What is the average weight of the three new team members who were rece

What is the average weight of the three new team members who were recently included in the team?

(1) The average weight of the team increases by 20 kg
(2) The three new men substitute the earlier members whose weights were 64 kg, 75 kg and 66 kg.

S=total weight of the team members; n=number of members; a+b+c=total weight of new team members
we need to find (a+b+c)/3
1) S/n+20= (S+a+b+c)/n or
S/n+20= (S+a+b+c)/(n+3) or anything else; it depends on how the team composition changed with coming and goings. Insufficient

2) The second information is also insufficient by itself because we need to know how the average weight is affected by replacing three members and number of members or
initial average weight.
Combining the two, we have S/n+20= (S-64-75-66+a+b+c)/n
S/n+20 = S/n - 205/n + (a+b+c)/n
20= (a+b+c)/n-205/n
We are left with two unknowns and one equation.
So the answer is E, IMO

Let the initial number of team members be N and their average weight = w
—> w = sum/N
—> Sum of all weights (S) = wN

(1) The average weight of the team increases by 20 kg
New weight = w + 20
New number of team members = N + 3
Let the weights of 3 new members = a,b & c
—> w + 20 = (S + a+b+c)/(N+3)
—> a + b + c = 20N + 3w + 60 —> Insufficient

(2) The three new men substitute the earlier members whose weights were 64 kg, 75 kg and 66 kg.

Since, there is no information on the new weights. We cannot find the average weight or any relation —> Insufficient

Combining (1) & (2),
—> Since replaced, total number will remain same = N and new average weight is increased by 20

New sum of weights = S + (a+b+c) - 205

—> w + 20 = (S + (a+b+c) - 205)/N
—> wN + 20N = S + (a+b+c) - 205 = wN + (a+b+c) - 205
—> a+b+c = 20N + 205
We don’t know the value of N —> Insufficient

IMO Option E

Posted from my mobile device

The only information we know from the question stem is that three new team members have been included in the team.
We are to determine the average weight of the three new members.

From Statement 1: we know that the average weight of the team increased by 20kg.
Not sufficient. We need to know the number of team members, and the average weight prior to the inclusion of the three new members, and there is no way of determining these extra pieces of information based on statement 1 and the information provided in the question stem.

From statement 2: we know the weights of the three previous members that the new members have replaced are 64kg, 75kg, and 66kg. From the information provided in statement 2, we need the total number of team members before we can arrive at a unique average weight for the three new team members. Statement 2 alone is therefore not sufficient.

1+2:
Statement 1 requires the number of team members and the average weight previously before the introduction of the three new members. Combining both statements does not provide us with the extra information required by statement 1.
Statement 2 requires the total number of team members in order to narrow down to the average weight of the three new team members. Combining both statements does not provide the information required by statement 2.

Therefore, taking both statements together is still not sufficient to answer the question.

The answer is E in my view.

What is the average weight of the three new team members who were recently included in the team?

Average of weight of the three new men = ?
Question is basically asking what is the sum of weights of three men.

(1) The average weight of the team increases by 20 kg
The team can have any number of members.
For example: There are 10 members with the average weight of 50 kg. After 3 new men are added(assuming that they are not replacing existing members) to the team the average becomes 70 kg.

Thus, difference in total weight of team = 70*13 - 50*10 = 410. Hence average weight of 3 new men = 410/3 kg

If there are 10 members with the average weight of 100 kg. After 3 new men are added(assuming that they are not replacing existing members) to the team the average becomes 120 kg.

Then, difference in total weight of team = 120*13 - 100*10 = 560. Hence average weight of 3 new men = 560/3 kg

INSUFFICIENT.

(2) The three new men substitute the earlier members whose weights were 64 kg, 75 kg and 66 kg.
Sum of replaced members = 64 + 75 + 66 = 205 kg.

If there are 10 members with the average weight of 50 kg. After 3 new men replace existing members of the team, weight of 7 members = 50*10 - 205 = 295 kg.
Then, total weight of 3 new men = Average of new team * 10 - 295 = ?

INSUFFICIENT.

Together 1 and 2

If there are 10 members with the average weight of 100 kg. After 3 new men replace existing members of the team the average becomes 120 kg.
Now, weight of 7 members = 100*10 - 205 = 795 kg.
Then, total weight of 3 new men = 120*10 - 795 = 405 kg.

OR

If there are 10 members with the average weight of 50 kg. After 3 new men replace existing members of the team the average becomes 70 kg.
Now, weight of 7 members = 50*10 - 205 = 295 kg.
Then, total weight of 3 new men = 70*10 - 295 = 405 kg.

SUFFICIENT.

IMO Answer C.
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What is the average weight of the three new team members who were recently included in the team?

(1) The average weight of the team increases by 20 kg
We do not know the average weight of the original team members before the increase nor their weights therefore, we can't calculate the total weight of the three new team members to get their average weight.
insufficient

(2) The three new men substitute the earlier members whose weights were 64 kg, 75 kg and 66 kg.
We do not have information of the additional men.
insufficient

Together is what we need. We know the average of the 3 earlier members and we know the increased average weight when the 3 men are substituted.

Therefore, C

\(\frac{(n_1+n_2+n_3)}{3}\)=???

Statement1: The average weight of the team increases by 20 kg.

Let's say that Average weight of the team is
--> \(\frac{(t_1+t_2+...t_n)}{n}\)=A
--> (n- # of people in the team)

if we add 3 people to the team,

--> \(\frac{(t_1+t_2+...t_n+n_1+n_2+n_3)}{(n+3)}\)=A+20

An+\(n_1+n_2+n_3\)=An+3A+20n+60

\(n_1+n_2+n_3\)=3A+20n+60

Clearly insufficient.

Statement2: The three new men substitute the earlier members whose weights were 64kg, 75kg and 66kg.
"...earlier members whose weights were 64kg, 75 kg and 66kg" -doesn't mean that weights of them match those of new three men.
Insufficient

Taken together 1&2,

\(\frac{(An-205+n_1+n_2+n_3)}{n}\)=A+20

An-205+\(n_1+n_2+n_3\)=An+20n

\(n_1+n_2+n_3\)=20n+205
Still no info about what n is.

Insufficient

The answer is E.

What is the average weight of the three new team members who were recently included in the team?

(1) The average weight of the team increases by 20 kg
(2) The three new men substitute the earlier members whose weights were 64 kg, 75 kg and 66 kg.[/quote]

Answer :

Question asks for the Arithmetic Mean Weight of 3 newly inducted team members.

From (1), it gives us only increase in average weight after induction. We don't know -

- what is the earlier total weight.
- how many members are there
Thus INSUFFICIENT.

From (2), we only know the members whose mentioned weights are substituted. We dont know whether, substituted weights are equal. INSUFFICIENT.

Combining both is INSUFFICIENT. Thus correct answer is E

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