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.

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:

The Discreet Charm of the DS [#permalink]
02 Feb 2012, 03:15

29

This post received KUDOS

Expert's post

35

This post was BOOKMARKED

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.

Re: The Discreet Charm of the DS [#permalink]
12 Dec 2012, 18:36

Bunuel wrote:

SOLUTIONS:

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?

Bonnie and Clyde when working together complete the painting of the car ins \frac{xy}{x+y} hours (sum of the rates equal to the combined rate or reciprocal of total time: \frac{1}{x}+\frac{1}{y}=\frac{1}{T} --> T=\frac{xy}{x+y}). Now, if x=y then the total time would be: \frac{x^2}{2x}=\frac{x}{2}, since x is odd then this time would be odd/2: 0.5 hours, 1.5 hours, 2.5 hours, ....

(1) x^2+y^2<12 --> it's possible x and y to be odd and equal to each other if x=y=1 but it's also possible that x=1 and y=3 (or vise-versa). Not sufficient.

(2) Bonnie and Clyde complete the painting of the car at 10:30am --> they complete the job in 3/4 of an hour (45 minutes), since it's not odd/2 then x and y are not equal. Sufficient.

Answer: B.

i didnot understand this..... the questions says if they are working independently.... why are you considering combined rate? my analysis I ended up at the same answer though

Re: The Discreet Charm of the DS [#permalink]
13 Dec 2012, 02:24

Expert's post

Amateur wrote:

Bunuel wrote:

SOLUTIONS:

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?

Bonnie and Clyde when working together complete the painting of the car ins \frac{xy}{x+y} hours (sum of the rates equal to the combined rate or reciprocal of total time: \frac{1}{x}+\frac{1}{y}=\frac{1}{T} --> T=\frac{xy}{x+y}). Now, if x=y then the total time would be: \frac{x^2}{2x}=\frac{x}{2}, since x is odd then this time would be odd/2: 0.5 hours, 1.5 hours, 2.5 hours, ....

(1) x^2+y^2<12 --> it's possible x and y to be odd and equal to each other if x=y=1 but it's also possible that x=1 and y=3 (or vise-versa). Not sufficient.

(2) Bonnie and Clyde complete the painting of the car at 10:30am --> they complete the job in 3/4 of an hour (45 minutes), since it's not odd/2 then x and y are not equal. Sufficient.

Answer: B.

i didnot understand this..... the questions says if they are working independently.... why are you considering combined rate? my analysis I ended up at the same answer though

Because they are working simultaneously and independently to paint the same car. _________________

Re: The Discreet Charm of the DS [#permalink]
29 Dec 2012, 06:37

Bunuel wrote:

2. Is xy<=1/2?

(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.

(2) x^2-y^2=0 --> |x|=|y|. Clearly insufficient.

Answer: A.

Dear Bunuel

Why haven't we used (x+y)^2 instead of (x-y)^2 in statement 1 ? Sorry if its a silly question _________________

Re: The Discreet Charm of the DS [#permalink]
11 Feb 2013, 02:38

Bunuel wrote:

SOLUTIONS:

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?

Bonnie and Clyde when working together complete the painting of the car ins \frac{xy}{x+y} hours (sum of the rates equal to the combined rate or reciprocal of total time: \frac{1}{x}+\frac{1}{y}=\frac{1}{T} --> T=\frac{xy}{x+y}). Now, if x=y then the total time would be: \frac{x^2}{2x}=\frac{x}{2}, since x is odd then this time would be odd/2: 0.5 hours, 1.5 hours, 2.5 hours, ....

(1) x^2+y^2<12 --> it's possible x and y to be odd and equal to each other if x=y=1 but it's also possible that x=1 and y=3 (or vise-versa). Not sufficient.

(2) Bonnie and Clyde complete the painting of the car at 10:30am --> they complete the job in 3/4 of an hour (45 minutes), since it's not odd/2 then x and y are not equal. Sufficient.

Answer: B.

Hi,

Just a question, Only if we get a answer which is "odd/2" then x & y are considered to be equal . rite?

Re: The Discreet Charm of the DS [#permalink]
11 Feb 2013, 04:45

Expert's post

FTG wrote:

Bunuel wrote:

SOLUTIONS:

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?

Bonnie and Clyde when working together complete the painting of the car ins \frac{xy}{x+y} hours (sum of the rates equal to the combined rate or reciprocal of total time: \frac{1}{x}+\frac{1}{y}=\frac{1}{T} --> T=\frac{xy}{x+y}). Now, if x=y then the total time would be: \frac{x^2}{2x}=\frac{x}{2}, since x is odd then this time would be odd/2: 0.5 hours, 1.5 hours, 2.5 hours, ....

