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 =
= 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,
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.
Let the initial number of persons be x. Then,
16x + 20 * 15 = 15.5 (x + 20) <=> 0.5x = 10 <=> x = 20.
Average speed = (2xy) /(x + y) km/hr
= (2 * 50 * 30) / (50 + 30) km/hr.
37.5 km/hr.
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
Manager's monthly salary Rs. (1600 * 21 - 1500 * 20) = Rs. 3600.
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
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