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

Question

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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chetan2u · Answer

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

minustark · Answer

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.
C is the correct answer.

AnirudhaS · Answer

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

Papist · Answer

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

TarunKumar1234 · Answer

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

TychoBrahe147 · Answer

1) insufficient - we don't know anything about female members
2) insufficient - we don't know how who is ambidextrous, who is not

Combining 1&2:
consider M=100, therefore 50 M are ambidextrous
F=100/2-1= 49
Even if none of the F is ambidextrous we get that 50/149 members are ambidextrous.
50/149 > 50/150 = 1/3
Sufficient

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

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