(1) x^2+y^2<12 --> it's possible x and y to be odd and equal to each other if x=y=1 but it's also possible that x=1 and y=3 (or vise-versa). Not sufficient.

(2) Bonnie and Clyde complete the painting of the car at 10:30am --> they complete the job in 3/4 of an hour (45 minutes), since it's not odd/2 then x and y are not equal. Sufficient.

Answer: B.

Hi,

Just a question, Only if we get a answer which is "odd/2" then x & y are considered to be equal . rite?

Re: The Discreet Charm of the DS [#permalink]
14 Feb 2013, 00:24

Bunuel wrote:

SOLUTIONS:

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?

Bonnie and Clyde when working together complete the painting of the car ins \frac{xy}{x+y} hours (sum of the rates equal to the combined rate or reciprocal of total time: \frac{1}{x}+\frac{1}{y}=\frac{1}{T} --> T=\frac{xy}{x+y}). Now, if x=y then the total time would be: \frac{x^2}{2x}=\frac{x}{2}, since x is odd then this time would be odd/2: 0.5 hours, 1.5 hours, 2.5 hours, ....

(1) x^2+y^2<12 --> it's possible x and y to be odd and equal to each other if x=y=1 but it's also possible that x=1 and y=3 (or vise-versa). Not sufficient.

(2) Bonnie and Clyde complete the painting of the car at 10:30am --> they complete the job in 3/4 of an hour (45 minutes), since it's not odd/2 then x and y are not equal. Sufficient.

Answer: B.

Bunuel, I did not understand the reasoning behind statement 2 being sufficient. Line marked in red is my doubt. what does it mean ?

Bunuel, Here's how I did . I don't know why it is wrong. Please advise :

6a=3b=7c

=> 12a=6b=14c

From 1) ac=6b => ac=12a => c=12.. a can be cancelled because it is not inequality. .

so c =12..

however,

ac-12a=0 --> a(c-12)=0 so a=0 OR c=12 .. Is this the reason why my soln is wrong?

Never reduce an equation by a variable (or expression with a variable), if you are not certain that the variable (or the expression with a variable) doesn't equal to zero. We can not divide by zero.

So, if you divide (reduce) ac=12a by a, you assume, with no ground for it, that a does not equal to zero thus exclude a possible solution (notice that both a=0 AND c-12=0 satisfy the equation).

Re: The Discreet Charm of the DS [#permalink]
14 Feb 2013, 01:16

Expert's post

thinktank wrote:

Bunuel wrote:

SOLUTIONS:

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?

Bonnie and Clyde when working together complete the painting of the car ins \frac{xy}{x+y} hours (sum of the rates equal to the combined rate or reciprocal of total time: \frac{1}{x}+\frac{1}{y}=\frac{1}{T} --> T=\frac{xy}{x+y}). Now, if x=y then the total time would be: \frac{x^2}{2x}=\frac{x}{2}, since x is odd then this time would be odd/2: 0.5 hours, 1.5 hours, 2.5 hours, ....

(1) x^2+y^2<12 --> it's possible x and y to be odd and equal to each other if x=y=1 but it's also possible that x=1 and y=3 (or vise-versa). Not sufficient.

(2) Bonnie and Clyde complete the painting of the car at 10:30am --> they complete the job in 3/4 of an hour (45 minutes), since it's not odd/2 then x and y are not equal. Sufficient.

Answer: B.

Bunuel, I did not understand the reasoning behind statement 2 being sufficient. Line marked in red is my doubt. what does it mean ?

From the stem we got that if x=y then the total time would be: \frac{x^2}{2x}=\frac{x}{2}, since x is odd then this time would be odd/2: 0.5 hours (1/2 hours = 0.5 hours), 1.5 hours (3/2 hours = 1.5 hours), 2.5 hours (5/2 hours = 2.5 hours), 3.5 hours (7/2 hours = 3.5 hours), 4.5 hours (9/2 hours = 4.5 hours), ....

Now, from the second statement we got that they complete the job in 0.75 hours, since the total time (0.75 hours) is NOT odd/2 (0.5 hours, 1.5 hours, 2.5 hours, 3.5 hours, 4.5 hours, ....), then x and y are not equal.

Bunuel, Here's how I did . I don't know why it is wrong. Please advise :

6a=3b=7c

=> 12a=6b=14c

From 1) ac=6b => ac=12a => c=12.. a can be cancelled because it is not inequality. .

so c =12..

however,

ac-12a=0 --> a(c-12)=0 so a=0 OR c=12 .. Is this the reason why my soln is wrong?

Never reduce an equation by a variable (or expression with a variable), if you are not certain that the variable (or the expression with a variable) doesn't equal to zero. We can not divide by zero.

