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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, 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]

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24 Feb 2012, 13:52
I have seen references to use your guides in many threads, Bunuel. Now I can see why. I will be sure to use your challenge sets in the coming weeks to hopefully boost my quant score into the 48-50 range. Thank you so much for your invaluable contributions for relative GMAT newbies, like myself.
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Re: The Discreet Charm of the DS [#permalink]

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25 Feb 2012, 04:27
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,
Vids
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Re: The Discreet Charm of the DS [#permalink]

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25 Feb 2012, 04:53
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,
Vids

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

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24 Mar 2012, 08:38
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?

(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 neither an integer nor integer/2 then $$x$$ and $$y$$ are not equal. Sufficient.

please consider adding "working together" to the stmt 2- as I deciphered they both worked separately and ended at the same time so cool enough (though same answer but got carried away by the wording)

Thanks once again for nice collection!!
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Re: The Discreet Charm of the DS [#permalink]

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24 Mar 2012, 14:08
yogesh1984 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?

(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 neither an integer nor integer/2 then $$x$$ and $$y$$ are not equal. Sufficient.

please consider adding "working together" to the stmt 2- as I deciphered they both worked separately and ended at the same time so cool enough (though same answer but got carried away by the wording)

Thanks once again for nice collection!!

Thank you for the suggestion.
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Re: The Discreet Charm of the DS [#permalink]

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14 May 2012, 15:40
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.

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

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

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14 May 2012, 22:19
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.

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

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!

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

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

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15 May 2012, 10:39
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.

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

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15 May 2012, 10:43
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.

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?

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

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15 May 2012, 11:16
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.

Hi bunuel,
Did not got how u solved option 2.Could you please explain in detail.
thanks.
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Re: The Discreet Charm of the DS [#permalink]

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16 May 2012, 00:39
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piyushksharma 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,
Did not got how u solved option 2.Could you please explain in detail.
thanks.

(2) 2x - 3 < 3y - 4 --> $$x<1.5y-\frac{1}{2}$$ --> $$x<y+(0.5y-\frac{1}{2})$$. Now, since $$y$$ is a negative number then $$0.5y-\frac{1}{2}=negative$$ so, we have that: $$x<y+negative$$. $$y+negative$$ is less then $$y$$ and if $$x$$ is less than $$y+negative$$ then it must also be less than $$y$$ itself: $$x<y$$.

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

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19 May 2012, 07:13
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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 also 2 and 3: $$2+3=5=prime$$. $$x$$ can be 12 or 100. Not sufficient.

(1)+(2) $$x$$ can only be 100. Sufficient.

hey..., can sm1 pls explain how primes of 100 can be 2 and 3?...(2nd last line),,..
thanx..
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Re: The Discreet Charm of the DS [#permalink]

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19 May 2012, 07:20
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vivekdhawan 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 also 2 and 3: $$2+3=5=prime$$. $$x$$ can be 12 or 100. Not sufficient.

(1)+(2) $$x$$ can only be 100. Sufficient.

hey..., can sm1 pls explain how primes of 100 can be 2 and 3?...(2nd last line),,..
thanx..

It should be: "... distinct primes of 100 are 2 and 5: $$2+5=7=prime$$. $$x$$ can be 12 or 100".
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Re: The Discreet Charm of the DS [#permalink]

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21 May 2012, 05:32
I plugged in number for question 2 statement (1).

Any numbers that I could think of really met the inequasion.

But do you have an algebric way of showing this rule?
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21 May 2012, 05:35
ronr34 wrote:
I plugged in number for question 2 statement (1).

Any numbers that I could think of really met the inequasion.

But do you have an algebric way of showing this rule?

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.

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

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25 May 2012, 10:58
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.

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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25 May 2012, 11:29
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.

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'.

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

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27 May 2012, 02:21
Here is my approach :
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

Ans: x & y are odd integer.
statement1: x=1,y=1 or x=3,y=1 not sufficient
statement2: time of completion is 10.30am-9.45am =45min =3/4Hr; i.e rate and time consumed by both is same.
hence, statement B is sufficient.

Ans : B

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

Ans: statement 1: x^2+y^2=1,i choose the no: x=y=sqrt(1/2); hence sufficient
x=0,y=1 ;hence sufficient
x=1,y=0 ;hence sufficient

i didnt find any number which doesnot comply to statement 1.
so, sufficient.
statement2: x^2-y^2=0 ==> mod(x) = mod(y)
==> x=y
& x=-y not sufficient

Ans: A

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

Ans: a,b,c are integers,not in sequence.
statement 1: b is half way between a & c.
a=2,b=4,c=6 abc=48 even
a=2,b=3,c=6 abc=36 even
a=3,b=5,c=7 abc=105 odd

statement 1 Not sufficient
Statement 2: a=b-c ==> b=a+c ; we cant say that abc will be even or odd because we dont know whether a,b,c is odd or even.
Not sufficient
on combining both statement also, we cant say anything about abc.

