The average age of the members of a 10-person committee is the same as it was 4 years ago, because an old member has just been replaced by a new member.
How much younger is the new member than the old member who was replaced?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
I have a question, why isn't it (x+z)/10=(x+36+y)/10 because after 4 years, it must be (z+4) replaced by y.
Log in to reply
While this is true in this case your looking for the difference (z+4)-y which is still 40
Consider it this way.... (At Present)Let the sum of 9 members age= x and the new member = y. So average is (x+y)/10. (4 years ago) the sum of nine members age = x-36 and the old member age was= z-4. So avg =( x+z-40)/10. Therefore, (x+y)=(x+z-40) So y=z+40 Note: y and z are the present ages.
I agree with you!
I think the answer should be 36. Four years ago, let the sum of the nine who stayed be x, and the old member be z, so the average then was (x + z)/10. Now the present board consists of nine members of age x + 36 and a new member of age u, so the average is (x + 36 + u)10. Equating the two averages, (times 10), x + z = x + 36 +u, so z - u = 36. Ed Gray
The answer should be 36.
The answer should be 36
Non-math explanation:
Every year that passes, all the members are 1 year older, so the sum of all the ages would be 10 higher. Then in 4 years, the sum of ages would be 10*4=40 years higher.
Then if both 4 years ago average and the current average are the same, then the difference between the replaced member and the new member has to be 40 years less.
Only 9 of the members are 4 years older. The oldest is removed from the new average. The new younger member is X. (X1+X2+X3+X4+X5+X6+X7+X8+X9+X10) - (X1+X2+X3+X4+X5+X6+X7+X8+X9) -9*4 = X, thus, X10=X+36.
Let S be the sum of ages of all old members and let a and b be the old member who was replaced and the new member who was added respectively. We have (S-40)/10=(s-a+b)/10. Simplifying gives s-40=s-a+b, s cancels andwe rearrange to get a-b=40.
Problem Loading...
Note Loading...
Set Loading...
Let the sum of nine member (total) =x and the age of old one=z so its average 4 yrs before=(x+z)/10. after 4 yrs let z be replaced by y. so now avg=(x+4*10+y)/10
now (x+z)/10=(x+40+y)/10 so after solving it found z=y+40. so old person is 40yrs older than young one.