let A goes up and B comes down.
assume A takes 1 step in 1sec.
hence,B takes 5 steps in 1sec.
50 steps ll be moved by A in 50/1=50secs
125 steps ll be moved by B in 125/5=25 secs
hence speed of escalator=(difference of no of
steps)/(difference of time)
speed=(125-50)/(50-25)= 3 steps/sec
hence during upward journey of 50secs by A,total steps=50 by
A and 150 by escalator..total=150+50=200
during downward journey of 25 secs by B,total steps=125 by B
and 75 by escalator..total=125+75=200steps.
ANS:200 steps
12.5
x and y can be equal to 1..
If so,then he gets 1/1 of Rs.10 which is equal to 10 and
again 1/1 of Rs.10=Rs.10..so,he gets total of 20..and
returns 20..so,no gain and no loss..
If x=1,y=2,he gets, 5+20=25..and returns 20..so he may not
lose..
So,whatever be the values of x and y,only these two answers
are possible..
so its a)He never loses..
Total 55
Manisha is a girl name so 54 boys
1 girl
A
0
5
FASHION= FOIHSAN
It is F-ASHIO-N to F-OIHSA-N.
So PROBLEM will be P-ROBLE-M,
the answer is PELBORM.
list price – actual price markrd by the company
net price (average price of all items inclusive all discounts,breakage & so on.) which is 425 i.e. 50% of the marked price
hence the list price or marked price is twice the net sale price = 850
50%=425
100%=?
=100%*425/50=850
Bus started at 8:00
it travelled with 18mph…
distance of destination is 27 miles…
we know velocity=(distance)/time
i.e., time=27/18hours=3/2hours=90 min…
i.e., the bus reached destination at 9:30
the bus stayed for 30 min…
so the bus started return journey at 10:00
now the bus returned with velocity 18 + (18/2)=27mph
the taken to the bus to travel = 27/27=1 hour = 60 min…
so bus would be returned on 11:00…
I am looking for a qualified chartered accountant in my profile who is known for his excellence in the field.
6 cuts
5 trains
24000
625
CASE 1: First we should take six balls divided equally and
then it is placed on the two pans.three on one and three on
other..
if the two pans are balanced then the defective ball is not
in the six..then we should the two and keep them one ball
on each.
CASE2: Again We should take any of the six balls and
divided equally and then it is placed on the two pans.. if
any of the pan weighs less than the other.. We should take
the three balls seperately..Now from that three we should
take any two and placed one on each.. fi both the pan
balances the ball which is left over is the defective.. if
one ball weighes less than the other,while keeping one on
each,then it is the defective one….
How will you know the odd is in lighter one or heavier one from only one weighing. It will require 2 weighing to find the odd set and one weighing for odd coin in that set i.e total 3 weighings.
A rude person who doesn’t respect me