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# The Discreet Charm of the DS

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Manager
Joined: 10 Mar 2014
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Re: The Discreet Charm of the DS  [#permalink]

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01 Jun 2014, 21:18
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.

Hi Bunnel,

One question here.

in st2 i can consider 8-0 also then why we are considering only least value. in st1 we are considering least value as well as other value. why this is different in st2?

Thanks
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02 Jun 2014, 01:23
PathFinder007 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.

Hi Bunnel,

One question here.

in st2 i can consider 8-0 also then why we are considering only least value. in st1 we are considering least value as well as other value. why this is different in st2?

Thanks

Because considering the least value of a (7), while knowing nothing about b is enough to say that the second statement is not sufficient. If b=8, then the answer is yes but if b is 0, then the answer is no.
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Re: The Discreet Charm of the DS  [#permalink]

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03 Jun 2014, 00:39
Hi Bunuel,
Referring to below:
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

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.

I didn't quite get the explanation on the second statement. Do you mind explaining with some numbers?

Thanks,
Mitesh
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03 Jun 2014, 01:49
mrvora wrote:
Hi Bunuel,
Referring to below:
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

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.

I didn't quite get the explanation on the second statement. Do you mind explaining with some numbers?

Thanks,
Mitesh

If mean=median=prime=2, then the 5 consecutive integers would be {0, 1, 2, 3, 4}. But this set is not possible since we are told that the set consists of 5 positive integers and 0 is neither positive nor negative integer.

Hence, mean=median=odd prime --> {Odd, Even, Odd prime, Even, Odd}. So, the set includes 2 consecutive even numbers. Out of 2 consecutive even integers only one is a multiple of 4.

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

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18 Sep 2014, 11:05
ajaym28 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.

Hi Bunnel,

Please clear my doubt regarding second statement:

assumtion 1 :

x = -1
y = -2
it means x > y

it will give me
2(-1) - 3 < 3(-2) - 4
-5 < -10
[False]

assumtion 2 :

x = -2
y = -1
it means x < y

it will give me
2(-2) - 3 < 3(-1) - 4
-7 < -7
[False]

How x < y is satisfying through numbers .. pls help

When you pick numbers, they must satisfy the statement you are testing and the info given in the stem. So, when picking numbers for (2) numbers must satisfy 2x - 3 < 3y - 4 and x and y must be negative (stem). Then you should try to get an YES and NO answer to the question to prove insufficiency.
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Re: The Discreet Charm of the DS  [#permalink]

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18 Sep 2014, 11:18
Bunuel wrote:
ajaym28 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.

Hi Bunnel,

Please clear my doubt regarding second statement:

assumtion 1 :

x = -1
y = -2
it means x > y

it will give me
2(-1) - 3 < 3(-2) - 4
-5 < -10
[False]

assumtion 2 :

x = -2
y = -1
it means x < y

it will give me
2(-2) - 3 < 3(-1) - 4
-7 < -7
[False]

How x < y is satisfying through numbers .. pls help

When you pick numbers, they must satisfy the statement you are testing and the info given in the stem. So, when picking numbers for (2) numbers must satisfy 2x - 3 < 3y - 4 and x and y must be negative (stem). Then you should try to get an YES and NO answer to the question to prove insufficiency.

Thanks Bunnel, just one more ques putting number in that kind of equations will be good idea or solving it by algebra(like you did) is a good idea ? i mean how to figure what to use and when ?
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Re: The Discreet Charm of the DS  [#permalink]

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18 Sep 2014, 11:29
ajaym28 wrote:
Bunuel wrote:
ajaym28 wrote:

Hi Bunnel,

Please clear my doubt regarding second statement:

assumtion 1 :

x = -1
y = -2
it means x > y

it will give me
2(-1) - 3 < 3(-2) - 4
-5 < -10
[False]

assumtion 2 :

x = -2
y = -1
it means x < y

it will give me
2(-2) - 3 < 3(-1) - 4
-7 < -7
[False]

How x < y is satisfying through numbers .. pls help

When you pick numbers, they must satisfy the statement you are testing and the info given in the stem. So, when picking numbers for (2) numbers must satisfy 2x - 3 < 3y - 4 and x and y must be negative (stem). Then you should try to get an YES and NO answer to the question to prove insufficiency.

Thanks Bunnel, just one more ques putting number in that kind of equations will be good idea or solving it by algebra(like you did) is a good idea ? i mean how to figure what to use and when ?

When you decide that a statement is sufficient based only on plug-in method you should make sure that you tried several different numbers (and saw some pattern maybe), and even in this case you may not be 100% sure that the answer would be correct. Though if several numbers give the same answer and you are able to see some pattern, then you can make an educated guess that a statement is sufficient and move-on.

Generally on DS questions when plugging numbers, your goal is to prove that the statement is NOT sufficient. So you should try to get an YES answer with one chosen number(s) and a NO with another.

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

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20 Oct 2014, 20:40
Dear Bunuel,

If every one will have 4 then the difference will be 0 which is not even.

Posted from my mobile device
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21 Oct 2014, 00:56
Thoughtosphere wrote:
Dear Bunuel,

If every one will have 4 then the difference will be 0 which is not even.

Posted from my mobile device

0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

Check for more here: http://gmatclub.com/forum/number-proper ... 74996.html

Hope it helps.
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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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The Discreet Charm of the DS  [#permalink]

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

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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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06 Nov 2016, 05:18
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.

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

Can you explain further..??
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06 Nov 2016, 06:06
1
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.

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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15 Dec 2016, 11:16
Hello Bunuel,

For question 9, option B -- What if we assign the -100 for x and -99 for y? x<y, but the equation (rearranged) 3y-2x>1 does not hold true. Please help!
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15 Dec 2016, 11:23
Shru91 wrote:
Hello Bunuel,

For question 9, option B -- What if we assign the -100 for x and -99 for y? x<y, but the equation (rearranged) 3y-2x>1 does not hold true. Please help!

(2) says that 2x - 3 < 3y - 4, so when plugging numbers for x and y they must satisfy this inequality. After that you should check whether you get x < y or not. Not vise-versa.
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10 Jun 2017, 11:50
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.

Hi Bunuel, a small doubt instead of $$2x+8\geq{0}$$ --> $$x\geq{-4}$$ this logic if we take |x+10| to be both positive and negative we get 2 vale by solving equation x=2 and x=-6. however by putting values back in eqtn we can see that only for x=2 equation is satisfying. Is this correct approach?
Re: The Discreet Charm of the DS   [#permalink] 10 Jun 2017, 11:50

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