Crestline coach ltd. Interview Questions, Process, and Tips

I think it’s none of these because Sunita is the grandmother of Renu so that makes the girl her mother in law ?

90 degree

3hour 45 min = 13500 sec
In 1 sec he covers = 12 m
In 13500 sec he covers = 12×13500 m = (12×13500)/1000 km = 162 km

Ans : 162 km

C. Monday
A year has 52 weeks and every week has 7 days.
We just add up one day to the present day of the year and that would automatically act as the day of the next year.
In case of leap years in between, one extra day gets added in the entire year making it 366 days instead of 365 days in an ordinary year and we add two days in that case.

Relative velocity of trains = 49kmph -25 kmph = 24kmph
=6.67m/s
Time taken to cross the first train = 10 sec
Length of the second train = Relative Velocity * Time taken
to cross the first Train = 6.67 *10 = 66.7 metres

Form IV is the register of receipt and issue of raw materials which. With the replacement of private records in lieu of statutory records, no separate Form V is required provided you maintain private accounts for receipt, issue and closing stock of raw materials.

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Question is not completed
The letters A, B, C, D, E, F and G, not necessarily in that order, stand for seven consecutive integers from 1 to 10
D is 3 less than A
B is the middle term
F is as much less than B as C is greater than D
G is greater than F

1. The fifth integer is
(a) A
(b) C
(c) D
(d) E
(e) F
ans:a
2.A is as much greater than F as which integer is
less than G
(a) A
(b) B
(c) C
(d) D
(e) E
ans:a

3. If A = 7, the sum of E and G is
(a) 8
(b) 10
(c) 12
(d) 14
(e) 16

4. An integer T is as much greater than C as C is
greater than E. T
can be written as A + E. What is D?
(a) 2
(b) 3
(c) 4
(d) 5
(e) Cannot be determined
ans:a

perimeter of rectangle = 2(l+b)
given l abd b as 18 cm and 26 cm
therefore perimeter of rectangle = 2(44)
perimeter of circle = 2*pi*r
given 2*pi*r=2(l+B)
i.e, 2*(22/7)*r= 2(44)
(22/7)*r=44
r=14 cms
area of the circle = pi*r*r
=(22/7)*14*14
=616 sqcm is the answer

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….