Last visit was: 09 Jul 2025, 03:12 It is currently 09 Jul 2025, 03:12
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
stonecold
Joined: 12 Aug 2015
Last visit: 09 Apr 2024
Posts: 2,249
Own Kudos:
3,462
 [5]
Given Kudos: 893
GRE 1: Q169 V154
GRE 1: Q169 V154
Posts: 2,249
Kudos: 3,462
 [5]
1
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
avatar
Vianand4
Joined: 17 May 2016
Last visit: 13 Feb 2017
Posts: 5
Own Kudos:
1
 [1]
Given Kudos: 7
WE:Information Technology (Computer Software)
Posts: 5
Kudos: 1
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
stonecold
Joined: 12 Aug 2015
Last visit: 09 Apr 2024
Posts: 2,249
Own Kudos:
3,462
 [1]
Given Kudos: 893
GRE 1: Q169 V154
GRE 1: Q169 V154
Posts: 2,249
Kudos: 3,462
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
stonecold
Joined: 12 Aug 2015
Last visit: 09 Apr 2024
Posts: 2,249
Own Kudos:
3,462
 [1]
Given Kudos: 893
GRE 1: Q169 V154
GRE 1: Q169 V154
Posts: 2,249
Kudos: 3,462
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
User avatar
kanusha
Joined: 25 Mar 2013
Last visit: 03 Aug 2017
Posts: 159
Own Kudos:
146
 [1]
Given Kudos: 101
Location: United States
Concentration: Entrepreneurship, Marketing
GPA: 3.5
Products:
Posts: 159
Kudos: 146
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
User avatar
kanusha
Joined: 25 Mar 2013
Last visit: 03 Aug 2017
Posts: 159
Own Kudos:
146
 [1]
Given Kudos: 101
Location: United States
Concentration: Entrepreneurship, Marketing
GPA: 3.5
Products:
Posts: 159
Kudos: 146
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
User avatar
stonecold
Joined: 12 Aug 2015
Last visit: 09 Apr 2024
Posts: 2,249
Own Kudos:
Given Kudos: 893
GRE 1: Q169 V154
GRE 1: Q169 V154
Posts: 2,249
Kudos: 3,462
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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%



User avatar
stonecold
Joined: 12 Aug 2015
Last visit: 09 Apr 2024
Posts: 2,249
Own Kudos:
Given Kudos: 893
GRE 1: Q169 V154
GRE 1: Q169 V154
Posts: 2,249
Kudos: 3,462
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
User avatar
Lucy Phuong
Joined: 24 Jan 2017
Last visit: 12 Aug 2021
Posts: 119
Own Kudos:
343
 [1]
Given Kudos: 106
GMAT 1: 640 Q50 V25
GMAT 2: 710 Q50 V35
GPA: 3.48
Products:
GMAT 2: 710 Q50 V35
Posts: 119
Kudos: 343
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
stonecold
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
User avatar
gmatexam439
User avatar
Moderator
Joined: 28 Mar 2017
Last visit: 18 Oct 2024
Posts: 1,066
Own Kudos:
2,130
 [1]
Given Kudos: 200
Location: India
Concentration: Finance, Technology
GMAT 1: 730 Q49 V41
GPA: 4
Products:
GMAT 1: 730 Q49 V41
Posts: 1,066
Kudos: 2,130
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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 .
Show SpoilerExplanation
Since 2,3,5 are the only prime factors of p=> The general expression for p will be 2a∗3b∗5c2a∗3b∗5c where a,b,c are positive integers.
Statement 1->p>100.
There exist infinite such cases.
In-fact there are only a few cases where p will be less than 100.
Hence not sufficient.
Statement 2->
Factors =12
Hence (a+1)*(b+1)*(c+1)=12=2*2*3
NOTE-> We can't arrange 12 as 1*1*12 or 2*6*1 or 4*3*1 as a,b,c are positive integers.
Hence a,b,c => (1,1,2) or (1,2,1) or (2,1,1)
Posible values of p=> 150,90,60
Hence not sufficient.
Combining the two statements => p must be 150.
Hence C.

My answer is
Show Spoilermy answer
B

Show SpoilerMy understanding
The option B explicitly states that 12 is a factor of p. So, 2 must have a power of 2 and 3 must have power 1. Now since number of factors is 12, 5 should have power=1.
Thus only feasible solution is p=60.

Please point out the issue in my understanding
User avatar
Luckisnoexcuse
User avatar
Current Student
Joined: 18 Aug 2016
Last visit: 16 Apr 2022
Posts: 520
Own Kudos:
655
 [1]
Given Kudos: 198
Concentration: Strategy, Technology
GMAT 1: 630 Q47 V29
GMAT 2: 740 Q51 V38
Products:
GMAT 2: 740 Q51 V38
Posts: 520
Kudos: 655
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
12 is not a factor of 150 or 90...so U know the answer


