It is currently 03 Mar 2021, 00:20 |

Customized

for You

Track

Your Progress

Practice

Pays

- Mar
**08**### Free GRE Prep Hour

08:00 PM CDT

-09:00 PM CDT

Strategies and techniques for approaching featured GRE topics, 1-hour of live, online instruction. Join March 8th - Mar
**09**### It's time to save on Magoosh!

09:00 AM AKDT

-10:00 AM AKDT

Take 20% off the plan of your choice, now through midnight on 3/9: Code GMAT20Sale - Mar
**11**### Free Workshop for the GRE

07:00 PM CDT

-08:30 PM CDT

Join a Free GRE Workshop with Kaplan Thursday, March 11th at 7 PM Eastern - Mar
**12**### How to Approach GMAT Combinatorics: Key Principles

08:00 PM AKDT

-09:00 PM AKDT

Combinatorics is the art of counting. You’ll need to understand this art to do well on the GRE and GMAT.

Retired Moderator

Joined: **07 Jun 2014 **

Posts: **4804**

WE:**Business Development (Energy and Utilities)**

Set A has 50 members and set B has 53 members. At least 2 of
[#permalink]
03 Jul 2016, 16:18

4

Expert Reply

1

Bookmarks

Question Stats:

Set A has 50 members and set B has 53 members. At least 2 of the members in set A are not in set B. Which of the following could be the number of members in set B that are not in set A ?

Indicate all such numbers.

A. 3

B. 5

C. 13

D. 25

E. 50

F. 53

_________________

Indicate all such numbers.

A. 3

B. 5

C. 13

D. 25

E. 50

F. 53

Practice Questions

Question: 13

Page: 121-122

Question: 13

Page: 121-122

_________________

Sandy

If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

Re: Set A has 50 members and set B has 53 members. At least 2 of
[#permalink]
13 Aug 2017, 06:04

11

Expert Reply

Explanation

Use the Double Matrix Method.

This technique can be used for most questions featuring a population in which each member has two characteristics associated with it (aka overlapping sets questions).

Here, we have a population of numbers, and the two characteristics are:

- In set A or NOT in set A

- In set B or NOT in set B

With the given information, we can set up our Matrix as follows:

Our goal is to determine which values can go in the bottom left corner (marked with a star)

Let's start by seeing what happens if we minimize the number of values that are in the TOP RIGHT corner (the number of values that are in set A BUT NOT in set B)

Since at least 2 values must be in this box, let's see what happens when there are 2 numbers there.

We get:

In this case, there are 5 numbers that are in set B but not in set A (bottom left box)

Now let's see what happens when we place a 3 in the box representing the number of values that are in set A BUT NOT in set B (i.e., the TOP RIGHT corner box):

In this case, there are 6 numbers that are in set B but not in set A (bottom left box)

Now let's see what happens when we place a 4 in the box representing the number of values that are in set A BUT NOT in set B (i.e., the TOP RIGHT corner box):

In this case, there are 7 numbers that are in set B but not in set A (bottom left box)

Now let's see what happens when we MAXIMIZE the box representing the number of values that are in set A BUT NOT in set B (i.e., the TOP RIGHT corner box):

The biggest value that can go here is 50, since numbers in the top row of boxes must add to 50.

In this case, there are 53 numbers that are in set B but not in set A (bottom left box)

So, the possible numbers that can be in the bottom left box range from 5 to 53 INCLUSIVE

Answer: B, C, D, E, F

RELATED VIDEO

_________________

Use the Double Matrix Method.

This technique can be used for most questions featuring a population in which each member has two characteristics associated with it (aka overlapping sets questions).

Here, we have a population of numbers, and the two characteristics are:

- In set A or NOT in set A

- In set B or NOT in set B

With the given information, we can set up our Matrix as follows:

Our goal is to determine which values can go in the bottom left corner (marked with a star)

Let's start by seeing what happens if we minimize the number of values that are in the TOP RIGHT corner (the number of values that are in set A BUT NOT in set B)

Since at least 2 values must be in this box, let's see what happens when there are 2 numbers there.

We get:

In this case, there are 5 numbers that are in set B but not in set A (bottom left box)

Now let's see what happens when we place a 3 in the box representing the number of values that are in set A BUT NOT in set B (i.e., the TOP RIGHT corner box):

In this case, there are 6 numbers that are in set B but not in set A (bottom left box)

Now let's see what happens when we place a 4 in the box representing the number of values that are in set A BUT NOT in set B (i.e., the TOP RIGHT corner box):

In this case, there are 7 numbers that are in set B but not in set A (bottom left box)

Now let's see what happens when we MAXIMIZE the box representing the number of values that are in set A BUT NOT in set B (i.e., the TOP RIGHT corner box):

The biggest value that can go here is 50, since numbers in the top row of boxes must add to 50.

