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GMAT question! Need help, can someone take me through this!

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by myworld316 » Sat Jul 18, 2009 12:04 pm
Two members of a club are to be selected to represent the club at a national meeting. If there are 190 different possible selections of the 2 members, how many members does the club have?

A) 20
B) 27
C) 40
D) 57
E) 95
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Source: — Problem Solving |

by zenithexe » Sat Jul 18, 2009 3:01 pm
How about using elimination?
since its XC2=190
There is no doubt that B) D) are incorrect as they both end with 7, and 7*6 won't produce unit digit of 0 when divided by 2.

So lets just plug in 40 and see, 40*39= way too big, therefore answer is A)

or you can solve algebratically

since XC2=190, this can be deduced to
(X)(X-1)(X-2)(X-3)..../2!*(X-2)(X-3)... =190
(X)(X-1)/2=190
(X^2-X)/2=190
X^2-x=380
X^2-x-380=0
(X-20)(x+19)=0
X=20,-19

hence Answer is A), 20
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The answer is A - > 20.

Let there be N members to choose from. So the total possible selections are N choose 2 which is given to be 190.
So, 190 = N choose 2
190 = N! / ( 2! * (N-2)!)
190 = N * (N-1) * (N-2)! / ( 2! * (N-2)! )

Solving it further you shall get that N * (N-1) = 380.

This means that we need to find two consecutive numbers whose product is 380. At this point just plug in the number. Starting with A, N = 20 and hence N-1 = 19. This gives us the desired product which is 380.

Here we get lucky with the first option. But say the first option was wrong, then based on how far you are from the needed product of 380, use educated guess to select the next option.

Let me know if this helps.

Thanks.
myworld316 wrote:Two members of a club are to be selected to represent the club at a national meeting. If there are 190 different possible selections of the 2 members, how many members does the club have?

A) 20
B) 27
C) 40
D) 57
E) 95
Last edited by ankitns on Sat Jul 18, 2009 8:21 pm, edited 1 time in total.
Attempt 1: 710, 92% (Q 42, 63%; V 44, 97%)
Attempt 2: Coming soon!
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Thanks

by myworld316 » Sat Jul 18, 2009 6:23 pm
Thanks to both replies, they were both very helpful.
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