Sent from my iPhone using GMAT Club Forum
User avatar
gmatexam439
User avatar
Moderator
Joined: 28 Mar 2017
Last visit: 18 Oct 2024
Posts: 1,066
Own Kudos:
2,130
 [1]
Given Kudos: 200
Location: India
Concentration: Finance, Technology
GMAT 1: 730 Q49 V41
GPA: 4
Products:
GMAT 1: 730 Q49 V41
Posts: 1,066
Kudos: 2,130
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Luckisnoexcuse
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
User avatar
stonecold
Joined: 12 Aug 2015
Last visit: 09 Apr 2024
Posts: 2,249
Own Kudos:
Given Kudos: 893
GRE 1: Q169 V154
GRE 1: Q169 V154
Posts: 2,249
Kudos: 3,462
Kudos
Add Kudos
Bookmarks
Bookmark this Post
gmatexam439
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 .
Show SpoilerExplanation
Since 2,3,5 are the only prime factors of p=> The general expression for p will be 2a∗3b∗5c2a∗3b∗5c where a,b,c are positive integers.
Statement 1->p>100.
There exist infinite such cases.
In-fact there are only a few cases where p will be less than 100.
Hence not sufficient.
Statement 2->
Factors =12
Hence (a+1)*(b+1)*(c+1)=12=2*2*3
NOTE-> We can't arrange 12 as 1*1*12 or 2*6*1 or 4*3*1 as a,b,c are positive integers.
Hence a,b,c => (1,1,2) or (1,2,1) or (2,1,1)
Posible values of p=> 150,90,60
Hence not sufficient.
Combining the two statements => p must be 150.
Hence C.

My answer is
Show Spoilermy answer
B

Show SpoilerMy understanding
The option B explicitly states that 12 is a factor of p. So, 2 must have a power of 2 and 3 must have power 1. Now since number of factors is 12, 5 should have power=1.
Thus only feasible solution is p=60.

Please point out the issue in my understanding

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
User avatar
gmatexam439
User avatar
Moderator
Joined: 28 Mar 2017
Last visit: 18 Oct 2024
Posts: 1,066
Own Kudos:
2,130
 [1]
Given Kudos: 200
Location: India
Concentration: Finance, Technology
GMAT 1: 730 Q49 V41
GPA: 4
Products:
GMAT 1: 730 Q49 V41
Posts: 1,066
Kudos: 2,130
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
stonecold
gmatexam439
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 .
Show SpoilerExplanation
Since 2,3,5 are the only prime factors of p=> The general expression for p will be 2a∗3b∗5c2a∗3b∗5c where a,b,c are positive integers.
Statement 1->p>100.
There exist infinite such cases.
In-fact there are only a few cases where p will be less than 100.
Hence not sufficient.
Statement 2->
Factors =12
Hence (a+1)*(b+1)*(c+1)=12=2*2*3
NOTE-> We can't arrange 12 as 1*1*12 or 2*6*1 or 4*3*1 as a,b,c are positive integers.
Hence a,b,c => (1,1,2) or (1,2,1) or (2,1,1)
Posible values of p=> 150,90,60
Hence not sufficient.
Combining the two statements => p must be 150.
Hence C.

My answer is
Show Spoilermy answer
B

Show SpoilerMy understanding
The option B explicitly states that 12 is a factor of p. So, 2 must have a power of 2 and 3 must have power 1. Now since number of factors is 12, 5 should have power=1.
Thus only feasible solution is p=60.

Please point out the issue in my understanding

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
avatar
Bobzi
Joined: 10 Feb 2017
Last visit: 12 Mar 2020
Posts: 9
Own Kudos:
7
 [1]
Given Kudos: 98
Posts: 9
Kudos: 7
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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!
User avatar
gmatexam439
User avatar
Moderator
Joined: 28 Mar 2017
Last visit: 18 Oct 2024
Posts: 1,066
Own Kudos:
2,130
 [3]
Given Kudos: 200
Location: India
Concentration: Finance, Technology
GMAT 1: 730 Q49 V41
GPA: 4
Products:
GMAT 1: 730 Q49 V41
Posts: 1,066
Kudos: 2,130
 [3]
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bobzi
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 !! :)
User avatar
stonecold
Joined: 12 Aug 2015
Last visit: 09 Apr 2024
Posts: 2,249
Own Kudos:
3,462
 [1]
Given Kudos: 893
GRE 1: Q169 V154
GRE 1: Q169 V154
Posts: 2,249
Kudos: 3,462
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bobzi
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

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
User avatar
loserunderachiever
Joined: 26 Feb 2018
Last visit: 14 Aug 2018
Posts: 36
Own Kudos:
Given Kudos: 24
WE:Sales (Internet and New Media)
Posts: 36
Kudos: 8
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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 ?
User avatar
stonecold
Joined: 12 Aug 2015
Last visit: 09 Apr 2024
Posts: 2,249
Own Kudos:
Given Kudos: 893
GRE 1: Q169 V154
GRE 1: Q169 V154
Posts: 2,249
Kudos: 3,462
Kudos
Add Kudos
Bookmarks
Bookmark this Post
loserunderachiever
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
User avatar
loserunderachiever
Joined: 26 Feb 2018
Last visit: 14 Aug 2018
Posts: 36
Own Kudos:
Given Kudos: 24
WE:Sales (Internet and New Media)
Posts: 36
Kudos: 8
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Stone ,

Can you give me a solution of these two ?
Attachments

Screen Shot 2018-03-03 at 12.57.18 PM.png
Screen Shot 2018-03-03 at 12.57.18 PM.png [ 104.42 KiB | Viewed 4736 times ]

   1   2   3   4   5   
Moderator:
Math Expert
102604 posts