Edreams odigeo Interview Questions, Process, and Tips

% error= (new no-actual no)/actual no *100
% error= [(x/7)-7x]/7x * 100
% error= 48x/49x * 100
% error= 0.9795*100=97.95

5

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);

MEDICINE will become EOJDJEFM in the same code.

According to the question:
⇒(x+5)(y−10)=300
⇒xy+5y−10x−50=xy
⇒5×300​−10x−50=0
⇒−150+x2+5x=0
⇒x2+15x−10x−150=0
⇒x(x+15)−10(x+15)=0
⇒x=10 or −15
Price cannot be negative
So, As x>0,x=10.

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

The number of ways of selecting a group of eight is
5 men and 3 women=5C5​×6C3​ =20
4 men and 4 women=5C4​×6C4​ =75
3 men and 5 women=5C3​×6C5​=60
2 men and 6 women=5C2​×6C6​=10
Thus the total possible cases is 20+75+60+10=165.

15%

Explanation:

The number of ways of selecting three men, two women and three children is:

= ⁴C₃ * ⁶C₂ * ⁵C₃
= (4 * 3 * 2)/(3 * 2 * 1) * (6 * 5)/(2 * 1) * (5 * 4 * 3)/(3 * 2 * 1)
= 4 * 15 * 10
= 600 ways.

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