am I missing something? : gmat MATH tough problems.doc

[gmattarget700]

hi there..

I had downloaded from this forum "gmat MATH tough problems.doc" sometime back..

now, very first question and explanation thats put in the document is this:

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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...
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I am not convinced, as for me n = 159

Here is how I think -

the arithmetic series: 2, 4, 6, 8....... m [note that m = n-1, as n is odd number and so, even number before n would be n-1)

lets assume, n1 = total numbers in the series

Now, with the same approach as mentioned above,
SUM = n1/2 [2*2 + (n1-1)*2] = 79*80
=> n1[n1+1) = 79*80
=> n1 = 79

m = a + (n1-1) d = 2 + (79-1) *2 = 2 (1+78) = 2*79

i.e. m = 2*79

=> m = n-1 = 2 *79 = 158
=> n = 158 +1 = 159

can you verify? Thanks

[Ian Stewart]

Yes, you are entirely correct. There is a serious problem with the solution provided - the 'n' in that solution does not represent the same value as the 'n' the question asks you to find. In the solution you pasted in, the 'n' is the number of terms we are adding. The question does not ask how many terms we are adding; it asks what we will get if we add 1 to the largest term. The answer should be 159, not 79, as you have correctly demonstrated.

[gmattarget700]

Thanks Ian for confirming...

I just went back and found that it was uploaded years back by Eric

https://www.beatthegmat.com/difficult-gm ... 20problems

so, I am sure it was overlook...

May I request Eric or someone to review and correct if any such issues in doc.....I wanted to fall back on those questions, to check if I need to spend good time on Maths or I can be fine with just refresh...

[piyush.sinha.2007]

Hi! From where did you download the questions??

also can somebody let me know which resource should be the best for QA
i have gone through OG and Manhattan QA book....questions in both of them looked very trivial...

[gmattarget700]

hey buddy...i have given the URL above

[ullaas]

Hi,

I am new here but I have a question regarding a solution in the doc (question 51):

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51. How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3?

A.14 B.15. C.16 D.17 E.18

Soln: if we arrange this in AP, we get
4+7+10+.......+49

so 4+(n-1)3=49: n=16
C is my pick
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I think the solution is D(17). The solution is overlooking integer '1' which also yields a remainder of 1 when divided by 3. Hence the solution should be 16+1 = 17.
Please can someone verify if this is correct or am I being obtuse.

Thank you very much.