32400 students login everyday to elitmus website. And stay on the website for 9 minutes. If the access to website is only 18 hours in a day, then how many students can be found online at a point of time?

Options : A) 324 B) 300 C) 270 D) 200

Answer is : C

Solution :

Given :-
1. 32400 students login everyday
2. every student stays for 9 minutes.
3. accessible times is 18 hours

What To Find:- 
Total Students can be found at any sec ?

Procedure :-
1, accessible time is 18 hours = 18 * 60 minutes
2. how many 9 minutes exist in 18 hours = 18 * 60 / 9 = 120  (say this is period)
3. So total students  have to access the site  in one of  these 120 periods.
4. at any point total students can be found = 32400 / 120 = 270.

So.... Simple...... :-)

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1. A and B started from the same point on a circular track. The ratio of their speed is n:1 (n is a whole number). If 5th and 17th time they meet at the same point, what can not be the value of n? 

Options : A) 2 b) 3 c) 4 d) 6

Answer is : D

Solution :

If speed of two bodies moving in same direction in circular path are in the simplest ratio of a:b, then no. of pts they will meet = |a-b|
Let circumference of circle be 'L'.
Time for 1st meeting at any of the |a-b| pts = L/(a-b).
Time for 1st meeting at Starting Pt = LCM of (L/a, L/b).
Time for nth meeting at Starting Pt = n*LCM of (L/a, L/b).

Checking with the options :- 
1 . Check Option A :
When n = 2 :
a=2,b=1.
No:of meeting pts = (2-1)=1pt.
Only meeting pt is L.
Time for first meeting = L/1.
Time for first meeting at starting pt = LCM of (L/2, L/1) = L.
Time for 5th meeting = 5L.
Time for 17th meeting = 17L.

Both tyms, they meet at pt L.
So, n=2 is valid.

Check Option B :

When n = 3 :
a=3,b=1.
No:of meeting pts = (3-1)=2pts.
Meeting pts are : L/2, L .
Time for first meeting = L/2.
Time for first meeting at starting pt = LCM of (L/3, L/1) = L.
Time for 5th meeting = 5*L/2.
Time for 17th meeting = 17*L/2.

Both tyms, they meet at pt L/2.
So, n=3 is valid.

Check Option C :

When n = 4 :
a=4,b=1.
No.of meeting pts = (4-1)=3pts.
Meeting pts are : L/3, L/2, L .
Time for first meeting = L/3.
Time for first meeting at starting pt = LCM of (L/4, L/1) = L.
Time for 5th meeting = 5*L/3.
Time for 17th meeting = 17*L/3.

Both times, they meet at pt L/3.
So, n=4 is valid.

Check Option D : 

When n = 6 :
a=6, b=1.
No:of meeting pts = (6-1)=5pts.
Meeting pts are : L/5, L/4, L/3, L/2, L .
Time for first meeting = L/5.
Time for first meeting at starting pt = LCM of (L/5, L/1) = L.
Time for 5th meeting = 5*L/5 = L.
Time for 17th meeting = 17*L/5.

Both tyms, they meet at different pts namely L and L/5.

So, n=6 is not valid.

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