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If G^2 < G, which of the following could be G?
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Updated on: 09 Jan 2023, 10:37
Updated on: 09 Jan 2023, 10:37
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If \(G^2 < G\), which of the following could be G?
(A) 1
(B) 23/7
(C) 7/23
(D) −4
(E) -2
(A) 1
(B) 23/7
(C) 7/23
(D) −4
(E) -2
Originally posted by GeminiHeat on 02 Feb 2021, 03:26.
Last edited by Carcass on 09 Jan 2023, 10:37, edited 1 time in total.
Last edited by Carcass on 09 Jan 2023, 10:37, edited 1 time in total.
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Re: If G^2 < G, which of the following could be G?
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02 Feb 2021, 07:20
02 Feb 2021, 07:20
2
GeminiHeat wrote:
If \(G^2 < G\), which of the following could be G?
(A) 1
(B) 23/7
(C) 7/23
(D) −4
(E) -1
(A) 1
(B) 23/7
(C) 7/23
(D) −4
(E) -1
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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Re: If G^2 < G, which of the following could be G?
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16 Dec 2021, 04:59
16 Dec 2021, 04:59
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
\(G^2\) < 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
