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The Discreet Charm of the DS

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

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New post 25 Feb 2012, 05:53
vidhya16 wrote:
All,

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

In respect to the question above, I assumed that any two of those 19 customer might have bought 5 & 3 oranges and hence I, marked the option insufficient. Bunnel have equated and treated the option in totally different way. I ,lack the skill to convert these sort of condition in to equation.

please can some post or point to the list of similar Word translation sentences and how to convert them in to equation. Im very new to GMAT club so please forgie me if this is the repeated posting.

Thanks,
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I did not use any equation for this question.

Statement (2) says: the difference between the number of oranges bought by ANY two customers is even --> in order the difference between ANY number of oranges bought to be even, either all customers must have bought odd number of oranges or all customers must have bough even number of oranges.

Now, the sum of 19 odd integers is odd and we have that fruit stand sold total of 76, so even number of oranges, which means that the case where all customers buy odd number of oranges is not possible. And since 1 is odd then no one bought only one orange. Sufficient.

As for word translation check this: word-problems-made-easy-87346.html

Hope it helps.
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New post 14 May 2012, 16:40
Bunuel wrote:
5. What is the value of integer x?

(1) 2x^2+9<9x --> factor qudratics: \((x-\frac{3}{2})(x-3)<0\) --> roots are \(\frac{3}{2}\) and 3 --> "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) --> since there only integer in this range is 2 then \(x=2\). Sufficient.

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(2x+8\geq{0}\) --> \(x\geq{-4}\), for this range \(x+10\) is positive hence \(|x+10|=x+10\) --> \(x+10=2x+8\) --> \(x=2\). Sufficient.

Answer: D.

Check this for more on solving inequalities like the one in the first statement:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486



Hope it helps.



Hey Bunuel,

Two questions.

1) How did you factor 2x^2+9<9x (ie 2x^2-9x+9<0) so quickly? I always struggle with factoring polynomials in which a coefficient other than 1 is on the x^2. Did you use the quadratic formula? I am interested in knowing if there is a quicker way than the quadratic formula method.

2) Once you determined that 1.5 and 3 were the roots of the equation, how did you figure that the solution was in between 1.5 and 3 from just looking at the sign "<"??
I used the dumb method of just plugging values that lie from (-infinity , 1.5), (1.5, 3) and (3, +infinity). How did you know the sign "<" told you the solution was in in between (1.5,3)?

Many thanks Bunuel! Your my hero dude!
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New post 14 May 2012, 23:19
alphabeta1234 wrote:
Bunuel wrote:
5. What is the value of integer x?

(1) 2x^2+9<9x --> factor qudratics: \((x-\frac{3}{2})(x-3)<0\) --> roots are \(\frac{3}{2}\) and 3 --> "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) --> since there only integer in this range is 2 then \(x=2\). Sufficient.

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(2x+8\geq{0}\) --> \(x\geq{-4}\), for this range \(x+10\) is positive hence \(|x+10|=x+10\) --> \(x+10=2x+8\) --> \(x=2\). Sufficient.

Answer: D.

Check this for more on solving inequalities like the one in the first statement:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486

Hope it helps.



Hey Bunuel,

Two questions.

1) How did you factor 2x^2+9<9x (ie 2x^2-9x+9<0) so quickly? I always struggle with factoring polynomials in which a coefficient other than 1 is on the x^2. Did you use the quadratic formula? I am interested in knowing if there is a quicker way than the quadratic formula method.

2) Once you determined that 1.5 and 3 were the roots of the equation, how did you figure that the solution was in between 1.5 and 3 from just looking at the sign "<"??
I used the dumb method of just plugging values that lie from (-infinity , 1.5), (1.5, 3) and (3, +infinity). How did you know the sign "<" told you the solution was in in between (1.5,3)?

Many thanks Bunuel! Your my hero dude!


