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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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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.

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4. How many numbers of 5 consecutive positive integers is divisible by 4?

(1) The median of these numbers is odd --> the median of the set with odd number of terms is just a middle term, thus our set of 5 consecutive numbers is: {Odd, Even, Odd, Even, Odd}. Out of 2 consecutive even integers only one is a multiple of 4. Sufficient.

(2) The average (arithmetic mean) of these numbers is a prime number --> in any evenly spaced set the arithmetic mean (average) is equal to the median --> mean=median=prime. Since it's not possible that median=2=even, (in this case not all 5 numbers will be positive), then median=odd prime, and we have the same case as above. Sufficient.

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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.
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6. If a and b are integers and ab=2, is a=2?

Notice that we are not told that a and b are positive.

There are following integer pairs of (a, b) possible: (1, 2), (-1, -2), (2, 1) and (-2, -1). Basically we are asked whether we have the third case.

(1) b+3 is not a prime number --> rules out 1st and 4th options. Not sufficient.
(2) a>b --> again rules out 1st and 4th options. Not sufficient.

(1)+(2) Still two options are left: (-1, -2) and (2, 1). Not sufficient.

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8. If x=0.abcd, where a, b, c and d are digits from 0 to 9, inclusive, is x>7/9?

First of all 7/9 is a recurring decimal =0.77(7). For more on converting Converting Decimals to Fractions see: math-number-theory-88376.html

(1) a+b>14 --> the least value of a is 6 (6+9=15>14), so in this case x=0.69d<0.77(7) but a=7 and b=9 is also possible, and in this case x=0.79d>0.77(7). Not sufficient.

(2) a-c>6 --> the least value of a is 7 (7-0=7>6), but we don't know the value of b. Not sufficient.

(1)+(2) The least value of a is 7 and in this case from (1) least value of b is 8 (7+8=15>14), hence the least value of x=0.78d>0.77(7). Sufficient.

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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.

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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.

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11. If x and y are integers, is x a positive integer?

(1) x*|y| is a prime number --> since only positive numbers can be primes, then: x*|y|=positive --> x=positive. Sufficient

(2) x*|y| is non-negative integer. Notice that we are told that x*|y| is non-negative, not that it's positive, so x can be positive as well as zero. Not sufficient.

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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 --> $$5*14x=8*7x+4*6x$$ --> $$70x=80x$$ --> $$10x=0$$ --> $$x=0$$ --> $$a=b=c=0$$ --> $$a+b+c=0$$. Sufficient.

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

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Hey Bunuel can I ask a question for 12?

We know 6a=3b

And for statement one:

ac =6b. Can't 6b =12a

Then it becomes ac=12a ==> c=12. I know it's wrong since if a is 0 then they will be equal regardless, but can you explain why what I did was wrong?
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Re: The Discreet Charm of the DS  [#permalink]

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kys123 wrote:
Hey Bunuel can I ask a question for 12?

We know 6a=3b

And for statement one:

ac =6b. Can't 6b =12a

Then it becomes ac=12a ==> c=12. I know it's wrong since if a is 0 then they will be equal regardless, but can you explain why what I did was wrong?

Also,

6a = 3b = 7c

Can we say a/b= 1/2, b/c = 7/3, and a/c = 7/6
a) ac = 6b, therefore c = 6b/a
substituting this in b/c => b / (6b/a) = 7/3 => a =14, b=28, c = 12

Isnt A also sufficient? Am I ignoring something?
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kys123 wrote:
Hey Bunuel can I ask a question for 12?

We know 6a=3b

And for statement one:

ac =6b. Can't 6b =12a

Then it becomes ac=12a ==> c=12. I know it's wrong since if a is 0 then they will be equal regardless, but can you explain why what I did was wrong?

ac=12a (here you can not reduce by a and write c=12 as you exclude possibility of a=0) --> a(c-12)=0 --> either a=0 OR c=12. So, we get either a=b=c=0 or a=14, b=28 and c=12.

nhemdani wrote:
Also,

6a = 3b = 7c

Can we say a/b= 1/2, b/c = 7/3, and a/c = 7/6
a) ac = 6b, therefore c = 6b/a
substituting this in b/c => b / (6b/a) = 7/3 => a =14, b=28, c = 12

Isnt A also sufficient? Am I ignoring something?

Your doubt is partially addressed above, though there is another thing: from 6a = 3b you can not write a/b=1/2 because b can be zero and we can not divide by zero. The same for other ratios you wrote.

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

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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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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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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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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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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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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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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. Re: The Discreet Charm of the DS   [#permalink] 15 May 2012, 12:16

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