Hi Stonecold ! First of all thanks for sharing these awesome conceptual questions. I have one doubt - Why "Z" cannot be -ve (Negative integer) for question no 1 in Mock test1

1)If z is divisible by all the integers from 1 to 5 inclusive,what is the smallest value of z?
Possible minimum values for Z also include -60,-120 etc. 60 should be the smallest positive integer.

Please correct me if i am wrong.

Hi.
Thanks for the feedback on the Quiz.

Coming back to your question.
Technically you are right.

But the Bottom line is -> Factors and multiples on the GMAT are always positive integers.
Hence z would always be 60.


Regards
Stone Cold
_________________

Mock Test 4 (600-700) Updated.



http://gmatclub.com/forum/stonecold-s-mock-test-217160.html#p1768052
_________________

1)If z is divisible by all the integers from 1 to 5 inclusive,what is the smallest value of z?

Without any options how can we think and try z = 60?
Yes I understand 60 is divisible by all numbers 1 to 5
Assume If z is divisible by all integers from 1 to 7 inclusive, what is the smallest value of z? without knowing any options?
Thanks

2)Data Sufficiency-> Is the positive integer z divisible by 12?
A)z is divisible by 3.
B)z is not divisible by 2.

1, z / 3 = integer
z could be 3,6,9,12,15,24...
Insufficient
BCE
2, z / 2 # integer
z must be an odd number
Sufficient

3)Data Sufficiency-> Is integer z divisible by 21?
A)z is divisible by 7 --- Insufficient
B)z is divisible by 3 ---- Insufficient
Z = 21, divisible by both 3 and 7
C

4, E

A quadrilateral has a perimeter of 16. Which of the following alone would provide sufficient information to determine the area of the quadrilateral.

A)The quadrilateral contains equal sides.

B)The quadrilateral is formed by combining two isosceles right triangles.

C) Two pairs of congruent angles are in a 2:1 ratio.

D) The width is 4o% of the length and all angles are of equal measure.

E) If the perimeter was decreased by 50%, the area would decrease by 25%

Show: ::

D.

_________________

The difference between the digits of a two digit number is 3.How many such numbers are possible ?
A)12
B)13
C)14
D)15
E)16
_________________

Hello, I am new in your topic. Is there anyone still discussing this?

The following is my take. Hope to receive your comment on it. Thank you.

Difference between 2 digits is 3 => Smaller digit must range from 0 to 6 (because if the smaller is 7, then the larger will be 10 -> this cannot happen).
The difference is constant, so we have 7 numbers
Tens digit will be either the larger or smaller, so we have 7x2=14 number
Eliminate the case [03], because 0 cannot play the role of tens digit.

Therefore, we have 14-1=13 numbers => (B) is correct

Sorry I am unable to find the existing thread for this question:

If 2,3,5 are the only prime factors of a positive integer p,what is the value of p?
A)p>100
B)p has exactly 12 factors including 1 and 12.

Why is the answer .


My answer is

12 is not a factor of 150 or 90...so U know the answer


Sent from my iPhone using GMAT Club Forum
_________________

Hi bro .. how is prep?

Please have a look at the official explanation. i think that is wrong, hence my doubt

My BAD.!
You are absolutely correct.
The OA should definitely be B.

if 12 is the factor the only possible combination would be 2^2*3*5


Best
Stone
_________________

hehe finally i succeeded in finding one wrong .....

Regards

Hi,

Thank you for posting awesome test. I have a question though. For Mock 3 question # 2, I am not able to follow the second stem and wondering if you could help me clarify my lack of understanding.


2)Data Sufficiency->If n is a positive integer,is n divisible by 2?
A)7n-8 is divisible by 20.
B)3n^2+2n+5 is a prime number.

I get that A is sufficient but I can't figure out B. Answer is D. Can you please stonecold help me figure this out. Thank you!

Hi Bobzi,

Statement 2:
\(3n^2+2n+5\) is a prime number. Now since n is a positive integer, it means that n >=1. If that is the case then, \(3n^2+2n+5\) would always be greater than 2. So for \(3n^2+2n+5\) to be prime, it ought to be odd.

Start with n=1: \(3n^2+2n+5\)=10 --Even; Thus \(3n^2+2n+5\) can never be prime when n=odd
With, n=2: \(3n^2+2n+5\)=21 --Odd; Thus \(3n^2+2n+5\) will always be odd when n=even and thus there will exist a number which will be prime [we need not find that number].

So we can conclude that n=even and even numbers are always divisible by 2. --Sufficient

I hope that clears your doubt !!

Hi Bobzi

Although gmatexam439 is spot on with his method, I would like to use a more fundamental approach for statement 2.

We need to use the basics.
Remember all primes >2 are odd? Yes, we will use that here.

We are told 3n^2+2n+5 is a prime number
Since n is a positive integer the least possible value of n is one.

So the least possible value of 3n^2+2n+5 is 3+3+5 => 11
Hence 3n^2+2n+5 is a prime number ≥11

What is the common thing about all primes greater than 2? They are odd.

Hence 3n^2+2n+5 is odd.

So 3n^2+ even + odd = odd
3n^2 + odd = odd
3n^2 = odd-odd = even

3n^2 = even

Hence n = even.

Hence sufficient
Smash that D.



Best
Stone
_________________

Excellent mate , Can we have answer explanations for these ? To be honest I'm getting many errors , would like to have some explanations . If you could help ?
_________________

Hi,
Which questions do you want me to discuss?

I am happy to help with any quant questions you have for me.
Just post in your question below and I will post in the solution.

Best
Stone
_________________

Stone ,

Can you give me a solution of these two ?

_________________