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.

3. If a, b and c are integers, is abc an even integer?

In order the product of the integers to be even at leas on of them must be even

(1) b is halfway between a and c --> on the GMAT we often see such statement and it can ALWAYS be expressed algebraically as \(b=\frac{a+c}{2}\). Now, does that mean that at leas on of them is be even? Not necessarily, consider \(a=1\), \(b=3\) and \(c=5\). Of course it's also possible that \(b=even\), for example if \(a=1\) and \(b=7\). Not sufficient.

(2) a = b - c --> \(a+c=b\). Since it's not possible that the sum of two odd integers to be odd then the case of 3 odd numbers is ruled out, hence at least on of them must be even. Sufficient.

4. How many numbers of 5 consecutive positive integers is divisible by 4?

(1) The median of these numbers is odd --> the median of the set with odd number of terms is just a middle term, thus our set of 5 consecutive numbers is: {Odd, Even, Odd, Even, Odd}. Out of 2 consecutive even integers only one is a multiple of 4. Sufficient.

(2) The average (arithmetic mean) of these numbers is a prime number --> in any evenly spaced set the arithmetic mean (average) is equal to the median --> mean=median=prime. Since it's not possible that median=2=even, (in this case not all 5 numbers will be positive), then median=odd prime, and we have the same case as above. Sufficient.

(1) 2x^2+9<9x --> factor qudratics: \((x-\frac{3}{2})(x-3)<0\) --> roots are \(\frac{3}{2}\) and 3 --> "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) --> since there only integer in this range is 2 then \(x=2\). Sufficient.

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(2x+8\geq{0}\) --> \(x\geq{-4}\), for this range \(x+10\) is positive hence \(|x+10|=x+10\) --> \(x+10=2x+8\) --> \(x=2\). Sufficient.

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 --> this basically means that all customers bought exactly 4 oranges (76/19=4), because if even one customer bought less than 4, the sum will be less than 76. Hence, no one bought only one orange. Sufficient.

(2) The difference between the number of oranges bought by any two customers is even --> in order the difference between ANY number of oranges bought to be even, either all customers must have bought odd number of oranges or all customers must have bough even number of oranges. But the first case is not possible: the sum of 19 odd numbers is odd and not even like 76. Hence, again no one bought only one=odd orange. Sufficient.

8. If x=0.abcd, where a, b, c and d are digits from 0 to 9, inclusive, is x>7/9?

First of all 7/9 is a recurring decimal =0.77(7). For more on converting Converting Decimals to Fractions see: math-number-theory-88376.html

(1) a+b>14 --> the least value of a is 6 (6+9=15>14), so in this case x=0.69d<0.77(7) but a=7 and b=9 is also possible, and in this case x=0.79d>0.77(7). Not sufficient.

(2) a-c>6 --> the least value of a is 7 (7-0=7>6), but we don't know the value of b. Not sufficient.

(1)+(2) The least value of a is 7 and in this case from (1) least value of b is 8 (7+8=15>14), hence the least value of x=0.78d>0.77(7). Sufficient.

(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.

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 2 and 5: \(2+5=7=prime\). \(x\) can be 12 or 100. Not sufficient.

11. If x and y are integers, is x a positive integer?

(1) x*|y| is a prime number --> since only positive numbers can be primes, then: x*|y|=positive --> x=positive. Sufficient

(2) x*|y| is non-negative integer. Notice that we are told that x*|y| is non-negative, not that it's positive, so x can be positive as well as zero. Not sufficient.

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.

Notice that we are not told that a and b are positive.

There are following integer pairs of (a, b) possible: (1, 2), (-1, -2), (2, 1) and (-2, -1). Basically we are asked whether we have the third case.

(1) b+3 is not a prime number --> rules out 1st and 4th options. Not sufficient. (2) a>b --> again rules out 1st and 4th options. Not sufficient.

(1)+(2) Still two options are left: (-1, -2) and (2, 1). Not sufficient.

Answer: E.

Hi, I was looking at this solution and had a question. If b+3 is not a prime number, only when a number in our set equals 1, the sum is 4 (1+3= 4 (Prime), 2+3, -1+3, -2+3 are all prime numbers), but option (1,2) means b+3 = 5 (Prime number) and A tells us that b+3 is not Prime so the only set is (2,1), therefore YES a=2

???

Last edited by nhk13579 on 06 Feb 2012, 07:40, edited 1 time in total.

Notice that we are not told that a and b are positive.

There are following integer pairs of (a, b) possible: (1, 2), (-1, -2), (2, 1) and (-2, -1). Basically we are asked whether we have the third case.

(1) b+3 is not a prime number --> rules out 1st and 4th options. Not sufficient. (2) a>b --> again rules out 1st and 4th options. Not sufficient.

(1)+(2) Still two options are left: (-1, -2) and (2, 1). Not sufficient.

Answer: E.

Hi, I was looking at this solution and had a question. If b+3 is not a prime number, only when a number in our set equals 1, in this case the sum os 4 (1+3= 4 (Prime), 2+3, -1+3, -2+3 are all prime numbers), shouldnt the answer be A, as we are only left with option (1,2) so YES a=2

???

1 is not a prime number, so the case of (-1, -2) is still possible: -2+3=1 not a prime.

Notice that we are not told that a and b are positive.

There are following integer pairs of (a, b) possible: (1, 2), (-1, -2), (2, 1) and (-2, -1). Basically we are asked whether we have the third case.

(1) b+3 is not a prime number --> rules out 1st and 4th options. Not sufficient. (2) a>b --> again rules out 1st and 4th options. Not sufficient.

(1)+(2) Still two options are left: (-1, -2) and (2, 1). Not sufficient.

Answer: E.

Hi, I was looking at this solution and had a question. If b+3 is not a prime number, only when a number in our set equals 1, in this case the sum os 4 (1+3= 4 (Prime), 2+3, -1+3, -2+3 are all prime numbers), shouldnt the answer be A, as we are only left with option (1,2) so YES a=2

???

1 is not a prime number, so the case of (-1, -2) is still possible: -2+3=1 not a prime.

Then it becomes ac=12a ==> c=12. I know it's wrong since if a is 0 then they will be equal regardless, but can you explain why what I did was wrong?

ac=12a (here you can not reduce by a and write c=12 as you exclude possibility of a=0) --> a(c-12)=0 --> either a=0 OR c=12. So, we get either a=b=c=0 or a=14, b=28 and c=12.

nhemdani wrote:

Also,

6a = 3b = 7c

Can we say a/b= 1/2, b/c = 7/3, and a/c = 7/6 a) ac = 6b, therefore c = 6b/a substituting this in b/c => b / (6b/a) = 7/3 => a =14, b=28, c = 12

Isnt A also sufficient? Am I ignoring something?

Your doubt is partially addressed above, though there is another thing: from 6a = 3b you can not write a/b=1/2 because b can be zero and we can not divide by zero. The same for other ratios you wrote.

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.

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...