Replacement of the Old Guard

Let the total age of the group be a and the total average $\frac {a}{10}$ .

If each original member stayed in their job, each of the 10 would have aged by 4 years, since we have to add 40 years to the groups total age after 4 years.

x = is the number subtracted from the total age of the group after 4 years on account of the new member being younger.

We then have $\frac {a}{10}= \frac {a+40-x}{10}$ )

For both sides to be equal it must mean that x=40, hence this how much younger the new member was.

Let the sum of nine member (total) =x

and the age of old one=z

so its average 4 yrs ago=(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 we find

z=y+40.

so the old person is 40yrs older than the young one

$Let\quad { x }_{ 1 },\quad { x }_{ 2 },\quad ...,\quad { x }_{ 10 }\quad be\quad the\quad ages\quad of\quad the\quad 10\quad people.\quad \\ Now,\quad one\quad of\quad the\quad membres\quad is\quad replaced\quad by\quad a\quad younger\quad member.\quad \\ Let\quad { x }_{ 10 }\quad is\quad replaced\quad by\quad new\quad member\quad { x }_{ 11 }.\quad \\ Now\quad four\quad years\quad ago,\quad theirs\quad ages\quad are\quad ({ x }_{ 1 }-4),\quad { (x }_{ 2 }-4),\quad ...,\quad ({ x }_{ 9 }-4)\quad and\quad ({ x }_{ 11 }-4).\\ \\ \frac { { x }_{ 1 }+...+{ x }_{ 9 }+{ x }_{ 10 } }{ 10 } =mean\quad age\\ \frac { ({ x }_{ 1 }-4)+...+({ x }_{ 9 }-4)+({ x }_{ 11 }-4) }{ 10 } =mean\quad age\\ or,\quad \frac { { x }_{ 1 }+...+{ x }_{ 9 }+{ x }_{ 11 }-40 }{ 10 } =mean\quad age\\ Hence,\quad { x }_{ 1 }+...+{ x }_{ 9 }+{ x }_{ 10 }\quad =\quad { x }_{ 1 }+...+{ x }_{ 9 }+{ x }_{ 11 }-40\\ or,\quad { x }_{ 11 }-{ x }_{ 10 }=40$

In 4 yrs now,age of 10 members of committee has increased by 10 X4 =40 yrs.Since average are is the same by replacing old guard by young member ,therefore difference of young member in age from old guard is 40 Yrs Ans. K.K.GARG,India

4 years ago

The age of the old man = x

the sum of ages of the other 9 members = y

the average = a

so

x + y = 10 a ................................... (1)

Now

The age of the old man = x + 4

the sum of ages of the other 9 members = y + 36

The age of the young man = z

the average = a

so

z + y + 36 = 10 a ........................(2)

from (1) , (2)

x = z + 36

x + 4 = z + 36 + 4

the age of the old man = the age of the young man + 40

let the set of ages of first 9 people be x and the age of the old person be y,

average (4 years ago)= (x+y)/10

after 4 years, age of 10 people increases by 4 each. that is, 4*10=40

let the age of the young person be z,

therefore, the average of ages now becomes (x+40+z)/10

it is given that the two averages are equal,

therefore, (x+y)/10=(x+40+z)/10

which gives y=z+40

therefore, the new member is younger to the old person by 40 years

Why so big solutions with a, x etc. If by old man vacating the average reduces to 4 year ago average for 10 ppl.. we need not care who replaces him! Just that 10+10+10+10 which happened in 4 years gets back with new member viz. 40!

Let y= Age of the younger member

Let x= CURRENT Age of the old member

*Since both generations are having 10 members each, then the sum of the ages of the current members is equal to the sum of the ages of the old members four years ago.

a+b+c+d+e+f+g+h+i+Y=(a-4)+(b-4)+(c-4)+(d-4)+(e-4)+(f-4)+(g-4)+(h-4)+(i-4)+(X-4)

a+b+c+d+e+f+g+h+i+Y=a+b+c+d+e+f+g+h+i+X-40

By addition property/transposition method:

Y=X-40

.:X-Y=40

Total 10 persons in the committee , avg remains same even after 4 years because of replacing oldest member by youngest member, in order to compensate the avg the age difference between the old and young member is=(years gap * total number of members) = (4*10) = 40.

