Quantitative Aptitude - Arithmetic Ability Questions

At 3 o'clock, the minute hand is 15 min. spaces apart from the hour hand. 

 To be coincident, it must gain 15 min. spaces. 

 55 min. are gained in 60 min.

  Then 15 min spaces are gained in  = 6055×15 min. 

Therefore,  The hands are coincident at  16411 min. past 3 o'clock.

distance covered in 1 revolution = 2πr = 2 x (22/7) x 20 = 880/7 cm
required no of revolutions = 17600 x (7/880) = 140

Let length = x meters, then breadth = 0.6x 

Given that  perimeter = 800 meters 

=> 2[ x + 0.6x] = 800 

=> x = 250 m 

Length = 250m and breadth = 0.6 x  250 = 150m 

Area = 250 x 150 = 37500 sq.m

Quantity 1:
Let the first pipe alone takes x hours to fill the
tank.
⇒The second and third pipes will take (x-5) and
(x-9) hours respectively.
According to the given information:
∴ 1x-9

⇒ (x-9)(2x-5) = x2 – 5x
⇒ 2x2 – 5x – 18x + 45 = x2 – 5x
⇒ x
2 -18x + 45 = 0
⇒ (x-15) (x-3) = 0
⇒ x = 15, 3

The first pipe can take 15 hours to fill the kund.
∵ 3 hours doesn’t satisfy the statement.
Quantity 2:
∴ Time taken by second pipe = x-5
⇒ Time taken by second pipe = 15-5 = 10hours
∴ Time taken by third pipe = x -9
⇒ Time taken by third pipe = 15- 9 = 6 hours
Now,

Net part filled in 1 hour = 115+110+16
⇒ Net part filled in 1 hour = 4+6+1060
⇒Net part filled in 1 hour = 2060=13
∴The Kund will be full in 3/1 hours if all the pipes are opened simultaneously
Now, comparing
15 > 3
Thus, Quantity 1 > quantity 2

 = 

   =         where  

B ---> speed of boat in still water

R---->speed of current

    ( By componendo and dividendo)

A leap year has 52 weeks and two days
Total number of cases = 7
Number of favourable cases = 1
i.e., {Friday, Saturday}

Required Probability = 1/7