The average of A and B is 20 years. If C were to replace A, the average would be 19 and if C were to replace B, the Average would be 21. What are the ages of A, B and C?

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