Just set up scenarios.

Rules: M>T

Your goal: Fill up Monday through Thursday with 20 parents while Friday has 0.

Statement 1)

Additional rule: No day is greater than 5, which means each day is equal to or less than 5.

So if monday=5, Tuesday=4, Wednesday=5, Thursday=5 -->Friday = 1 in order for total to be 20.

Sufficient. At least one parent has to choose friday for this to work.

Statement 2)

Additional rule: M>W

so if Monday=5, Tuesday=4, Wednesday=4, Thursday=5 --> Friday =2 in order for total to be 20.

Sufficient. At least one parent has to choose friday for this to work.

D

I was thinking A.

For B

You can use a number higher than 5 and make a number of different combinations of 20.

IMO A

to minimize on friday, maximize for other days.


1)In order from monday to friday

5 5 5 5 0
now On monday the number is greater than on tuesday

5 4 5 5 1, at least one parent will chose Friday

SUFFICIENT

2) 6 5 5 4 0
and
6 5 2 1 6

INSUFF

answer: A

my ans is A.

Hi,

We have the following facts.

- 20 parents
- 5 days to choose from (Mo, Tu, We, Th, Fr)
- Each parent will select maximum 1 day
- More people chose "Mo" than "Tu".
- No restriction on the parent/day distribution or the maximum parents per day!
- There is a possibility that some days were not chosen by any parents (since the question already asks whether Friday was selected).

Now, if i interpret statement 1, "None of the five days was chosen by more than 5 parents", then IMO it will mean that:

- More than 5 parents didn't choose any of these five days
- More than five means at least 6.
- That leaves us with a maximum of 14 parents.
- 14 Parents will choose from 5 days.
- Option 1: We can distribute 14 parents accross 4 days without including Friday and still conform to the fact that more poeple chose "Mo" than "Tu": Mo (6), Tu (5), We (2), Th (1), Fr(0)
- Option 2: We can also eliminate "Th" and add 1 more vote to "We": Mo (6), Tu (5), We (3), Th (0), Fr (0)
- Option 3: we can move that vote to Friday instead: Mo (6), Tu (5), We (1), Th (1), Fr (1)
- So, my conclusion is that statement 1 is not sufficient.

For statement 2, more people also chose Monday than Wednesday, then we will distribute 20 parents across 5 days:
- Option 1: Mo (10), Tu (5), We (3), Th (2), Fr(0) -> Friday can be ignored
- Option 2: Mo (10), Tu (5), We (3), Th (1), Fr(1) -> Friday can be selected

.. then again.. statement 2 is not sufficient.

Combining both statements will tell us that at most 14 parents will chose from 5 days and Monday will be chosen more than Tuesday and more than Wednesday separately .. not more than both of them combined. So, options 2 and 3 from statement 1 can still give us the same result.


Hence, IMO it should be E.

Each of 20 parents chose one of five days from Monday through Friday to attend parent-teacher conferences. If more parents chose Monday than Tuesday, did at least one of the parents choose Friday?
(1) None of the five days was chosen by more than 5 parents.
(2) More parents chose Monday than Wednesday.


OA is A

The reason D is a mistake is because you were still considering statement 1 when working with statement 2.. you need to consider each one individually.. and judge each one on it's own merits!

Yup. This is a very common mistake made. I used to make this same one when i used condition 1 to evaluate condition 2 instead of evaluating it independently and ended up with different answers. careful!!