Average Numbers

[mksreeram]

If the average of 6 numbers is 20, how many of the numbers are equal to 21?

(1) Three of the numbers are equal to 19
(2) Three of the numbers are less than 21

[gmatrant]

is the answer D

[sudhir3127]

IMO E

[parallel_chase]

I think the answer is E.


Statement I

3 numbers are 19 each

remaining 3 can be 21, 22, 20 or 23, 20, 20

Insufficient.

Statement II

3 numbers are less than 21, remaining 3 can be any number, endless possibilities!!!

Insufficient


Combining I & II
Three numbers are 19, three numbers are less than 21, still dont know anything about remaining 3.

Hence Insufficient.


E is the answer.

Whats the OA?

[pepeprepa]

I think it is C.
I agree with statements 1 and 2 alone are insufficients.


(1) Three of the numbers are equal to 19
(2) Three of the numbers are less than 21

19 19 19 X Y Z
We must choose X,Y,Z to get an average of 20.
If X=21 Y=21 Z=21 the average is 20, we agree on that.

If X=22 we have to reach 120. We already have 19*3+22=79
So Y+Z=41
We can see that if you choose 22 for one of the numbers you therefore have to choose an integer less than 21, and you contradict the hypothesis.

If X=21
We have 19*3+21=78
So Y+Z=42
That is the same reason, to get 42 either 21 21 or 20 22 and you are wrong because you choose a number less than 21.

So C

[parallel_chase]

if you look at both the statements, arent they saying the same thing?

3 numbers are 19
3 numbers are less than 21


3 numbers less than 21 are 3 numbers equal to 19 (if this answer is E)

or

3 numbers less than 21 apart from 3 numbers that are 19 (if this answer isC)

We dont know thats why E is the answer.

Let me know what you think.

[pepeprepa]

I believe that the question tells us when you consider 1)and2):
You have 3 of the 6 integers which are under 21 and they are 19 each

Your second hypothesis is impossible to get an average at 20.

[parallel_chase]

I believe that the question tells us when you consider 1)and2):
You have 3 of the 6 integers which are under 21 and they are 19 each

Your second hypothesis is impossible to get an average at 20.

Nice work Pepeprepa, answer is C with 3 numbers 21 each.

[sudhir3127]

yeah i think i shud be C.

i missed the second point which says the number of numbers less than 21 ( 19 as said by first statement ) shud be only 3.

so in order to meet the average of 20... the next 3 numbers shud be 21 21 and 21...

thanks pepeprepa Nice one ... ( IIMB waiting!!!!!!!!)

[pepeprepa]

Yep sudhir IIMB waiting

[nitya34]

whats the OA?
I got C

[doclkk]

If the average of 6 numbers is 20, how many of the numbers are equal to 21?

(1) Three of the numbers are equal to 19
(2) Three of the numbers are less than 21

Answer must be C

AD/BCE

Statement 1. PICK NUMBERS 19,19,19 20, 22, 21 or 21,21,21

CLEARLY NOT A

Statement 2. Pick numbers 19, 18, 17, 21, 22, 23 or 3 19's and 3 21's.

CLEARLY NOT B

TOGETHER

19, 19, 19 and 21, 21, 21.

There's no other way.

Answer is C.

[cool_sni]

Hello doclkk,
How did you arrive with

19, 19, 19 and 21, 21, 21.

Why can't it be
19,19,19 and 20,20,23
or
19,19,19 and 21,21,22