GATE Questions

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Let Mohan present age = p

ATQ,

5/6 x p = p - 5

p = 30

His wifes age = 30 - 3 = 27 yrs.

For 1 hour, 20 minutes is slower, then for
06 miuntes - 2 minutes is slower
30 minutes - 10 minutes
Actual time | false time
12 12
01 PM 12:40 PM (20 minutes slower)
02 PM 01:20 PM
03 PM 02:00 PM

So, when actual time is 3pm, the false time is 2 pm.
So, it loses 60 minutes when it shows 2 pm.

For this we have to find the LCM of 24, 36 and 30

LCM of 24, 36 and 30 = 360 sec

360/60 min = 6 minutes.

Given A,B,C can complete a work in 15,20 and 30 respectively.

The total work is given by the LCM of 15, 20, 30 i.e, 60.

A's 1 day work = 60/15 = 4 units

B's 1 day work = 60/20 = 3 units

C's 1 day work = 60/30 = 2 units

(A + B + C) worked for 2 days = (4 + 3 + 2) 2 = 18 units

 Let B + C worked for x days = (3 + 2) x = 5x units

C worked for 2 days = 2 x 2 = 4 units

Then, 18 + 5x + 4 = 60

22 + 5x = 60

5x = 38

x = 7.6

Therefore, total number of days taken to complete the work = 2 + 7.6 + 2 = 11.6 = 11 3/5 days.

729 = 9 x 9 x 9

=> Cube root of 729 = 9

Now, required square of 9 = 9 x 9 = 81.

Indira Gandhi the first Prime Minister of India is known as the Iron Lady of India.

Given total number of students in the class = 21

So each student will have 20 greeting cards to be send or receive (21 - 1(himself))

Therefore, the total number of greeting cards exchanged by the students = 20 x 21 = 420.

Given a secret is told to 3 persons in 3 minutes.
And if the process continues with each person saying 3 other persons,
In 30 minutes => the rocess for 10 times because each person time is 3 minutes.
Required number of persons = 3 + 32 + 33 + .... + 39 + 310

Now it forms a geometric progression,

Sum to n terms is S = arn - 1r - 1

Here a = 3, r = 3

 
=> 59048 x 3 /2

= 88572.