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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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03 Feb 2012, 13:57
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12. If 6a=3b=7c, what is the value of a+b+c?
(1) ac=6b
(2) 5b=8a+4c

Stmt 1 = I get the values of a,b and c but a,b,c can be 0 also. So insufficient.
Stmt 2 - Insufficient because of same reason.
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03 Feb 2012, 14:00
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sourabhsoni 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

The catch is that they are working independently.

Answer B

Just to clarify more, working independently does not mean they are painting different cars. They are still painting the same car. Only that, the events are mutually exclusive, as you say in probabilistic terms, so that rates are not affected when they work simultaneously.

Another way :
1) more than one solution possible.
2) let's say 1)x=y=1. Work will be completed in 1/2 hour i.e. 10.15 am.
2)x=y=3. Work will be completed in 3/2 hour i.e. 11.15 am.
Not possible. B.
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Re: The Discreet Charm of the DS [#permalink]

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03 Feb 2012, 16:37
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Bunuel wrote:
sourabhsoni wrote:
2. Is xy<=1/2?
(1) x^2+y^2=1
(2) x^2-y^2=0

My funda - Area of square is largest among all the quadilateral with same perimeter.
Stmt 1 - Only possible values of x and y are 1/Sqrt(2). So sufficient as xy = 1/2
Stmt 2 - Only says x and y are equal. Not sufficient
Answer A

You are close to correct reasoning for (1), though from it you can not say that xy=1/2 and the only possible value for x and y are 1/Sqrt(2). Consider the following example: 0^1+1^2=1.

As for (2): x^2-y^2=0 doesn't mean that x=y.

1) x2 + y2 = 1 is a circle with radius 1, and origin as center. We are only concerned with 1st and 3rd quad (think why? ), Draw a line y=x which intersects circle at x=y=1/sqrt(2). We can observe, as x becomes greater than 1/sqrt2, y gets lesser than 1/sqrt2 moving on the circle (below y=x line). Hence xy<= 1/2. Similarly, as y becomes greater than 1/sqrt2, x gets lesser than 1/sqrt2 moving on the circle (above y=x line). Hence xy<= 1/2. Hence sufficient.

For more mathematically inclined, draw graph of xy=1/2. It has only one point of contact with the circle, tangential at x=y=1/sqrt(2).
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Re: The Discreet Charm of the DS [#permalink]

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03 Feb 2012, 17:08
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vailad wrote:
IMO, if you can imagine drawing the circle and y=x line, it takes less than 30 sec. to figure it out. It only takes so much longer to explain in text. And ofcourse, no need to draw the xy graph.
Will wait though for the even faster method, if any.

If you answered this question in less than 30 sec using this approach then all I can say is great job!

As for the easier/faster solution, little hint: it involves simplest algebraic manipulation.
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03 Feb 2012, 17:22
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vailad wrote:
Bunuel wrote:
If you answered this question in less than 30 sec using this approach then all I can say is great job!

As for the easier/faster solution, little hint: it involves simplest algebraic manipulation.

F***ing good !

(x-y)2 = 1-2xy >=0 => xy>=1/2.

Now this should take 15sec !

That's it. +1 again.

Now tell me which one is easier/faster?

Just a little typo there: xy<=1/2.
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sourabhsoni wrote:
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.

prime # are always positive

Stmt 1 - |y| will be always positive to x*|y| to be prime z has to be positve. Sufficient
Stmt 2 - Same concept x will be positive number Sufficient

Answer is D

Statement 2: x can be zero or +. ( x*|y| is a non negative integer. i.eit can be 0 as well.. NS)
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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.

Answer: A.
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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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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.

Answer: B.

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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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.

Answer: C.

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

Answer: C.

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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10 Aug 2013, 23:33
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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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Re: The Discreet Charm of the DS [#permalink]

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22 Oct 2014, 06:03
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sagarjain90 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.

Hi Bunuel,
I considered $$(x+y)^2$$ $$>=$$ 0 and arrived at $$xy >=$$ $$\frac{-1}{2}$$.
And hence concluded that the statement is insufficient.
Please correct me.

Yes, you need to consider $$(x-y)^2\geq{0}$$ to get sufficiency.
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17 Apr 2016, 08:05
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bikographer 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.

Considering (2), and rearranging it,

2x - 3y < -1 .......... (i)

substituting x=-2 and y=-1 in (i), we get

-4 + 3 = -1 which is not <-1

This means that the condition doesn't hold good, and hence x < y doesn't hold good.

Could you please point out where exactly I am making a mistake in my process?

The statements in DS questions are true. So, if you pick numbers you should pick such that they satisfy the statement. x and y cannot be -2 and -1 respectively because they do not satisfy 2x - 3 < 3y - 4.
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Re: The Discreet Charm of the DS [#permalink]

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06 Nov 2016, 05:06
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hardik0491 wrote:
Bunuel wrote:
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.

Answer: A.

I think the option should be E
Option 1 -x * -y will result in positive, where x is negative.

Can you explain further..??

If x is negative, then x*|y| = negative*non-negative = non-positive, which cannot be prime, since only positive numbers are primes.
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Re: The Discreet Charm of the DS [#permalink]

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02 Feb 2012, 05:55
1. B
2. A
3. B
4. D
5. D
6. E
7. D
8. C
9. B
10. C
11. D
12. E
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Re: The Discreet Charm of the DS [#permalink]

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02 Feb 2012, 06:48
time to get cracking... thanks for posting
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Re: The Discreet Charm of the DS [#permalink]

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02 Feb 2012, 10:55
1-b
2-a
3-b
4-a
5-b
6-e
7-d
8-b
9-e
10-a
11-d
12-a
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Re: The Discreet Charm of the DS [#permalink]

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02 Feb 2012, 18:01
Ans:
1-B
2-A
3-A
4-E
5-D
6-E
7-C
8-A
9-C
10-A
11-D
12-D
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Re: The Discreet Charm of the DS [#permalink]

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03 Feb 2012, 02:36
vailad wrote:
1.B
2.A
3.B
4.D
5.A
6.E

5 correct answers out of 6.

sourabhsoni wrote:
1. B
2. A
3. B
4. D
5. D
6. E
7. D
8. C
9. B
10. C
11. D
12. E

10 correct answers out of 12.

khaadu wrote:
1-b
2-a
3-b
4-a
5-b
6-e
7-d
8-b
9-e
10-a
11-d
12-a

7 correct answers out of 12.

vinayaerostar wrote:
Ans:
1-B
2-A
3-A
4-E
5-D
6-E
7-C
8-A
9-C
10-A
11-D
12-D

4 correct answers out of 12.

Good job everyone! By the way it's better if you post the solutions along with the answers: others will benefit with your approaches and you'll get 1 Kudos point per correct solution.

Will post explanations in couple of days, so that to give some more time to those who want to participate.
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