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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

After posting some 700+ questions, I've decided to post the problems which are not that hard. Though each question below has a trap or trick so be careful when solving. I'll post OA's with detailed solutions after some discussion. Good luck.

1. There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four socks are picked at random what is the probability of getting two socks of the same color? A. 1/5 B. 2/5 C. 3/4 D. 4/5 E. 1

2. If x is an integer and 9<x^2<99, then what is the value of maximum possible value of x minus minimum possible value of x? A. 5 B. 6 C. 7 D. 18 E. 20

3. Fanny and Alexander are 360 miles apart and are traveling in a straight line toward each other at a constant rate of 25 mph and 65 mph respectively, how far apart will they be exactly 1.5 hours before they meet? A. 25 miles B. 65 miles C. 70 miles D. 90 miles E. 135 miles

4. If -3<x<5 and -7<y<9, which of the following represent the range of all possible values of y-x? A. -4<y-x<4 B. -2<y-x<4 C. -12<y-x<4 D. -12<y-x<12 E. 4<y-x<12

5. The angles in a triangle are x, 3x, and 5x degrees. If a, b and c are the lengths of the sides opposite to angles x, 3x, and 5x respectively, then which of the following must be true? I. c>a+b II. c^2>a^2+b^2 III. c/a/b=10/6/2

A. I only B. II only C. III only D. I and III only E. II and III only

6. Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colors. How many different arrangements are possible? A. 30 B. 60 C. 120 D. 240 E. 480

7. After 2/9 of the numbers in a data set A were observed, it turned out that 3/4 of those numbers were non-negative. What fraction of the remaining numbers in set A must be negative so that the total ratio of negative numbers to non-negative numbers be 2 to 1? A. 11/14 B. 13/18 C. 4/7 D. 3/7 E. 3/14

8. There are 15 black chips and 5 white chips in a jar. What is the least number of chips we should pick to guarantee that we have 2 chips of the same color? A. 3 B. 5 C. 6 D. 16 E. 19

9. Julie is putting M marbles in a row in a repeating pattern: blue, white, red, green, black, yellow, pink. If the row begins with blue marble and ends with red marble, then which of the following could be the value of M? A. 22 B. 30 C. 38 D. 46 E. 54

11. The numbers {1, 3, 6, 7, 7, 7} are used to form three 2-digit numbers. If the sum of these three numbers is a prime number p, what is the largest possible value of p? A. 97 B. 151 C. 209 D. 211 E. 219

12. If \({-\frac{1}{3}}\leq{x}\leq{-\frac{1}{5}}\) and \({-\frac{1}{2}}\leq{y}\leq{-\frac{1}{4}}\), what is the least value of \(x^2*y\) possible? A. -1/100 B. -1/50 C. -1/36 D. -1/18 E. -1/6

4. If -3<x<5 and -7<y<9, which of the following represent the range of all possible values of y-x? A. -4<y-x<4 B. -2<y-x<4 C. -12<y-x<4 D. -12<y-x<12 E. 4<y-x<12

To get max value of y-x take max value of y and min value of x: 9-(-3)=12; To get min value of y-x take min value of y and max value of x: -7-(5)=-12;

Hence, the range of all possible values of y-x is -12<y-x<12.

Answer: D.

This answer is not 100% right, because there is not the sign <= but only <. therefore (assuming that X and Y are integers the answer is: (-6-(-4))<Y-X<(8-(-2))

The answer IS 100% correct.

First of all we are not told that x and y are integers.

Next, consider the following approach, we have -3<x<5 and -7<y<9,

Add y<9 and -3<x --> y-3<9+x --> y-x<12; Add -7<y and x<5 --> -7+x<y+5 --> -12<y-x;

Notice that most of the problems have short, easy and elegant solutions, since you've noticed a trick/trap hidden in the questions.

1. There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four socks are picked at random what is the probability of getting two socks of the same color? A. 1/5 B. 2/5 C. 3/4 D. 4/5 E. 1

No formula is need to answer this one. The trick here is that we have only 3 different color socks but we pick 4 socks, which ensures that in ANY case we'll have at least one pair of the same color (if 3 socks we pick are of the different color, then the 4th sock must match with either of previously picked one). P=1.

Answer: E.

Bunuel, I was wondering... if the question asked: if 2 socks are picked randomly, what is the probability of getting 2 socks of the same color, am I correct in going \(\frac{10}{20}*\frac{9}{19} + \frac{6}{20}*\frac{5}{19}+\frac{4}{20}*\frac{3}{19}\)?

