Brilliant Technology Staffing Interview Questions, Process, and Tips

13

Mm

A. Tuesday

Lets assume total LCM(5,8) = 40units.
As, 5 men or 8 women do equal amount of work in a day,
1 Man does 8units/day and 1 Woman does 5units/day.
3M and 5W in 10 days do (3*8 + 5*5)*10 = 490units
To do 490 units in 14 days, number of Women required = 490/(14*5) = 7

30/100*142.85 = approx. 100
so ans is 42.85

4) QEOKTWG

2400

3606

127.179(app)
given distance of the train along the wind is 695
and againt the wind is 498
and time = distance/speed
as we know that time is equal in both the cases hence equate
695/s1=498/s2———–(1);
where s1=speed of the plane + speed of the wind
and s2=speed of the plane -speed of the wind
given that speed of the wind is 21k/h
s1=sp+21
s2=sp-21
substitu in eq 1
we get the answer as 27.17(app)

When A runs 1000 meters, B runs 900 meters and when B runs 800 meters, C runs 700 meters.

Therefore, when B runs 900 meters, the distance that C runs = (900 x 700)/800 = 6300/8 = 787.5 meters.

So, in a race of 1000 meters, A beats C by (1000 – 787.5) = 212.5 meters to C.

So, in a race of 600 meters, the number of meters by Which A beats C = (600 x 212.5)/1000 = 127.5 meters.

speed of the train respect to man
= (63 – 3) km/hr
= 60 km/hr
= 60 * 1000 / 3600 m/sec
= 50/3 m/sec

time
= distance/speed
= 500 * 3/ 50
= 30 sec