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What percent of the 300 students are girls who had taken a course in

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What percent of the 300 students are girls who had taken a course in  [#permalink]

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New post 20 Nov 2019, 06:01
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75% (01:24) correct 25% (01:58) wrong based on 28 sessions

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What percent of the 300 students are girls who had taken a course in writing?

(1) 1/2 of the girls and 2/5 of the boys had taken a course in writing.

(2) The number of girls who had taken a course in writing is 6 more than the number of the boys who had taken a course in writing.
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Re: What percent of the 300 students are girls who had taken a course in  [#permalink]

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New post 22 Nov 2019, 19:46
rohan2345 wrote:
What percent of the 300 students are girls who had taken a course in writing?

(1) 1/2 of the girls and 2/5 of the boys had taken a course in writing.

(2) The number of girls who had taken a course in writing is 6 more than the number of the boys who had taken a course in writing.



So we have 4 categories.,
Girls taken course G and not taken Gn, and similarly B and ABN.
We have to find G/300*100..
Statement I.
0.5(G+Gn)=G....G=Gn
0.4(B+Bn)=B......2Bn=3B....Bn=3B/2
Statement II
G=B+6

Combined all terms are interconnected
G+B+Gn+Bn=300=2G+5B/2
2G+(5/2)(G-6)=300......9G/2-15=300....G=315*2/9=70
%=100*70/300=70/3%

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Re: What percent of the 300 students are girls who had taken a course in  [#permalink]

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New post 23 Nov 2019, 01:28
rohan2345 wrote:
What percent of the 300 students are girls who had taken a course in writing?

(1) 1/2 of the girls and 2/5 of the boys had taken a course in writing.

(2) The number of girls who had taken a course in writing is 6 more than the number of the boys who had taken a course in writing.

Lets no of girls be g and no of boys be b
Now, g+b = 300 -----(i)

(1) No of girls taking course = g/2
No of boys taking course = 2b/5
Insufficient

(2) Insufficient

(1) + (2)
g/2 = 6 + 2b/5 ----- (2)

from equation (i) and (ii) value of g can be calculated.
Sufficient
C is correct
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Re: What percent of the 300 students are girls who had taken a course in   [#permalink] 23 Nov 2019, 01:28
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