11. The numbers {1, 3, 6, 7, 7, 7} are used to form three 2-digit numbers. If the sum of these three numbers is a prime number p, what is the largest possible value of p? A. 97 B. 151 C. 209 D. 211 E. 219

This is how i solved the ques:

We have to deal with 3 two digit numbers whose sum is as given in the ques. Considering the unit's digit in the various answer choices we have 9,7,1

Next look from the given set of integers (1, 3, 6, 7, 7, 7) , by a way of chosing any 3 values we get a possiblity of only "1" (--3+--7+--1)as a the unit digit. Hence our options are narowed to B/D i.e 151 , 211

Since we need to find the largest possible value check for option 211 which can be writen as 67+73+71=211

Also can be thought about in this way : If i add only unit digits of --3+--7+--1 i get the sum with 1 as the unit digit and a carry of 1 to the tens digit . So we have to see if there are any 3 numbers which can add to a sum of 20 , so that i can add the carry forward of "1" to make the number 21. A quick scan leaves with the possibility of 6,7,7 to make up for the 10's digit....

Hope this technique of looking at problems helps...

2. If x is an integer and 9<x^2<99, then what is the value of maximum possible value of x minus minimum possible value of x? A. 5 B. 6 C. 7 D. 18 E. 20

Also tricky. Notice that \(x\) can take positive, as well as negative values to satisfy \(9<x^2<99\), hence \(x\) can be: -9, -8, -7, -6, -4, 4, 5, 6, 7, 8, or 9. We asked to find the value of \(x_{max}-x_{min}\), ans since \(x_{max}=9\) and \(x_{min}=-9\) then \(x_{max}-x_{min}=9-(-9)=18\).

Answer: D.

X(max) should be 9.99 and X(min) should be -9.99

\sqrt{99} = 9.99

19.98 will be the correct answer. _________________

2. If x is an integer and 9<x^2<99, then what is the value of maximum possible value of x minus minimum possible value of x? A. 5 B. 6 C. 7 D. 18 E. 20

Also tricky. Notice that \(x\) can take positive, as well as negative values to satisfy \(9<x^2<99\), hence \(x\) can be: -9, -8, -7, -6, -4, 4, 5, 6, 7, 8, or 9. We asked to find the value of \(x_{max}-x_{min}\), ans since \(x_{max}=9\) and \(x_{min}=-9\) then \(x_{max}-x_{min}=9-(-9)=18\).

Answer: D.

X(max) should be 9.99 and X(min) should be -9.99

\sqrt{99} = 9.99

Notice that we are told that x is an integer. _________________

2. If x is an integer and 9<x^2<99, then what is the value of maximum possible value of x minus minimum possible value of x? A. 5 B. 6 C. 7 D. 18 E. 20

Also tricky. Notice that \(x\) can take positive, as well as negative values to satisfy \(9<x^2<99\), hence \(x\) can be: -9, -8, -7, -6, -4, 4, 5, 6, 7, 8, or 9. We asked to find the value of \(x_{max}-x_{min}\), ans since \(x_{max}=9\) and \(x_{min}=-9\) then \(x_{max}-x_{min}=9-(-9)=18\).

Answer: D.

X(max) should be 9.99 and X(min) should be -9.99

\sqrt{99} = 9.99

Notice that we are told that x is an integer.

Sorry boss, these silly mistakes will kill me in exam.

7. After 2/9 of the numbers in a data set A were observed, it turned out that 3/4 of those numbers were non-negative. What fraction of the remaining numbers in set A must be negative so that the total ratio of negative numbers to non-negative numbers be 2 to 1? A. 11/14 B. 13/18 C. 4/7 D. 3/7 E. 3/14

If choose variable for set A there will be too many fractions to manipulate with, so pick some smart #: let set A contain 18 numbers.

"2/9 of the numbers in a data set A were observed" --> 4 observed and 18-4=14 numbers left to observe; "3/4 of those numbers were non-negative" --> 3 non-negative and 1 negative; Ratio of negative numbers to non-negative numbers to be 2 to 1 there should be total of 18*2/3=12 negative numbers, so in not yet observed part there should be 12-1=11 negative numbers. Thus 11/14 of the remaining numbers in set A must be negative.

Answer: A.

Hey Bunuel, Can you please advice how you chose 18 as the smart number? Thanks,

7. After 2/9 of the numbers in a data set A were observed, it turned out that 3/4 of those numbers were non-negative. What fraction of the remaining numbers in set A must be negative so that the total ratio of negative numbers to non-negative numbers be 2 to 1? A. 11/14 B. 13/18 C. 4/7 D. 3/7 E. 3/14

If choose variable for set A there will be too many fractions to manipulate with, so pick some smart #: let set A contain 18 numbers.

"2/9 of the numbers in a data set A were observed" --> 4 observed and 18-4=14 numbers left to observe; "3/4 of those numbers were non-negative" --> 3 non-negative and 1 negative; Ratio of negative numbers to non-negative numbers to be 2 to 1 there should be total of 18*2/3=12 negative numbers, so in not yet observed part there should be 12-1=11 negative numbers. Thus 11/14 of the remaining numbers in set A must be negative.

