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.
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.
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
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 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.
Required average = [(76 * 16) - (75 * 10)] / 6 = (1216 - 750)/6 = 466/3 = 233 /3.
Clearly, to find the average, we ought to know the number of boys, girls or students in the class, neither of which has been given.
So, the data provided is inadequate.
Four tenths = 0.4
Five thousandths = 0.005
The average is (0.4 + 0.005)/2 = 0.2025
The 5 consecutive odd numbers whose average is k are (k-4), (k-2), k, (k+2), (k+4)
Again the average of (k-4), (k-2), k, (k+2), (k+4), (k+6), (k+8) is (k+2)
Clearly, we have : x = (3y + 3z ) / 6 or 2x = y + z.
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