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# The Discreet Charm of the DS

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Math Expert
Joined: 02 Sep 2009
Posts: 47920
Re: The Discreet Charm of the DS  [#permalink]

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10 Jun 2017, 12:47
KARISHMA315 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.

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?

Yes, that's also a correct way of solving. Good thing you did is that you did not forget to test both values after you got them.
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Re: The Discreet Charm of the DS  [#permalink]

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06 Sep 2017, 00:43
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.

Hi Bunuel,

I understood that x*|y|=positive; but if y<=0 then |y|=-y and in this case x can be negative.
Am I missing something?

Thanks.
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Joined: 02 Sep 2009
Posts: 47920
Re: The Discreet Charm of the DS  [#permalink]

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06 Sep 2017, 00:49
1
Yashodhan123 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.

Hi Bunuel,

I understood that x*|y|=positive; but if y<=0 then |y|=-y and in this case x can be negative.
Am I missing something?

Thanks.

Yes.

An absolute value of a number cannot be negative: |a| is positive or 0, no matter whether a itself is negative or not.

If y <= 0, then |y| = -y, yes, but even in this case -y = -negative = positive.
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Re: The Discreet Charm of the DS  [#permalink]

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08 Jun 2018, 15:04
12. If 6a=3b=7c, what is the value of a+b+c?
(1) ac=6b
(2) 5b=8a+4c

Let 6a=3b=7c=K (where K is a constant)
Thus, a=K/6
b= K/3
c=K/7

Therefore, we need to find the value of K/6 + K/3 + K/7 = 9K/14. OR if we find the value of just K it will give us the answer.

Stmt 1: ac=6b
$$K/6*K/7 = 6*K/3;$$

$$K^2/42-2K=0$$

$$K(K/42-2)=0$$

So, K = 0 or K= 84. NOT Sufficient.

Stmt 2:
5b=8a+4c
5K/3=8K/6 +4K/7
40K/21-5K/3=0
5K/3=0
K=0 SUFFICIENT

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

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09 Jun 2018, 01:36
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.

Dear sir,
Why are not you taking (x+y)^2 >=0 i.e. x^2+y^2+2xy>=0?
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Re: The Discreet Charm of the DS  [#permalink]

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10 Jun 2018, 00:16
BelalHossain046 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.

Dear sir,
Why are not you taking (x+y)^2 >=0 i.e. x^2+y^2+2xy>=0?

The point is that if you consider $$(x+y)^2 \ge 0$$, you'll get $$xy\geq{-\frac{1}{2}}$$, which is not helpful at all. Actually since both $$xy\geq{-\frac{1}{2}}$$ (from $$(x+y)^2 \ge 0$$) and $$xy \le \frac{1}{2}$$ (from $$(x-y)^2 \ge 0$$) are true, then we get that $$-\frac{1}{2} \leq{xy} \leq{\frac{1}{2}}$$. But only the approach given in the solution above gives you the answer we are looking for, while another one gives you an inequality which is not helpful.

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

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18 Jul 2018, 05:26
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.

Shouldn't we consider zero as positive ?
2 is not sufficient as |y| can be zero so x can be negative or positive
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Posts: 47920
Re: The Discreet Charm of the DS  [#permalink]

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18 Jul 2018, 05:29
1
teaserbae 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.

Shouldn't we consider zero as positive ?
2 is not sufficient as |y| can be zero so x can be negative or positive

ZERO.

1. 0 is an integer.

2. 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.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.
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Re: The Discreet Charm of the DS &nbs [#permalink] 18 Jul 2018, 05:29

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