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 DS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers. Good luck!

1. Bonnie can paint a stolen car in x hours, and Clyde can paint the same car in y hours. They start working simultaneously and independently at their respective constant rates at 9:45am. If both x and y are odd integers, is x=y? (1) x^2+y^2<12 (2) Bonnie and Clyde complete the painting of the car at 10:30am

4. How many numbers of 5 consecutive positive integers is divisible by 4? (1) The median of these numbers is odd (2) The average (arithmetic mean) of these numbers is a prime number

7. A certain fruit stand sold total of 76 oranges to 19 customers. How many of them bought only one orange? (1) None of the customers bought more than 4 oranges (2) The difference between the number of oranges bought by any two customers is even

10. The function f is defined for all positive integers a and b by the following rule: f(a,b)=(a+b)/GCF(a,b), where GCF(a,b) is the greatest common factor of a and b. If f(10,x)=11, what is the value of x? (1) x is a square of an integer (2) The sum of the distinct prime factors of x is a prime number.

1. Bonnie can paint a stolen car in x hours, and Clyde can paint the same car in y hours. They start working simultaneously and independently at their respective constant rates at 9:45am. If both x and y are odd integers, is x=y? (1) x^2+y^2<12 (2) Bonnie and Clyde complete the painting of the car at 10:30am

The catch is that they are working independently.

Answer B

Just to clarify more, working independently does not mean they are painting different cars. They are still painting the same car. Only that, the events are mutually exclusive, as you say in probabilistic terms, so that rates are not affected when they work simultaneously.

Another way : 1) more than one solution possible. 2) let's say 1)x=y=1. Work will be completed in 1/2 hour i.e. 10.15 am. 2)x=y=3. Work will be completed in 3/2 hour i.e. 11.15 am. Not possible. B. _________________

My funda - Area of square is largest among all the quadilateral with same perimeter. Stmt 1 - Only possible values of x and y are 1/Sqrt(2). So sufficient as xy = 1/2 Stmt 2 - Only says x and y are equal. Not sufficient Answer A

You are close to correct reasoning for (1), though from it you can not say that xy=1/2 and the only possible value for x and y are 1/Sqrt(2). Consider the following example: 0^1+1^2=1.

As for (2): x^2-y^2=0 doesn't mean that x=y.

1) x2 + y2 = 1 is a circle with radius 1, and origin as center. We are only concerned with 1st and 3rd quad (think why? ), Draw a line y=x which intersects circle at x=y=1/sqrt(2). We can observe, as x becomes greater than 1/sqrt2, y gets lesser than 1/sqrt2 moving on the circle (below y=x line). Hence xy<= 1/2. Similarly, as y becomes greater than 1/sqrt2, x gets lesser than 1/sqrt2 moving on the circle (above y=x line). Hence xy<= 1/2. Hence sufficient.

For more mathematically inclined, draw graph of xy=1/2. It has only one point of contact with the circle, tangential at x=y=1/sqrt(2).

Attachments

gmatclub exp.png [ 10.25 KiB | Viewed 18171 times ]

IMO, if you can imagine drawing the circle and y=x line, it takes less than 30 sec. to figure it out. It only takes so much longer to explain in text. And ofcourse, no need to draw the xy graph. Will wait though for the even faster method, if any.

If you answered this question in less than 30 sec using this approach then all I can say is great job!

As for the easier/faster solution, little hint: it involves simplest algebraic manipulation. _________________

(1) 3x + 4 < 2y + 3 --> \(3x<2y-1\). \(x\) can be some very small number for instance -100 and \(y\) some large enough number for instance -3 and the answer would be YES, \(x<y\) BUT if \(x=-2\) and \(y=-2.1\) then the answer would be NO, \(x>y\). Not sufficient.

Hi bunuel, Did not got how u solved option 2.Could you please explain in detail. thanks.

(2) 2x - 3 < 3y - 4 --> \(x<1.5y-\frac{1}{2}\) --> \(x<y+(0.5y-\frac{1}{2})\). Now, since \(y\) is a negative number then \(0.5y-\frac{1}{2}=negative\) so, we have that: \(x<y+negative\). \(y+negative\) is less then \(y\) and if \(x\) is less than \(y+negative\) then it must also be less than \(y\) itself: \(x<y\).

10. The function f is defined for all positive integers a and b by the following rule: f(a,b)=(a+b)/GCF(a,b), where GCF(a,b) is the greatest common factor of a and b. If f(10,x)=11, what is the value of x?

Notice that the greatest common factor of 10 and x, GCF(10,x), naturally must be a factor of 10: 1, 2, 5, and 10. Thus from f(10,x)=11 we can get four different values of x:

(1) x is a square of an integer --> \(x\) can be 1 or 100. Not sufficient.

(2) The sum of the distinct prime factors of x is a prime number ---> distinct primes of 12 are 2 and 3: \(2+3=5=prime\), distinct primes of 45 are 3 and 5: \(3+5=8\neq{prime}\) and distinct primes of 100 are also 2 and 3: \(2+3=5=prime\). \(x\) can be 12 or 100. Not sufficient.

(1)+(2) \(x\) can only be 100. Sufficient.

Answer: C.

hey..., can sm1 pls explain how primes of 100 can be 2 and 3?...(2nd last line),,.. thanx..

