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# If 6 students, representing 16 2/3% of the class, failed algebra, how

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Joined: 02 Sep 2009
Posts: 44293
If 6 students, representing 16 2/3% of the class, failed algebra, how [#permalink]

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01 Mar 2018, 00:15
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If 6 students, representing 16 2/3% of the class, failed algebra, how many students passed the course?

(A) 48

(B) 42

(C) 36

(D) 30

(E) 32
[Reveal] Spoiler: OA

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If 6 students, representing 16 2/3% of the class, failed algebra, how [#permalink]

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01 Mar 2018, 00:24
Bunuel wrote:
If 6 students, representing 16 2/3% of the class, failed algebra, how many students passed the course?

(A) 48

(B) 42

(C) 36

(D) 30

(E) 32

If 6 is $$16\frac{2}{3}$$% = $$\frac{50}{3}$$%, then x -> 100%

Total number of students(x) = $$100*6 = \frac{50x}{3}$$ -> x = $$100*6*\frac{3}{50} = 36$$

Therefore, 30 out of 36 students passed the course(Option D)
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If 6 students, representing 16 2/3% of the class, failed algebra, how [#permalink]

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01 Mar 2018, 20:07
Bunuel wrote:
If 6 students, representing 16 2/3% of the class, failed algebra, how many students passed the course?

(A) 48

(B) 42

(C) 36

(D) 30

(E) 32

Let C = total students

6 = 16 2/3 % of the class
6 = 16 2/3 % C

16 2/3 % C=$$\frac{16\frac{2}{3}}{100}=\frac{\frac{50}{3}}{100}=\frac{50}{300}=\frac{1}{6}C$$

$$6=\frac{1}{6}C$$
$$(6*\frac{6}{1}) = C = 36$$

C = 36
6 failed
(36-6) = 30 passed

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Re: If 6 students, representing 16 2/3% of the class, failed algebra, how [#permalink]

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03 Mar 2018, 00:09
Bunuel wrote:
If 6 students, representing 16 2/3% of the class, failed algebra, how many students passed the course?

(A) 48

(B) 42

(C) 36

(D) 30

(E) 32

IMO D
6=16 2/3% x where x is total number of students in class
6=48+2/3*100 x
6=50/3*100 x
6=1*x/6
x=36
failed is 6 then passes is 36-6=30
Re: If 6 students, representing 16 2/3% of the class, failed algebra, how   [#permalink] 03 Mar 2018, 00:09
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