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Of the students in a certain class, 55% of the female and [#permalink]
 29 Mar 2009, 10:59
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Of the students in a certain class, 55% of the female and 35% of the male passed an exam. Did more than half of the students in the class pass the exam?
(1) More than half of the students in the class are female.
(2) The number of the female students is 20 more than the number of the male students. Good Q show your calculations
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Re: DS-male/female in a class [#permalink]
29 Mar 2009, 11:38
nitya34 wrote: Of the students in a certain class, 55% of the female and 35% of the male passed an exam. Did more than half of the students in the class pass the exam?
(1) More than half of the students in the class are female.
(2) The number of the female students is 20 more than the number of the male students.
Good Q
show your calculations I'll venture a B. stmnt1 is clearly not sufic. stmnt2 - plugging in numbers, we can find the answer.
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Re: DS-male/female in a class [#permalink]
29 Mar 2009, 11:57
Suppose there are f% of the students are female: 55f/100 + 35(100-f)/100 = 35 + 20f/100 percent of the students passed. Thus we need to know whether f/5 > 15, i.e. whether f > 75
(1) f > 50 not suff
(2) only tells us that f > 50 not suff
(t) not suff
easier drawing a number line:
(male)35-----15-----50--5--55 (female)
we need to know whether 15m < 5f i.e f > 3m
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Re: DS-male/female in a class [#permalink]
17 Apr 2009, 19:43
i got B. after picking several numbers, i got the answer that more than 1/2 of the class did not pass.
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Re: DS-male/female in a class [#permalink]
18 Apr 2009, 00:50
Can some one help on this question please.
Brief explaination is needed.
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Re: DS-male/female in a class [#permalink]
19 Apr 2009, 11:09
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nitya34 wrote: Of the students in a certain class, 55% of the female and 35% of the male passed an exam. Did more than half of the students in the class pass the exam?
(1) More than half of the students in the class are female.
(2) The number of the female students is 20 more than the number of the male students.
Good Q
show your calculations Lets say total students=100, Female x,then Male=100-x. As per question 0.55x+0.35(100-x) > 0.5 * 100 solving we get is x>75 ? St 1. not sufficient as we need to prove number of female > 75 St 2 : if x+y=100 and x-y=20 then solving we get x=60 (number of female). therefore we can say Number of female < 75 and therefore SUFFICIENT. hence IMO B
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Re: DS-male/female in a class [#permalink]
19 Apr 2009, 23:40
Ok I got B as well but did it in a different way. I like Bandit's way better though. None the less, here goes:
1 not sufficient..we all agree.
2: The minimum number of males must be 20 and this is because we need a whole number that can split into the proportions 35% and 65%. I arrived at this by computing the prime factorizing 35, 65 and 100. 35= 5*7; 65= 5*13 and 100 = 2*2*5*5. Thus the 5 of 100 will cancel but the 2*2*5 will not. Thus the number of males must be multiples of 20.
Now I used M=20 and F=40 (because females is simply 20 more)
At this the numbers passed = 29 < 30
We can stop here..we need not calculate for M=40, 60,etc because as the proportion of males increase, since the passing rate is low for males, we are all the more sure that less than half of the students have passed.
Thus Ans B
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Re: DS-male/female in a class [#permalink]
20 Apr 2009, 06:46
thank you bandit and shkusira
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Re: DS-male/female in a class [#permalink]
10 Jun 2009, 04:38
E for me.
When statements involve absolute relationship (i.e Stmt2 F-20=M) vs. relative (e.g. F/M=2/5) , you cannot say "lets assume F+M=100".
Let's look at the stem:
0.55F+0.35M>0.5(F+M)?
Simplify: 0.05F>0.15M? or F>3M or
F/M>3?
Stmt1: F>M or F/M>1
This is not sufficient to answer the question.
Stmt2: F-20=M
This statement does not give a relative value of F to M. It does not matter whether F-1=M or F-100000=M. We need total number of students to be able to use this statement. Hence, it is not sufficient.
Together, still we cannot answer the question. Hence, E.
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Re: DS-male/female in a class [#permalink]
10 Jun 2009, 09:50
seofah wrote: Stmt2: F-20=M
This statement does not give a relative value of F to M. It does not matter whether F-1=M or F-100000=M. We need total number of students to be able to use this statement. Hence, it is not sufficient.
We cannot get 55% and 35% for any F and M. This is a trap here. In other words, M cannot be 1 or any number up to 19, because 35% * M will not be an integer value. So, M can be 20 and more. But for M>=20, the total number of students who passed exam will be always less than a half.
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Re: DS-male/female in a class [#permalink]
10 Jun 2009, 19:42
Let F = number of females,
M = number of males
\Rightarrow (F+M) = number of people in class.
Question is whether \frac{(0.55) F + (0.35) M}{(F+M)} \hspace{20}\geq \hspace{20} \frac{1}{2}
Simplifying, we get:
F \hspace{10}> \hspace{10}3M
(1)
It is NOT SUFFICIENT because the statement only says that F \hspace{10}> \hspace{10}M
(2)
F \hspace{5}= \hspace{5}M + 20
Substituting statement-2 in F \hspace{10}> \hspace{10}3M
\Rightarrow M + 20 \hspace{10}> \hspace{10}3M
\Rightarrow 10 \hspace{10}> \hspace{10}M
Clearly, the above statement can be disproved using the fact that 35% \hspace{5}of\hspace{5} M \geq\hspace{5} 20, in order for it to be an integer.
Thus, statement-2 is SUFFICIENT.
Answer B.
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Re: DS-male/female in a class [#permalink]
10 Jun 2009, 23:12
B.
Stmt 1 - We agree that it is insuff.
Stmt 2 -
No. of Boys = X
No. of girls = X + 20
So total students = X + X + 20 = (2X + 20)
We can assume that the number of students be 100 (it will not make a diff. as X will change accordingly)
So Boys = 40
Girls = 60
%Boys passed = 35% of 40 = 14
%Girls passed = 55% of 60 = 33
Total passed = 47 which is less then 50.
Hence suff.
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Re: DS-male/female in a class
[#permalink]
10 Jun 2009, 23:12
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