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slower train – 48 kmph = 40/3 m/s
say faster train, v m/s
therefore, {v-(40/3)}*180 = 600, => v= 60 kmph
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….
400
P+R=200
Q+R=350
+
—————-
P+Q+2R=550
P+Q+R=500
–
____________
R=50
A. Rs 140
7+8+11=26
B: 7/26*520= 140
Because he is very short for his height he can press upto 18th floor button only
( a ) BQDCJCMF
“TERMINAL” split it into the half we get “TERM” “INAL”
Now the first half is decreased by one and the next half is
increased by one, so we get:
“SDQLJOBM” (pls note in the ques we have “SDQIJOBM” where
‘L’ shud have come instead of ‘I’)
so “CREDIBLE” is to “BQDCJCMF”
Let ‘x’ be the number of perons in the group and let ‘y’ be
amount per head which they have to pay.
Then xy = 2400.
or y = 2400/x
Since two friends have forget the purse, x-2 persons should
share the total amount(2400).
If they share, they have to make an extra contribution of
Rs 100 to pay up the bill
That is x-2 persons should pay y+100 each
or (x-2)(y+100) = 2400
or y+100 = 2400/(x-2)
or y = 2400/(x-2) – 100
Therefore we have got two equations namely,
y= 2400/x and y = 2400/(x-2) – 100
Comparing these two, we get
2400/x = 2400/(x-2) – 100
Solving this we get
x^2 – 2x -48 = 0
or (x-8) (x+6) = 0
or x = 8 or x = -6
Since ‘x’ denote the number of persons, it should be
positive.
So, x = 8.
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