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

Hi, the Answer is C.

For your question why negative numbers cannot be a value of G.

we have the inequality as \(G^2\)<G.

When simplified we get G<0 or G<1.

G can never be less than zero, since \(G^2\) is less than G.

If G is negative, for example G= -2 then \(G^2\) becomes 4 which contradicts the statement \(G^2\)<G.

So choose the value which is less than 1 and by POE we get \(7/23\)


Hope it helps

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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 \(0<G<1\), then \(G^2 < G\). For example, \((\frac{1}{2})^2<\frac{1}{2}\).

Answer: C.


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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
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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
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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


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