the
trains A & B is 19/30th of the dist between stations X & y.how much time would it take
each train (A & B) to cover the distance X to Y,if train A reaches half an hr ltr to its
dest as compared to B.(Ans 10 hrs,9 hrs)

3.A train aftr travelling 50 km frm A meets with an accident and proceeds at 4/5th of the
former speed and reaches B,45 min late.had the accident happened 20 km further on ,it would
hav arrived 12 min sooner.find the original speed & the distance.(Ans 25 km/hr,125 km)

[aspirehigh]

[QUOTE=rainbowsparks]hey ppl thanks a lot fr replying.here r a few more bouncers.
i hav got answers fr all the following questions but I just don't understand the
explanation given in my book .plz help me out.plz explain me each & every step.
make it as simple as possible.

1.A car which was driven in fog passed a man walking at 3 km/hr in the same direction.
he could see the car for 4 min & upto a distance of 100 m.what was the speed of the car.
(Ans 4&1/2 km/hr)
2.Two trains A & B start frm stations X & y towards each other.B leaves station Y half an
hour aftr train A leaves station X.two hrs aftr train A has strtd,the dist between the
trains A & B is 19/30th of the dist between stations X & y.how much time would it take
each train (A & B) to cover the distance X to Y,if train A reaches half an hr ltr to its
dest as compared to B.(Ans 10 hrs,9 hrs)

3.A train aftr travelling 50 km frm A meets with an accident and proceeds at 4/5th of the
former speed and reaches B,45 min late.had the accident happened 20 km further on ,it would
hav arrived 12 min sooner.find the original speed & the distance.(Ans 25 km/hr,125 km)


Dear Rainbow,

1. Let the speed of the car be 'x'. therefore the relative speed = (x-3)kmph.
given, he could see the car upto 100 metres, in 4 minutes (that means the car's relatively travelled 100metres in 4 min, or 0.1 km in 40 minutes.
therefore w.r.t. 0.1 --- 40min.; a(suppose) --- 60min., which comes to 1.5 kmph.
now calculate x. (x-3)=1.5, so x=4.5kmph

2. I can't recall the formula, but its something like root of b/a. plz confirm.
3. sorry.

Regards,
aspirehigh

[aspirehigh]

[QUOTE=rainbowsparks]hey ppl thanks a lot fr replying.here r a few more bouncers.
i hav got answers fr all the following questions but I just don't understand the
explanation given in my book .plz help me out.plz explain me each & every step.
make it as simple as possible.

1.A car which was driven in fog passed a man walking at 3 km/hr in the same direction.
he could see the car for 4 min & upto a distance of 100 m.what was the speed of the car.
(Ans 4&1/2 km/hr)
2.Two trains A & B start frm stations X & y towards each other.B leaves station Y half an
hour aftr train A leaves station X.two hrs aftr train A has strtd,the dist between the
trains A & B is 19/30th of the dist between stations X & y.how much time would it take
each train (A & B) to cover the distance X to Y,if train A reaches half an hr ltr to its
dest as compared to B.(Ans 10 hrs,9 hrs)

3.A train aftr travelling 50 km frm A meets with an accident and proceeds at 4/5th of the
former speed and reaches B,45 min late.had the accident happened 20 km further on ,it would
hav arrived 12 min sooner.find the original speed & the distance.(Ans 25 km/hr,125 km)


Dear Rainbow,

1. Let the speed of the car be 'x'. therefore the relative speed = (x-3)kmph.
given, he could see the car upto 100 metres, in 4 minutes (that means the car's relatively travelled 100metres in 4 min, or 0.1 km in 40 minutes.
therefore w.r.t. 0.1 --- 40min.; a(suppose) --- 60min., which comes to 1.5 kmph.
now calculate x. (x-3)=1.5, so x=4.5kmph

2. I can't recall the formula, but its something like root of b/a. plz confirm.
3. sorry.

Regards,
aspirehigh

[msksent]

Hi
This is msksent. To your question 3:
Let d be the distance and s be the speed. Then we can write
(50/s) + ((d-50)/(4s/5)) = (d/s) + (3/4) -----(1) and
(70/s) + ((d-70)/(4s/5)) = (d/s) + (33/60) --- (2)
Substituting 1 in 2, we will get as (20/s) -(20/(4s/5)) + (d/s) +(3/4) = (d/s) + (33/60)
Which on simplification will give u s = 25. Substituting back in 1, we get d = 125 km.