GATE Questions

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Speed of the current = 24-16/2

= 8/2

= 4 km/hr.

First is produced by the action of the second.

Hence, Bread : Yeast :: Curd : Bacteria.

Given A alone can fill the tank of capacity 240 lit in 16 hrs.

=> A can fill in 1 hr = 240/16 = 15 lit

=> B alone can fill the tank of capacity 240 lit in 12 hrs.

=> B can fill in 1 hr = 240/12 = 20 lit

Now, (A + B) in 1 hr = 15 + 20 = 35 lit

But they are opened for 2 hrs

=> 2 x 35 = 70 lit rae filled

Remaining water to be filled in tank of 240 lit = 240 - 70 = 170 lit.

      It is a case of shifting as given below:

      1  2  3  4  5                    2  1  3  5  4

      O  F  T  E  N   →  F  O  T  N  E

---------------------------------------------------------------------------

       1  2  3  4  5                   5  4  3  1  2

1.    H  E  A  R  T       →      T  R  A  H  E    

2.    O  P  E  N  S  →  S  N  E  O  P

        1  2  3  4  5         2  1  3  5  4

3.    R  I  S  K  Y  →   I  R  S  Y  K

4.    F  I  R  S  T  →   I  F  R  S  T

Hence option 3 is correct.

Double Difference pattern is followed in the above series. +2, +5, + 10, +17, +26

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

Let work done by 1 man in i day be m

and Let work done by 1 boy in 1 day be b

From the given data,

4(5m + 3b) = 23

20m + 12b = 23....(1)

2(3m + 2b) = 7

6m + 4b = 7 ....(2)

By solving (1) & (2), we get

m = 1, b = 1/4

Let the number of required boys = n

6(7 1 + n x 1/4) = 45

=> n = 2.

Given numbers are 1, 2 and 3

Factors of 1 = 1

Factors of 2 = 1, 2

Factors of 3 = 1, 2 and 3

Common factors = 1

Hence, the HCF of 1, 2 and 3 = 1.