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.
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 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 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.
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
Let the third subject marks be 'x'
=> Second subject marks = 2x
=> Third subject marks = 4x
Given avg = 224
x + 2x + 4x = 224 x 3
=> 7x = 224 x 3
=> x = 96
Hence, Second subject marks = 2x = 2 x 96 = 192.
Given that,
Number of windows = 50
Each window covering covers 15 windows
=> 50 windows requires 50/15 window coverings
= 50/15 = 3.333
Hence, more than 3 window coverings are required. In the options 4 is more than 3.
Hence, 4 window coverings are required to cover 50 windows of each covering covers 15 windows.
Required average = [(76 * 16) - (75 * 10)] / 6 = (1216 - 750)/6 = 466/3 = 233 /3.
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