If G^2

Question

If G^2 < G, which of the following could be G?

(A) 1
(B) 23/7
(C) 7/23
(D) -4
(E) -2

I do not understand this basic and I need clarification on this question...

Show Spoiler

G raised to the power of 2 < G
=> G raised to the power of 2 -G < 0
=> G(G-1) < 0
=> Either G < 0 OR G-1 < 0
=> G < 0 or G < 1
=> G must be less than 0.

So, shouldn't the negative numbers be the possible values of G?

What I am missing, please?

Gnpth · Answer

Hi, the Answer is C. For your question why negative numbers cannot be a value of G. we have the inequality as G^2

Also Please read the rules for posting before posting any question

Bunuel · Answer

One can solve this question algebraically (check the links at the end of the post) or simply go through the options. First of all notice that G cannot possibly be negative, because if G is negative then G^2=positive and in this case (G^2=positive)>(G=negative) , which contradicts the given condition that G^2 < G . Discard D and E. Clearly, G cannot be 1 either: in this case G^2 = G . Discard A. The same way, G cannot be greater than 1: in this case G^2 > G . Discard B. Only option C remains: if 0x2-4x-94661.html#p731476 inequalities-trick-91482.html data-suff-inequalities-109078.html range-for-variable-x-in-a-given-inequality-109468.html everything-is-less-than-zero-108884.html graphic-approach-to-problems-with-inequalities-68037.html All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184 All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189 700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html

ScottTargetTestPrep · Answer

In order for G^2 to be less than G, G must be a value between 0 and 1. Thus, G could be 7/23.

Answer: C

BrentGMATPrepNow · Answer

APPROACH #1 : Properties of exponents
 case i : If x > 1, then 0 < x < x²
 case ii : If 0 < x < 1, then 0 < x² < x 
 case iii : If -1 < x < 0, then x < 0 < x²
 case iv : If x < -1, then x < 0 < x²
 
Since G² < G, then we're dealing with case ii, which means 0 < x < 1
Answer: C

APPROACH #2 : Process of elimination
(A) If G = 1, then G² = 1. These values don't satisfy the condition that G² < G. Eliminate A. 
(B) If G = 23/7, then G² = 23²/7². These values don't satisfy the condition that G² < G. Eliminate B. 
(C) If G = 7/23, then G² = 7²/23². These values DO satisfy the condition that G² < G 
(D) If G = -4, then G² = 16. These values don't satisfy the condition that G² < G. Eliminate D. 
(E) If G = -1, then G² = 1. These values don't satisfy the condition that G² < G. Eliminate E. 
Answer: C

BrushMyQuant · Answer

Given that G^2 < G and we need to find a possible value of G from the answer choice G^2 < G => G^2 -G<0 => G (G-1)<0 Product of two values < 0 => these two values will have OPPOSITE Signs Now this will give us two cases Case 1 G < 0 and G-1 > 0 => G < 0 and G > 1 Intersection will give us NO SOLUTION in this case Case 2 G > 0 and G-1 < 0 => G > 0 and G < 1 => 0 < G < 1 Only possible answer in this range is C So, Answer will be C Hope it helps! Watch the following video to learn the Basics of Inequalities https://www.youtube.com/watch?v=1THrnWJtOqc

bumpbot · Answer

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If G^2 < G, which of the following could be G?

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