Bytek Automobiles Interview Questions, Process, and Tips

1/24-1/40
(5-3)/120
2/120
1/60
60 minutes

3630

20 litres
cos
140 280 320 liters oil are exactly fittable in 20 litred tin

Dear friends i ofcourse confused like you when i come
across to solve this problem. Really friends its very
simple if you understand the question clearly. First thing
is what is mean by “as many as” means its called ‘idiom and
phrase’ in english and it means “the same number of”. now
read the question “how many pairs of letters in STAINLESS
which has same number of letters between them in the word
as they have in english alphabet”.
In the alphabetical order, A-Z can be numbered as 1-26.
In A(INL)E which is same as in the alphabetical order A
(BCD)E. In both the cases E is in the Fourth position. so
we got one pair.
And in ST, there are no letters between them in the word
stainless. In alphabetical orer from A-Z also there is no
letters between them..so we got the second pair…
In STAINLESS it has two pairs ST and AE

9/32

1:2

1500

Take one fruit from box with label mixture. If we see
orange, because the basket lies (it cant have a mixture),
then it has only oranges. The other 2 are labeled apples and
oranges. The one labeled apples,
cannot have oranges inside, cos they are allready been
identified, and because it lies, it cannot have apples either.
So it has a mixture. And we are left with the one labeled
oranges that lies and has apples.

Let 10’splace digit is x and unit’s place digit y

First milestone : 10x+y
Second milestone : 10y+x
Third milestone: 100x+y

Since the speed is uniform so
Distance covered in first Hr = Distance covered in Second Hr
(10y+x)-(10x+y) = (100x+y)-(10y+x)
After solving, we get —-> y=6x but since x and y are digits so only possible combination is x=1 and y=6,
So average speed = 45 KM/HR

3 miutes

A. 14th

135 is wrong because from starting difference between 7 after that 9 after 11 ,13,17 and so on

None

7 months

TOTAL NUMBER OF BOOKS 40+40=80
ISSUED BOOKS 30
REMAINING BOOKS 80-30=50
RATE=50/80=5/8

One day’s work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10.

C leaves 4 days before completion of the work, which means only A and B work during the last 4 days.

Work done by A and B together in the last 4 days = 4 (1/24 + 1/30) = 3/10.

Remaining Work = 7/10, which was done by A,B and C in the initial number of days.

Number of days required for this initial work = 7 days.

Thus, the total numbers of days required = 4 + 7 = 11 days.