New Questions to solve.?

[sbhaskaran2k]

1) Padmaja left her residence between 4 and 5 'o clock and returned back between 5 and
6 'o clock . She notices that the hour hand and minute hand of the clock interchanged
theie position with what they were when she went out. Wht time did she go out and when
diod she return ?


2) A bus B1 left a town T1 for another town T2 at 6:00 A.M . At 7:30 A.M another bus
B2 also left T1 for T2 and the speed of the bus B2 was 5 kmph higher than that of B1.
At 10:30 P.M on the same day the busses were 21 km apart . Find the speed of the bus
B1 ?

Help me to solve these questions?

[assignus]

Originally Posted by sbhaskaran2k

1) Padmaja left her residence between 4 and 5 'o clock and returned back between 5 and
6 'o clock . She notices that the hour hand and minute hand of the clock interchanged
theie position with what they were when she went out. Wht time did she go out and when
diod she return ?


Help me to solve these questions?

left at 4 hr 26 min 51.19 sec
returned at 5 hr 22 min 14.27 sec

the way i worked out is very cumbersome to post.....
lemme know if this is the right answer,..

[sbhaskaran2k]

Originally Posted by assignus

left at 4 hr 26 min 51.19 sec
returned at 5 hr 22 min 14.27 sec

the way i worked out is very cumbersome to post.....
lemme know if this is the right answer,..

Ya it is right.... Can u tell me the answers in fraction ...

and the steps too...

[cat_cracker1]

is 24km/h the answer for the 2nd one.

check out if its the right one,
then i will post the way in
which i solved that.

[sbhaskaran2k]

Originally Posted by cat_cracker1

is 24km/h the answer for the 2nd one.

check out if its the right one,
then i will post the way in
which i solved that.

Actually we shd get 2 answers ...

1) when the buses r not crossed each other and the distance is 21 km
2) when the buses r crossed each other and the distance is 21 km


u missed these points....dude

[assignus]

Originally Posted by sbhaskaran2k

Ya it is right.... Can u tell me the answers in fraction ...

and the steps too...

its easy to deduce that time at which xyz leaves is between 4:25-4:30
and time when xyz returns is 5:20-5:25

now...Hours hand moves 5 degrees in 60 min..
say left time is 4:X minutes...... so in X min Hours hand moves (5xX)/60 degrees

Now returm time in terms of X is 5 hours [20+(5xX)/60] minutes

now calculate how many degrees hours hand moves in [20+(5xX)/60] minutes .....this added to 25 minutes will the position of hours hand when xyz returns...... that is also equal to X ..i.e

X=25+(5/60) [20+(5xX)/60]

solving u get X=26.8531 = 26 min 51.18 sec

hope it helps..

ps: is there any shortcut..or a simpler way of solving?

[wippybaba]

Originally Posted by sbhaskaran2k

Actually we shd get 2 answers ...

1) when the buses r not crossed each other and the distance is 21 km
2) when the buses r crossed each other and the distance is 21 km


u missed these points....dude

Yes.. you are right...

let speed of bus B1 be x, and speed of bus B2 == x+5

at 10:30, bus B1 has travelled 16.5 hours and bus B2 has travelled 15 hours.

==> 16.5x - 15(x+5) = 21 (for the first case when B1 is still ahead of B2)

or 15(x+5) - 16.5x = 21 (for the second case when B2 has overtaken B1)

solving, we get x = 64 for the first case and x = 36 for second case??? is it right or am i missing something?

[sbhaskaran2k]

Originally Posted by wippybaba

Yes.. you are right...

let speed of bus B1 be x, and speed of bus B2 == x+5

at 10:30, bus B1 has travelled 16.5 hours and bus B2 has travelled 15 hours.

==> 16.5x - 15(x+5) = 21 (for the first case when B1 is still ahead of B2)

or 15(x+5) - 16.5x = 21 (for the second case when B2 has overtaken B1)

solving, we get x = 64 for the first case and x = 36 for second case??? is it right or am i missing something?

ya u got it right ............

Wht abt the first one ???

[cat_cracker1]

actally made the mistake by taking it 10.30 a.m.
so made it 4.5 hrs.
that's why i got 24km/h as ans.

and also missed the second case.

and yes the answer shud b 64 and 36

ciao

[DesiGuru]

Originally Posted by assignus

its easy to deduce that time at which xyz leaves is between 4:25-4:30
and time when xyz returns is 5:20-5:25

now...Hours hand moves 5 degrees in 60 min..
say left time is 4:X minutes...... so in X min Hours hand moves (5xX)/60 degrees

Now returm time in terms of X is 5 hours [20+(5xX)/60] minutes

now calculate how many degrees hours hand moves in [20+(5xX)/60] minutes .....this added to 25 minutes will the position of hours hand when xyz returns...... that is also equal to X ..i.e

X=25+(5/60) [20+(5xX)/60]

solving u get X=26.8531 = 26 min 51.18 sec

hope it helps..

ps: is there any shortcut..or a simpler way of solving?

I dont get the need for this complication
She went out at 4:25 and came back at 5:20
Isnt the above correct too?

K