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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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05 Feb 2012, 03:59
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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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05 Feb 2012, 04:21
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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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05 Feb 2012, 04:32
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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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05 Feb 2012, 04:33
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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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05 Feb 2012, 04:43
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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 --> this basically means that all customers bought exactly 4 oranges (76/19=4), because if even one customer bought less than 4, the sum will be less than 76. Hence, no one bought only one orange. Sufficient.

(2) 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. But the first case is not possible: the sum of 19 odd numbers is odd and not even like 76. Hence, again no one bought only one=odd orange. Sufficient.

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05 Feb 2012, 04:56
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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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05 Feb 2012, 05:35
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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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05 Feb 2012, 06:11
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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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05 Feb 2012, 06:38
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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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05 Feb 2012, 06:49
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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 --> $$5*14x=8*7x+4*14x$$ --> $$70x=80x$$ --> $$10x=0$$ --> $$x=0$$ --> $$a=b=c=0$$ --> $$a+b+c=0$$. Sufficient.

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

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05 Feb 2012, 06:52
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Just posted the answers. Kudos points given to everyone with correct solution. Let me know if I missed someone.
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Re: The Discreet Charm of the DS [#permalink]

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06 Feb 2012, 06:36
Bunuel wrote:
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.

Hi, I was looking at this solution and had a question. If b+3 is not a prime number, only when a number in our set equals 1, the sum is 4 (1+3= 4 (Prime), 2+3, -1+3, -2+3 are all prime numbers), but option (1,2) means b+3 = 5 (Prime number) and A tells us that b+3 is not Prime so the only set is (2,1), therefore YES a=2

???

Last edited by nhk13579 on 06 Feb 2012, 06:40, edited 1 time in total.
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Re: The Discreet Charm of the DS [#permalink]

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06 Feb 2012, 06:40
nhemdani wrote:
Bunuel wrote:
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.

Hi, I was looking at this solution and had a question. If b+3 is not a prime number, only when a number in our set equals 1, in this case the sum os 4 (1+3= 4 (Prime), 2+3, -1+3, -2+3 are all prime numbers), shouldnt the answer be A, as we are only left with option (1,2) so YES a=2

???

1 is not a prime number, so the case of (-1, -2) is still possible: -2+3=1 not a prime.

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

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06 Feb 2012, 06:42
Bunuel wrote:
nhemdani wrote:
Bunuel wrote:
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.

Hi, I was looking at this solution and had a question. If b+3 is not a prime number, only when a number in our set equals 1, in this case the sum os 4 (1+3= 4 (Prime), 2+3, -1+3, -2+3 are all prime numbers), shouldnt the answer be A, as we are only left with option (1,2) so YES a=2

???

1 is not a prime number, so the case of (-1, -2) is still possible: -2+3=1 not a prime.

Hope it's clear.

Sorry I got excited

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

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06 Feb 2012, 07:43
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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06 Feb 2012, 08:24
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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Re: The Discreet Charm of the DS [#permalink]

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06 Feb 2012, 08:54
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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06 Feb 2012, 19:52
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.

I did this for option 1 -

ac=6b
ac=2 * 3 * b

since 6a=3b=7c

a c = 2 * 6a
c = 12
now we can find a and b also, so seems sufficient.

so where am i going wrong?
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Re: The Discreet Charm of the DS [#permalink]

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06 Feb 2012, 23:11
Well cause there is 2 options for statement 1.

ac = 3b ==>
ac = 12a
Now if a was 0 then 0*12 = 0*c ==> 6*0=3*0=7*0
a+b+c= 0
or

12a = ca
c=12
7*12(c)=3*28(b)=6*14(a)
a+b+c= 54.

There is 2 possible solutions, so you do not know if it's 0 or 54
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Re: The Discreet Charm of the DS [#permalink]

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09 Feb 2012, 01:12
HI Bunuel,

I personally dont feel very comfortable with your solution for Q9. Just not very intuitive for me.

Note: I really enjoyed doing these set of questions. U r taking GMAT club Quant practice questions to a next level alltogether. Thanks
Re: The Discreet Charm of the DS   [#permalink] 09 Feb 2012, 01:12

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