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If a total of 84 students are enrolled in two sections of a [#permalink ]
02 Jul 2004, 21:28

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85% (01:42) correct

15% (00:57) wrong

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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.

Last edited by

Bhai on 03 Jul 2004, 07:05, edited 1 time in total.

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question looks incomplete, but i guess "sufficient" enough to derive( guess

my answer as C.

- ash

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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.

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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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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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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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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

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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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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.

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Your right. The OA is E.

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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.

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Yea E it is we havent been told the class sizes

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Re: DS - Percentages / Sets - OG 12 [#permalink ]
31 Mar 2009, 19:14

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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Re: DS - Percentages / Sets - OG 12 [#permalink ]
31 Mar 2009, 19:18

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.

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Re: DS - Percentages / Sets - OG 12 [#permalink ]
02 Apr 2009, 14:59

E we dont know how sections are divided.

Re: DS - Percentages / Sets - OG 12
[#permalink ]
02 Apr 2009, 14:59