GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 21 Oct 2019, 21:33 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

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.  The Ultimate Q51 Guide [Expert Level]

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8025
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

Show Tags

For a positive integer n, when 12n is divided by 15, which of the following cannot be the remainder?

A. 0
B. 3
C. 5
D. 6
E. 9

==>From 12n=15Q+r, 12 and 15 are the multiples of 3, so remainder=r must also be the multiple of 3. Therefore, 5 cannot be the remainder.

The answer is C.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8025
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

Show Tags

G(x) is the greatest integer less than or equal to x and L(x) is the least integer greater than or equal to x. When x is not an integer, which of the following is the value of L(x)-G(x)?

A. -2
B. -1
C. 0
D. 1
E. 2

==> You get G(x)=round down and L(X)=round up. Then, x≠integer, so if you substitute x=1.2, you get L(1.2)-G(1.2)=2-1=1.

The answer is D.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8025
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

Show Tags

If the average (arithmetic mean) of a, b, and c is (a+2b)/3, what is the value of b?

1) a=1
2) c=2

==> If you modify the original condition and the question and check the question again, from is (a+2b)/3=(a+b+c)/3, you get a+2b=a+b+c, then b=c. In order to find b, you only need to know c.

Therefore, the answer is B.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8025
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

Show Tags

If two integers x and y such x>y are selected at random between -8 and 11, inclusive, how many cases are possible?

A. 150
B. 180
C. 190
D. 210
E. 240

==> Since two integers from -8 to 11 are being randomly selected and x>y, you use combination. Thus, the number of integers from -8 to 11 becomes 11-(-8)+1=20, so 20, then 20C2=(20)(19)/2!=190.

The answer is C.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8025
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

Show Tags

Which of the following points reflect to y=-x at (-3,2)?

A. (-2, 3)
B. (2,-3)
C. (3,2)
D. (3,-2)
E. (2,3)

==> In order to become symmetrical to y=-x, you need to substitute –y value on x coordinate, and –x value on y coordinate. Thus, you get (-3,2)-->(-2,3).

The answer is A.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8025
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

Show Tags

There are 6 red balls and 4 blue balls in a jar. If 2 balls are selected from the jar, what is the probability that 2 balls selected are red balls?

A. 1/10
B. 1/9
C. 3/10
D. 3/15
E. 2/15

==> You get 4C2/10C2==2/15. The answer is E.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8025
GMAT 1: 760 Q51 V42 GPA: 3.82
The Ultimate Q51 Guide [Expert Level]  [#permalink]

Show Tags

5 people including A and B line up in a row. How many possible cases are there such that at least one person stands between A and B?

A. 24
B. 36
C. 48
D. 60
E. 72

==> Since it is the number of cases such that at least one person stands between A and B when 5 people are lined up in a row, you need to subtract the number of cases where A and B stands next to each other from the total number of cases. Then, you get ABCDE-(AB)CDE, which becomes 5!-4!(2)=72. You get 4!(2) because there are cases where A and B switch the order in the line. Therefore, the answer is E.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8025
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

Show Tags

If x and y are integers, is x+y an odd number?

1) y=3x+1
2) y=2x+3

==> In the original condition, there are 2 variables (x,y) and in order to match the number of variables to the number of equations, there must be 2 equations. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2), from 3x+1=2x+3, you get x=2 y=7. Then, you get x+y=2+7=9=odd, hence yes, it is sufficient. The answer is C. However, this is an integer question, one of the key questions, so you apply CMT 4 (A: if you get C too easily, consider A or B). For con 1), from y=3x+1, you get (x,y)=(even,odd) or (odd,even), which always becomes x+y=odd, hence yes, it is sufficient. For con 2), you get (x,y)=(1,5) no, (2,7) yes, it is not sufficient. Therefore, the answer is A.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8025
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

Show Tags

Is kr<0?

1) k^2r^3<0
2) |k+r|<|k|+|r|

==> For con 1), you ignore the square, so t<0, and for con 2), you get kr<0, hence yes, it is sufficient. The reason is that from (|k+r|)^2<(|k|+|r|)^2, you get k^2+r^2+2kr<k^2+r^2+2|kr|, and if you get rid of k^2+r^2 from both sides, you get 2kr<2|kr|, then kr<|kr|, which becomes kr<0. The answer is B.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8025
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

Show Tags

If a and b are 1-digit positive integers, is 100a+10b+2 divisible by 4?

1) a=2.
2) b=3.

