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# If a total of 84 students are enrolled in two sections of a

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SVP
Joined: 16 Oct 2003
Posts: 1766
If a total of 84 students are enrolled in two sections of a [#permalink]

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02 Jul 2004, 22:28
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Difficulty:

25% (medium)

Question Stats:

79% (00:36) correct 21% (00:39) wrong based on 77 sessions

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If a total of 84 students are enrolled in two sections of a calculus course, how many of the 84 students are female?

(1) $$\frac{2}{3}$$ of the students in Section 1 are female.
(2) $$\frac{1}{2}$$ of the students in Section 2 are male.
[Reveal] Spoiler: OA

Last edited by Bhai on 03 Jul 2004, 08:05, edited 1 time in total.
Senior Manager
Joined: 21 Mar 2004
Posts: 445
Location: Cary,NC

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03 Jul 2004, 00:25
question looks incomplete, but i guess "sufficient" enough to derive( guess my answer as C.

- ash
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ash
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SVP
Joined: 16 Oct 2003
Posts: 1766

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03 Jul 2004, 08:06
ashkg wrote:
question looks incomplete, but i guess "sufficient" enough to derive( guess my answer as C.

- ash

Sorry about that. I changed the question.
Manager
Joined: 10 Jun 2004
Posts: 81
Re: DS-3 [#permalink]

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03 Jul 2004, 08:18
Bhai wrote:
12. If a total of 84 students are enrolled in two sections of a calculus course, how many of the 84 students are female?

(1) 2/3 of the students in Section 1 are female.

(2) 1/2 of the students in Section 2 are male

I choose E.

For both 1,2 We don't have the number of students enrolled in one section at least. Therefore, both are insuffecient.
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Senior Manager
Joined: 21 Mar 2004
Posts: 445
Location: Cary,NC
Re: DS-3 [#permalink]

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03 Jul 2004, 09:39
dr_sabr wrote:
Bhai wrote:
12. If a total of 84 students are enrolled in two sections of a calculus course, how many of the 84 students are female?

(1) 2/3 of the students in Section 1 are female.

(2) 1/2 of the students in Section 2 are male

I choose E.

For both 1,2 We don't have the number of students enrolled in one section at least. Therefore, both are insuffecient.

Did you try C ?

if section 1 has x students, section 2 has 84-x students.
now work on the ratios using A and B.

Its a linear equation with one variable.

C should be sufficient.
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ash
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Manager
Joined: 10 Jun 2004
Posts: 81
Re: DS-3 [#permalink]

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03 Jul 2004, 09:46
ashkg wrote:
Did you try C ?

if section 1 has x students, section 2 has 84-x students.
now work on the ratios using A and B.

Its a linear equation with one variable.

C should be sufficient.

You got me ash, I should look deeply to the question next time.
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Intern
Joined: 17 Dec 2003
Posts: 19
Location: Russia

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05 Jul 2004, 15:36
To ashkg,

Ans. E

We should know at least the number of students enrolled in one of the sections.

Let's say there are 42 students in each section, then the number of female in both equals to 49, whereas if there are 78 in the first and 6 in the second, then the number of female in both is 55.

My regards
Senior Manager
Joined: 21 Mar 2004
Posts: 445
Location: Cary,NC
Re: DS-3 [#permalink]

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05 Jul 2004, 15:47
hi there,

yes, I think the ans should be E.

the linear equation doesnt get solved, bcoz there is nothing to equate to !

bummed
- ash

dr_sabr, u were right about this one ........i should take a closer look at the problems !
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ash
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Manager
Joined: 11 Jan 2005
Posts: 57
Location: Mexico City

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15 Mar 2005, 09:05
i have it down to c or e and I will got with E.

I cannot think of a way to figure it out. The only thing i can thing of is that the number 84 has some propretie sthat I am not thinking about.

I tried dividing the two groups as 48 and 48 which works getting 32 females and 24 females equalling 56

then i used 60 and 24 getting 40 females and 20 females which of course gives a different answer.

has to be E.
Intern
Joined: 23 Oct 2003
Posts: 4
Location: Atlanta

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15 Mar 2005, 10:28
Your right. The OA is E.
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5021
Location: Singapore

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15 Mar 2005, 11:21
It's (E). There's no way we can find out.

Let's assume there are n students in section 1, then there will be 84-n students in section 2.

Using (1), we're told 2/3n = number of female students in section 1.
not sufficient

Using (2), we're told 1/2(84-n) = number of female students in section 2
not sufficient

Using (1)+(2),
Doesn't help us either, we still cant' find out what's n.

So E it is.
Manager
Joined: 15 Feb 2005
Posts: 245
Location: Rockville

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15 Mar 2005, 14:33
Yea E it is we havent been told the class sizes
SVP
Joined: 29 Aug 2007
Posts: 2457
Re: DS - Percentages / Sets - OG 12 [#permalink]

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31 Mar 2009, 20:14
mrsmarthi wrote:
If a total of 84 students are enrolled in two sections of a calculus course, how many of the 84 students are female?

1) 2/3 of the students in Section 1 are female.
2) 1/2 of the students in Section 2 are male.

total = 84
students in Section 1 = n
students in Section 2 = 84 - n

1: female = 2n/3
male = n - (2n/3) = n/3 ......................nsf

2: male = (84-n)/2
female = (84-n)/2 ...............nsf

1&2: doesnot give any solution for n..... nsf...

E.
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Manager
Joined: 19 Aug 2006
Posts: 233
Re: DS - Percentages / Sets - OG 12 [#permalink]

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31 Mar 2009, 20:18
I also think it's E.

x+y=84

stmnt 1 - 2/3x - female, 1/3x - male
not suffic.

stmnt 2 - 1/3y - male, 2/3y - female
not suffic.

Combining them would still result in x+y=84 - not suffic.
Current Student
Joined: 13 Jan 2009
Posts: 353
Location: India
Re: DS - Percentages / Sets - OG 12 [#permalink]

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02 Apr 2009, 15:59
E we dont know how sections are divided.
Re: DS - Percentages / Sets - OG 12   [#permalink] 02 Apr 2009, 15:59
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# If a total of 84 students are enrolled in two sections of a

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