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# Is d > 1 (1) Median of d, -d, 1/d is d. (2). Median of d d^2 d^3 i

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Is d > 1 (1) Median of d, -d, 1/d is d. (2). Median of d d^2 d^3 i  [#permalink]

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18 Mar 2019, 10:43
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Difficulty:

65% (hard)

Question Stats:

51% (01:39) correct 49% (02:01) wrong based on 37 sessions

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Is $$d>1$$?

(1) Median of $$d$$, $$-d$$, $$\frac{1}{d}$$ is $$d$$.

(2) Median of $$d$$, $$d^2$$, $$d^3$$ is $$d$$

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Re: Is d > 1 (1) Median of d, -d, 1/d is d. (2). Median of d d^2 d^3 i  [#permalink]

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29 Mar 2019, 08:06
PriyankaPalit7 wrote:
Is $$d>1$$?

(1) Median of $$d$$, $$-d$$, $$\frac{1}{d}$$ is $$d$$.

(2) Median of $$d$$, $$d^2$$, $$d^3$$ is $$d$$

Since we have three numbers, and since the median is just the middle of 3, then translating out statements into equations is easy.
We'll do this and simplify, a Precise approach.

(1) Since d is the median then 1/d ≤ d ≤ -d or -d ≤ d ≤ 1/d. In the first case d ≤ -d implies d ≤ 0 so it is definitely smaller than 1. In the second case d ≤ 1/d implies that d ≤ 1 (just multiply by d) so it is once again smaller than 1.
Then the answer is NO: Sufficient.

(2) Our two options are d^2 ≤ d ≤ d^3 or d^3 ≤ d ≤ d^2. In the first case d^2 ≤ d implies d(d - 1) ≤ 0 so d is between 0 and 1. In the second case d^3≤d^2 implies d^2(d-1) ≤ 0 so, since d^2 is nonnegative, then d - 1 ≤ 0 and therefore d ≤ 1.
Then the answer is once again NO: Sufficient.

Note that another solution would be to notice that d = 1 works in both cases, and then to see that if d > 1 then 1/d must be the median of (-d, 1/d, d) and d^2 must be the median of (d, d^2, d^3) in contradiction to the given data. This is a more 'Logical' approach, as it revolves around seeing the logical patterns behind the number choices.
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Re: Is d > 1 (1) Median of d, -d, 1/d is d. (2). Median of d d^2 d^3 i  [#permalink]

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29 Mar 2019, 09:06
PriyankaPalit7 wrote:
Is $$d>1$$?

(1) Median of $$d$$, $$-d$$, $$\frac{1}{d}$$ is $$d$$.

(2) Median of $$d$$, $$d^2$$, $$d^3$$ is $$d$$

#1
value of d=0 only sufficinet
#2
value d= 1 , 0
so is not >1 ; sufficient
IMO D
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Re: Is d > 1 (1) Median of d, -d, 1/d is d. (2). Median of d d^2 d^3 i  [#permalink]

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03 Apr 2019, 02:00
DavidTutorexamPAL wrote:
PriyankaPalit7 wrote:
Is $$d>1$$?

(1) Median of $$d$$, $$-d$$, $$\frac{1}{d}$$ is $$d$$.

(2) Median of $$d$$, $$d^2$$, $$d^3$$ is $$d$$

Since we have three numbers, and since the median is just the middle of 3, then translating out statements into equations is easy.
We'll do this and simplify, a Precise approach.

(1) Since d is the median then 1/d ≤ d ≤ -d or -d ≤ d ≤ 1/d. In the first case d ≤ -d implies d ≤ 0 so it is definitely smaller than 1. In the second case d ≤ 1/d implies that d ≤ 1 (just multiply by d) so it is once again smaller than 1.
Then the answer is NO: Sufficient.

(2) Our two options are d^2 ≤ d ≤ d^3 or d^3 ≤ d ≤ d^2. In the first case d^2 ≤ d implies d(d - 1) ≤ 0 so d is between 0 and 1. In the second case d^3≤d^2 implies d^2(d-1) ≤ 0 so, since d^2 is nonnegative, then d - 1 ≤ 0 and therefore d ≤ 1.
Then the answer is once again NO: Sufficient.

Note that another solution would be to notice that d = 1 works in both cases, and then to see that if d > 1 then 1/d must be the median of (-d, 1/d, d) and d^2 must be the median of (d, d^2, d^3) in contradiction to the given data. This is a more 'Logical' approach, as it revolves around seeing the logical patterns behind the number choices.

-d ≤ d ≤ 1/d. "In the second case d ≤ 1/d implies that d ≤ 1 (just multiply by d) so it is once again smaller than 1."
Here you just multiplied both sides by d assuming it is >0. If d<0, your derivation just doesn't hold. Try -2 for example:
-(-2)<=-2<=1/-2 or 2<=-2<=-0.5 isnt correct see.
Re: Is d > 1 (1) Median of d, -d, 1/d is d. (2). Median of d d^2 d^3 i   [#permalink] 03 Apr 2019, 02:00
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