If $a$ and $b$ are negative, and $c$ is positive, which of the following statement(s) is/are true?
I) $a - b < a - c$
II) if $a < b$, then $\dfrac{a}{c} < \dfrac{b}{c}$
III) $\dfrac{a}{b} > \dfrac{a}{c}$

Question

Solve the inequality: $ \log_2(x^2 - 5x + 6) >1 $

Answer 1

A quick way to check these statements is to plug in real numbers that fit the given signs and see which claims hold up, then confirm with a second set of numbers so the pattern is not a fluke.

Test with $a = -4, b = -1, c = 2$: For I, $a - b = -4-(-1) = -3$ and $a - c = -4-2 = -6$. Is $-3 < -6$? No, so I fails here. For II, since $a < b$ is true ($-4 < -1$), check $\dfrac{a}{c} = -2$ and $\dfrac{b}{c} = -0.5$; $-2 < -0.5$ holds, so II works. For III, $\dfrac{a}{b} = \dfrac{-4}{-1} = 4$ and $\dfrac{a}{c} = -2$; $4 > -2$ holds, so III works.
Test with $a = -1, b = -5, c = 10$: For I, $a - b = -1-(-5) = 4$ and $a - c = -1-10 = -11$; $4 < -11$ is false, so I fails again. For III, $\dfrac{a}{b} = \dfrac{-1}{-5} = 0.2$ and $\dfrac{a}{c} = -0.1$; $0.2 > -0.1$ holds again.

Statement I keeps failing because subtracting a negative number acts like adding, which always pushes the result up, while subtracting a positive number always pushes it down, so $a-b$ is always more than $a-c$, never less. Statement II is guaranteed by the rule that dividing an inequality by a positive number keeps its direction. Statement III is guaranteed because a negative divided by a negative is always positive, while a negative divided by a positive is always negative, and positive always beats negative.

Let's summarize:

I is false, II is always true, III is always true, so the answer is II and III only.

This matches option D.

Answer 2

A quick way to check these statements is to plug in real numbers that fit the given signs and see which claims hold up, then confirm with a second set of numbers so the pattern is not a fluke.

Test with $a = -4, b = -1, c = 2$: For I, $a - b = -4-(-1) = -3$ and $a - c = -4-2 = -6$. Is $-3 < -6$? No, so I fails here. For II, since $a < b$ is true ($-4 < -1$), check $\dfrac{a}{c} = -2$ and $\dfrac{b}{c} = -0.5$; $-2 < -0.5$ holds, so II works. For III, $\dfrac{a}{b} = \dfrac{-4}{-1} = 4$ and $\dfrac{a}{c} = -2$; $4 > -2$ holds, so III works.
Test with $a = -1, b = -5, c = 10$: For I, $a - b = -1-(-5) = 4$ and $a - c = -1-10 = -11$; $4 < -11$ is false, so I fails again. For III, $\dfrac{a}{b} = \dfrac{-1}{-5} = 0.2$ and $\dfrac{a}{c} = -0.1$; $0.2 > -0.1$ holds again.

Statement I keeps failing because subtracting a negative number acts like adding, which always pushes the result up, while subtracting a positive number always pushes it down, so $a-b$ is always more than $a-c$, never less. Statement II is guaranteed by the rule that dividing an inequality by a positive number keeps its direction. Statement III is guaranteed because a negative divided by a negative is always positive, while a negative divided by a positive is always negative, and positive always beats negative.

Let's summarize:

I is false, II is always true, III is always true, so the answer is II and III only.

This matches option D.

If \(a\) and \(b\) are negative, and \(c\) is positive, which of the following statement(s) is/are true?
I) \(a - b < a - c\)
II) if \(a < b\), then \(\dfrac{a}{c} < \dfrac{b}{c}\)
III) \(\dfrac{a}{b} > \dfrac{a}{c}\)

Show Hint

The Correct Option is D

Solution and Explanation

Top Questions on inequalities

Questions Asked in SNAP exam

If \(a\) and \(b\) are negative, and \(c\) is positive, which of the following statement(s) is/are true?I) \(a - b < a - c\)II) if \(a < b\), then \(\dfrac{a}{c} < \dfrac{b}{c}\)III) \(\dfrac{a}{b} > \dfrac{a}{c}\)