• 7 CATs FREE!
If you earn 100 Forum Points

Engage in the Beat The GMAT forums to earn
100 points for \$49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS
Earn 10 Points Per Post
Earn 10 Points Per Thanks
Earn 10 Points Per Upvote

## Difficult GMAT Math Problems

##### This topic has expert replies
Senior | Next Rank: 100 Posts
Posts: 32
Joined: 13 Jun 2009
Location: Mumbai, India
by pg850 » Sun Jul 05, 2009 8:13 am
hey Eric, I was wondering if you could tell me where these qs are from? I plan to do the MGMAT tests and if these qs are from the same pool I dont want to solve them now and then know what the answers are when I m doing the tests..

Newbie | Next Rank: 10 Posts
Posts: 2
Joined: 15 Jul 2009

### help

by nikhilvinci » Wed Jul 15, 2009 3:52 am
hi
can someone please mail me the collection of tough math problems .i am not able to download it
my email id is nikhilvinci@gmail.com
thankyou

Junior | Next Rank: 30 Posts
Posts: 11
Joined: 05 Jun 2008
Thanked: 2 times
by Jake Blues » Thu Jul 23, 2009 1:18 am
hey Eric, I was wondering if you could tell me where these qs are from? I plan to do the MGMAT tests and if these qs are from the same pool I dont want to solve them now and then know what the answers are when I m doing the tests..

Fellas,any idea about that? I'm in the same situation,just do not want to spoil my CAT experience in case these questions are taken from CAT's pool.
Can anybody shed the light on the origin of these questions?

Newbie | Next Rank: 10 Posts
Posts: 1
Joined: 30 Jan 2009
by busybee1 » Wed Jul 29, 2009 7:20 am
Hi All,

I believe you are right tohellandback.
For this question the answer shld be 159 or the question shld be stating 79*80 to be the sum of first n even numbers.

-----------------------------------------------
are you sure the answers are correct..I did the first problem and I think it is done wrong

1. The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?
Sol: First term a=2, common difference d=2 since even number

therefore sum to first n numbers of Arithmetic progression would be

n/2(2a+(n-1)d)

= n/2(2*2+(n-1)*2)=n(n+1) and this is equal to 79*80

therefore n=79 which is odd...
now the n here is the number of terms in the series 2,4,6..
the real n being asked in the question should be 79*2+1=159        Newbie | Next Rank: 10 Posts
Posts: 1
Joined: 04 Feb 2010
by charlesdiss » Thu Feb 04, 2010 11:54 pm
I thought on the GMAT we didn't need to memorize formulas? Looking at the first problems it seems that I needed to have the formulas for arthimetic progressions memorized. Any thoughts?

Master | Next Rank: 500 Posts
Posts: 109
Joined: 12 Oct 2009
Thanked: 12 times
Followed by:1 members
GMAT Score:720
by jeffedwards » Fri Feb 05, 2010 12:24 pm
Thanks so much for posting this. Awesome! I've been looking for difficult questions....on a budget Newbie | Next Rank: 10 Posts
Posts: 1
Joined: 18 May 2010
by guillermogallicc » Tue May 18, 2010 9:10 am
your right man, just came to the same conclusion..159

Junior | Next Rank: 30 Posts
Posts: 17
Joined: 01 Jul 2010
Location: Houston
Thanked: 1 times
by TheCloakedMonk » Wed Sep 01, 2010 12:32 pm
I have a much better approach for question #3. (The current explanation is confusing to me)

You can look at this as and "OR" problem.

If a leap year happens every 4 years, then a person has a 1/4 chance of being born in that year because he/she can only possibly be born in 1 out of the 4 given years.

Therefore, the probability of a person being born out of the four given years is 1/4 = 25%. Therefore the probability the person A "OR" person B was born {P(A or B)} in the leap year is 1/4 + 1/4 = 2/4 = 50%.

The question asks for the number of people that would yield a greater than >50% probability. Therefore, since 2 people yields 50%, 3 people are needed to surpass 50%. There is a 75% chance {1/4 + 1/4 +1/4} that one of three people {P(A or B or C)} was born in the leap year.

Am I correct or have I just gotten lucky?

Enjoy!
Anything is possible if you believe in yourself and have faith in your actions.

Junior | Next Rank: 30 Posts
Posts: 17
Joined: 01 Jul 2010
Location: Houston
Thanked: 1 times
by TheCloakedMonk » Wed Sep 01, 2010 12:55 pm
Question #1

The easiest way to do this is:

Sum of Numbers = (#of Numbers)(Middle Number)
Sum of Even Numbers {Where n is odd...like in this problem} = ((n-1)/2)*((n-1)/2+1) = 79*80
Therefore, 79 = (n-1)/2, so n = 159

What you must see:
# of Even Numbers = (n-1)/2 = 79
Middle # = (n-1)/2 +1 = 80

You can easily devise this because the only difference between the two terms is "+1" which is obviously 80 NOT 79.

Hope that clears the slate.
Anything is possible if you believe in yourself and have faith in your actions.

Junior | Next Rank: 30 Posts
Posts: 19
Joined: 14 Aug 2010
by boysangur » Tue Sep 07, 2010 10:25 pm
Man, this problem set got me very concerned. Are there really some problems like these on the actual GMAT? Some of them are not very hard but, although it is subjective, some seem a little too hard. Even looking at the solutions, some of them are pretty darn complex and lengthy. Like the solution to number 11 is given as follows:

Since 12w=3x=4y,

w:x=3:12=1:4 and x:y=4:3

so, w = 1
x = 4
y = 3

the fractional part of the square is shaded:
{(w+x)^2 - [(1/2)wx + (1/2)wx +(1/2)xy + (1/2)w(2w)]}/(w+x)^2

= {(w+x)^2 - [wx + (1/2)xy + w^2)]}/[(w+x)^2]

=[(5^2) -(4+6+1)]/(5^2)

= (25 - 11)/25

= 14/25

I mean, really, there is no way I could figure this out and solve it in anywhere near 2 minutes. I thought I was getting close to the finish line in my math review, as I solved most of the OG 12 problems easily but now I'm really discouraged.

Legendary Member
Posts: 1255
Joined: 07 Nov 2008
Location: St. Louis
Thanked: 312 times
Followed by:90 members
by Tani » Wed Sep 15, 2010 8:38 am
Looks to me as though the answer to the first problem is wrong.
Tani Wolff

Newbie | Next Rank: 10 Posts
Posts: 2
Joined: 25 Oct 2010
by ecmpec » Mon Oct 25, 2010 6:27 am
Can any one mail me the the file along with answers at rahul.mechies@gmail.com?

Legendary Member
Posts: 1255
Joined: 07 Nov 2008
Location: St. Louis
Thanked: 312 times
Followed by:90 members
by Tani » Mon Oct 25, 2010 7:01 am
Re: leap year problem (#3).

If you add the probabilities that a single person would be born in a leap year (1/4), then with four people you would have a 100% possibility that one would be born in leap year. But, clearly all four could have been born in the same, non-leap year, so the possibility is not 100%. The explanation that came with the problem is correct.
Tani Wolff

Newbie | Next Rank: 10 Posts
Posts: 4
Joined: 10 Jul 2011
by rohitkalla » Sun Jul 10, 2011 8:11 pm
Can somebody please mail the math tough questions doc to me ? it's not opening in my system..

bpositive09@gmail.com.

Thanks.

Senior | Next Rank: 100 Posts
Posts: 33
Joined: 27 Nov 2007
by dev.gavande » Mon Sep 12, 2011 9:20 pm
Thanks!!!