Average Questions

Weight of the teacher = (35.4 x 25 - 35 x 24) kg = 45 kg.

Correct Sum = (36 * 50 + 48 - 23) = 1825.

Correct mean = = 1825/50 = 36.5

Middle numbers  =  [(10.5 x 6 + 11.4 x 6) - 10.9 x 11] = (131.4 - 119-9) = 11.5.

 Since the month begins with a Sunday, so there will be five Sundays in the month,

Required average = (510 * 5 + 240 * 25) / 30 = 8550/30 = 285.

One way to deal with fractions is to convert them all to decimals. 

In this case all you would need to do is to see which is greater than 0.5.

Otherwise to see which is greater than ½, double the numerator and see if the result is greater than the denominator. In B doubling the numerator gives us 8, which is bigger than 7. 

M + Tu + W + Th = 4 x 48 = 192
Tu + W + Th + F = 4 x 46 = 184
M = 42
Tu + W + Th = 192 - 42 = 150
F = 184 – 150 = 34

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.

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.