Ajay who is a King, at his eleventh hour, called his three queens and distributed his gold in the following way: He gave 50% of his wealth to his first wife, 50% of the rest to his second wife and again 50% of the rest to his third wife. If their combined share is worth1,30,900 kilograms of gold, find the quantity of gold King Ajay was having initially?

let us assume the ten's digit number be x and unit's place digit be y.
Then two digit Original Number = 10x + y
According to question ? y = 2x - 1 ................(1)
When digits are interchanged then new number = 10y + x
If the digits at unit's and ten's place are interchanged, the difference between the new and the original number is less than the original number by 20,
New Number - Original Number = Original number - 20
? 10y + x - (10x + y) = 10x + y - 20
? 10y + x - 10x - y - 10x - y = - 20
? 8y - 19x = - 20
? 19x - 8y = 20 ...................(2)
Put the value of y from equation (1) in above equation (2).
? 19x - 8(2x - 1) = 20
? 19x - 16x - 8 = 20
? 7x = 20 + 8
? 7x = 28
? x = 28/7
? x = 4
Put the value of y in equation (1)
From (i) y = 2 x 4 -1 = y = 7
? original number = 10x + y =10 x 4 + 7 = 47.

Let the number be N
According to the question N x ( N x 3)/4 = 10800
? 3N² / 4 = 10800
? N² = (10800 x 4) / 3= 14400
N = ?14400 = 120

? A runs 100 m while B runs 95 m
? A runs 400 m while B runs 95 x 400 / 100 = 380 m

Again, B runs 200 m while C runs 185 m
? B runs 380 m while C runs 185 x 200 / 380 = 3515 / 10 = 351.5 m

Hence, A runs 400 m while C runs 351.5 m
So, A can beat C by (400 ? 351.5) = 48.5 m

? Relative speed of the train = (40 - 25) km/hr = 15 x (5/18) = 25/6 m/sec
? Length of the train = (48 x 25) / 6 = 200 meters .

From above given triangle , It is clear that the altitudes AL, BM and CN of ?ABC intersect at H. Then H is the orthocentre of ?ABC.
In ?ABC, HL ? BC and BN ? CH.
Thus, the two altitudes HL and BN of ?HBC, intersects at A.
Hence the orthocentre of ?HBC is M .

Let the radius of circular field = r m.
Speed of person in m/s = 30/60 = 1/2m/s
According to the question,
[(2?r) /(1/2)] - [(2r)/(1/2)] = 30
? 4?r - 4r = 30
? [4 x (22/7) - 4]r =30
? (125 - 4)r = 30
? (8.5)r = 30
? r = 30/8.5 = 3.5 m

Let P(B) = x
Given, P(A?B) = 0.8 and P(A) = 0.3
? P(A) + P(B) - P(A?B) = 0.8
? P(A) + P(B) - P(A) P(B) = 0.8 {?A and B are independent}
? 0.3 + x - 0.3x = 0.8
? 0.7x = 0.5
? x = 5/7

From given figure , we have
In quadrilateral ABCD , we can see that
Diagonal AB = Diagonal CD ......... ( ? )
AO = OB ......... ( ? )
DO = CO ......... ( ? )
?DOB = 90° ......... ( ? )
From ( ? ) , ( ? ) , ( ? ) and ( ? ) , we get
From above , it is clear that In quadrilateral ABCD , diagonals are equal and they bisect each other at 90° .
? Quadrilateral ABCD is square.

Let x km . be covered in y hrs.
then, 1st speed = (x / y) km/hr.

2nd speed = [(x/2) / 2y)] km/hr.
= (x/4y) km/hr.

? Ratio of speed = x/y : x/4y = 1 :1/4 = 4: 1