Radiant technologies Interview Questions, Process, and Tips

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240.

X : (3000 x 8) + (2000 x 4) = 32000
Y : (4000 x 8) + (5000 x 4) = 52000
X:Y = 32000:52000 = 8:13
8k + 13k = 630
k = 30
X:8 x 30 = 240

30

Given:

In a group of 15 students,

7 have studied Latin,

8 have studied Greek,

3 have not studied either.

To find:

The number of students who studied both Latin and Greek.

Solution:

In a group of 15 students, have studied Latin, 8 have studied Greek, 3 have not studied either.

Therefore,

n(A∪B) = 15 – 3

n(A∪B) = 12

7 have studied Latin,

n(A) = 7

8 have studied Greek,

n(B) = 8

n(A∩B) is the number of students who studied both Latin and Greek.

n(A∩B) = n(A) + n(B) – n(A∪B)

n(A∩B) = 7 + 8 – 12

n(A∩B) = 15 – 12

n(A∩B) = 3

The number of students who studied both Latin and Greek is 3

Final answer:

3 of them studied both Latin and Greek.

Thus, the correct answer .3

1506

6*6=36 medals
6 days

1st day=1+35/7=6 remaining 30 medals
2nd day=2+28/7=6 remaining 24 medals
3rd day=3+21/7=6 remaining 18 medals
…
6th day 6 medals

2637
4815
OR
2817
4635

A. 80 : 81 : 84

C. 20

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​

if rs 12 and rs 15 per meter then 750*12+750*15

1056/(2*pi*r)=1056/(2*3.14*14)=1059/88 =12 revolution