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In a club for left-handed people that also admits the ambidextrous, ar

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In a club for left-handed people that also admits the ambidextrous, ar  [#permalink]

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04 Nov 2019, 02:43
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65% (hard)

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54% (01:53) correct 46% (01:39) wrong based on 68 sessions

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In a club for left-handed people that also admits the ambidextrous, are more than 1/3 of the members ambidextrous?

(1) Exactly 50% of the male members of the club are ambidextrous.

(2) The number of females in the club is exactly 1 fewer than half the number of male members.

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Re: In a club for left-handed people that also admits the ambidextrous, ar  [#permalink]

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04 Nov 2019, 19:49
In a club for left-handed people that also admits the ambidextrous, are more than 1/3 of the members ambidextrous?

So if the total is x, is the number of ambidextrous >x/3 or >33.33%

(1) Exactly 50% of the male members of the club are ambidextrous.
We do not know the % of male overall or females who are ambidextrous.
Say there are 10 males and 40 females and no female is ambidextrous, so only 5 out 50 or 10%<33%
But if there are any p males and no females, ans is yes as 50% is the answer then.

(2) The number of females in the club is exactly 1 fewer than half the number of male members.
f=(m/2)-1

Combined.
Total =f+m=(m/2)-1+m=(3m/2)-1
Number of ambidextrous will be at least m/2

So let m be 10, so f=(10/2)-1=4.....total =10+4=14
Number of ambidextrous is at least 5..
Fraction of ambidextrous $$\geq{\frac{5}{14}}$$, hence >1/3
Sufficient
C
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Re: In a club for left-handed people that also admits the ambidextrous, ar  [#permalink]

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04 Nov 2019, 21:19
Bunuel wrote:
In a club for left-handed people that also admits the ambidextrous, are more than 1/3 of the members ambidextrous?

(1) Exactly 50% of the male members of the club are ambidextrous.

(2) The number of females in the club is exactly 1 fewer than half the number of male members.

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Statement 1:
50% of male members are ambidextrous, but the number or ratio of total, male or female members are not given. Thus Not Sufficient.
Statement 2:
Female members = (50% of male members) -1. No information about ambidextrous members are given. Not sufficient.
Combining 1 & 2:
Lets say, total members are 20. From statement 2, we can say among them 14 are male members and 6 are female members. As per statement 1, ambidextrous members are 7, which is SUFFICIENT to find out the answer.
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Re: In a club for left-handed people that also admits the ambidextrous, ar  [#permalink]

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30 Dec 2019, 06:45
Question stem: Let the total members be t. And ambidextrous be a. And Males = m, Females = f (say)
Is $$a>\frac{t}{3}$$ ?
(1) Exactly 50% of the male members of the club are ambidextrous.
$$\frac{m}{2} = a$$
Insufficient as the split between male and female is not given.

(2) The number of females in the club is exactly 1 fewer than half the number of male members.
$$f=\frac{m}{2}-1$$
Again,
$$m+f=t;$$
$$or, m+ \frac{m}{2}-1=t;$$
$$or, m = \frac{2(t+1)}{3};$$
Insufficient as we don't have any information about how many ambidextrous.

(1) + (2)
Putting the value of m found in 2, into 1
$$a = \frac{1}{2}*m = \frac{1}{2}*\frac{2(t+1)}{3} = \frac{t+1}{3} = \frac{t}{3} + \frac{1}{3} >\frac{t}{3}$$
Hence sufficient.

Ans: C
Re: In a club for left-handed people that also admits the ambidextrous, ar   [#permalink] 30 Dec 2019, 06:45
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