GMATprep - quant Q indicative of real exam?

Fei

Hi
I just took the GMAT prep and found that there was little overlap. I got 720. I wish the solutions were provided though because I came across a few hairy questions.

Q3 ( for the third question, I don't consider this easy, and I am a bit worried about running into this on the real thing so early on).
h(n) is the product of even integers from 2 to n, if p is the smallest prime factor of h(100) +1, then p is
A. between 2&10, B between 10&20, C between 20&30, D between 30&40, E greater than 40.

Q18. For which of the following functions is f(a+b) = f(a) +f(b) for all positive numbers a & b?
A. f(x)=x squared
B. f(x)=x+1,
C. f(x)=square root x
D f(x)=2/x,
E.f(x) = -3x

Can anyone help me on these? nothing similar was in the OG, and definitely not covered in the theory.

also I found two mistakes in my tests, is this possible as everyone says that the GMATprep uses recycled questions. I really wish they provided the solutions. Are the questions above considered easy?

I appreciate anyone's thought on the above questions, thanks beatthegmat.

Fei · Posted: Sat Jun 17, 2006 8:16 am Post subject:

thanks to my friend - here is the solution to Q18 above.

Very Happy

Just have to work through one by one
A: a2+b2 ≠ (a+b)2

B: (a+1)+(b+1) = (a+b+2) ≠ (a+b)+1

C: a1/2+b1/2 ≠ (a+b)1/2

D: 2/a + 2/b = (2a + 2b)/ab ≠ 2/(a+b)

E: -3a + -3b = -3(a+b) CORRECT

If anyone can provide a solution for Q3, that would be greatly appreciated. My big day is next Thursday.

okrachokra · Posted: Mon Aug 07, 2006 4:46 pm Post subject: Solution for Q3

let k(n) be product of all integers from 2 to n.
In which case h(100) = 2 * k(50).

Now take the number 2*k(50) + 1;

Let p be the smallest prime that divides it.

if p is a prime divisor that divides 2*k(50) then it cannot divide 2*k(50) + 1

if k is a prime divisor that divides 2*k(50) then k belongs to the set primes <50.

Therefore the smallest prime p must be > 50.

The option E fits the answer best.

Let me know if this answer sounds right.

katwilson09 · Posted: Tue Nov 28, 2006 1:54 pm Post subject: hmmm...don't think so

I do not think that you are correct with the reasoning

h(100) = 2*k(50)

2*(2*4*6*....50) != (2*4*6*....50*52*54*.....98*100)

That is like saying 2*50! == 100! which is incorrect

katwilson09

I missed something in your answer but still

2*(2*3*4*5...50) != (2*4*6*8...100)

But I am still unsure of the correct answer. My guess is that the lowest prime should be in the smallest range.

Tried to pop the numbers into a spreadsheet and the result was too large to get a meaningful answer.

Mark Dabral · Posted: Wed Nov 29, 2006 9:12 pm Post subject: One possible solution

Please see my approach to this problem. The solution is attached an image and you will have to be logged on to see the image.

Cheers,
Mark

Mark Dabral

GMATPrep Quantitative is significantly harder than the actual test. To give you an idea, one can make upto six mistakes in Quantitative section in the GMATPrep and still get a full score of Q51 (99%). My guess is that on the real exam you can at most do two wrong to get a full score.

I don't know why Pearson Vue wrote GMATPrep Quant to be harder than the actual test.

So don't despair if you timing and the number of incorrect questions are high. The scaled score will be the same and takes the into account the difficulty of the test.

I hope this helps.

Cheers,
Mark

Guest · Posted: Tue Jun 26, 2007 2:08 pm Post subject:

Mark, I am logged in, but I do not see your attachment.

Guest · Posted: Tue Jun 26, 2007 3:09 pm Post subject:

You're not logged in--your last post shows as 'Guest'

mamo · Posted: Mon Jul 02, 2007 8:29 am Post subject: Re: One possible solution

That was a smart approach!!