Birla White Interview Questions, Process, and Tips

Speed = (45×518)m/sec=252m/sec

Total distance covered = (360+140) m = 500 m.

∴Required time= 500×225sec = 40 sec.

when know that 13 is factor of 39 so it will also not be able to divide it differently. so the remainder will be same 20 but as we know that 20 is greater then 13 so we can divide it once again and in the casE of 13 the remainder will be 20-13=7

600

let no of boys in group is = x
then total sum = 30 * x
after joining one more boys with a weight of 35 kg the total sum is = 30x + 35
after joining the new student the weight will increase 1 kg
so total sum
30x + 35 = 31(x+1)
x = 4

240 m

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

G.c.d=num1 *num2/lcm
Num1=20
Gcd=5
Lcm=60
5=(20*num2)/60
Num2= 5*60/20
Num2=15

South-West

Suppose the lengthier arm of weighing pan is of x cm and other arm is y cm .Also let weight if each melon be m kg.
so applying equilibrium of torque principles ,we get
case 1:-
1x-8my=0
case 2:-
2mx-1y=0
using case 1 equation , we substitute value of x into case 2 equation;
16mmy-1y=0
(16mm-1)y=0
y is length of weighing arm and cannot be 0,
therefore ,
16mm-1 =0
16mm=1
mm=1/16
m= square root (1/16)
m=+-1/4
m is weight of melon and cant be negative.
Hence m, weight of one melon is 1/4 kg

995

24.
exp: 0 is formed by multiplying 5 with 2. so first we find
how many 5’s and 2’s are there in 100!.
No of 5’s : 100/5=20/5=4/5.
20 +4 =24.
No of 2’s : 100/2=50/2=25/2=12/2=6/2=3/2=1/2.
50 +25 +12 +6 +3 +1.
It has 24 5’s and 2’s.
so the no. of zeros=24.