Ted rogers school of management Interview Questions, Process, and Tips

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11

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

Length of train = 125Mtrs
Speed of man = 5Km/Hr
=(5*1000)/(1*3600)
=1.4M/s
Time taken for train to pass = 10s

Spd = D/T
Total Distance (Man distance+ Train Length)
#Distance covered by Man in 10s
D=Spd*T
= (1.4M/s*10s)
= 14Mtrs

#Train length = 125Mtrs

Total D = 14Mtrs + 125Mtrs
= 139Mtrs
Time taken = 10s

Sp = D/t
=139M/10s
=13.9M/s

answer is 28
>2+8=10
>reverse of 28 is 82
>subtract 28 from 82
>82-28=54
the number is 28

264

740

DISSENTIAL

NTFS — New Technology File System
Fat — file allocation table
NTFS having a quota, commpress system with securites base as
administrator giving, multipal user, groups, to set permission

47.5

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.

(x-2)^3==x^3-6x^2+12x-8
(x-2)^3==(2^(2/3)+2^(1/3))^3
therefore,by solving
Ans: x^3-6x^2+x== -8-5*2^(2/3)-5*2^(1/3)

5/9

Ans- 6,6 litres
Let x Litres from can1 (where, water = x/4 litres and milk = 3x/4 litres)
Now
12-x liters from can 2 ( where, water = 6-x/2, milk = 6-x/2)

Now given
water/milk = 3/5
0.25x+6-0.5x / 0.75 +6-0.5x = 3/5
6-0.25x/6+0.25 = 3/5
Solve it you get
x = 6 litres from can1
Then 12-x= 12-6=6 litres from can2

9s

area is doubled.
actual area = 1/2 bh
after increase area = 1/2 *4b* h /2
=bh
1/2 *bh *2 = bh
therefore area is doubled