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The Discreet Charm of the DS

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

Solution: the-discreet-charm-of-the-ds-126962-20.html#p1039633

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

Solution: the-discreet-charm-of-the-ds-126962-20.html#p1039634

3. If a, b and c are integers, is abc an even integer?
(1) b is halfway between a and c
(2) a = b - c

Solution: the-discreet-charm-of-the-ds-126962-40.html#p1039637

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

Solution: the-discreet-charm-of-the-ds-126962-40.html#p1039645

5. What is the value of integer x?
(1) 2x^2+9<9x
(2) |x+10|=2x+8

Solution: the-discreet-charm-of-the-ds-126962-40.html#p1039650

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

Solution: the-discreet-charm-of-the-ds-126962-40.html#p1039651

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

Solution: the-discreet-charm-of-the-ds-126962-40.html#p1039655

8. If x=0.abcd, where a, b, c and d are digits from 0 to 9, inclusive, is x>7/9?
(1) a+b>14
(2) a-c>6

Solution: the-discreet-charm-of-the-ds-126962-40.html#p1039662

9. If x and y are negative numbers, is x<y?
(1) 3x + 4 < 2y + 3
(2) 2x - 3 < 3y - 4

Solution: the-discreet-charm-of-the-ds-126962-40.html#p1039665

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.

Solution: the-discreet-charm-of-the-ds-126962-40.html#p1039671

11. If x and y are integers, is x a positive integer?
(1) x*|y| is a prime number.
(2) x*|y| is non-negative integer.

Solution: the-discreet-charm-of-the-ds-126962-40.html#p1039678

12. If 6a=3b=7c, what is the value of a+b+c?
(1) ac=6b
(2) 5b=8a+4c

Solution: the-discreet-charm-of-the-ds-126962-40.html#p1039680
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Re: The Discreet Charm of the DS [#permalink] New post 03 Feb 2012, 07:52
Bunuel wrote:
sourabhsoni wrote:
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.

-----------------------------

yes but isn't it correct to say that when they are working independently and starting at same time (as per the question) and ending at same time as per stmt 2 then they must be working at same rate - i.e. X = Y...

As stmt 2 doesn't say they are not working together.
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Re: The Discreet Charm of the DS [#permalink] New post 03 Feb 2012, 07:59
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sourabhsoni wrote:
yes but isn't it correct to say that when they are working independently and starting at same time (as per the question) and ending at same time as per stmt 2 then they must be working at same rate - i.e. X = Y...

As stmt 2 doesn't say they are not working together.


No, that's not correct. Again when: two or more entities (machines, people, ...) are working together they all stop working when the job is done, no matter what their respective rates are. I think that you are thrown away by the phrase "they start working simultaneously and independently", which simply means that they start at the same time and work together (obviously they will also end the work at the same time, when the work is done).

Hope it's clear.
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Re: The Discreet Charm of the DS [#permalink] New post 03 Feb 2012, 08:08
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kys123 wrote:
1)b
2)a
3)b
4)d
5)d
6)e
7)d
8)c
9)e
10)a
11)d
12)a


8 correct answers out of 12.

Well done.

kys123 wrote:
1) statement 2 means

rate = 1/x + 1/y = 4/3 (1/[45 mins/ 60 mins]).

The only integer that would work is 1 n 3. Therefore x =/=y. Since X has to be 1 or 3 and Y is whatever X isn't.


That's correct, though there is another way of doing this.
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Re: The Discreet Charm of the DS [#permalink] New post 03 Feb 2012, 13:52
7.A
8.C
9.B
10.C
11.D
12.B
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Re: The Discreet Charm of the DS [#permalink] New post 03 Feb 2012, 13:56
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.
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Re: The Discreet Charm of the DS [#permalink] New post 03 Feb 2012, 13:56
9. If x and y are negative numbers, is x<y?
(1) 3x + 4 < 2y + 3
(2) 2x - 3 < 3y - 4

Given x is -ve and y is -ve

Stmt 1 - 3x + 4 < 2y + 3 which can be written as
3x < 2y - 1
suppose x < y
then 3x < 3y

You cannot dedude from here. So insufficient.

Stmt 2 - 2x - 3 < 3y - 4 which cane be written as
2x + 1 < 3y
which can be furhter written as 2y > 3y

therefore 2x + 1 < 2y even after 1 2x is smaller than 2y that proves x < y.

So answer B
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Re: The Discreet Charm of the DS [#permalink] New post 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.

Combining stmt 1 and stmt 2 we get x = 100.

So answer is C
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Re: The Discreet Charm of the DS [#permalink] New post 03 Feb 2012, 13:57
11. If x and y are integers, is x a positive integer?
(1) x*|y| is a prime number.
(2) x*|y| is non-negative integer.

prime # are always positive

Stmt 1 - |y| will be always positive to x*|y| to be prime z has to be positve. Sufficient
Stmt 2 - Same concept x will be positive number Sufficient

Answer is D
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Re: The Discreet Charm of the DS [#permalink] New post 03 Feb 2012, 13:57
12. If 6a=3b=7c, what is the value of a+b+c?
(1) ac=6b
(2) 5b=8a+4c

Stmt 1 = I get the values of a,b and c but a,b,c can be 0 also. So insufficient.
Stmt 2 - Insufficient because of same reason.
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Re: The Discreet Charm of the DS [#permalink] New post 03 Feb 2012, 15:53
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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.
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Re: The Discreet Charm of the DS [#permalink] New post 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.
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Re: The Discreet Charm of the DS [#permalink] New post 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. :)
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Re: The Discreet Charm of the DS [#permalink] New post 03 Feb 2012, 17:22
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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.
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Re: The Discreet Charm of the DS [#permalink] New post 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.
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Re: The Discreet Charm of the DS [#permalink] New post 05 Feb 2012, 06:52
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Re: The Discreet Charm of the DS [#permalink] New post 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.
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Re: The Discreet Charm of the DS [#permalink] New post 06 Feb 2012, 06:40
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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.

Hope it's clear.
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Re: The Discreet Charm of the DS [#permalink] New post 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.

Hope it's clear.


Sorry I got excited :)

Makes sense
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Re: The Discreet Charm of the DS [#permalink] New post 06 Feb 2012, 07:43
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?
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Re: The Discreet Charm of the DS [#permalink] New post 06 Feb 2012, 08:24
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?


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?
Re: The Discreet Charm of the DS   [#permalink] 06 Feb 2012, 08:24
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