Given,
mean of 100 observations is 40
=> Total of 100 observations = 40 x 100 = 4000
84 is misread as 48
=> Difference = 84 - 48 = 36
=> Now, new total of 100 observations = 4000 + 36 = 4036
Correct Mean = 4036/100 = 40.36
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.
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 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
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
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.
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.
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.
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