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

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

Let's assume there are 100 people in the club.

S1: We are told 50% of the males in the club are ambidextrous. However we have no way of determining what proportion of the club members are male. NOT SUFFICIENT.

S2: Using this, we can find the number of males in our sample using the equation 100 = M + F or 100 = M + (M/2 - 1) and solving for M. However, we are not given any information about how many students are ambidextrous. NOT SUFFICIENT.

S1+S2: We can multiply 50% by the value of M we got from S2 and compare that number to the total number of students to find the solution. SUFFICIENT.

ANSWER: C

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

Stat1: Exactly 50% of the male members of the club are ambidextrous.
So, L= M/2, but what about female members. Not sufficient.

Stat2: The number of females in the club is exactly 1 fewer than half the number of male members.
But, we don't know, how many males or females are ambidextrous. Not sufficient.

Combining both,
L= M/2 and F= M/2 -1, It means ratio = (M/2)/ (M/2+M/2 -1 ) = (M/2)/ (M -1) > 1/3 sufficient.

So, Ans. C.

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