If a < b < 0, which of the following must be true?
A. a^2 < b^2
B. b − 10 < a
C. b + a > a
D. ab < b^2
E. ab < a^2
OA E
Source: Veritas Prep
If a < b < 0, which of the following must be true?
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Given that a < b < 0, we know that both a and b are negative and the absolute value of a is greater than that of b; thus, though a < b, we have |a| > |b|.BTGmoderatorDC wrote:If a < b < 0, which of the following must be true?
A. a^2 < b^2
B. b − 10 < a
C. b + a > a
D. ab < b^2
E. ab < a^2
OA E
Source: Veritas Prep
We can take suitable values for a and b. Say a = - 3 and b = -2.
Lets' take each option one by one.
A. a^2 < b^2
=> (-3)^2 < (-2)^2 => 9 < 4. This is incorrect since 9 > 4.
B. b - 10 < a
=> -2 - 10 < -3 => -12 < -3. This is correct; however is this must be correct?
Let's take another extremem example; say b = -2 and a = -15. Thus, from b - 10 < a, we have -2 - 10 < -15 => -12 < -15. This is incorrect since -15 < -12.
C. b + a > a
Say a = - 3 and b = -2. Thus, from b + a > a => b > 0; cancelling a from both sides. This is correct since b < 0.
D. ab < b^2
Say a = - 3 and b = -2. Thus, from ab < b^2 => -2 *- 3 < (-2)^2 => 6 < 4. This is correct since 6 > 4.
E. ab < a^2
Option E must be the correct answer. Let's check.
Say a = - 3 and b = -2. Thus, from ab < a^2 => -2 *- 3 < (-3)^2 =>6 < 9. This is correct. We can cross check this logically too.
We have ab < a^2 => b > a; cancelling a from both sides, note the sign reversal of the inequality. Since a is negative, upon canceling a, the sign of the inequality will change.
So, we are left with b > a, which is correct as per the information a < b < 0.
The correct answer: E
Hope this helps!
-Jay
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Since a is less than b and both a and b are negative, ab will always be less than a^2 (for example, if a = -2 and b = -1, ab = 2 and a^2 = 4).BTGmoderatorDC wrote:If a < b < 0, which of the following must be true?
A. a^2 < b^2
B. b − 10 < a
C. b + a > a
D. ab < b^2
E. ab < a^2
OA E
Source: Veritas Prep
Answer: E
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