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:

I'm posting the next set of medium/hard PS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers. Good luck!

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000 B. 810/100,000 C. 858/100,000 D. 860/100,000 E. 1530/100,000

3. For the past k days the average (arithmetic mean) cupcakes per day that Liv baked was 55. Today Bibi joined and together with Liv they baked 100 cupcakes, which raises the average to 60 cupcakes per day. What is the value of k? A. 6 B. 8 C. 9 D. 10 E. 12

5. There are 7 red and 5 blue marbles in a jar. In how many ways 8 marbles can be selected from the jar so that at least one red marble and at least one blue marble to remain in the jar? A. 460 B. 490 C. 493 D. 455 E. 445

6. A pool has two water pumps A and B and one drain C. Pump A alone can fill the whole pool in x hours, and pump B alone can fill the whole pool in y hours. The drain can empty the whole pool in z hours, where z>x. With pumps A and B both running and the drain C unstopped till the pool is filled, which of the following represents the amount of water in terms of the fraction of the pool which pump A pumped into the pool? A. \(\frac{yz}{x+y+z}\)

7. Metropolis Corporation has 4 shareholders: Fritz, Luis, Alfred and Werner. Number of shares that Fritz owns is 2/3 rd of number of the shares of the other three shareholders, number of the shares that Luis owns is 3/7 th of number of the shares of the other three shareholders and number of the shares that Alfred owns is 4/11 th of number of the shares of the other three shareholders. If dividends of $3,600,000 were distributed among the 4 shareholders, how much of this amount did Werner receive? A. $60,000 B. $90,000 C. $100,000 D. $120,000 E. $180,000

8. A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is -185 what is the sum of the 5 smallest integers of set A? A. -165 B. -175 C. -195 D. -205 E. -215

11. In an infinite sequence 1, 3, 9, 27, ... each term after the first is three times the previous term. What is the difference between the sum of 13th and 15th terms and the sum of 12th and 14th terms of the sequence? A. 10*3^11 B. 20*3^11 C. 10*3^12 D. 40*3^11 E. 20*3^12

12. x, y and z are positive integers such that when x is divided by y the remainder is 3 and when y is divided by z the remainder is 8. What is the smallest possible value of x+y+z? A. 12 B. 20 C. 24 D. 29 E. 33

13. If \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}\), what is the product of the tens and the units digits of \(\frac{x}{(8!)^3}-39\)? A. 0 B. 6 C. 7 D. 12 E. 14

-y = +y and |y| no matter what, comes out as positive hence Sqrt (-y*|y| ) comes out as possitive y and sqrtx^2 =|x|= positive x , of course from your tone i get that i'm going wrong some where but

What I meant is that if y<0 then |y|=-y and -1-|y|=-1-(-y)=-1+y. _________________

-y = +y and |y| no matter what, comes out as positive hence Sqrt (-y*|y| ) comes out as possitive y and sqrtx^2 =|x|= positive x , of course from your tone i get that i'm going wrong some where but

What I meant is that if y<0 then |y|=-y and -1-|y|=-1-(-y)=-1+y.

isn't -1 -|y|=-1-y for the same reason |x|/x became -1 because numerator x was positive and denominator x was negative . sqrt(negative -y *|y|) so negative -y =postive y and |y| = positive hence when y comes out it comes as positive y and because of the negative sign -1 - y=-1-y please help bunuel

-y = +y and |y| no matter what, comes out as positive hence Sqrt (-y*|y| ) comes out as possitive y and sqrtx^2 =|x|= positive x , of course from your tone i get that i'm going wrong some where but

What I meant is that if y<0 then |y|=-y and -1-|y|=-1-(-y)=-1+y.

isn't -1 -|y|=-1-y for the same reason |x|/x became -1 because numerator x was positive and denominator x was negative . sqrt(negative -y *|y|) so negative -y =postive y and |y| = positive hence when y comes out it comes as positive y and because of the negative sign -1 - y=-1-y please help bunuel

sqroot of (negative number ^2) = (negative number ^2)^1/2=negative number

for example sqroot (-5*|5|)=sqroot (-5^2)=((-5)^2)^1/2=-5 _________________

Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth

Bunuel, could u please tell us which question is still unanswered or answered wrong by all of us? I want to think on such questions more, before u post solutions.

and btw, do u read our solutions or just only answers? I wonder whether my way of thinking was ok. in some cases I tried to use another method not to repeat others. I just wonder whether it worked, or it was just coincidence _________________

Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth

then why is |x|/x =-1 , should not it be 1 as |x|=-x and denominator is negative as well

You are right that |x|=-x since x is negative but tell me, what is -x, negative or positive? Negative of negative gives you positive, right? So -x must be positive. Now, if x is negative, -x/x must be positive/negative giving you -1.

You are confusing yourself too much with negatives and positives. Just think of it this way:

|x|/x = -x/x (By definition, since |x| = -x when x < 0) -x/x = -1 ( x and x get canceled here leaving you with -1) _________________

9. If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}\)? A. 1+y B. 1-y C. -1-y D. y-1 E. x-y

Note that \(\sqrt{a^2}=|a|\). Next, since \(x<0\) and \(y<0\) then \(|x|=-x\) and \(|y|=-y\).

