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# Let x = 0.99, y = 0.99, and z = 0.99^2. Then which of the following

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Math Expert
Joined: 02 Sep 2009
Posts: 58453
Let x = 0.99, y = 0.99, and z = 0.99^2. Then which of the following  [#permalink]

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09 Apr 2019, 21:44
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Let x = 0.99, $$y = \sqrt{0.99}$$, and z = 0.99^2. Then which of the following is true?

(A) x < z < y
(B) z < y < x
(C) z < x < y
(D) y < x < z
(E) y < z < x

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Joined: 17 May 2018
Posts: 140
Location: India
Re: Let x = 0.99, y = 0.99, and z = 0.99^2. Then which of the following  [#permalink]

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09 Apr 2019, 22:40
2
For numbers between 0 and 1 we always follow the rule x^2<x<√x
Based on this rule
IMO C

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Intern
Joined: 17 Sep 2018
Posts: 42
Re: Let x = 0.99, y = 0.99, and z = 0.99^2. Then which of the following  [#permalink]

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09 Apr 2019, 23:00
1
If you forgot the above rule you could also solve 0.99^2 as (1-0.01)^2 = 0.9801. And can remember that the square root of a decimal X is higher than decimal X (and that, effectively, square rooting, is the reverse of squaring for decimals where, as we established above, the result is smaller).
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Joined: 26 Mar 2013
Posts: 2345
Re: Let x = 0.99, y = 0.99, and z = 0.99^2. Then which of the following  [#permalink]

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10 Apr 2019, 06:56
4
Bunuel wrote:
Let x = 0.99, $$y = \sqrt{0.99}$$, and z = 0.99^2. Then which of the following is true?

(A) x < z < y
(B) z < y < x
(C) z < x < y
(D) y < x < z
(E) y < z < x

Do not stick to the number. Simple change the numbers in the question to an easier number.

Let $$x=\frac{1}{4}$$ , $$y=\frac{1}{2}$$, $$z = \frac{1}{16}$$

From above Z < X < Y

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Re: Let x = 0.99, y = 0.99, and z = 0.99^2. Then which of the following  [#permalink]

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13 Apr 2019, 18:43
Bunuel wrote:
Let x = 0.99, $$y = \sqrt{0.99}$$, and z = 0.99^2. Then which of the following is true?

(A) x < z < y
(B) z < y < x
(C) z < x < y
(D) y < x < z
(E) y < z < x

Since the square root of a number between 0 and 1 will increase its value, and the square of that same number will decrease its value:

z < x < y

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Re: Let x = 0.99, y = 0.99, and z = 0.99^2. Then which of the following   [#permalink] 13 Apr 2019, 18:43
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