1. Solving and Factoring Quadratics:
http://www.purplemath.com/modules/solvquad.htm
http://www.purplemath.com/modules/factquad.htm

2. Solving inequalities:
x2-4x-94661.html#p731476 (Check this first)
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486

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

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New post 15 May 2012, 11:39
Bunuel wrote:
5. What is the value of integer x?

(1) 2x^2+9<9x --> factor qudratics: \((x-\frac{3}{2})(x-3)<0\) --> roots are \(\frac{3}{2}\) and 3 --> "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) --> since there only integer in this range is 2 then \(x=2\). Sufficient.

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(2x+8\geq{0}\) --> \(x\geq{-4}\), for this range \(x+10\) is positive hence \(|x+10|=x+10\) --> \(x+10=2x+8\) --> \(x=2\). Sufficient.

Answer: D.

Hope it helps.


Hi bunuel,
Isn't |x+10|=2x+8 be written as
Either x+10=2x+8 or x+10=-(2x+8) ? and then this should be solved?
Please help on this one.
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New post 15 May 2012, 11:43
piyushksharma wrote:
Bunuel wrote:
5. What is the value of integer x?

(1) 2x^2+9<9x --> factor qudratics: \((x-\frac{3}{2})(x-3)<0\) --> roots are \(\frac{3}{2}\) and 3 --> "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) --> since there only integer in this range is 2 then \(x=2\). Sufficient.

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(2x+8\geq{0}\) --> \(x\geq{-4}\), for this range \(x+10\) is positive hence \(|x+10|=x+10\) --> \(x+10=2x+8\) --> \(x=2\). Sufficient.

Answer: D.

Hope it helps.


Hi bunuel,
Isn't |x+10|=2x+8 be written as
Either x+10=2x+8 or x+10=-(2x+8) ? and then this should be solved?
Please help on this one.


We goth that x is more than or equal to 4. Now, for this range x+10>0 so |x+10| expands only as x+10 (|x+10|=x+10).
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New post 15 May 2012, 12:16
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.

Answer: B.



Hi bunuel,
Did not got how u solved option 2.Could you please explain in detail.
thanks.
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New post 25 May 2012, 11:58
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: \(a=1\), \(b=5\) and \(c=3\), of course it's also possible that for example \(b=even\), for \(a=1\) 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.


What about the case when all a,b,c are zero. In this case, abc = 0 and 0 is neither odd nor even. Hence 'E'.
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New post 25 May 2012, 12:29
avinash2603 wrote:
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: \(a=1\), \(b=5\) and \(c=3\), of course it's also possible that for example \(b=even\), for \(a=1\) 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.


What about the case when all a,b,c are zero. In this case, abc = 0 and 0 is neither odd nor even. Hence 'E'.


Welcome to GMAT Club. Below is an answer to your question.

Notice that zero is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder. Since 0/2=0=integer then zero is even.

For more on this subject please check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.
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New post 04 Jun 2012, 21:39
Bunuel wrote:
5. What is the value of integer x?

(1) 2x^2+9<9x --> factor qudratics: \((x-\frac{3}{2})(x-3)<0\) --> roots are \(\frac{3}{2}\) and 3 --> "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) --> since there only integer in this range is 2 then \(x=2\). Sufficient.



\((x-\frac{3}{2})(x-3)<0\)

I agree one solution of this inequality is

\((x-\frac{3}{2})\) > 0 , (x-3)<0 => \((\frac{3}{2})\) < x < 3

However, Don't u think this can also resort to

\((x-\frac{3}{2})\) < 0 , (x-3)>0 => x < \((\frac{3}{2})\) , x > 3

and in that case x can have infinite values.

and if that is the case Stmt-1 alone would not be sufficient.
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New post 05 Jun 2012, 00:51
anordinaryguy wrote:
Bunuel wrote:
5. What is the value of integer x?

(1) 2x^2+9<9x --> factor qudratics: \((x-\frac{3}{2})(x-3)<0\) --> roots are \(\frac{3}{2}\) and 3 --> "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) --> since there only integer in this range is 2 then \(x=2\). Sufficient.



