Of the following, which is greater than ½ ?

Indicate ALL such fractions.

Let the average bill paid by twenty members = 'x'

But 19 men paid each = Rs. 70

20th man paid Rs. 90.25 more than the avg bill of 20 = x + 90.25

20x = 19(70) + x + 90.25
19x = 1330 + 90.25
19x = 1420.25
x = 1420.25/19 = Rs. 74.75

But the total bill = 20 x 74.75 = Rs. 1495.

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.

Number of runs scored more to increse the ratio by 1 is 26 - 14 = 12
To raise the average by one (from 14 to 15), he scored 12 more than the existing average.
Therefore, to raise the average by five (from 14 to 19), he should score 12 x 5 = 60 more than the existing average. Thus he should score 14 + 60 = 74.

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.

Assuming Sripad has scored the least marks in subject other than science,
Then the marks he could secure in other two are 58 each.
Since the average mark of all the 3 subject is 65.
i.e (58+58+x)/3 = 65
116 + x = 195
x = 79 marks.

Therefore, the maximum marks he can score in maths is 79.

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 total distance to be covered is 84 kms.
Time taken to cover the distance without stoppage = 84/42 hrs = 2 hrs
Time taken to cover the distance with stoppage = 84/28 = 3 hrs.
Thus, he takes 60 minutes to cover the same distance with stoppage.
Therefore, in 1 hour he stops for 20 minutes.

Actual average salary of 12 officers = 400

 Total salary of 12 officers = 400 x 12 = 4800

Average of these 12  officers as a whole =60

Total salary of 12 officers as a whole = 60 x 12 = 720

The total salary of rest  = 4800 - 720 = 4080

Difference in average of the rest of officers  = 60 - 56 = 4

Rest of the officers = 

Total number of employees = 1020 + 12 = 1032