In this case, there are 53 numbers that are in set B but not in set A (bottom left box)

So, the possible numbers that can be in the bottom left box range from 5 to 53 INCLUSIVE

Answer: B, C, D, E, F

RELATED VIDEO

_________________

Re: Set A has 50 members and set B has 53 members. At least 2 of
[#permalink]
18 Sep 2018, 11:05

4

Attachment:

Untitled.png [ 8.87 KiB | Viewed 18342 times ]

another slightly different approach with venn diagram.

clearly from the given conditions, A-B must be greater than or equal to 2. as such, their intersection must be less than or equal to 48. consequently B-A must be greater than 5 and any value satisfying this interval is correct. needless to say, the upper limit is 53, since n(B)=53.

Re: Set A has 50 members and set B has 53 members. At least 2 of
[#permalink]
19 Aug 2019, 12:40

1

It seems like one could also use a hybrid approach of the ones shown above, see attached (but maybe not).

We would start with A not B is >=2. Given A is 50, it would follow that A and B must be <=48.

Then, since we know A and B is <=48, and it's given that B is 53, then B not A must be be >=5.

It seems to make sense.

We would start with A not B is >=2. Given A is 50, it would follow that A and B must be <=48.

Then, since we know A and B is <=48, and it's given that B is 53, then B not A must be be >=5.

It seems to make sense.

hybridVenn.png [ 74.08 KiB | Viewed 13515 times ]

Re: Set A has 50 members and set B has 53 members. At least 2 of
[#permalink]
22 May 2020, 05:37

1

Had to read this like 3 times to understand the wording...my brain just refused to understand, at first.

Also, took a different approach because I kinda had to visualize this to make sense of what it was saying with the "not A" and the "not B". Here's how I did this.

If at least 2 are in ONLY A, then 48 are in BOTH, and 5 are ONLY B...I did a quick table to visualize

ONLY A BOTH ONLY B

2 48 53-48=5

3 47 53-47=6

... ... ...

50 0 53-0=53

So, B would have between 5 and 53 members not in A (or only in B).

Also, took a different approach because I kinda had to visualize this to make sense of what it was saying with the "not A" and the "not B". Here's how I did this.

If at least 2 are in ONLY A, then 48 are in BOTH, and 5 are ONLY B...I did a quick table to visualize

ONLY A BOTH ONLY B

2 48 53-48=5

3 47 53-47=6

... ... ...

50 0 53-0=53

So, B would have between 5 and 53 members not in A (or only in B).

Re: Set A has 50 members and set B has 53 members. At least 2 of
[#permalink]
05 Jun 2020, 05:16

1

This is a tricky question to grasp, indeed. Needs a lot of reasoning and inference than mathematical calculation.

Set A: 50 members

Set B: 53 members

At least 2 members in Set A but not in Set B. That means 2 to 50 members in A are not in B. The remaining members would be both in A and B.

So (50-2) to (50-50) are common members in A and B. (48 to 0)

If 0 to 48 members in B are in A, then (53-0) to (53 - 48) members will be in B but NOT in A. So the range is 53 to 5.

5 to 53 members can be in Set B but not in A.

Only 3 from the options can be excluded from this range.

Set A: 50 members

Set B: 53 members

At least 2 members in Set A but not in Set B. That means 2 to 50 members in A are not in B. The remaining members would be both in A and B.

So (50-2) to (50-50) are common members in A and B. (48 to 0)

If 0 to 48 members in B are in A, then (53-0) to (53 - 48) members will be in B but NOT in A. So the range is 53 to 5.

5 to 53 members can be in Set B but not in A.

Only 3 from the options can be excluded from this range.

Re: Set A has 50 members and set B has 53 members. At least 2 of
[#permalink]
12 Dec 2020, 12:51

1

What about the case where no numbers in A and B overlap?

For example, Set A could contain numbers 1-50 and set B could contain numbers 51-103 for all we know. It fulfills the requirement "At least 2 of the members in set A are not in set B" because in this case all 50 numbers in set A are not in B. In that case, "number of members in set B that are not in set A" are numbers 51-103 so it should be A,B,C,D,E,F.

Official answer not including A but don't understand why since it is possible for B to have no intersecting numbers from A.

For example, Set A could contain numbers 1-50 and set B could contain numbers 51-103 for all we know. It fulfills the requirement "At least 2 of the members in set A are not in set B" because in this case all 50 numbers in set A are not in B. In that case, "number of members in set B that are not in set A" are numbers 51-103 so it should be A,B,C,D,E,F.

Official answer not including A but don't understand why since it is possible for B to have no intersecting numbers from A.

gmatclubot

Moderators:

Multiple-choice Questions — Select One or More Answer Choices |
||

## Hi Guest,Here are updates for you:## ANNOUNCEMENTSEmail Newsletter is Here!
Learn More | Subscribe New Features are Here!
See the full details here ## GRE Prep Club REWARDS |