Just simple 10 ×the difference between =10×4=40

Let the outgoing member's age TODAY be X and the incoming member's age be Y . We will be looking for the value of X - Y .

Further, et the sum of the ages of the 9 constant members of the committee TODAY be A .

Then the sum of their ages 4 YEARS AGO would be A - 36 (each member is younger by 4 years, all nine of them together by (4 x 9) = 36).

And the outgoing member's age 4 YEARS AGO would be X - 4 .

It is given that the average age of the group today, with the new member, is the same as that of 4 years ago, with the old member.

The average age of the group today is $\frac{A + Y}{10}$

The average age of the group 4 years ago was $\frac{(A - 36) + (X - 4)}{10}$ = $\frac{A + X - 40}{10}$

Therefore, $A + Y = A + X - 40$

(the two fractions above are equal, and since the denominators are also equal, the numerators must be equal)

Re-arranging, we get $X - Y = 40$ , Q.E.D.

Let the old member be Z, the young member be A and the sum of the ages of all the other members be Y.

As stated in the questions $\frac{ Y+Z}{10}$ - 4 = $\frac{Y+A}{10}$

Then Y + Z - 40 = Y + A

Therefore, Z - A = 40

• t = ages diffrence between old and new member
• x = todays all members age sum
• $\frac{x-t}{10}$ = $\frac{x-40}{10}$
• 10x-10t = 10x-400
• -10t=-400
• t = 40

Average age of committee at $t_y = 0$ consisting of 9 members of the age $x_1 .. x_9$ and the old member of the age $x_{A,0}$ :

$f(t_y = 0) = \frac{(\sum\limits_{i=1}^{9} x_i) + x_A }{10}$

Average age at $t_y = 4$ after replacement of old member of the age $x_{A,t_y=4}$ by the young member of the age $x_B$ :

$f(t_y=4) = \frac{(\sum\limits_{i=1}^{9} x_i + t_y) + x_B}{10} = \frac{(\sum\limits_{i=1}^{9} x_i) + 9 \cdot t_y + x_B}{10}$

Equal average age at $t_y = 0$ and $t_y = 4$ :

$f(t_y = 0) = f(t_y=4) \implies \frac{(\sum\limits_{i=1}^{9} x_i) + x_{A,0}}{10} = \frac{(\sum\limits_{i=1}^{9} x_i) + 9 \cdot t_y + x_B}{10} \implies x_{A,0} = 9 \cdot t_y + x_B$

At $t_y = 4$ the old member has aged 4 years since $t_y = 0$ , so:

$x_{A,t_y=4} = x_{A,0} + t_y = x_{A,0} + 4$

The absolute difference in age between the old member and the young member:

$|x_{A,t_y} - x_B| = |x_{A,0} + t_y - x_B| = |9 \cdot t_y + x_B + t_y - x_B| = |10 \cdot t_y| = |10 \cdot 4| = \boxed{40}$

That is, how many years younger is the new member than the old member is NOW (not how many years younger is the new member than the old member was four years ago).

Let's let S = the sum of the ages of the unchanged members of the committee now, Y = age of new member, E = age of old member. We can set up the average comparison as follows:

(S + Y)/10 = (S-36 + E-4)/10

Solving for E-Y,

S + Y = S - 36 + E - 4

40 = E - Y

As we can see, the younger member was 40 years younger than the older one when he was replaced today, so the answer is 40.

It should be 36.

Previous average age = $\frac {\displaystyle \sum_{n=1}^{10}a_i}{10} =\frac {\displaystyle \sum_{n=1}^{9}(a_i+4)+b}{10}$
where b=age of new guy. $=\frac {\displaystyle \sum_{n=1}^{9}a_i + 36 + b }{10}$

Simplifying: $\displaystyle \sum_{i=1}^{9}a_i + a_{old}=\displaystyle \sum_{i=1}^{9}a_i + 36 + b$

$a_{old} = 36+b$
Hence, new guy is 36 years younger than retired guy.

