Clearly, we have :

A = B - 3 ...(i)

D + 5 = E ...(ii)

A+C = 2E ...(iii)

B + D = A+C = 2E ...(iv)

A+B + C + D + E=150 ...(v)

From (iii), (iv) and (v), we get: 5E = 150 or E = 30.

Putting E = 30 in (ii), we get: D = 25.

Putting E = 30 and D = 25 in (iv), we get: B = 35.

Putting B = 35 in (i), we get: A = 32.

Putting A = 32 and E = 30 in (iii), we get: C = 28.

When Ravi's brother was born, let Ravi's father's age = x years and mother's age = y years.

Then, sister's age = (x - 28) years. So, x - 28 = 4 or x = 32.

Ravi's age = (y - 26) years. Age of Ravi's brother = (y - 26 + 3) years = (y - 23) years.

Now, when Ravi's brother was born, his age = 0 i.e. y - 23 = 0 or y = 23.

We can get this by any of two explanations.

Answer 1: 41

7 = 3 + 4  = 34

Similarly, 5 = 4 + 1  = 41

Answer 2 :  41

7 - 3 = 4  => 34

Similarly, 5 - 4 = 1  => 41

let 'b' be the Length of bridge from cow to the near end of the bridge and 'a' be the distance of the train from the bridge.
'x' be speed of cow => '4x' speed of train
Then the total length of the bridge 2b + 10.
(a-2)/4x = b/x
=> a-2 = 4b........(1)
Now if it had run in opposite direction
(a+2b+10-2)/4x = (b+10-2)/x
=> a - 2b = 24......(2)
Solving (1) and (2)
b = 11 ,
Therefore length of the bridge is 2 x 11 + 10 = 32mts.

Let the total number of workers be x. Then, 

Number of women = x/3 and number of men = x - x/3  = 2x/3

Number of women having children = x/18

Number of men having children = x/3

Number of workers having children = (x/18) + (x/3) = 7x/18

Therefore, workers having no children = x - 7x/18 = 11/18 of all workers.

Clearly, the required number would be such that it leaves a remainder of 1 when divided by 3, 4, 5, or 6 and no remainder when divided by 7. Thus, the number must be of the form (L.C.M of 3, 4, 5, 6) x + 1 i.e., (60x + 1 ) and a multiple of 7. Clearly, for x = 5, the number is a multiple of 7. So the number is 301.

Candidates passed in atleast four subjects

= (Candidates passed in 4 subjects) + (Candidates Passed in all 5 subjects)

= (Candidates failed in only 1 subject ) + ( Candidates passed in all subjects)

= (78 + 275 + 149 + 147 + 221) + 5685 = 870 + 5685 = 6555    

If Vijay gives 'x' marbles to Ajay then Vijay and Ajay would have V - x and A + x marbles.
V - x = A + x --- (1)
If Ajay gives 2x marbles to Vijay then Ajay and Vijay would have A - 2x and V + 2x marbles.
V + 2x - (A - 2x) = 30 => V - A + 4x = 30 --- (2)
From (1) we have V - A = 2x
Substituting V - A = 2x in (2)
6x = 30 => x = 5.

Let 'x' be the number of new Rs.2000 notes.

Given for one new 2000 note there are three old 500 notes.

i.e, 3x

Given 10 more new Rs.2000 notes are added to the collection
and the ratio of new Rs.2000 notes to old Rs.500 notes 

x + 10 3 x = 1 2 ? x = 20

Therefore, number of new Rs.2000 notes =  x+10 = 30.

old Rs.500 notes = 3x = 60.

Thus, the total number of notes in the collection  = 30 + 60 = 90.

3T+12C = 48,250----(1)

21T+36C = ?

multiply eq (1) with 3
i.e,, 3(7T+12C) = 48,250*3
21T+36C = 1,44,750/-