If x, y, z 0 and x y z which of the following could be true?

Question

If x, y, z > 0 and x > y > z which of the following  could  be true?

I.  \sqrt{x} >\sqrt{y} > \sqrt{z}

II.  \sqrt{z} > \sqrt{y} > \sqrt{x}

III.  \sqrt{x} > \sqrt {z} > \sqrt{y}

A. Only I 
B. Only II
C. I and II
D. I and III
E. I, II and III

Taking quite some time to solve it by trial and error, Any efficient workaround to do this?

chetan2u · Accepted Answer

If you know the way the numbers behave when raised to successive powers, you will get your answer without getting into hit and trial.

x>y>z>0

If you take same power for all of them, the order will remain same.

So, I.  \sqrt{x} >\sqrt{y} > \sqrt{z}  is correct, but 
II.  \sqrt{z} > \sqrt{y} > \sqrt{x}  is not.

let us see the third one now as z and y are raised to different powers, and both x and y raised to SAME power follow the order, that is  \sqrt{x} >\sqrt{y}

In z and y, we are taking the third root and second root respectively. 
So, a value greater than 1, will surely give a smaller value when raised to third root as compared to second root, but a value between 0 and 1, will do opposite:  It will increase the value .

So if z and y are between 0 and 1, we can find a value satisfying the given order.

say  y=\frac{1}{16};z=\frac{1}{27}........y>z......\frac{1}{16}>\frac{1}{27}

But  \sqrt {z}=\sqrt {\frac{1}{27}}=\frac{1}{3}  and  \sqrt{y}=\sqrt{\frac{1}{16}}=\frac{1}{4} .
Here,  \sqrt {z}>\sqrt{y} 
 Hence III can also be true.

D

terminatork07 · Answer

chetan2u  Are there any resources/tips/tricks for these sort of questions. Putting smart-numbers always leaves room for errors and doesn't give me confidence.

egmat · Answer

terminatork07 Here's a decision framework that eliminates guesswork. See if it helps you! The Critical Divider: n = 1 This is the only number you need to remember. Numbers behave oppositely on either side of 1 when taking roots/powers: Rule 1: For numbers > 1 Taking any root decreases the value Higher roots create smaller results Example: 16^{1/2} = 4 but 16^{1/3} = 2.52 Rule 2: For numbers between 0 and 1 Taking any root increases the value Higher roots create larger results Example: (1/16)^{1/2} = 1/4 but (1/16)^{1/3} = 0.397 Your 3-Second Decision Process: When you see root comparisons: 1. First check: Are the roots the same? → If yes, order stays same 2. Second check: Different roots? → Consider if values could be < 1 3. Apply rule: If < 1 possible, higher roots give larger values For This Specific Problem: Statement III ( \sqrt{x} > \sqrt {z} > \sqrt{y} ) works because: - We can choose 0 < z < y < 1 - The cube root of z "jumps higher" than the square root of y - While x > 1 keeps \sqrt{x} largest overall Pattern Recognition: This logic appears in: - Data Sufficiency with inequalities and roots - PS problems comparing expressions like x^2 vs x^3 - Work/rate problems where rates are fractional You can practice similar questions here (these are OG questions) - select Exponents & Roots under Problem Solving and choose Medium/Hard level questions (as per your current ability levels for the topic).

If x, y, z > 0 and x > y > z which of the following could be true?

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