At first take any 10 types of ages that should be (20 and above) and then add them. Find out the average. Then add 4 years to all members ages which will increase the total of their ages by 40 years which is your answer.

The correct answer would be 40 because we compare the age of the two in the same time: 36 would be right if he we compare the older persons age at 4 years ago with the younger persons present age.

Eg: let O=Older person 4years ago Let A=the other nine 4years ago Let X=average age Let Y=Younger person present day Let Ot=Older person present day

(O+A)/10 = X "4years ago" (Y+A+4*9)/10=X "Present day" Also Ot=O+4 "Aged 4 years"

Thus X=X

O+A=Y+A+4 9 Ot-4=Y+4 9 Ot = Y+40

Thus the older person today is 40 years older thank the younger man today. Remember you can't compare the 4 years ago age of the older person to the present day age of the younger person. It should be in the same time.

Shouldn't it be 36?

The correct answer is 36, because everyone is multiplying 4 years to all 10 members, but the it should be 9, one is changing

The sum of all ages increases by 40 in 4 years. So to maintain the same average after 4 years we should subtract 40 from the new sum of all ages. Therefore is obvious that the new members age must be 40 years less!

let n=new members age x = average age of one member

since one member left, it is 9x (9 original members). then "n" is added, as it is the new members age which is a variable by itself

9x+n/10 = 10(x-4)/10

n + 9x = 10x - 40

thus n = x-40

the difference would be 40

We don't need to worry about the average (division by 10). The sum of the ages of the 10 members must be the same now and 4 years ago:

Let the sum of their ages 4 years ago be: X1+X2+X3+.........+O
and the sum of their ages now be: (X1+4) +(X2+4)+(X3+4)+........(O+4) where O stands for the oldest member X1+X2+X3+.........+O ≠ (X1+4) +(X2+4)+(X3+4)+........(O+4) ≠ X1+X2+X3+.........+O+40 To make both sides equal, we need to replace O by O - 40 , Let is call it Y (for youngest) X1+X2+X3+.........+O = X1+X2+X3+.........+Y where Y = O - 40

(number of members)X(number of years ago)=(number of years in between the old member and the new member)

10X4=40

Let x represent all members excluding old member and new member

Let y represent old member being replaced

Let z represent new member replacing.

9(x-4)+ (y-4) = 9x + z

9x - 36 + y - 4 = 9x + z

y - 40 = z

OR

y = z + 40

Let the average = x y = the age of the young (new member) subtracted from the age of the old (replaced)

In 4 years, 10x = 9(x+4) + (x+4-y) 10x = 9x+36 + x+4-y y = 40

Suppose the committee's total age 4 years ago is x. By now it would be x + 40. So to make it x again by replacing one member, the new member should be 40 years younger.

I think answer will be 36 if the new guy replaces the old one after the 4years. But if it is at the start like he replaced old guy 4 years ago then answer is 40.

Question is imprecise. It should ask "how many years younger" not "how much younger". The correct answer as written is "40 years", not "40" ("4 decades" is also a correct answer as written).

X = old, Y = young, A = sum others today. So, we have : (A + X - 40)/10 = (A + Y)/10 Hence, X - 40 = Y

Obviously the age of each member of the committee is not relevant! Let assume that all old members have the same age, let it be "b"! Let "a" the common age (and the average) 4 years ago!
b = a + 4 Let n - the age of the newcomer. The sum of ages must be the same now and 4 years ago: 10 x a = 9 x (a + 4) + n ==> n = a - 36 Warning: "a" is the age they had 4 years ago. Nowadays the common age of the members is b = a+4 and: n = b - 40 . Answer: younger with 40 years

Let the age of all the members be denoted as x1,x2,......x10. Now,we have that {x1 + x2 +......+x9 + y}/10 = {(x1 -4) + (x2 - 4)+......+(x10 - 4)}/10 Therefore, Y = X10 + 40 or Y - X10 = 40.So,the age difference is 40 years.

age of commitee 4 years ago = y

age of commitee now = y

age of the old member = x

age of the young member = t

$y= y+40 - x +t$

$0=40-x+t$

$x-t=40$