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# The Discreet Charm of the DS

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02 Feb 2012, 04:15
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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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03 Nov 2014, 13:57
Hi Bunnel,
for 1st condition if we take (x+y) square which will = x^2+2xy+y^2>=0 and since x^2+y^2=1 then: 1-2xy\geq{0} --> xy\leq{\frac{1}{2}}. Sufficient.
will be right or wrong.

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04 Nov 2014, 04:39
taleesh wrote:
Hi Bunnel,
for 1st condition if we take (x+y) square which will = x^2+2xy+y^2>=0 and since x^2+y^2=1 then: 1-2xy\geq{0} --> xy\leq{\frac{1}{2}}. Sufficient.
will be right or wrong.

Check here: the-discreet-charm-of-the-ds-126962-20.html#p1039634
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04 Nov 2014, 10:30
Hi Bunnel,

I am sorry to say i did read the link before and has query that taking (x+y)^2 or (x-y)^2 ,both are correct or not. And if the first scenario is not right than why is it so? Your reply do not help me to clarify my doubt?
Please bear with me if i am asking silly question but understand that i need to know it.

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04 Nov 2014, 10:42
taleesh wrote:
Hi Bunnel,

I am sorry to say i did read the link before and has query that taking (x+y)^2 or (x-y)^2 ,both are correct or not. And if the first scenario is not right than why is it so? Your reply do not help me to clarify my doubt?
Please bear with me if i am asking silly question but understand that i need to know it.

You need to consider $$(x-y)^2\geq{0}$$ to get sufficiency.

Using $$(x+y)^2\geq{0}$$ leads to $$xy\geq{\frac{1}{2}}$$, which is no use for us.
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04 Nov 2014, 11:33
Thanks Bunnel,

Its clear now.

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27 Jan 2015, 16:39
Dear Bunuel, I answerd the above question by this way:

1- X^2+Y^2=1 so X^2=1-Y^2 X=(1-Y) replace X by (1-y)

(1-y)y<1/2 so any number used gave that A is the answer

2- (x-y)(x+y)=0 so x=y or x=-y insuff

is that correct?
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28 Jan 2015, 01:55
23a2012 wrote:
Dear Bunuel, I answerd the above question by this way:

1- X^2+Y^2=1 so X^2=1-Y^2 X=(1-Y) replace X by (1-y)

(1-y)y<1/2 so any number used gave that A is the answer

2- (x-y)(x+y)=0 so x=y or x=-y insuff

is that correct?

How does x = 1 - y from x^2 = 1 - y^2? That's not correct.
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28 Jan 2015, 06:41
Oh, Thank you Bunuel I see my mistake so I think it should be

X^2+Y^2=1

X^2=1-Y^2

Now I square the both sides of xy≤1/2

x^2 y^2≤1/4 and i replaced x^2 by 1-y^2

1-y^2*y^2≤1/4

so A will be the answer.

is that correct?
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28 Jan 2015, 07:31
23a2012 wrote:
Oh, Thank you Bunuel I see my mistake so I think it should be

X^2+Y^2=1

X^2=1-Y^2

Now I square the both sides of xy≤1/2

x^2 y^2≤1/4 and i replaced x^2 by 1-y^2

1-y^2*y^2≤1/4

so A will be the answer.

is that correct?

That's also wrong.

We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative.

We don't know whether xy is positive, hence we cannot square xy≤1/2.

The correct way of solving this question is here: the-discreet-charm-of-the-ds-126962-20.html#p1039634

For more on inequalities check here: inequalities-tips-and-hints-175001.html

Hope it helps.
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28 Jan 2015, 08:17
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.

So, Bunuel could you please show me how we can get (x-y)^2 from x^2+y^2=1? I faced difficulty in

understand this point.

Thank you
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28 Jan 2015, 08:31
23a2012 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.

So, Bunuel could you please show me how we can get (x-y)^2 from x^2+y^2=1? I faced difficulty in

understand this point.

Thank you

$$x^2+y^2=1$$

$$x^2-2xy+y^2\geq{0}$$ --> $$(x^2+y^2)-2xy\geq{0}$$ --> $$1-2xy\geq{0}$$.
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The Discreet Charm of the DS [#permalink]

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01 Oct 2015, 04:28
Hi Bunuel,

I like your compilation of questions, but some are poorly written.

