Adgm Interview Questions, Process, and Tips

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.

Let numbers be 2x, 3x
Given: (2x)^3 + (3x)^3 = 945
8x^3 + 27x^3 = 945
x = 3
Difference between numbers = 3x – 2x = x
Hence difference = 3

For optimal size of a project team..
the % increase in staff size should be “Zero”
reason:
additional member directly proportional to increase in
staff size..

4/9

speed = distance / time.
240/16=15m/sec=54km/hr

let the average of 25 person be x

total age of a person=average×no.of person

total age of 25 person=X×25

=25X

when a new person of 46kg come average decrease by 5kg(X-5)

Total age of 26 person =25x+46

Average age of 26 person=X-5

A/Q

.25X+46=26(X-5)

25X+46=26X-130

26X-25X=130+46

X =176

hence the av. of 25 person be 176kg

time (dist covrd by husband) (dist covrd by wife)
0.5hr 20 miles 0 miles
1 hr 40 miles 25 miles
1.5 hr 60 miles 50 miles
2 hrs 80 miles 75 miles
2.5 hrs 100 miles 100 miles
Therefore wife catches up after 2.5 hrs..

Sunday
Monday
Tuesday

27

let a = 9 * 9 * 9;
let b = 12 * 12 * 12;
let c = 15 * 15 * 15;

sum = a + b + c;

edge = Math.cbrt(sum);

console.log(edge);

b

P1/P2 = C1*T1/ C2*T2
where P1 = Partner 1’s Profit.
C1 = Partner 1’s Capital.
T1 = Time period for which Partner 1 contributed his capital.
P2 = Partner 2’s Profit.
C2 = Partner 2’s Capital.
T2 = Time period for which Partner 2 contributed his capital
6000/3000= 20,000*6/x*12
x=5000

They’ll never meet unless some of them stop and other continue to run