Total age of 4 members, 10 years ago = (24 x 4) years = 96 years.
Total age of 4 members now = [96 + (10 x 4)] years = 136 years.
Total age of 6 members now = (24 x 6) years = 144 years.
Sum of the ages of 2 children = (144 - 136) years = 8 years.
Let the age of the younger child be x years.
Then, age of the elder child = (x+2) years.
So, x+(x+2) =8 <=> x=3
Age of younger child = 3 years.
Let the first number be x,
Then, sum of the four numbers = x + 4x = 5x.
so, 5x/4 = 60 or x = (60 * 4) / 5 = 48.
Correct Sum = (36 * 50 + 48 - 23) = 1825.
Correct mean = = 1825/50 = 36.5
Let there be x pupils in the class.
Total increase in marks = (x * 1/2) = x/2.
x/2 = (83 - 63) => x/2 = 20 => x = 40.
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.
Let the average expenditure be Rs. x Then,
9x=[(8*30)+(x+20)]<=>9x =x+260 <=> x =32.50
Total money spent = 9x = Rs. (9 x 32.5O) = Rs 292. 50
Let the number of papers be x. Then, 63x + 20 + 2 = 65x or 2x = 22 or x = 11.
Middle numbers = [(10.5 x 6 + 11.4 x 6) - 10.9 x 11] = (131.4 - 119-9) = 11.5.
Total age increased = (8 * 2) years = 16 years.
Sum of ages of two new men = (21 + 23 + 16) years = 60 years
Average age of two new men = (60/2) years = 30 years.
Sum of the present ages of husband, wife and child = (23 * 2 + 5 * 2) + 1 = 57 years.
Required average = (57/3) = 19 years.
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