Clearly, we have : x = (3y + 3z ) / 6 or 2x = y + z.
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)
Four tenths = 0.4
Five thousandths = 0.005
The average is (0.4 + 0.005)/2 = 0.2025
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.
Required average = [(76 * 16) - (75 * 10)] / 6 = (1216 - 750)/6 = 466/3 = 233 /3.
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.
It is the same as a person with 20 years more age replaces an existing person of the group ( or village)
Since the total age of the village having n persons, is being increased by 20 years and the average age of village is being increased by 1 year, hence there are total 20 people in the village.
Alternatively : ( n x 42 ) + 20 = ( n x 43 )
=> n=20
Let the fixed expenditure of the hotel be Rs.x and the variable expenditure ( which is dependent on the guest ) is Rs.y , then
x + 10y = 600 ---------(1)
x + 20y = 800 ----------(2)
From (1) & (2)
10y = 200
=> y = Rs. 20 and x= 400
Hence the total expenditure when there are 40 guests = 400 + 40 x 20 = 1200
Therefore, average expenditure = 1200/40 = Rs. 30
Let the average expenditure per head be Rs. p
Now, the expenditure of the mess for old students is Rs. 44p
After joining of 15 more students, the average expenditure per head is decreased by Rs. 3 => p-3
Here, given the expenditure of the mess for (44+15 = 59) students is increased by Rs. 33
Therefore, 59(p-3) = 44p + 33
59p - 177 = 44p + 33
15p = 210
=> p = 14
Thus, the expenditure of the mess for old students is Rs. 44p = 44 x 14 = Rs. 616.
Let 'K' be the total number of sweets.
Given total number of students = 112
If sweets are distributed among 112 children,
Let number of sweets each student gets = 'L'
=> K/112 = L ....(1)
But on that day students absent = 32 => remaining = 112 - 32 = 80
Then, each student gets '6' sweets extra.
=> K/80 = L + 6 ....(2)
from (1) K = 112L substitute in (2), we get
112L = 80L + 480
32L = 480
L = 15
Therefore, 15 sweets were each student originally supposed to get.
Manager's monthly salary
= Rs. (1900 x 25 - 1500 x 24) = Rs. 11,500
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