If \(0 < a < b < c,\) which of the following statements must be true?

VJesus12 wrote: ↑

Thu Jun 04, 2020 7:51 am

If \(0 < a < b < c,\) which of the following statements must be true?

I. \(2a > b + c\)
II. \(c – a > b - a\)
III. \(\dfrac{c}{a} < \dfrac{b}{a}\)

A) I only
B) II only
C) III only
D) I and II
E) II and III

[spoiler]OA=B[/spoiler]

Source: Official Guide

I. 2a > b + c
Consider this scenario: a = 1, b = 2 and c = 3. This meets the given condition that 0 < a < b < c.
HOWEVER, if we plug these values into statement I, we see that it is NOT the case that 2a > b + c
So, statement I NEED NOT BE TRUE

II. c – a > b - a
It's already given that c > b
If we subtract ANY VALUE (such as a) from both sides, the inequality remains valid.
So, statement II MUST BE TRUE

III. c/a < b/a
Consider this scenario: a = 1, b = 2 and c = 3. This meets the given condition that 0 < a < b < c.
HOWEVER, if we plug these values into statement III, we see that it is NOT the case that c/a < b/a
So, statement III NEED NOT BE TRUE

Answer: B

Cheers,
Brent

VJesus12 wrote: ↑

Thu Jun 04, 2020 7:51 am

If \(0 < a < b < c,\) which of the following statements must be true?

I. \(2a > b + c\)
II. \(c – a > b - a\)
III. \(\dfrac{c}{a} < \dfrac{b}{a}\)

A) I only
B) II only
C) III only
D) I and II
E) II and III

[spoiler]OA=B[/spoiler]

Source: Official Guide

Solution:

Let’s analyze each inequality in the given Roman numerals.

I. 2a > b + c

If a = 1, b = 2 and c = 3, we see that 2a = 2 and b + c = 5. However, since 2 is not greater than 5, we see that I is not true.

II. c - a > b - a

Adding a to both sides, we have c > b. Since we are given that b < c, II is true.

III. c/a < b/a

Multiplying both sides by a, we have c < b (notice that we don’t have to switch the inequality sign since a is positive). However, we are given that b < c, so III is not true.

Answer: B