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# The Discreet Charm of the DS

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The Discreet Charm of the DS [#permalink]

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02 Feb 2012, 04:15
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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]

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15 Nov 2012, 23:10
gmatDeep 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.

Bunuel, while solving (1), how do you know which nos to plugin in and test? Can you suggest some approach please? I got this one wrong because for the values I plugged in, I was always getting $$x<y$$

I'm quite confused by the fundamental concept here. I sub in x = -1.1 and y = -1 (which makes x<y), and equation 2x - 3 < 3y - 4 doesn't work, but then if you switch them around, x = -1 and y = -1.1, the equation still doesn't work.

What are we trying to find with substituting in numbers here? I feel like I have tied a knot in my head. Someoone please help? Thanks.

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Re: The Discreet Charm of the DS [#permalink]

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06 Dec 2012, 23:19
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Bunuel 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

Responding to a pm:

Time taken by Bonnie to complete one work = x hrs
Time taken by Clyde to complete one work = y hrs
x and y are odd integers i.e. they could take values such as 1/3/5/7/9/11...

Question: Is x = y? i.e. is the time taken by Bonnie equal to time taken by Clyde? i.e. is the speed of Bonnie equal to the speed of Clyde?

(1) x^2+y^2<12
This info is not related to work concepts. It's just number properties. x and y are odd integers.
If x = y = 1, this inequality is satisfied.
If x = 1 and y = 3, this inequality is satisfied.

This means x may or may not be equal to y. Not sufficient.

(2) Bonnie and Clyde complete the painting of the car at 10:30 am.r
Together, they take 45 mins to complete the painting of the car. This means, if their rate of work were the same, each one of them would have taken 1.5 hrs working alone. But their time taken is an integer value. We can say that they do not take the same time i.e. x is not equal to y. Hence this statement is sufficient to say $$x \neq y$$

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Re: The Discreet Charm of the DS [#permalink]

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07 Dec 2012, 08:41
VeritasPrepKarishma wrote:
Bunuel 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

Responding to a pm:

Time taken by Bonnie to complete one work = x hrs
Time taken by Clyde to complete one work = y hrs
x and y are odd integers i.e. they could take values such as 1/3/5/7/9/11...

Question: Is x = y? i.e. is the time taken by Bonnie equal to time taken by Clyde? i.e. is the speed of Bonnie equal to the speed of Clyde?

(1) x^2+y^2<12
This info is not related to work concepts. It's just number properties. x and y are odd integers.
If x = y = 1, this inequality is satisfied.
If x = 1 and y = 3, this inequality is satisfied.

This means x may or may not be equal to y. Not sufficient.

(2) Bonnie and Clyde complete the painting of the car at 10:30 am.r
Together, they take 45 mins to complete the painting of the car. This means, if their rate of work were the same, each one of them would have taken 1.5 hrs working alone. But their time taken is an integer value. We can say that they do not take the same time i.e. x is not equal to y. Hence this statement is sufficient to say $$x \neq y$$

Thanks a lot for your explanation and response to my pm
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Re: The Discreet Charm of the DS [#permalink]

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10 Dec 2012, 05:58
Bunnel,
Can you please explain once again why in 1st question statement2 is sufficient...If you dont mind.Thanks in adavance
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Re: The Discreet Charm of the DS [#permalink]

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10 Dec 2012, 06:01
skamal7 wrote:
Bunnel,
Can you please explain once again why in 1st question statement2 is sufficient...If you dont mind.Thanks in adavance

You should tell me what didn't you understand in the solution provided.

Check other approaches here: the-discreet-charm-of-the-ds-126962-80.html#p1151803

Hope it helps.
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Re: The Discreet Charm of the DS [#permalink]

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12 Dec 2012, 07:00
Amazing collection bunuel, thanks a ton. Helped me a lot.

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Re: The Discreet Charm of the DS [#permalink]

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12 Dec 2012, 19:36
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]

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13 Dec 2012, 03:24
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]

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29 Dec 2012, 07:37
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]

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11 Feb 2013, 03:38
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]

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11 Feb 2013, 05:45
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]

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14 Feb 2013, 00:31
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]

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14 Feb 2013, 01:24
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]

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14 Feb 2013, 02:08
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]

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14 Feb 2013, 02:16
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]

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14 Feb 2013, 02:44
Crystal Clear..Thanks Bunuel

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Re: The Discreet Charm of the DS [#permalink]

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14 Feb 2013, 02:48
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]

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23 Feb 2013, 16:17
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]

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23 May 2013, 15:59
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]

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23 May 2013, 16:09
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] 23 May 2013, 16:09

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