So, \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y\)

Answer: D.

Hi Bunuel,

I didn't quite understand this ..

X and Y are odd then how can |x| = -x it has to be +ive.

x and y are negative numbers, not odd numbers.

If \(x\geq{0}\) then \(|x|=x\); If \(x<{0}\) then \(|x|=-x\). So if \(x\) is negative then \(|x|=-x=-negative=positive\). For example, if \(x=-2\) then \(|x|=|-2|=2=-x\).

Got 9 correct with @ 2.5 mins/Q ... how about others ? (2 Q left without going to deep as killed more than 4 mins on those Q4,7- this proves that not to waste time on something you don't get idea in 1.5-2 mins time frame and missed 2 narrowly- Q9, 10: 10 I din't read 2y but only y so...)..will post in details some explanations, esp. those which are little different from already posted by the very own legend Mr B!! _________________

6. A pool has two water pumps A and B and one drain C. Pump A alone can fill the whole pool in x hours, and pump B alone can fill the whole pool in y hours. The drain can empty the whole pool in z hours, where z>x. With pumps A and B both running and the drain C unstopped till the pool is filled, which of the following represents the fraction of the pool which pump A filled? A. \(\frac{yz}{x+y+z}\)

B. \(\frac{yz}{yz+xz-xy}\)

C. \(\frac{yz}{yz+xz+xy}\)

D. \(\frac{xyz}{yz+xz-xy}\)

E. \(\frac{yz+xz-xy}{yz}\)

With pumps A and B both running and the drain unstopped the pool will be filled in a rate \(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{yz+xz-zy}{xyz}\) pool/hour. So, the pool will be filled in \(\frac{xyz}{yz+xz-xy}\) hours (time is reciprocal of rate).

In \(\frac{xyz}{yz+xz-xy}\) hours pump A will do \(\frac{1}{x}*\frac{xyz}{yz+xz-xy}=\frac{yz}{yz+xz-xy}\) part of the job.

Answer: B.

hi , i feel the ans to this Q is wrong. as we have to tell what fraction of the pool was filled up by pump A. it will not depend on the drain C as the filling was done only by pump A and B. ans shud be y/(x+y). the given ans would be correct if we were to find "the amount of water pumped by A in that duration"

Actually that was the intended meaning of the question. I edited it so to avoid ambiguity: "which of the following represents the amount of water in terms of the fraction of the pool which pump A pumped into the pool?"

Now, if you analyze each option you'll see that only \(52^4=2^4*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000 B. 810/100,000 C. 858/100,000 D. 860/100,000 E. 1530/100,000

Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.

# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.

\(P=\frac{favorable}{total}=\frac{810}{10^5}\)

Answer: B.

Hello Bunnel....

Can you kindly clarify this question and elaborate on the solution!!

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000 B. 810/100,000 C. 858/100,000 D. 860/100,000 E. 1530/100,000

Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.

# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.

\(P=\frac{favorable}{total}=\frac{810}{10^5}\)

Answer: B.

Hello Bunnel....

Can you kindly clarify this question and elaborate on the solution!!

Thanks in advance.

Can you please specify which part didn't you understand? Thanks. _________________

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000 B. 810/100,000 C. 858/100,000 D. 860/100,000 E. 1530/100,000

Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.

# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.

\(P=\frac{favorable}{total}=\frac{810}{10^5}\)

Answer: B.

Hey Bunuel,

Sorry to trouble you. If you could, I would appreciate if you could clarify more about 5C3. So this is how I approached the problem:

We have three - 6's. With two slots that can be filled by 9 numbers.

666 99 . Now the question is how many ways can we rearrange these codes of three 6's and 2 non 6's? Well, have 5 elements 3 identical (because we have 3 identical 6's and two non-6's) Hence we have 5!/(3!2!) ways of rearranging the numbers. Is this why we have 5C3 in the problem?

But the two non-6's can be identical or different. For example 66611 66622 66633 66644 66655 ... 66699... etc all satisfy the condition of only three 6's. The number of ways to rearrange the letters {aaabb} is 5!/(3!2!). Moreover, we can also have the two non-6's be different, 66612 66654 66624, etc . the number of ways to rearrange the letters {aaabc} is 5!/(3!).

What do we do in this situation? Do we look at this problem as (#6)(#6)(#6)(not #6)(not #6) therefore 5!/(3!2!). Three repeating elements of #6 and two repeating elements of not #6.

OR

Do we look at the problem as (#6)(#6)(#6)(x)(y), where x and y two numbers that are not 6, that may or may not be equal, therefore 5!/3! (3! for the 3 repeating 6's but we cannot say anything about whether x and y are equal to each other)

Bunuel, as always, thank you so much!! And again, sorry to trouble you.

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

HBS alum talks about effective altruism and founding and ultimately closing MBAs Across America at TED: Casey Gerald speaks at TED2016 – Dream, February 15-19, 2016, Vancouver Convention Center...

By Libby Koerbel Engaging a room of more than 100 people for two straight hours is no easy task, but the Women’s Business Association (WBA), Professor Victoria Medvec...