For example: "How many numbers of 5 consecutive positive integers is divisible by 4?"
What is supposed to be divisible by 4? the "5 consecutive integers" ? or the "numbers"? Either way it makes no sense. Do you mean each of these 5 numbers? the product of these 5 integer? their sum? This sentence would just never fly in SC.

Other example:
"A and B ran, at their respective constant rates, a race of 480m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat etc."
This one is confusing. You say the protagonist run ONE race of 480m, and then say they run TWO heats. So.. do they run 480m or 2*480m? Moreover the terms "heat" and "head start" are unclear. I assume you mean "lap" and "A leaves when B has run 48m". But I am not sure. I rarely have to guess on OG questions.

I am thinking of subscribing to the Quant database. Is someone proof-reading the questions?

Then again, great compilation and answers.

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10 Jan 2016, 13:31
Bunuel wrote:
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 --> 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.

HI Bunuel! Why do you assume that as per statement 1 the value of b cannot be 6? If that were the case both would be insufficient together. Is it a typo and abcd are meant to be consecutive?

Otherwise I can imagine the set of possibilities for a,b as per statement 1 would be : (9,6), (8,7), (7,8) & (6,9) in which case we have a b >6

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10 Jan 2016, 13:39
kamakazizuru 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?

(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.

HI Bunuel! Why do you assume that as per statement 1 the value of b cannot be 6? If that were the case both would be insufficient together. Is it a typo and abcd are meant to be consecutive?

Otherwise I can imagine the set of possibilities for a,b as per statement 1 would be : (9,6), (8,7), (7,8) & (6,9) in which case we have a b >6

Not following you. b can be 6 for (1). Also, in (2) it's a-c>6, not a-b>6.
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10 Jan 2016, 14:04
Bunuel wrote:
kamakazizuru 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?

(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.

HI Bunuel! Why do you assume that as per statement 1 the value of b cannot be 6? If that were the case both would be insufficient together. Is it a typo and abcd are meant to be consecutive?

Otherwise I can imagine the set of possibilities for a,b as per statement 1 would be : (9,6), (8,7), (7,8) & (6,9) in which case we have a b >6

Not following you. b can be 6 for (1). Also, in (2) it's a-c>6, not a-b>6.

My bad! I realize that if a is 7 then b has to be atleast 8 (i.e. the case where b is 6 becomes irrelevant)

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20 Jan 2016, 04:12
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.

My line of reasoning:

We are told that x and y are integers, and are asked to ascertain if x is positive. Consider x*|y|. |y| = y, when y > 0; |y| = -y, when y < 0
Case A: y > 0 => xy is a prime number, prime numbers are positive, hence x HAS TO BE positive
Case B: y < 0 => -xy is a prime number, prime numbers are positive, to make positive the quantity, the negative y has to be multiplied by a negative x, and so x is negative

How could Statement 1 alone then, be sufficient? Kindly clarify

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20 Jan 2016, 06:18
BigFatBassist 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.

My line of reasoning:

We are told that x and y are integers, and are asked to ascertain if x is positive. Consider x*|y|. |y| = y, when y > 0; |y| = -y, when y < 0
Case A: y > 0 => xy is a prime number, prime numbers are positive, hence x HAS TO BE positive
Case B: y < 0 => -xy is a prime number, prime numbers are positive, to make positive the quantity, the negative y has to be multiplied by a negative x, and so x is negative

How could Statement 1 alone then, be sufficient? Kindly clarify

In case B: if y is negative, then -y is positive, thus -xy = x*(-y)= x*positive = prime = positive --> x = positive.
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04 Feb 2016, 11:37
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.

Why did you switch the sign at the end? Could you explain, if we add 2xy to both sides, then the sign would not change? But if we minus 1 from both sides, then divide by a negative, the sign does change?

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05 Feb 2016, 01:33
ZaydenBond 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.

Why did you switch the sign at the end? Could you explain, if we add 2xy to both sides, then the sign would not change? But if we minus 1 from both sides, then divide by a negative, the sign does change?

$$1-2xy\geq{0}$$;

$$1\geq{2xy}$$;

$$\frac{1}{2} \geq{xy}$$;, which is the same as $$xy\leq{\frac{1}{2}}$$.
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16 Apr 2016, 04:45
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.

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?

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