Difficult GMAT Math Problems
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..
 nikhilvinci
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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
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

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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?
Thanks in advance!
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+(n1)d)
= n/2(2*2+(n1)*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
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+(n1)d)
= n/2(2*2+(n1)*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

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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?
 jeffedwards
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Thanks so much for posting this. Awesome! I've been looking for difficult questions....on a budget

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 TheCloakedMonk
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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!
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.
 TheCloakedMonk
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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} = ((n1)/2)*((n1)/2+1) = 79*80
Therefore, 79 = (n1)/2, so n = 159
What you must see:
# of Even Numbers = (n1)/2 = 79
Middle # = (n1)/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.
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} = ((n1)/2)*((n1)/2+1) = 79*80
Therefore, 79 = (n1)/2, so n = 159
What you must see:
# of Even Numbers = (n1)/2 = 79
Middle # = (n1)/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.
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.
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.
 Tani
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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, nonleap year, so the possibility is not 100%. The explanation that came with the problem is correct.
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, nonleap year, so the possibility is not 100%. The explanation that came with the problem is correct.
Tani Wolff

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Can somebody please mail the math tough questions doc to me ? it's not opening in my system..
bpositive09@gmail.com.
Thanks.
bpositive09@gmail.com.
Thanks.

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