Brandywine global investment management Interview Questions, Process, and Tips

EMNXDS

let the amount of money be x

cloths 1/3 X x=rs.x/3

balance = x- x/3 = 2x/3

food = 1/5 X 2x/3 = 2x/15

balance = 2x/3 – 2x/15= 8x/15

travel = 1/4 X 8x/15 = 2x/15

now he has 100 rupees

2x/5 = 100
2x= 500
x = 500/2
x = 250

let first digit be ‘X’
then 5th digit is ‘3X’
let 2nd digit be ‘Y’
then 3rd digit is ‘Y-3’
and 4th digit is ‘Y+4’
then the no is ‘(X)(Y)(Y-3)(Y+4)(3X)’
from the above we can say 3X<=9 so X<=3 and any of the digit in the number is <=9 and also given that 3 pairs sum is 11... so make trial and error.. if X=1...any of the no is 10 which is wrong trial.... if X=2...then let Y+4=9 ==> Y=5
then no is 25296
first pair 2+9=11
second pair 2+9=11
third pair 5+6=11
now the answer is 25296

7000

p = prob(1st digit not 7)*prob(2nd digit not 7)*prob(3rd
digit not 7)
=(8/9)*(9/10)*(9/10)
=0.72
Sandhya had it right except that there are 900 numbers
between 100 and 999 inclusive (not 899).

835,sum(digits)= 16
734 = 14
642=12
751= 13,
853=16,
981=18,
532=10

answer(c)751

1min = 60 seconds
6 min=6*60=360seconds
4 sec=10
360sec=360*10 /4=900

10000

12%

30.5

D

Here is the solution to the given version of the puzzle (9 balls, one is heavier, need to identify oddball), where we label the balls A, B, …, I:

1. Weigh ABC versus DEF.

Scenario a: If these (1) balance, then we know the oddball is one of G, H, I.

2. Weigh G versus H.

Scenario a.i: If these (2) balance, the oddball is I.

Scenario a.ii: If these (2) do not balance, the heavier one is the oddball.

Scenario b: If these (1) do not balance, then the oddball is on the heavier side. For simplicity, assume the ABC side is heavier, so the oddball is one of A, B, C.

2. Weigh A versus B.

Scenario b.i: If these (2) balance, the oddball is C.

Scenario b.ii: If these (2) do not balance, the heavier one is the oddball.

answer is maximum of 2.

8400

How it is possible ? Please write in detail !!! This answer
s not a use !!!

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