The average age of 10 members of a committee is the same as it was 4 years ago, because an old member has been replaced by a young member.
Find how much younger is the new member?
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I am so stupid that I computed for 4 years from now
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.
You caught me. I forgot to add 4 years into the old member's current age
Let the ages 4 years ago be this:
The 9 members who stayed: i = 1 ∑ 9 x i
The old member: a
Let the ages today be this:
The 9 members who stayed: i = 1 ∑ 9 ( x i + 4 ) = i = 1 ∑ 9 x i + i = 1 ∑ 9 4 = i = 1 ∑ 9 x i + 3 6
The old member: a + 4
The new member: b
The average age of the members then and now are the same. Therefore, the sum of their ages then and now are also the same:
i = 1 ∑ 9 x i + a = i = 1 ∑ 9 x i + 3 6 + b a − b = 3 6 ( a + 4 ) − b = 4 0
Therefore, the old member is 4 0 years older than the young member
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But I suppose there were 3 tries so you should have been able to catch the mistake. Bad luck!
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I guessed the last two XD
i too did it the same way . by the do you know ayushrai
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Yes I chat with him on slack
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by the way i am his cousin
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@Abhishek Alva – Why dont you come on slack? And in which class are you?
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Let the average four years ago and today be a , the age of the old member and young member be Y and y respectively. Then the different in age Δ y = Y − y . Since the average age is the same today and four years ago. The total age today after the replacement is 1 0 a . The total age before the replacement is 1 0 ( a + 4 ) . Therefore, we have:
1 0 ( a + 4 ) − Y + y 1 0 a + 4 0 − Δ y ⟹ Δ y = 1 0 a = 1 0 a = 4 0
The new member is 4 0 years younger.