==> If you modify the original condition and the question, the remainder when an integer n is divided by 4 is equal to the remainder when only the units digit and the tens digit are divided by 4, and so you only need to find b. Thus, for con 2), you get b=3, which becomes 10(3)+2=32, and so it is always divisible by 4, hence yes, it is sufficient. The answer is B.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8025
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

Show Tags

[p] is the greatest integer less than or equal p. What is the value of [π-1]?

A. 2
B. 3
C. 4
D. 5
E. 6

==>You get π-1=3.14-1=2.14, and thus [π-1]=[2.14]=2. The answer is A.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8025
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

Show Tags

If $$n=s^at^b$$, where r, a, b and s are integers, is $$\sqrt{n}$$ an integer?

1) a+b is an even number
2) a is an even number

==> In the original condition, there are 5 variables (n,s,t,a,b) and 1 equation ($$n=s^at^b$$). In order to match the number of variables to the number of equations, there must be 5 equations. Since there is 1 for con 1) and 1 for con 2), E is most likely to be the answer.
By solving con 1) and con 2), s=t=2 and a=b=2 yes, but s=t=2 and a=b=-2 no, hence it is not sufficient.

Therefore, the answer is E.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8025
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

Show Tags

How many possible 6-digit codes can be formed from three a’s, two b’s, and one c?

A. 40
B. 50
C. 60
D. 70
E. 80

==> Since it is the number of ways of listing a,a,a,b,b,c, you get $$\frac{6!}{(3!)(2!)(1)}=60$$.
The answer is C.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8025
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

Show Tags

If a and b are positive integers, what is the number of the different prime factors of a^b?

1) a has different 2 prime factors.
2) b has different 3 prime factors.

==> If you modify the original condition and the question, the number of prime factors doesn’t affect the exponent. In other words, the number of prime factors of a^b is equal to the number of prime factors of a.

Therefore, the answer is A.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8025
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

Show Tags

Is xyz>0?

1) xy>0
2) yz>0

==> In the original condition, there are 3 variables (x,y,z) and in order to match the number of variables to the number of equations, there must be 3 equations. Since there is 1 for con 1) and 1 for con 2), E is most likely to be the answer. By solving con 1) and con 2), if x=y=z=1, yes, but if x=y=x=-1, no, hence it is not sufficient.

Therefore, the answer is E.
_________________
Intern  B
Joined: 06 Feb 2016
Posts: 47
Location: Poland
Concentration: Finance, Accounting
GMAT 1: 730 Q49 V41 GPA: 3.5
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

Show Tags

Thank you MathRevolution. I got 49 and the types of questions you mentioned are exactly what I am the weakest at.
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8025
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

Show Tags

For an investment, you start off with $1,000, the interest rate of 5%, and you deposited two years as compound interest. What is the total interest earned in 2 years? A.$96.5
B. $98.5 C.$100.5
D. $102.5 E.$105.2

==> You get $$1,000(1+0.05)^2-1,000=102.5.$$

The answer is D.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8025
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

Show Tags

What is the median of the consecutive multiples of 7 in the first 50 positive integers?

A. 14
B. 21
C. 28
D. 35
E. 42

==> Since it is the first 50 positive integers, you need to find the median of the consecutive multiples of 7, you get 7, 14, 21, 28, 35, 42, 49, and the median is 28.

The answer is C.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8025
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

Show Tags

If [x] is the least integer greater than or equal to x, and [x/2]=3, what is the scope of x？

A. 3<x≤4
B. 8<x≤9
C. 8<x≤10
D. 9<x≤12
E. 4<x≤6

==> For [x], you get [1.2]=2, which means rounded up. In order to get [x/2]=3, you get 2<x/2≤3 and then 4<x≤6.

Therefore, the answer is E.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8025
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

Show Tags

What is the number of multiples of 6 from -35 to 69?

A. 14
B. 16
C. 17
D. 20
E. 21

==> 0 becomes the multiple of all integers. From 1~69, 11 of them are multiples of 6, from -1~-35, 5 of them are multiples of 6, hence you get 11+5+1=17.

The answer is C.
_________________ Re: The Ultimate Q51 Guide [Expert Level]   [#permalink] 15 Jun 2017, 18:13

Go to page   Previous    1  ...  10   11   12   13   14   15   16  ...  44    Next  [ 870 posts ]

Display posts from previous: Sort by

The Ultimate Q51 Guide [Expert Level]

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderator: DisciplinedPrep

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  