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# At a certain fitness center, are more than 1/6 of the members over 30

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Senior Manager
Joined: 17 Oct 2016
Posts: 318
Location: India
Concentration: Operations, Strategy
GPA: 3.73
WE: Design (Real Estate)
At a certain fitness center, are more than 1/6 of the members over 30  [#permalink]

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Updated on: 27 Nov 2017, 00:32
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25% (medium)

Question Stats:

85% (01:37) correct 15% (01:35) wrong based on 53 sessions

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At a certain fitness center, are more than 1/6 of the members over 30 years of age?

(1) Exactly 60 percent of male members are over 20 years of age, and of these, 3⁄5 are over 30 years of age.

(1) Exactly 15 female members are over 30 years of age.

_________________

Help with kudos if u found the post useful. Thanks

Originally posted by Sasindran on 27 Nov 2017, 00:27.
Last edited by Bunuel on 27 Nov 2017, 00:32, edited 1 time in total.
Renamed the topic, edited the question and edited the tags.
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Re: At a certain fitness center, are more than 1/6 of the members over 30  [#permalink]

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27 Nov 2017, 05:35
Sasindran wrote:
At a certain fitness center, are more than 1/6 of the members over 30 years of age?

(1) Exactly 60 percent of male members are over 20 years of age, and of these, 3⁄5 are over 30 years of age.

(1) Exactly 15 female members are over 30 years of age.

Statement 1: let there me $$x$$ males. so number of males over $$20=60$$%$$*x=0.6x$$

number of males over 30 $$= \frac{3}{5}*0.6x=0.36x$$

but nothing mentioned about female. So insufficient

Statement 2: let there me $$y$$ females. out of $$y$$, $$15$$ are over 30. but nothing mentioned about male. Insufficient

Combining 1 & 2 total member $$= x+y$$

total member over 30 $$=0.36x+15$$

Hence the ratio of member over 30 $$=\frac{(0.36x+15)}{x+y}$$. so we have two unknown variables. Insufficient

Option E
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Re: At a certain fitness center, are more than 1/6 of the members over 30  [#permalink]

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05 Dec 2017, 10:16
Sasindran wrote:
At a certain fitness center, are more than 1/6 of the members over 30 years of age?

(1) Exactly 60 percent of male members are over 20 years of age, and of these, 3⁄5 are over 30 years of age.

(1) Exactly 15 female members are over 30 years of age.

why is A not the answer? when we know that 1/6 = .16

and if we take 1 as total number of members in gym

So, 60 percent male above 20 age = .60

and 3/5 of .60 = .36

so 0.36>0.16

which means THERE ARE MORE THAN 1/6 people in the gym who are 30 year old or above...ONLY MALE CAN PROVIDE ENOUGH INFO

Please let me know where I am wrong
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Re: At a certain fitness center, are more than 1/6 of the members over 30  [#permalink]

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05 Dec 2017, 10:40
rocko911 wrote:
Sasindran wrote:
At a certain fitness center, are more than 1/6 of the members over 30 years of age?

(1) Exactly 60 percent of male members are over 20 years of age, and of these, 3⁄5 are over 30 years of age.

(1) Exactly 15 female members are over 30 years of age.

why is A not the answer? when we know that 1/6 = .16

and if we take 1 as total number of members in gym

So, 60 percent male above 20 age = .60

and 3/5 of .60 = .36

so 0.36>0.16

which means THERE ARE MORE THAN 1/6 people in the gym who are 30 year old or above...ONLY MALE CAN PROVIDE ENOUGH INFO

Please let me know where I am wrong

Hi rocko911

The highlighted portion is wrong. As you have assumed total member to be 1, so you cannot again assume that total male member is also 1.

60%of male => 0.6*something less than 1 (male+female=1)
so you will have to assume some other variable for male/female
Re: At a certain fitness center, are more than 1/6 of the members over 30 &nbs [#permalink] 05 Dec 2017, 10:40
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# At a certain fitness center, are more than 1/6 of the members over 30

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