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]

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

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]

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]

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.

[mamo]

That was a smart approach!!

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

[rara]

Mark I have looked at your answer over and over again and read other ones, but it is not clicking. I think I am missing a core concept that is being tested. Do you have any recommendations for what I should go back and review to fully understand what this question is asking/testing?

[navalpike]

Mark, I respect your opinion but couldn't disagree more. I have taken the Gmat as well as done the GmatPrep quant many times. The actual test was much more difficult than the prep test.

[outreach]

nice explanation by @Mark Dabral

[GMATGuruNY]

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.

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

Here's the number property rule that's being tested with this problem:

If x is a positive integer, the only factor common both to x and to x+1 is 1; they share no other factors. Any factor of x (other than 1) will NOT be a factor of x+1.

Let's examine why:

If x is a multiple of 2, what's the next largest multiple of 2? If x=4, the next largest multiple of 2 is 4+2=6. So the next largest multiple of 2 is x+2.

If x is a multiple of 3, what's the next largest multiple of 3? If x=6, the next largest multiple of 3 is 6+3=9. So the next largest multiple of 3 is x+3.

Using this logic, if we go from x to x+1, we get only to the next largest multiple of 1. So 1 is the only factor common both to x and to x+1.

Thus, in the problem above, we know that 1 is the only factor common both to h(100) and to h(100) + 1. They share no other factors.

h(100) = 2 * 4 * 6 *....* 94 * 96 * 98 * 100

Factoring out 2, we get:

h(100) = 2^50 (1 * 2 * 3 *... * 47 * 48 * 49 * 50)

Looking at the set of parentheses on the right, we can see that every prime number between 1 and 50 is a factor of h(100). This means that none of the prime numbers between 1 and 50 is a factor of h(100) + 1, because h(100) and h(100) + 1 share no factors other than 1.

So the smallest prime factor of h(100) + 1 must be greater than 50.

The correct answer is E.

This is a hard and strange question. Most test-takers would be wise to guess and move on to more familiar questions. The good news is that if you get all of the other questions right, skipping this question will have little impact on your score. And remember: a quarter of the questions are experimental and do not affect your score at all.

[ankur73]

cn someone explain Q no -18