Two trains running at 36 km/hr and 54 km/hr are running parallel in the opposite direction. If their lengths are 200m and 300m respectively. Then how much time will they take to cross each other?

Let OP be the tower of height h (say) and A and B be the two positions on the horizontal line through O, such that
?OAP = ?, ?OBP = ? and OB = x
In ?OBP, Use the trigonometry formula
Tan? = P/B = Perpendicular distance / Base distance
Tan? = OP/OB
? OB = OP/Tan?
? OB = OP Cot?
Put the value of OB and OP , We will get
x = h Cot ?...............(1)
In ?OAP, Similarly
Tan? = OP/OA
? OA = OP/ Tan?
? OA = OP Cot ?
Put the value of OA and OP
? a + x = h Cot ?
? x = h Cot ? - a ............(2)
From equation (1) and (2)
? h Cot ? = h Cot ? - a
? a = h Cot ? - h Cot ?
? a = h (Cot ? - Cot ?)
? a = h (Cos ?/ Sin ? - Cos ? / Sin ? )
? a = h( (Cos ? Sin ? - Cos ? Sin ? ) /Sin ? Sin ? )
? a = h( Sin(? - ?) / Sin ? Sin ?)
? h = a Sin ? Sin ?/ Sin(? - ?)

There are 3A's 2N's and one B. We have to find the total number of arrangements of 6 letters out of which 3 are alike of one kind, 2 are alike of second kind, thus the total number of words
= 6! / (3! 2!) = 60

Since 39 persons can repair a road by working 5 hours in a day in 12 days.
? 1 persons can repair a road by working 5 hours in a day in 39 x 12 days.
? 1 persons can repair a road by working 1 hours in a day in 39 x 12 x 5 days.
? 30 persons can repair a road by working 1 hours in a day in 39 x 12 x 5 / 30 days.
? 30 persons can repair a road by working 6 hours in a day in 39 x 12 x 5 / 30 x 6 days.
? 30 persons can repair a road by working 6 hours in a day in 39 x 2 x 1 / 6 days.
? 30 persons can repair a road by working 6 hours in a day in 39 x 1 x 1 / 3 days.
? 30 persons can repair a road by working 6 hours in a day in 13 days.

Let the required weight be w kg.
Less length, Less weight (Direct proportion)
45 : 12 :: 171 : w
? w = (12 x 171)/45 = 45.6 kg

Let the sum be Rs 'y' , so amount = 2y
Simple interest =Rs y
Let R be the rate of interest,
R= (100 x SI)/(P x T) = (100 x y) / (y x 8) = 12.5 %
where SI= Simple Interest
P = Principal
T = Time

now, the needed amount = Rs 4y
since SI = Rs (4x-x)= Rs 3y
since T= (100 x SI)/(P x R)
= (100 x 3y )/(y x 125)= 24 yr

Let us assume the first Odd number is a.
then 2nd odd number = a + 2
and 3rd odd number = a + 4
According to question,
Sum of three consecutive odd numbers is 20 more than the first of these numbers;
a + (a + 2) + (a + 4) = a + 20
? a + a + 2 + a + 4 - a = 20
? 2a + 6 = 20
? 2a + 6 = 20
? 2a = 20 - 6
? 2a = 14
? a = 7
The First odd number a = 7
The second odd number = a + 2 = 7 + 2 = 9
Second odd number is middle number = 9