So, if you divide (reduce) ac=12a by a, you assume, with no ground for it, that a does not equal to zero thus exclude a possible solution (notice that both a=0 AND c-12=0 satisfy the equation).

Hope it's clear.

thanks alot. .you rock man!! r u a phd in maths ? _________________

hope is a good thing, maybe the best of things. And no good thing ever dies.

Re: The Discreet Charm of the DS [#permalink]
23 Feb 2013, 15:17

Bunuel wrote:

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.

Answer: B.

1) b = a+c/2 i.e. a+c = even (as it is divisible by 2) and an even# divided by another even# can be odd or even (e.g. 46/2 = 23 an odd, but 48/2 = 24 an even). now if a + c = odd + odd = even and if a+c/2 = odd then all 3 numbers are odd (e.g. a=21, b=23 and c=25) and abc = odd but if a+c = even+even then a+c/2 = odd and abc = even (a=22, b=24 and c=26). so insufficient 2) a = b - c i.e. a + c = b. from number properties we know that i) odd+odd = even, ii) even+odd = odd iii) even+even = even so in any of the 3 cases you will end up with atleast one number that is even and hence abc = even. sufficient. correct ans. B _________________

___________________________________________ Consider +1 Kudos if my post helped

Re: The Discreet Charm of the DS [#permalink]
23 May 2013, 14:59

Hi Brunnel, I have trouble with question 9 for the second statement, 2x-3<3y-4, when I substitute x=-5; y= -1; the equation holds but when I substitute x=-2; y=-5; the equation collapsed. So, wouldn't the answer be E? Please help explain, Thanks.

Re: The Discreet Charm of the DS [#permalink]
23 May 2013, 15:09

Expert's post

smartyman wrote:

Hi Brunnel, I have trouble with question 9 for the second statement, 2x-3<3y-4, when I substitute x=-5; y= -1; the equation holds but when I substitute x=-2; y=-5; the equation collapsed. So, wouldn't the answer be E? Please help explain, Thanks.

The question does not ask whether 2x - 3 < 3y - 4, it asks whether x<y. While 2x - 3 < 3y - 4 is given to be true by the second statement. _________________

Re: The Discreet Charm of the DS [#permalink]
05 Jun 2013, 13:19

Bunuel wrote:

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.

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

Answer: C.

I think the answer to the above questions should be "A" not "C". Since the GCD for 10 and x can only be 1,2,5 and 10, the corresponding value for x can be 1, 12, 45 and 100. Given this information, the first clause leaves only 100 as the correct answer, which is a square of 10. Please let me know if there is any flaw in my reasoning.

Re: The Discreet Charm of the DS [#permalink]
05 Jun 2013, 22:17

Bunuel wrote:

9. If x and y are negative numbers, is x<y?

(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.

However, the questions is whether x<y. If x< y-(any negative term) doesn't mean that X< Y. For example: if x= -5 y= -3 (Here x<Y) and according to the above equation -5<-3 but -5<-3 + (a negative term, say -3) will make the questions incorrect.

Please tell me where I am going wrong with this. Thanks!

Re: The Discreet Charm of the DS [#permalink]
05 Jun 2013, 23:20

Expert's post

narulajasneet wrote:

Bunuel wrote:

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.

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

Answer: C.

I think the answer to the above questions should be "A" not "C". Since the GCD for 10 and x can only be 1,2,5 and 10, the corresponding value for x can be 1, 12, 45 and 100. Given this information, the first clause leaves only 100 as the correct answer, which is a square of 10. Please let me know if there is any flaw in my reasoning.

x could be 1, 12, 45 or 100.

(1) says that x is a square of an integer --> x could be 1^2=1 or 10^2=100. Two answers, thus the statement is insufficient.

Re: The Discreet Charm of the DS [#permalink]
05 Jun 2013, 23:27

Expert's post

narulajasneet wrote:

Bunuel wrote:

9. If x and y are negative numbers, is x<y?

(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.

I have a little doubt around the explanation given for Q9 Clause 2. You mentioned the following as your explanation: (2) 2x - 3 < 3y - 4 --> x<1.5y-\frac{1}{2} --> x<y+(0.5y-\frac{1}{2})=y+negative --> x<y (as y+negative is "more negative" than y). Sufficient.

However, the questions is whether x<y. If x< y-(any negative term) doesn't mean that X< Y. For example: if x= -5 y= -3 (Here x<Y) and according to the above equation -5<-3 but -5<-3 + (a negative term, say -3) will make the questions incorrect.

Please tell me where I am going wrong with this. Thanks!

It's the other way around: if x and y are negative numbers and IF x<y+negative, then x<y. _________________