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# Each of 30 parents chose one of five days from Monday throug

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Manager
Joined: 10 Feb 2011
Posts: 112

Kudos [?]: 380 [1], given: 10

Each of 30 parents chose one of five days from Monday throug [#permalink]

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14 Mar 2011, 16:31
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Difficulty:

75% (hard)

Question Stats:

45% (01:24) correct 55% (01:22) wrong based on 119 sessions

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Each of 30 parents chose one of five days from Monday through Friday to attend parent-teacher conferences. Was at least one of the five days chosen by more than 10 parents?

(1) 15 parents chosen Monday or Tuesday.
(2) Each of the five days was chosen by at least 5 parents.
[Reveal] Spoiler: OA

Kudos [?]: 380 [1], given: 10

Manager
Joined: 24 Nov 2010
Posts: 197

Kudos [?]: 94 [1], given: 7

Location: United States (CA)
Concentration: Technology, Entrepreneurship
Schools: Ross '15, Duke '15

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14 Mar 2011, 17:49
1
KUDOS
_ _ _ _ _
M T W T F

a. 15 parents chose M/T.. its possible that 8 chose M and 7 chose T ---> insufficient

b. each day was chosen by atleast 5 --> 5*5 = 25, the rest 5 can be assigned to any one of the days making the number = 10. but it cannot be more than 10. -->sufficient

ans b

Kudos [?]: 94 [1], given: 7

Math Expert
Joined: 02 Sep 2009
Posts: 42528

Kudos [?]: 135176 [1], given: 12669

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14 Mar 2011, 17:50
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banksy wrote:
284. Each of 30 parents chose one of five days from Monday through Friday to attend parent-teacher conferences. Was at least one of the five days chosen by more than 10 parents?
(1) 15 parents chosen Monday or Tuesday.
(2) Each of the five days was chosen by at least 5 parents.

Each of 30 parents chose one of five days from Monday through Friday to attend parent-teacher conferences. Was at least one of the five days chosen by more than 10 parents?

(1) 15 parents chosen Monday or Tuesday --> we can have YES answer, (for example: 14-1-5-5-5) as well as NO answer, (for example: 10-5-5-5-5). Not sufficient.

(2) Each of the five days was chosen by at least 5 parents --> even if we distribute parents: 5-5-5-5-10, we still won't have any day which was chosen by more than 10 parents (minimizing # of parents who chose each of 4 days to maximize # of parents who chose the fifth day). Sufficient.

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Kudos [?]: 135176 [1], given: 12669

Senior Manager
Joined: 27 May 2012
Posts: 426

Kudos [?]: 90 [0], given: 487

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22 Aug 2013, 00:10
Bunuel wrote:
banksy wrote:
284. Each of 30 parents chose one of five days from Monday through Friday to attend parent-teacher conferences. Was at least one of the five days chosen by more than 10 parents?
(1) 15 parents chosen Monday or Tuesday.
(2) Each of the five days was chosen by at least 5 parents.

Each of 30 parents chose one of five days from Monday through Friday to attend parent-teacher conferences. Was at least one of the five days chosen by more than 10 parents?

(1) 15 parents chosen Monday or Tuesday --> we can have YES answer, (for example: 14-1-5-5-5) as well as NO answer, (for example: 10-5-5-5-5). Not sufficient.

(2) Each of the five days was chosen by at least 5 parents --> even if we distribute parents: 5-5-5-5-10, we still won't have any day which was chosen by more than 10 parents (minimizing # of parents who chose each of 4 days to maximize # of parents who chose the fifth day). Sufficient.

Missed the word more and got the wrong answer .
If it was " Was at least one of the five days chosen by at least 10 parents ?" then the answer could be E

Then
(1) 15 parents chosen Monday or Tuesday --> case 1: 10, 5 answer is yes ( showing Monday, Tuesday only )
--------------------------------------------------- -- case 2: 8,7 answer is no
2) Each of the five days was chosen by at least 5 parents -->

case1: 6 6 6 6 6 ,Answer No
case2: 10 5 5 5 5,Answer Yes

1+2

10 5 5 5 5, Answer Yes
8 7 5 5 5, Answer No

Still not sufficient

with the word " more than 10 parents " then of course the answer is B
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- Stne

Kudos [?]: 90 [0], given: 487

Current Student
Joined: 20 Jan 2014
Posts: 175

Kudos [?]: 73 [0], given: 120

Location: India
Concentration: Technology, Marketing
Re: Each of 30 parents chose one of five days from Monday throug [#permalink]

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25 Sep 2014, 06:43
Each of 30 parents chose one of five days from Monday through Friday to attend parent-teacher conferences. Was at least one of the five days chosen by more than 10 parents?

(1) 15 parents chosen Monday or Tuesday.
(2) Each of the five days was chosen by at least 5 parents.

The catch here is at least more than 10 parents/.
From B we can find that max 10 can be assigned to one day but not in any condition we can assign more than 10 to any day. Hence B is the answer.
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Kudos [?]: 73 [0], given: 120

Re: Each of 30 parents chose one of five days from Monday throug   [#permalink] 25 Sep 2014, 06:43
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# Each of 30 parents chose one of five days from Monday throug

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