At a certain corporation, the planning committee has 23 members

solve using 2x2 matrix
---P----NP----total
F--- ----- b---20
NF-- a---- ----
total--23-- ---- --
given that a+b = 21
so
20-b= 23-a
we get
a-b=3
a+b=21
solve a = 12 and b = 9

---P----NP----total
F---11 ----- 9---20
NF-- 12---- ----
total--23-- ---- --

OPTION B 11

Can anyone please help me understand what I am missing here?

It would be

23 - x + 20 - x = 21
x=11
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If you take 23 + 20 - BOTH = 21, you will find that BOTH = 22. The trick on this problem the testmakers want you to fall for is not accounting for the redundant count.

the group of 20 finance members have BOTH members and the group of 23 planning committee members have BOTH members counted for. Therefore the answer you receive here counts the BOTH amount twice. You have to divide by 2 to get the actual answer.

This is an official GMAT question. I wasted a solid 4.6 minutes on this and got it wrong and rest of the mock went under the bus! Am I really bad reading or is this a poorly worded question? Would be grateful to have your opinions on this Bunuel KarishmaB
I interpreted this text as: 21 people are members of only one of the two committees, i.e., they are either members of planning committee or members of finance committee. Since only 20 members belong to finance committee, 21 people cannot be the members of only finance committee (since 21>20), therefore, these 21 must be members of ONLY the planning committee.

Finally, I had deduced these numbers:
N(P) = # of Planning committee = 23 (given)
N(P only) = # Planning committee ONLY = 21
N(F) # of Finance committee = 20 (given)

And hence, # of members in both committees must be = N(P)-N(P only) = 23-21 = 2 (NOT AN ANSWER CHOICE)

I even made this table and checked:
­

KarishmaB How would you solve it?

One easy way is discussed by Bunuel above.

Another is using the concept of instances and people. A person in only one set adds one instance to the total instances whereas a person in 2 sets adds 2 instances to the total instances (and a person in 3 sets adds 3 instances)
23 + 20 = 43 is the total number of instances. 21 of these come from people in only one set and account for 21 instances.
22 instances are left unaccounted for now. These must come from people in both sets and since each person here contributes 2 instances, there must be 11 people in both sets.

Answer (B)

Here are some interesting sets questions:
https://youtu.be/rnqM1YCaLp8
https://youtu.be/18ZDLj7KMkc
https://youtu.be/1RZdJTKCDYs

1) Let a be the no. of members in only one team and b in both teams. We have to find b.
a+2b=20+23
2b=43-21
b=11. Hence, answer.

2) Let a be the number of members in only planning team and b only in finance.
a+b=21
23-a=20-b
a-b=3
Solving these equations we get, 2a=24 a=12. Hence, 23-12=11. Answer B

3) Use options.
A) we get 20+17 not equal to 21
B)11 we get 12+9=21. Hence, the answer.