Total weight increased = (8 x 2.5) kg = 20 kg.
Weight of new person = (65 + 20) kg = 85 kg.
Average of 20 numbers = 0.
Sum of 20 numbers = (0 * 20) = 0.
It is quite possible that 19 of these numbers may be positive and if their sum is a, then 20th number is (- a).
Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.
Required sale = Rs.[(6500 x 6) - 34009]
= Rs. (39000 - 34009)
= Rs. 4991.
Average = total runs / no.of innings = 32
So, total = Average x no.of innings = 32 x 10 = 320.
Now increase in avg = 4runs. So, new avg = 32+4 = 36runs
Total runs = new avg x new no. of innings = 36 x 11 = 396
Runs made in the 11th inning = 396 - 320 = 76
Let the numbers be x, x + 2, x + 4, x + 6 and x + 8.
Then [x + (x + 2) + (x + 4) + (x + 6) + (x + 8) ] / 5 = 61.
or 5x + 20 = 305 or x = 57.
So, required difference = (57 + 8) - 57 = 8
Weight of the teacher = (35.4 x 25 - 35 x 24) kg = 45 kg.
Let the numbers be x, x + 1, x + 2, x + 3, x + 4, x + 5 and x + 6,
Then (x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6)) / 7 = 20.
or 7x + 21 = 140 or 7x = 119 or x =17.
Latest number = x + 6 = 23.
i) Let the ages of the five members at present be a, b, c, d & e years.
And the age of the new member be f years.
ii) So the new average of five members' age = (a + b + c + d + f)/5 ------- (1)
iii) Their corresponding ages 3 years ago = (a-3), (b-3), (c-3), (d-3) & (e-3) years
So their average age 3 years ago = (a + b + c + d + e - 15)/5 = x ----- (2)
==> a + b + c + d + e = 5x + 15
==> a + b + c + d = 5x + 15 - e ------ (3)
iv) Substituting this value of a + b + c + d = 5x + 15 - e in (1) above,
The new average is: (5x + 15 - e + f)/5
Equating this to the average age of x years, 3 yrs, ago as in (2) above,
(5x + 15 - e + f)/5 = x
==> (5x + 15 - e + f) = 5x
Solving e - f = 15 years.
Thus the difference of ages between replaced and new member = 15 years.
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