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

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Expert's post

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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]
03 Feb 2012, 13:56

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.

Given 10 + x / GCF (10,x) = 11. Possible values of GCF(10,x) = 1,2,5,10. So 10+x can be 11, 22, 55, 110 ; which will result in x = 1,12, 45, 100

Stmt 1 - x can be 1 and 10 so not sufficient. Stmt 2 - x can be 12 , 100 so not sufficient.

Re: The Discreet Charm of the DS [#permalink]
03 Feb 2012, 15:53

Expert's post

Solutions for: 4, 5, 7, 8, 9, and 10 are correct. +1 for each.

sourabhsoni wrote:

6. If a and b are integers and ab=2, is a=2? (1) b+3 is not a prime number (2) a>b

Possible values of a and b are : (1,2), (2,1), (-1,-2), (-2,-1) we have to find out if a = 2 or get if b is 1.

Stmt 1 - b+3 is prime number. Possible values of b 2 and -1 Not sufficient.

Stmt 2 a > b Pssible values of b are 1, -2. Not sufficient

Combining stmt 1 and stmt 2 no common answers. So insufficient.

Answer E.

As for this one: first of all (1) says that b+3 is NOT a prime number.

Next, on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other. So, we can not have the case when (1) says that b is either 2 or -1 and (2) says that b is either 1 or -2, because in this case statements would contradict each other.

Hope it's clear.

P.S. Funny thing: answer is still E. _________________

Re: The Discreet Charm of the DS [#permalink]
03 Feb 2012, 16:47

Expert's post

vailad wrote:

Bunuel wrote:

sourabhsoni wrote:

2. Is xy<=1/2? (1) x^2+y^2=1 (2) x^2-y^2=0

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).

Yes, the answer to the question is indeed A. +1.

Though let me point out that there are at least two other solutions that are shorter and easier. I'll post them in couple of days, after some more discussion. _________________

Re: The Discreet Charm of the DS [#permalink]
03 Feb 2012, 16:55

Bunuel wrote:

vailad wrote:

Bunuel wrote:

2. Is xy<=1/2? (1) x^2+y^2=1 (2) x^2-y^2=0

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).

Yes, the answer to the question is indeed A. +1.

Though let me point out that there are at least two other solutions that are shorter and easier. I'll post them in couple of days, after some more discussion.

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. _________________

Re: The Discreet Charm of the DS [#permalink]
04 Feb 2012, 03:59

Bunuel wrote:

vailad wrote:

Bunuel wrote:

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.

F***ing good !

(x-y)2 = 1-2xy >=0 => xy>=1/2.

Now this should take 15sec !

That's it. +1 again.

Now tell me which one is easier/faster?

Just a little typo there: xy<=1/2.

Haha. I said, then and there, this should take 15 sec. i.e. double the time I took.

P.S. : As we see, the co-ordinate geometry method might not be the best way to solve this particular problem. Having said that I must say one should keep this approach at the back of the mind. Solving complex inequalities can be monumental at times, and certainly more prone to errors while co-ordinate geometry is graphical, clean and quick. _________________

Re: The Discreet Charm of the DS [#permalink]
06 Feb 2012, 06:36

Bunuel wrote:

6. If a and b are integers and ab=2, is a=2?

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, 06:40, edited 1 time in total.

Re: The Discreet Charm of the DS [#permalink]
06 Feb 2012, 06:40

Expert's post

nhemdani wrote:

Bunuel wrote:

6. If a and b are integers and ab=2, is a=2?

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.

Re: The Discreet Charm of the DS [#permalink]
06 Feb 2012, 06:42

Bunuel wrote:

nhemdani wrote:

Bunuel wrote:

6. If a and b are integers and ab=2, is a=2?

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.

Re: The Discreet Charm of the DS [#permalink]
06 Feb 2012, 08:54

Expert's post

kys123 wrote:

Hey Bunuel can I ask a question for 12?

We know 6a=3b

And for statement one:

ac =6b. Can't 6b =12a

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.