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.
47/10000 = .0047
.02 =(2/100) x 100% =2%
Let N / 11 = 233
Then, N = 233 x 11 = 2563
? Missing digit is 5.
121012 = 12 x 10084 + 4
? remainder = 4
We have the important relation, More work, More time (days)
? A piece of work can be done in 6 days.
? Three times of work of same type can be done in 6 x 3
= 18 days
? = 750.0003 ÷ 19.999
? ? ? 750 ÷ 20
? ? ? 375 ? 38
Subtract 20, 25, 30, 35, 40, 45 from successive numbers. So 0 is wrong.
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