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  1. The average age of a board of 8 trustees remains the same as it was 3 years ago, when one of them is replaced by a new member. The new member is younger than the trustee in whose place he has been replaced by

Options:
A .  16 years
B .  24 years
C .  32 years
D .  none of these
Answer: Option B
Let the average age of the board of 8 trustees be x.
Then, the sum of the ages of the 8 trustees would be 8x.
Three years ago, the sum of their ages would have been (8x - 8*3) = (8x - 24).
Let the age of the trustee who has been replaced be y, and the age of the new member be z.
We know that z < y, since the new member is younger than the trustee he has replaced.
After the replacement, the sum of the ages of the 8 trustees remains the same as it was 3 years ago. Therefore, we have:
(8x - y + z) = (8x - 24)
Simplifying this expression, we get:
z - y = -24
Since z < y, we can rewrite this as:
y - z = 24
This means that the difference in ages between the trustee who has been replaced and the new member is 24 years.
Therefore, the answer is option B, 24 years.

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4 Comments

Lets assume that the sum of the age of the unchanged trustees 3 years ago be x. Also the age of the replaced trustee and the new trustee, 3 years ago be y and z respectively.
Now since the average age of the current trustees and the previous trustees is same,
(x + y)/8=(x + z + 24)/8
The 24 value accounts for the time lapse of 3 years for each of the 8 trustees.
Thus we get y - z = 24
Therefore the age gap is 24
8*3=24
Let the total age of seven unchanged trustees be x and the replacing trustee age be y and z
Now
x+y/8 = x+z+24
Here 24 is the total age of 8 trustees in 3 yrs
So y-z = 24
The average age of a board of 8 trustees remains the same as it was 3 years ago, when one of them is replaced by a new member. The new member is younger than the trustee in whose place he has been replaced
8x3=24

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