Age of the 15th student = [15 * 15 - (14 * 5 + 16 * 9)] = (225-214) = 11 years.

Let the speeds of the car, train and bus be 5x, 9x and 4x km/hr respectively.
Average speed = 5x + 9x + 4x/3 = 18x /3 = 6x km/hr.
Also, 6x = 72 => x = 12 km/hr
Therefore, the average speed of the car and train together is = 5 x + 9 x 2= 7x = 7 x 12 = 84 km/hr.

Let the three numbers be x, y, z.

From the gien data,

x = 2y ....(1)

x = z/2 => z = 2(2y) = 4y .....(From 1)  ...........(2)

Given average of three numbers = 56

Then,

$$\begin{gathered} x + y + z 3 = 56 \\ 2 y + y + 4 y 3 = 56 ( From 1 & 2 ) \\ 7 y = 56 x 3 \\ y = 24 \end{gathered}$$

Now,

x = 2y => x = 2 x 24 = 48

z = 4y = 4 x 24 = 96

Now, the highest number is z = 96 & smallest number is y = 24

Hence, required sum of highest number and smallest number

= z + y

= 96 + 24

= 120.

Manager's monthly salary Rs. (1600 * 21 - 1500 * 20) = Rs. 3600.

Let avg ages of 5 members at present is G

And age of new member is m and an older is n

Required = (5G + m - n)/5 = G - 3

m - n = 15 years

Let the ratio of initial quantity of oils be 'x' => 4x, 5x & 8x.
Let k be the quantity of third variety of oil in the final mixture.
Let the ratio of initial quantity of oils be 'y'
From given details,
4x + 5 = 5y ..... (1)
5x + 10 = 7y .....(2)
8x + k = 9y ......(3)

By solving (1) & (2), we get
x = 5 & y = 5

From (3) => k = 5

Therefore, quantity of third variety of oil was 9y = 9(5) = 45kg.

Let the side of the square plot be 'a' ft.
Given area of the plot (a x a) = 289 => a = 17
Length of the fence = Perimeter of the plot = 4a = 68 ft.
Cost of building the fence = 68 x 58 = Rs. 3944.

We need to find the total cost to send r messsages, r > 1000.

The first 1000 messsages will cost Rs.p each (Or) 

The total cost of first 1000 messsages = Rs.1000p

The remaining (r - 1000) messsages will cost Rs.q each (Or)

The cost of the (r - 1000) = Rs.(r - 1000)y

Therefore, total cost = 1000p + rq - 1000q

= 1000(p - q) + qr

Let the number of students in classes A, B and C be P, Q and R respectively.
Then, total score of A = 83P, total score of B = 76Q, total score of C = 85R.
Also given that,
(83P + 76Q) / (P + Q) = 79
=>4P = 3Q.
(76Q + 85R)/(Q + R) = 81
=>4R = 5Q,
=>Q = 4P/3 and R = 5P/3
Therefore, average score of A, B, C = ( 83P + 76Q + 85R ) / (P + Q + R) = 978/12 = 81.5