K G Somani & Co Interview Questions, Process, and Tips

FOR 15 = RS.20
FOR 20 = RS.18
FOR 25 = RS.16
FOR 30 = RS.14
FOR 35 = RS.12
FOR 40 = RS.10

Friends,
The Answer given by ‘Gaurav Sharma’ is correct and
the approach (bottom to top) suggested by ‘Shailesh’ is
good. But with minor correction we can arrive the solution
using this approach:
After 5th loot, No. of breads left = 3
after 4th loot, no. of breads left = (3+0.5)x2 = 7
after 3rd loot, no. of breads left = (7+0.5)x2 = 15
after 2nd loot, no. of breads left = (15+0.5)x2 = 31
after 1st loot, no. of breads left = (31+0.5)x2 = 63
So, before 1st loot, no. of breads left = (63+0.5)x2 = 127

6

31yrs

3 min

xy+2y-x=6
xy+2y-x-2=6-2
y(x+2)-1(x+2)=4
(y-1)(x+2)=4
so c is 4

-1

The last person covered 120.71 meters.
It is given that the platoon and the last person moved with
uniform speed. Also, they both moved for the identical
amount of time. Hence, the ratio of the distance they
covered – while person moving forward and backword – are
equal.
Let’s assume that when the last person reached the first
person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X)
meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-
X)] X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
Note that at first glance, one might think that the total
distance covered by the last person is 100 meters, as he
ran the total lenght of the platoon (50 meters) twice.
TRUE, but that’s the relative distance covered by the last
person i.e. assuming that the platoon is stationary.

1×2×…100=100!
Number of zeros in product of n numbers =[5n​]+[52n​]+[53n​]+…
Number of zeros in product of 100 numbers =[5100​]+[52100​]+[53100​]
where [.] is greatest integer function
=[20]+[4]+[0.8]=20+4=24

No loss no gain.

4

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