22 replies to this topic

[TrollMan]

There were 100 people present at a baseball-card show:

59 were men
72 were football- and baseball-card collectors
81 were college graduates
89 were right-handed

What is the least number of men who were both football- and baseball-car collectors and right-handed college graduates?

Show your reasoning

[IQ154]

There were 100 people present at a baseball-card show:

59 were men
72 were football- and baseball-card collectors
81 were college graduates
89 were right-handed

What is the least number of men who were both football- and baseball-car collectors and right-handed college graduates?

Show your reasoning

Am I missing something here? Isn't this simple arithmetic (deduction actually)? It's 31, right? I didn't show my reasoning because for some reason I think I've reasoned incorrectly. That is a reasonable assumption based on reasonable doubt. Come on, let's be reasonable, if for no other reason than ... that's it, it is unreasonable to go on with these puns (i.e. no rhyme nor reason).

[IQ154]

Oops, I just voted this down by mistake. Sorry Trollman. It was a slip of the click.

[sditty]

Spoiler for Final answer

There are 41 women. (100-59)

If all the women were dual card collectors, that means minimally 31 men are such collectors. (72-41)
If all women are college grads, then minimally 40 men are college grads. (81-41)

If all women are right handed, then minimally 48 men are right handed. (89-41)
If all left handed men are college grads, then minimally 29 men are right handed college grads. (40-(59-48))

So we have 31 men who are dual card collectors and 29 men who are right handed college grads. The minimum overlap of these two sets within a total of 59 men is 1.

So the answer is 1.

[SeaCalMaster]

This looks suspiciously like homework...

[TrollMan]

well, IQ154, with a name like mine who wouldnt downrate me

Homework?
Are you saying you didnt enjoy this puzzle!?

sditty, good job.

Edited by TrollMan, 20 January 2012 - 01:00 AM.

[mukulc]

There were 100 people present at a baseball-card show:

59 were men
72 were football- and baseball-card collectors
81 were college graduates
89 were right-handed

What is the least number of men who were both football- and baseball-car collectors and right-handed college graduates?

Show your reasoning

There are 41 women. Assuming all women are part of all groups. The least number of men that satisfy all the criteria is min of ((72-41), (81-41),(89-41)). Which makes it 31 men who were both football- and baseball-car collectors and right-handed college graduates.

[bhramarraj]

Spoiler for Least no. of Men

No. of total women = 100 - 59 = 41
If all the 41 women had been football- & baseball-card collectors, then least no. of men who are football & baseball collectors = 72 - 41 = 31.
Similarly least no. of men who are college graduates = 81 - 41 = 40.
------do------------------------------- right handed = 89 - 41 = 48

Then from above it is clear that 31 is the least no. of men who are both football- & baseball- card collectors and righthanded college graduates.

Edited by bhramarraj, 20 January 2012 - 07:57 AM.

[hhh3]

Spoiler for my go...

31

Edited by hhh3, 20 January 2012 - 08:00 AM.

[Dej Mar]

Spoiler for

59 men from 100 leaves 41 persons present who are not men (but may be women or children).
Assuming each of the 41 (women or children) were card collectors, college graduates and right-handed, then
72 - 41 = 31 card collectors who were men,
81 - 41 = 40 college graduates were men, and
89 - 41 = 48 men were right-handed.

If 27 men were right-handed college graduates but not card collectors, 20 men were right-handed card collectors but not college graduates, 10 men were card collectors and college graduates but not right-handed, and 1 man who was a college graduate but neither a card collector nor right-handed, then there would be 1 man who was a right-handed card collector and college graduate -- the minimum possible number (given the givens).