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The Discreet Charm of the DS

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The Discreet Charm of the DS [#permalink]  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]  12 Dec 2012, 06:00
Amazing collection bunuel, thanks a ton. Helped me a lot.
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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.

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
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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.

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.
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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.

Dear Bunuel

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

Hi,

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

Hi,

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

Yes, that's correct.
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Re: The Discreet Charm of the DS [#permalink]  13 Feb 2013, 23:31
Bunuel wrote:
12. If 6a=3b=7c, what is the value of a+b+c?

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.

(2) 5b=8a+4c --> $$5*14x=8*7x+4*14x$$ --> $$70x=80x$$ --> $$10x=0$$ --> $$x=0$$ --> $$a=b=c=0$$ --> $$a+b+c=0$$. Sufficient.

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?
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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.

Bunuel,
I did not understand the reasoning behind statement 2 being sufficient. Line marked in red is my doubt. what does it mean ?
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Re: The Discreet Charm of the DS [#permalink]  14 Feb 2013, 01:08
Expert's post
Sachin9 wrote:
Bunuel wrote:
12. If 6a=3b=7c, what is the value of a+b+c?

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.

(2) 5b=8a+4c --> $$5*14x=8*7x+4*14x$$ --> $$70x=80x$$ --> $$10x=0$$ --> $$x=0$$ --> $$a=b=c=0$$ --> $$a+b+c=0$$. Sufficient.

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.
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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.

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.

Hope it's clear.
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Re: The Discreet Charm of the DS [#permalink]  14 Feb 2013, 01:44
Crystal Clear..Thanks Bunuel
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Re: The Discreet Charm of the DS [#permalink]  14 Feb 2013, 01:48
Bunuel wrote:
Sachin9 wrote:
Bunuel wrote:
12. If 6a=3b=7c, what is the value of a+b+c?

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.

(2) 5b=8a+4c --> $$5*14x=8*7x+4*14x$$ --> $$70x=80x$$ --> $$10x=0$$ --> $$x=0$$ --> $$a=b=c=0$$ --> $$a+b+c=0$$. Sufficient.

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 ?
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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.

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
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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.
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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.
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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:

GCF(10,x)=1 --> $$f(10,x)=11=\frac{10+x}{1}$$ --> $$x=1$$;
GCF(10,x)=2 --> $$f(10,x)=11=\frac{10+x}{2}$$ --> $$x=12$$;
GCF(10,x)=5 --> $$f(10,x)=11=\frac{10+x}{5}$$ --> $$x=45$$;
GCF(10,x)=10 --> $$f(10,x)=11=\frac{10+x}{10}$$ --> $$x=100$$.

(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.

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.
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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.

(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.

Hi Bunuel,

First of all, great set of questions.

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!
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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:

GCF(10,x)=1 --> $$f(10,x)=11=\frac{10+x}{1}$$ --> $$x=1$$;
GCF(10,x)=2 --> $$f(10,x)=11=\frac{10+x}{2}$$ --> $$x=12$$;
GCF(10,x)=5 --> $$f(10,x)=11=\frac{10+x}{5}$$ --> $$x=45$$;
GCF(10,x)=10 --> $$f(10,x)=11=\frac{10+x}{10}$$ --> $$x=100$$.

(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.

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.

Hope it's clear.
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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.

(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.

Hi Bunuel,

First of all, great set of questions.

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.
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Re: The Discreet Charm of the DS [#permalink]  06 Jun 2013, 00:29
Bunuel wrote:
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.

(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.

Hi Bunuel,

First of all, great set of questions.

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.

Sorry for confusion. I was actually taking x<y to be true and thinking how it can prove x < y-(negative number). Thanks a lot!

Your posts are amazing and great learning!
Re: The Discreet Charm of the DS   [#permalink] 06 Jun 2013, 00:29

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