IMMO Information Technology Interview Questions, Process, and Tips

Let A=5x,B=3x,C not given so consider C=x
In triangle total angle=180°
A+B+C=180°
5x+3x+x=180°
9x=180°
X=180°/9
X=20°
A=5(20)=100°
B=3(20)=60°
C=20°
Therefore A=100° is the largest angle
Ans: 100°

1500

79.2

cannot be determined .

Let A, B and C be the three 6-faced dice.
Then, according to the question,
Since two dices has to be equal, that value can be any of the 6 faces, i.e., 6C1​ cases.
Now for each case, 2 equal dices can be selected from 3 dices in 3C2​ i.e., 3 ways.
And for each of the above, the third dice can have any of the 5 remaining faces
The possible outcomes are P(A)=61​,P(B)=61​,P(C)=65​,P(A)=61​,P(B)=65​,P(C)=61​ and P(A)=65​,P(B)=61​,P(C)=61​
Hence the required probability = 61​×61​×65​×6×3=21690​=125​

Question is wrong i think

2*3(15+4)=114
114*1/2=57 rs

brother

1/8

800 yards

area is given by Area of Rectangle = Length of Rectangle*(sqrt((Diagonal of Rectangle^2)-(Length of Rectangle^2)))
so substituting we get area=4*(sqrt((25-16)))
area=4*(sqrt(9))
since area cannot be negative hence we take sqrt of 9 as 3 (as sqrt of 9 is 3 and -3 as well)
area=4*3=12m^2

Out of 10 persons, 4 are graduates; so, (10 – 4) = 6 are under-graduates.

If there is no restriction, any three can be chosen from the ten in (10C3) = 120 ways.

Now, if all three chosen are under-graduates; it can take place in (6C3) = 20 ways.

Therefore, the probability that there will be no graduate among the three chosen = (20 / 120) = (1 / 6).

Therefore, the probability that there will be at least one graduate among the three chosen = {1 – (1 / 6)} = (5 / 6) = 0.8333.

The hour hand advances between individual numbers as the minute hand makes a full path around the hour. The hour hand was pointing directly at 3 at 3:00, but at 3:15 it has advanced 1/4 of the way between 3 and 4. The number of degrees between numbers 30 (because 360/12). That gives you 30/4 = 7.5. The number of degrees between the two hands is 7.5 degrees.

lcm (12,18,24,32,40) – 1
1440-1 = 1439

Here is the solution to the given version of the puzzle (9 balls, one is heavier, need to identify oddball), where we label the balls A, B, …, I:

1. Weigh ABC versus DEF.

Scenario a: If these (1) balance, then we know the oddball is one of G, H, I.

2. Weigh G versus H.

Scenario a.i: If these (2) balance, the oddball is I.

Scenario a.ii: If these (2) do not balance, the heavier one is the oddball.

Scenario b: If these (1) do not balance, then the oddball is on the heavier side. For simplicity, assume the ABC side is heavier, so the oddball is one of A, B, C.

2. Weigh A versus B.

Scenario b.i: If these (2) balance, the oddball is C.

Scenario b.ii: If these (2) do not balance, the heavier one is the oddball.

answer is maximum of 2.

2 Leap years
+2 to tuedsay
Thursday is the correct answer