\((x-\frac{3}{2})(x-3)<0\)

I agree one solution of this inequality is

\((x-\frac{3}{2})\) > 0 , (x-3)<0 => \((\frac{3}{2})\) < x < 3

However, Don't u think this can also resort to

\((x-\frac{3}{2})\) < 0 , (x-3)>0 => x < \((\frac{3}{2})\) , x > 3


and in that case x can have infinite values.

and if that is the case Stmt-1 alone would not be sufficient.


The second case of \((x-\frac{3}{2})<0\) and \((x-3)>0\) is not possible. This condition leads to \(x<\frac{3}{2}\) and \(x>3\), but \(x\) can not be simultaneously less than \(\frac{3}{2}\) (to make \(x-\frac{3}{2}\) negative) and more than 3 (to make \(x-3\) positive).

For more on how to solve such kind of inequalities check:
x2-4x-94661.html#p731476 (Check this first)
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486

Hope it helps.
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New post 27 Jun 2012, 08:03
Hi Bunuel,

Can we solve this question using graphical approach.

Since (x-a)^2+(y-b)^2 = r^2 ----(A)

Equation 1 represents equation of circle with radius 1.

Now, if we can maximize the value of both x and y, we can get the maximum value of xy and that could let us know whether statement (1) is sufficient or not.

Here, I require your help. Intuitively, I find that x=y will maximize the equation x^2+y^2=1. But I am not sure about my reasoning as well as the Mathematics behind. Could you please look into this.

Thanks

Bunuel wrote:

Please read the thread. Solutions to ALL the questions are given on the previous pages.

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.
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New post 27 Jun 2012, 08:17
imhimanshu wrote:
Hi Bunuel,

Can we solve this question using graphical approach.

Since (x-a)^2+(y-b)^2 = r^2 ----(A)

Equation 1 represents equation of circle with radius 1.

Now, if we can maximize the value of both x and y, we can get the maximum value of xy and that could let us know whether statement (1) is sufficient or not.

Here, I require your help. Intuitively, I find that x=y will maximize the equation x^2+y^2=1. But I am not sure about my reasoning as well as the Mathematics behind. Could you please look into this.

Thanks

Bunuel wrote:

Please read the thread. Solutions to ALL the questions are given on the previous pages.

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.


Useful property: For given sum of two numbers, their product is maximized when they are equal.

(1) says that \(x^2+y^2=1\). So, \(x^2y^2\) will be maximized when \(x^2=y^2\): \(x^2+x^2=1\) --> \(x^2=\frac{1}{2}\) --> the maximum value of \(x^2y^2\) thus is \(\frac{1}{4}\) and the maixmum value of \(xy\) is \(\frac{1}{2}\).

Hope it's clear.
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New post 28 Jul 2012, 07:36
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.


Why we recalled \((x-y)^2\geq{0}\) ?
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New post 29 Jul 2012, 01:55
AnanJammal wrote:
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.


Why we recalled \((x-y)^2\geq{0}\) ?


Because \((x-y)^2\geq{0}\) expands to \(x^2-2xy+y^2\geq{0}\), which leads to \(1-2xy\geq{0}\) (since \(x^2+y^2=1\)) and finally to \(xy\leq{\frac{1}{2}}\).
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New post 06 Dec 2012, 23:19
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 \(x \neq y\)

Answer (B)
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New post 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.

Answer: B.


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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New post 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.

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 ?
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New post 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.

Answer: B.


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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New post 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.

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.

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

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New post 22 Jul 2013, 23:53
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.

Answer: B.


Hi Bunuel,

In the 1st condition if we take ac = 6b and hence multiply the the given expression 6a=3b=7c by 2 we get 12a=6b=14c.

Now substituting 12a=ac=14c

I get 12a=ac --> c = 12 and similarly a = 14 and hence b= 28.

Can you please point where am I going wrong?

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Re: The Discreet Charm of the DS   [#permalink] 22 Jul 2013, 23:53

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