10. The function f is defined for all positive integers a and b by the following rule: f(a,b)=(a+b)/GCF(a,b), where GCF(a,b) is the greatest common factor of a and b. If f(10,x)=11, what is the value of x?

Notice that the greatest common factor of 10 and x, GCF(10,x), naturally must be a factor of 10: 1, 2, 5, and 10. Thus from f(10,x)=11 we can get four different values of x:

(1) x is a square of an integer --> \(x\) can be 1 or 100. Not sufficient.

(2) The sum of the distinct prime factors of x is a prime number ---> distinct primes of 12 are 2 and 3: \(2+3=5=prime\), distinct primes of 45 are 3 and 5: \(3+5=8\neq{prime}\) and distinct primes of 100 are also 2 and 3: \(2+3=5=prime\). \(x\) can be 12 or 100. Not sufficient.

(1)+(2) \(x\) can only be 100. Sufficient.

Answer: C.

hey..., can sm1 pls explain how primes of 100 can be 2 and 3?...(2nd last line),,.. thanx..

It should be: "... distinct primes of 100 are 2 and 5: \(2+5=7=prime\). \(x\) can be 12 or 100". _________________

Given: \(6a=3b=7c\) --> least common multiple of 6, 3, and 7 is 42 hence we ca write: \(6a=3b=7c=42x\), for some number \(x\) --> \(a=7x\), \(b=14x\) and \(c=6x\).

(1) ac=6b --> \(7x*6x=6*14x\) --> \(x^2=2x\) --> \(x=0\) or \(x=2\). Not sufficient.

(1) x^2+y^2=1. Recall that \((x-y)^2\geq{0}\) (square of any number is more than or equal to zero) --> \(x^2-2xy+y^2\geq{0}\) --> since \(x^2+y^2=1\) then: \(1-2xy\geq{0}\) --> \(xy\leq{\frac{1}{2}}\). Sufficient.

Hi Bunuel, I considered \((x+y)^2\) \(>=\) 0 and arrived at \(xy >=\) \(\frac{-1}{2}\). And hence concluded that the statement is insufficient. Please correct me.

Yes, you need to consider \((x-y)^2\geq{0}\) to get sufficiency. _________________

(1) 3x + 4 < 2y + 3 --> \(3x<2y-1\). \(x\) can be some very small number for instance -100 and \(y\) some large enough number for instance -3 and the answer would be YES, \(x<y\) BUT if \(x=-2\) and \(y=-2.1\) then the answer would be NO, \(x>y\). Not sufficient.

This means that the condition doesn't hold good, and hence x < y doesn't hold good.

Could you please point out where exactly I am making a mistake in my process?

The statements in DS questions are true. So, if you pick numbers you should pick such that they satisfy the statement. x and y cannot be -2 and -1 respectively because they do not satisfy 2x - 3 < 3y - 4. _________________

1. B 2. A 3. B 4. D 5. D 6. E 7. D 8. C 9. B 10. C 11. D 12. E

10 correct answers out of 12.

khaadu wrote:

1-b 2-a 3-b 4-a 5-b 6-e 7-d 8-b 9-e 10-a 11-d 12-a

7 correct answers out of 12.

vinayaerostar wrote:

Ans: 1-B 2-A 3-A 4-E 5-D 6-E 7-C 8-A 9-C 10-A 11-D 12-D

4 correct answers out of 12.

Good job everyone! By the way it's better if you post the solutions along with the answers: others will benefit with your approaches and you'll get 1 Kudos point per correct solution.

Will post explanations in couple of days, so that to give some more time to those who want to participate. _________________

1. Bonnie can paint a stolen car in x hours, and Clyde can paint the same car in y hours. They start working simultaneously and independently at their respective constant rates at 9:45am. If both x and y are odd integers, is x=y? (1) x^2+y^2<12 (2) Bonnie and Clyde complete the painting of the car at 10:30am

The catch is that they are working independently.

stmt 1 - no relation there are can be multiple values of x and y stmt 2 - both started at same time, finished at same time with no breaks means they have same working rate proves x = y sufficient

My funda - Area of square is largest among all the quadilateral with same perimeter. Stmt 1 - Only possible values of x and y are 1/Sqrt(2). So sufficient as xy = 1/2 Stmt 2 - Only says x and y are equal. Not sufficient Answer A

1. Bonnie can paint a stolen car in x hours, and Clyde can paint the same car in y hours. They start working simultaneously and independently at their respective constant rates at 9:45am. If both x and y are odd integers, is x=y? (1) x^2+y^2<12 (2) Bonnie and Clyde complete the painting of the car at 10:30am

The catch is that they are working independently.

stmt 1 - no relation there are can be multiple values of x and y stmt 2 - both started at same time, finished at same time with no breaks means they have same working rate proves x = y sufficient

Answer B

The logic for (2) is not correct, (though I'm not saying that (2) is insufficient). Even if two entities have different rates if they work together they both stop when the job is done. _________________

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

As you leave central, bustling Tokyo and head Southwest the scenery gradually changes from urban to farmland. You go through a tunnel and on the other side all semblance...

Ghibli studio’s Princess Mononoke was my first exposure to Japan. I saw it at a sleepover with a neighborhood friend after playing some video games and I was...