Dorchester collection Interview Questions, Process, and Tips

5c2*4c1/9c3 +5c3/9c3=25/42

brownish grey

Relative speed =65-45km/hr
Relative speed=20km/hr
Convert to m/s
S=20*5/18
S=5.555
Length of the faster train =5.55555*18seconds
=100meters

Ans 635.5
4000
1 year 5 % 200 rs int
Total 4200
2 year 5 % 210 ra int
Total 4410
4410-2210 = 2200
1 year 5% 110
Total 2310
2 year 5 % 115.5 rs int
Total 2425.5
Total interest paid
200 +210+ 110+115.5. = 635.5

12

5

SI = PTR / 100

37 SEATS….
because if passengers will occupy 3 seats, per head amount
payable will be 26.89.80.67 can not be divisible other than
3 and 26.89. so the,
Number of occupied seats will be 3
Number of unoccupied seats will be 40-3=37

33

let MP = 100x
25% discount = 75x -> SP
Profit% = (Profit/CP)x100
125 = (75x/CP)x100
.: CP = 60x

Now if discount = 10%
SP = 90x
.: Profit% = ((90x-60x)/60x)x100 = 50%

1208.9

Answer is 274 .

Let the required term be x.

Pattern :- 1) Find the difference between consecutive terms as :

2 – 1 = 1

4 – 2 = 2

13 – 4 = 9

31 – 13 = 18

112 – 31 = 81

2) Now write the above numbers obtained by differences in series as :

1, 2, 9, 18, 81, y

(Suppose y be next term after 81.)

3) Look the difference series carefully and it follows the below pattern as :

(9 * 2)/1 = 18

(18 * 9)/2 = 81

(81 * 18)/9 = 162 = y (the next term after 81)

Thus the required answer is :

x = 112 + y

x = 112 +162 =274

B. 100

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.

60*5/18*9=150