The range Interview Questions, Process, and Tips

1620

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

Assume his salary was Rs. X
He earns 5% raise. So his salary is (105*X)/100
A year later he receives 2.5% cut.
So his salary is ((105*X)/100)*(97.5/100)
which is Rs. 22702.68
So Ans is Rs.22176

let the total no of breads be x.
1st man 2nd man
x- (x/2) – 1/2 – 1/2((x-1)/2) – 1/2 ….. so on.
ans is 31.
first ate : 15.5 + .5 = 16 remaining 15
second ate : 7.5+0.5 = 8 remaining 7.
third ate : 3.5 +0.5 = 4 remaining 3.
fourth ate : 1.5 + 0.5 = 2 remaining 1
fifth ate : 0.5 + 0.5 = 1 remaining 0

C. 39

3*2+1 = 7
15*2+1 = *31*
31*2+1 = 63
63*2+1 = 127
127*2+1 = 255
255*2+1 = 511

9 because from 64 stubs 8 cigarettes and again from 8 stubs he’ll get 1 cigarette then his score will be 9

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)

A can give B (5 min – 4 1/2 min) = 30 sec start.
The distance covered by B in 5 min = 1000 m.
Distance covered in 30 sec = (1000 * 30)/300 = 100 m.
A can give B 100m start.

Son in law