Quantitative Aptitude - Arithmetic Ability Questions

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

If Rahul age is x, then Sachin age is x - 7,

so, (x-7)/x =7/9

 9x - 63 = 7x
 2x = 63
 x = 31.5

So Sachin age is 31.5 - 7 = 24.5

Speed of Man = 4.5 kmph
Speed of stream = 1.5 kmph
Speed in DownStream = 6 kmph
Speed in UpStream = 3 kmph
Average Speed = (2 x 6 x 3)/9 = 4 kmph.

NOTE: When a century year leaves a remiander 0, when divided by 400 then it is a leap year (366 days).

So, 1200 has 366 days.

Let the numbers be x, x + 2 and X + 4.

Then, x(x + 2)(x +4) / 8 = 720 => x(x + 2)(x +4) = 5760.

√x  *  √(x + 2) * √(x +4) = √x(x +2)(x + 4 ) = √5760  =  24√10.

Let the ratio be 'R'

Total number of students = 1050

Then,

9R + 5R = 1050

14R = 1050

=> R = 75

Hence, the number of boys = 9R = 9 x 75 = 675