Answer: A.

Hey Bunuel, Can you please advice how you chose 18 as the smart number? Thanks,

Choose a number which is a multiple of 9 (since we are told that "2/9 of the numbers in a data set..."). Try to choose a multiple so that you get another number ("3/4 of those numbers...") an integer too. _________________

Notice that most of the problems have short, easy and elegant solutions, since you've noticed a trick/trap hidden in the questions.

1. There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four socks are picked at random what is the probability of getting two socks of the same color? A. 1/5 B. 2/5 C. 3/4 D. 4/5 E. 1

No formula is need to answer this one. The trick here is that we have only 3 different color socks but we pick 4 socks, which ensures that in ANY case we'll have at least one pair of the same color (if 3 socks we pick are of the different color, then the 4th sock must match with either of previously picked one). P=1.

Answer: E.

Hi Bunnel, Isn't this is the probability of getting atleast 1 pair of socks? This will include cases where we have 2 pairs or all socks of same color. Question seems to be asking the probability of exactly 1 pair.

Notice that most of the problems have short, easy and elegant solutions, since you've noticed a trick/trap hidden in the questions.

1. There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four socks are picked at random what is the probability of getting two socks of the same color? A. 1/5 B. 2/5 C. 3/4 D. 4/5 E. 1

No formula is need to answer this one. The trick here is that we have only 3 different color socks but we pick 4 socks, which ensures that in ANY case we'll have at least one pair of the same color (if 3 socks we pick are of the different color, then the 4th sock must match with either of previously picked one). P=1.

Answer: E.

Hi Bunnel, Isn't this is the probability of getting atleast 1 pair of socks? This will include cases where we have 2 pairs or all socks of same color. Question seems to be asking the probability of exactly 1 pair.

In that case it would be asking about the probability of EXACTLY two socks of the same color. _________________

Thank you for the very high quality questions. Can you please explain your thinking behind the solution in question 8?

My confusion stems from the word "guarantee". Ideally the least number of chosen chips, that could result in 2 different colours, is indeed 3. However, the probability of this event is certainly not 1.

Thank you for the very high quality questions. Can you please explain your thinking behind the solution in question 8?

My confusion stems from the word "guarantee". Ideally the least number of chosen chips, that could result in 2 different colours, is indeed 3. However, the probability of this event is certainly not 1.

We are not interested in the probability in question 8: 8. There are 15 black chips and 5 white chips in a jar. What is the least number of chips we should pick to guarantee that we have 2 chips of the same color?

6. Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colours. How many different arrangements are possible? A. 30 B. 60 C. 120 D. 240 E. 480

Seems tough and complicated but if we read the stem carefully we find that the only way both conditions to be met for 5 red marbles, which are half of total marbles, they can be arranged only in two ways: R*R*R*R*R* or *R*R*R*R*R.

Here comes the next good news, in these cases BOTH conditions are met for all other marbles as well: no two adjacent marbles will be of the same color and the first and the last marbles will be of different colors.

Now, it's easy: 2 blue, 2 green and 1 yellow can be arranged in 5 empty slots in 5!/(2!*2!)=30 ways (permutation of 5 letters BBGGY out of which 2 B's and 2 G' are identical). Finally as there are two cases (R*R*R*R*R* and *R*R*R*R*R. ) then total # of arrangement is 30*2=60.

6. Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colours. How many different arrangements are possible? A. 30 B. 60 C. 120 D. 240 E. 480

Seems tough and complicated but if we read the stem carefully we find that the only way both conditions to be met for 5 red marbles, which are half of total marbles, they can be arranged only in two ways: R*R*R*R*R* or *R*R*R*R*R.

Here comes the next good news, in these cases BOTH conditions are met for all other marbles as well: no two adjacent marbles will be of the same color and the first and the last marbles will be of different colors.

Now, it's easy: 2 blue, 2 green and 1 yellow can be arranged in 5 empty slots in 5!/(2!*2!)=30 ways (permutation of 5 letters BBGGY out of which 2 B's and 2 G' are identical). Finally as there are two cases (R*R*R*R*R* and *R*R*R*R*R. ) then total # of arrangement is 30*2=60.

Answer: B.

what about R B G R Y R G R B R ?

This arrangement violates the condition that the first and the last marbles must be of different colours. _________________

Part 2 of the GMAT: How I tackled the GMAT and improved a disappointing score Apologies for the month gap. I went on vacation and had to finish up a...

So the last couple of weeks have seen a flurry of discussion in our MBA class Whatsapp group around Brexit, the referendum and currency exchange. Most of us believed...

This highly influential bestseller was first published over 25 years ago. I had wanted to read this book for a long time and I finally got around to it...