GATE Questions

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Thus, the derivative of sinx by first principle is cosx.

One day work of Raghu and Sam together = 1/12 + 1/15 = 9/60 = 3/20

Aru efficiency = 2/3 of (Raghu + Sam)

Number of days required for Aru to do thw work alone = 3/2 x 20/3 = 10 days. 

From statement A, we know that Pipe A can fill the tank in 40 hours. However, this information is not sufficient as we do not have the data for Pipe B. Hence, statement A alone cannot answer the given question.

From statement B, we know that Pipe B is one third as efficient as pipe A. However, we do not know the rate at which Pipe A fills the tank. Hence, we will not be able to find the rate at which Pipe B fills the cistern. Therefore, statement B alone is not sufficient to answer the question.

Now, if we combine the two statements, we know that Pipe A take 40 hours to fill the cistern.
Pipe B takes 120 hours to fill the cistern.

If they worked alternately, then either Pipe A could have started the cycle or Pipe B could have started the cycle.

If Pipe A started the sequence of filling alternately, then at the end of two hours, the two pipes together would have filled 1/40 + 1/120 = 1/30  th of the tank in an hour. Or the cistern will fill in 30 hours.

If Pipe B started the sequence, then at the end of 2 hours, the two pipes together would have filled 1/120 + 1/40 = 1/30 th of the tank in an hour. Or the cistern will fill in 30 hours.

As the answer obtained irrespective of which pipe started the sequence is the same, the correct answer is (3) - i.e., both the statement are sufficient to answer the question.

Given the salaries are in the ratio of 4 : 5 : 6

Now the salaries are increased by 50%, 60% and 50% respectively

The New ratio of salaries will be

4 x (150/100) : 5 x (160/100) : 6 x (150/100)

= 6 : 8 : 9

The new ratio of salaries of Akhil, Arun and Karthik is 9 : 8 : 6

Let 'd' be the distance between A and B
K -time = d/10 + d/9 = 19d/90 hours
L -time = 2d/12 = d/6 hours
We know that, 10 min = 1/6 hours
Thus, time difference between K and L is given as 10 minutes.
=> 19d/90 - d/6 = 1/6
=> (19d-15d)/90 = 1/6
=> 4d/90 = 1/6
Thus,
d= 15/4 km = 3.75 km.

hence the distance between A and B is 3.75 km.

The first person shakes hands with 22 different people, the second person also shakes hands with 22 different people, but one of those handshakes was counted in the 22 for the first person, so the second person actually shakes hands with 21 new people. The third person, 20 people, and so on...
So,
22 + 21 + 20 + 19 + 18 + 17 + 16 + 15 + 14 + 13 + 12 + 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1
= n(n+1)/2 = 22 x 23 /2 = 11 x 23 = 253.

Here I am very happy to start this Program is correct statement.

"Too" is not used with pleasant adjectives. It means 'more than required' and is usually used with unpleasant adjectives.

So, 'too healthy', 'too happy', 'too glad' etc are wrong.

Here total distance = 154 + 101 = 255 mts
Given speed = 120 kmph = 120x5/18 m/sec
Hence, Time = d/s = 255x18/120x5 = 7.65 sec