Number of books that can be placed in the larger box is given by Volume of Box / Volume of book.
Now, n = 25 x 42 x 60 / 7 x 6 x 5
= 63000 / 210
= 300.
Therefore, 300 books can be in that box.

Let initially 'X' ticket has been sold.
So now in 2nd week 20% increases, so
=> X x 120/100
In 3rd week 16% increases, so
=> X x (120/100) x (116/100)
In 4th week 20% decrease, so
=> X x (120/100) x (116/100) x (80/100) = 1392
X = 1250.

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.

Dr. Neelu Rohmetra from Jammu University has been appointed as one of the heads of IIMs. She became the first woman director of Indian Institute of Management Sirmaur, Himachal Pradesh. Dr. Neelu Rohmetra is the founder of the director of International Centre for Cross-Cultural Research and Human Resource Management (ICccR and HRM).

Indian Cricket team off-spin specialist Ravichandran Ashwin has become the fastest cricketer to take 250 wickets in the Test matches. Ashwin surpassed Australian fast bowler Dennis Keith Lillee to reach the milestone of 250 Test wicket. Thus Ravichandran Ashwin has become the sixth Indian bowler to reach 250 test wickets.

By trial and error method, we get
2880/3 = 960 is not a perfect square
2880/4 = 720 is not a perfect square
2880/5 = 576 which is perfect square of 24
Hence, 5 is the least number by which 2880 must be divided in order to make it into a perfect square.

From statement (b)
Average of 11 players = 11 x 28 = 308
Average of 10 players = 10 x 27.3 = 273
Age of captain = 308 - 273 = 35.

From the Statement (a) we can say that captain's age (C) = y + 11 ...(1) where y-youngers age
From the statement (c),the averge age of 3groups of 3players each was given,
It means total 9 players age was given indirectly,
1st group: 25 x 3 = 75
2nd group: 28 x 3 = 84
3rd group: 30 x 3 = 90
Total  is 75 + 84 + 90 = 249
Now {C + Y + (9 players age sum)}/11 = 28
Y + 11 + Y + (249) = 28 x 11=308
2Y+260=308
2Y = 308-260 = 40 ==> Y=24 ==> C(captain's age)= Y+11 = 24+11 = 35

Therefore, statement b alone or a & c only is the answer.

Let X = Time taken for each of Type B Problems(100 Problems)
And 2X = Time taken for each of Type A problems(50 problems)

Total time period = 3hrs = 3 x 60min = 180 minutes

100X + 50(2X) = 180
100X + 100X = 180
200X = 180 min

X = 180/200
X = 0.90 min

By convertiing into seconds,

X = 0.90 x 60 seconds
X = 54 sec

So, time taken for Part A problems is = 54 x 2 x 50 = 5400 seconds
= 5400/60sec = 90 minutes.