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Each of 20 parents chose one of five days from Monday

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Each of 20 parents chose one of five days from Monday [#permalink]

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New post 09 Oct 2008, 04:16
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
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Re: math--five days [#permalink]

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New post 09 Oct 2008, 07:00
The answer is (C)

Day: M T W R F
chosen by # of ppl 4 4 4 4 4 (evenly distributed)

S1: 5 5 5 5 0
OR it could be 4 4 4 4 4 (multiple possibilities, so Statement 1 is insufficient)

S2: 10 4 2 2 2
OR it could be 10 4 1 1 4 (multiple possibilities, so statement 2 is insufficient)

S1+ S2 5 4 3 4 4
OR it could be 5 4 4 3 4 etc (lots option)
but whatever option you select, theres always atleast one person who selected Friday. So S1+S2 sufficient.

Hope that helps.
Need more explanation? let me know...
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Re: math--five days [#permalink]

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New post 09 Oct 2008, 07:51
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Answer is (A) :twisted:

Dont forget: Mon > Tue is GIVEN.

1) no more than 5 on any days
Therefore, maximum number can be assigned on all other days except Friday is 19. Then, we know there will be at least 1 assigned to Friday. (SUFFICIENT)

5, 4, 5, 5, 1 = Yes (SUFFICIENT)

2) Mon > Wed
This means Mon > Tues > Wed

10, 9, 1, 0, 0 = No
9, 8, 2, 0, 1 = YES (INSUFFICIENT)

Therefore, the answer is A :P
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Re: math--five days [#permalink]

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New post 09 Oct 2008, 10:09
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albany09 wrote:
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.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.


Key to the solution: Try to allocate as many parents to Mon, Tue, Wed, and Thu and as possible. Or, try to minimize parents in Fri.

(1) Since the maximum parents allowed is 5 parents per day.
Mon = 5
Tue = 4 ---------- Since Mon > Tue
Try to minimize parents in Friday.
Wed = 5
Thu = 5
The minimum parents in Fri = 1.
Even though we try our best to make Fri = 0, the condition given here make Fri = 1.
So, at least one of the parents must choose Friday. Sufficient.

(2) Mon > Wed
Mon > Tue ---------- Given from the main question.

Case 1:
Mon = 17
Tue = 2
Wed = 1
Thu, Fri = 0.

Case 2:
Mon = 16
Tue = 2
Wed = 1
Thu = 0
Fri = 1.

Two answers. Insufficient.

The answer is A.
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Re: Each of 20 parents chose one of five days from Monday [#permalink]

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New post 28 Oct 2016, 19:08
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Re: Each of 20 parents chose one of five days from Monday   [#permalink] 28 Oct 2016, 19:08
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