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The Discreet Charm of the DS

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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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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?

Regards

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New post 22 Jul 2013, 23:56
Spaniard 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.


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?

Regards


ac=12a (here you cannot 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.

Does this make sense?
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Re: The Discreet Charm of the DS [#permalink]

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New post 22 Jul 2013, 23:58
Bunuel wrote:
Spaniard 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.


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?

Regards


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.

Does this make sense?


I had missed that possibility. Thanks a lot.

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New post 28 Jul 2013, 13:36
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.

Answer: B.


Hi Bunuel is there any reason to discard zero from statement #2, i mean if you consider that a could be zero b=c and then axbxc would be zero giving an additional result. It never says that these are positive integres. Thanks in advance i could be wrong it wouldn't be the first time :-)

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New post 28 Jul 2013, 13:39
duruindio 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, 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.

Answer: B.


Hi Bunuel is there any reason to discard zero from statement #2, i mean if you consider that a could be zero b=c and then axbxc would be zero giving an additional result. It never says that these are positive integres. Thanks in advance i could be wrong it wouldn't be the first time :-)


abc could be zero, but zero is even, so the answer is still YES.
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New post 02 Aug 2013, 03:42
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.


Can you please elaborate why you assumed X and Y to be integers?
Correct me if I am wrong, but are rational numbers not 'Numbers'?

Because in that case, the answer IMO is C

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New post 09 Aug 2013, 02:57
crash 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.

Answer: B.


Can you please elaborate why you assumed X and Y to be integers?
Correct me if I am wrong, but are rational numbers not 'Numbers'?

Because in that case, the answer IMO is C


Where in the second statement did I assume that the numbers are integers?
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New post 11 Aug 2013, 00:27
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.


I followed a different approach but I am getting that (1) also answers the question.

From 3b=7c => 6b=14c.

So ac=6b - equation 1)
14c=6b - equation 2)

Dividing 1) by 2) =>a/14= 1

So a=14!!!. Where am I committing a mistake in this logic? Bunuel, Please help!!

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pjagadish27 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.


I followed a different approach but I am getting that (1) also answers the question.

From 3b=7c => 6b=14c.

So ac=6b - equation 1)
14c=6b - equation 2)

Dividing 1) by 2) =>a/14= 1

So a=14!!!. Where am I committing a mistake in this logic? Bunuel, Please help!!


You cannot divide ac by 14c because c could be zero and division by zero is not allowed. The same applies to division of 6b by 6b.

Does this make sense?
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New post 08 Sep 2013, 08:44
Bunuel wrote:
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.

Answer: C.


Why have you taken value on rhs till second place of decimal. Thereby concentrating only on a and b? Please explain

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

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New post 08 Sep 2013, 08:55
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.


How did you decide which approach to use in A, one could have solved like the way you did in B
3X < 2y - 1 or x < .67y - 0.33 , so x < y + negative , so x is more negative, so x < Y.

Is this is a 700 question ?

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New post 08 Sep 2013, 10:55
ygdrasil24 wrote:
Bunuel wrote:
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.

Answer: C.


Why have you taken value on rhs till second place of decimal. Thereby concentrating only on a and b? Please explain


Please elaborate your question. Where did I do that?
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New post 08 Sep 2013, 11:02
ygdrasil24 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.

Answer: B.


How did you decide which approach to use in A, one could have solved like the way you did in B
3X < 2y - 1 or x < .67y - 0.33 , so x < y + negative , so x is more negative, so x < Y.

Is this is a 700 question ?


The approach used for the second statement does not work for the first:

\(3x + 4 < 2y + 3\) --> \(x<\frac{2y}{3}-\frac{1}{3}\) ---> \(x<y-\frac{y}{3}-\frac{1}{3}\) --> \(x<y+(-\frac{y}{3}-\frac{1}{3})\) --> we don't know whether \(-\frac{y}{3}-\frac{1}{3}=positive+negative\) is negative.

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

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New post 08 Sep 2013, 11:16
Is this is a 700 question ?[/quote]

The approach used for the second statement does not work for the first:

\(3x + 4 < 2y + 3\) --> \(x<\frac{2y}{3}-\frac{1}{3}\) ---> \(x<y-\frac{y}{3}-\frac{1}{3}\) --> \(x<y+(-\frac{y}{3}-\frac{1}{3})\) --> we don't know whether \(-\frac{y}{3}-\frac{1}{3}=positive+negative\) is negative.

Hope it's clear.[/quote]
No :(

We could have written similarily in B ? x < 1.5y - 0.5 or x < 2y +(-.5y - 0.5)

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New post 08 Sep 2013, 11:21
ygdrasil24 wrote:
Is this is a 700 question ?


The approach used for the second statement does not work for the first:

\(3x + 4 < 2y + 3\) --> \(x<\frac{2y}{3}-\frac{1}{3}\) ---> \(x<y-\frac{y}{3}-\frac{1}{3}\) --> \(x<y+(-\frac{y}{3}-\frac{1}{3})\) --> we don't know whether \(-\frac{y}{3}-\frac{1}{3}=positive+negative\) is negative.

Hope it's clear.[/quote]
No :(

We could have written similarily in B ? x < 1.5y - 0.5 or x < 2y +(-.5y - 0.5)[/quote]

I don;t understand what you mean.

You can write (2) 2x - 3 < 3y - 4 in many ways. Correct way will give you that the statement is sufficient.

Similarly you can write (1) 3x + 4 < 2y + 3 in many ways. None of which will give you that the statement is sufficient.
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Re: The Discreet Charm of the DS [#permalink]

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New post 31 Oct 2013, 16:05
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.


I don't understand something.
Sometimes, abs of "x" will be checked to equal -x if x is negative, or x if x is positive.
So why isn't there a check for x-10 and -(x-10)?

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

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New post 01 Nov 2013, 01:04
ronr34 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.


I don't understand something.
Sometimes, abs of "x" will be checked to equal -x if x is negative, or x if x is positive.
So why isn't there a check for x-10 and -(x-10)?


There can be more than one way to solve a question.
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Re: The Discreet Charm of the DS [#permalink]

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New post 22 Nov 2013, 06:36
Bunuel,
This
x^2+y^2=1 could not have been (x^2+y^2)^2?

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

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New post 22 Nov 2013, 08:17

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Re: The Discreet Charm of the DS   [#permalink] 22 Nov 2013, 08:17

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