Ans: E.

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

Ans: E ( No explanation)

5. What is the value of integer x?
(1) 2x^2+9<9x
(2) |x+10|=2x+8
Ans:
statement 1: 2x^2+9<9x ==>2x^2-9x+9<0
==>(2x-1)(x-3)<0
so, 1/2<x<3 or x>3&x<1/2
Not sufficient

Statement 2: |x+10| = 2x+8
if x>10;
x+10=2x+8 ==>x=2 but (x>10)
if x<10;
-x-10=2x+8 ==>x= -6 and (x<10)
so, x=-6
Sufficient

Ans B

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

statement 1: b+3 is not a prime number i.e
b+3=1,4,6,8 so, b could be = -2,1,3,5
Not sufficient
Statement 2: a>b
and ab=2 and a&b are integers..only possible value is
a=2 & b=1

Sufficient
Ans B

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
Ans:
total oranges =76
No of customer =19
how many bought only 1 oranges?
statement1:
if none bought more than 4, then,max no of oranges bought is 19x4 =76 oranges.
in short, each customer has bought 4 oranges.
sufficient
statement 2: customer can buy any no of oranges totaling 76. 4-4=0 even, 5-3=2 even,and many more.
not sufficient

Ans : A

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
Ans: x=0.abcd
7/9=0.777777

statement 1: a+b>14
(a,b) 7,8) ,(7,9),(8,9),(8,8),(9,9)
x=0.abcd ; replacing the value of a&b
x=0.78cd
x=0.79cd
x=0.89cd
x=0.88cd
x=0.99cd
all are greater then 0.77777 hence
Sufficient

statment 2: a-c>6
(a,c): (9,2) (7,0) and many more
x=0.92cd is >0.7777 ok
x=0.70cd is <0.7777 not ok

Not sufficient

Ans A

9. If x and y are negative numbers, is x<y?
(1) 3x + 4 < 2y + 3
(2) 2x - 3 < 3y - 4
Ans: x,y <0
statement 1: 3x+4<2y+3 ==>3x-2y+1<0 not sufficient

statement 2: 2x-3<3y-4 ==> 2x-3y +1<0 not sufficient

on combining both statement and solving for x& y
x< -1/5 & y< 1/5
so, y>x for interval (-1/5 to 1/5) since both are -ve so interval should be (-1/5 to 0)
and y=x for (-infinity to -1/5)
Not sufficient

Ans E

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.
Ans: f(10,x)=11 ==> (10+x)/GCF(10,x) =11 ==>x = GCF(10,x)-10

Statement 1: x could be =1,4,9,16,25..
GCF of (10,1) , (10,4),(10,9) will be different.
Not sufficient

Statement 2:
x= 2 , no of factor 2 (1&2) ok
x= 4 , no of factor 3 (1,2,4) ok
x= 10 , no of factor 4 (1,2,5,10) not ok
not sufficient

on combining I & II
we can get value like 1,4,25 which satisfy both the statement
but no unique value of x can be found.

Ans E

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.
Ans:
statement 1: x*|y| is prime no
no information about +ve or -ve no.
Not sufficient

Satement 2: for x*|y| has to non-ve integer both x& y has to -ve or +ve simultaneously
any value inside mode is always positive. mode(y) = positive
to make x*|y| +ve, X has to be positive.

hence sufficient.
Ans B

12. If 6a=3b=7c, what is the value of a+b+c?
(1) ac=6b
(2) 5b=8a+4c
Ans: 6a=3b=7c= k
a=k/6
b=k/3
c=k/7
a+b+c = (k/6)+(k/3)+k/7) if we can find the value of K, we wil have our answer.

Statement 1: ac=6b ==>(k/6)(k/7) = 6.k/3 ==>k=84
Sufficient

statement 2: 5b=8a+4c
==> 5.k/3 = (8k/6)+(4.k/7)
no value of k can be found.

Hence
Ans A.

please check my approach ans suggest if anything is missing or wrong.
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Re: The Discreet Charm of the DS [#permalink]

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04 Jun 2012, 20:39
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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Re: The Discreet Charm of the DS [#permalink]

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04 Jun 2012, 23:51
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

Hope it helps.
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Re: The Discreet Charm of the DS   [#permalink] 04 Jun 2012, 23:51

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# The Discreet Charm of the DS

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