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

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
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I. obvously we cannot say that it "must" be true.

II. c – a > b - a
Cancelling a on both sides
c > b
Definitely true as per question.

III. c/a < b/a
Multiply both sides by "a" (positive)
c < b
Definitely "not" true as per question.

So, only II is correct.

Hey,

For (1),
0<a<b<c, can't we write it as
a<b....i
a<c...ii

adding i and ii

2a<b+c ???? (addition property of inequalities)

Or
Is it like this property holds true only when abcd are different numbers??

Appreciate a clarification!

TIA.

Nothing indicates what a may be greater than, so I is wrong.
III is always wrong because the order should be reversed.

II is correct.

Answer choice B is correct.
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We are given that 0 < a < b < c and need to determine which statements are true. Let’s analyze each Roman numeral.

I. 2a > b + c

2a cannot be greater than the sum of b and c. Since a + a = 2a, a < b, and a < c, a + a < b + c, or 2a < b + c. Statement I is FALSE.

II. c – a > b - a

We can simplify the inequality to c > b. Since we are given that c is greater than b in the stem, c – a is greater than b - a. Statement II is TRUE.

III. c/a < b/a

We can multiply both sides by a and we have c < b (note: we don’t need to switch the inequality sign because a is positive). However, we are given that c is greater than b, so c/a can’t be less than b/a. Statement III is FALSE.

Thus, only Roman numeral II is true.

Answer: B
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A simple and quick approach is to assume values for a, b, c
If a=2, b=3, c=4
I. 2a > b + c = 4 > 7; this is wrong
II. c – a > b - a = 2 > 1; this is correct
III. c/a <b/a = 2 < 1.5; this is wrong

Therefore, II only is correct. Answer is B

[1] Assign Test Values

Must follow rule: 0 < a < b < c,

a = 2
b = 3
c = 4

[2] Plug In & Calculate

\(I. 2a > b + c \rightarrow 2(2) > 3 + 4 \rightarrow 4 > 7 \space (FALSE)\)

\(II. c – a > b - a \rightarrow 4 - 2 > 3 - 2 \rightarrow 2 > 1 \space (TRUE)\)

\(III. \frac{c}{a} < \frac{b}{a} \rightarrow \frac{4}{2} < \frac{3}{2} \rightarrow 2 < 1.5 \space (FALSE)\)

Answer: II ONLY (B)

Shouldn't the question say a,b,c are integers?

We can let smart numbers
a=2
so,b=3
c=4
Now, we'll apply on each options given☺
And we get option B(||) only matches

Answer is B

Posted from my mobile device

I. 2a > b + c
Taking numbers a = 1, b = 2 and c = 3 this is not true

II. c – a > b - a
In 0 < a < b < c, subtracting a from each
0 - a < a - a < b - a < c - a
Hence c - a > b - a True

III. \(\frac{c}{a}\) < \(\frac{b}{a}\)
As b < c
dividing by 'a' which is positive doesn't impact inequality
Hence \(\frac{b}{a} < \frac{c}{a}\)
not true

Answer B.
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Answer: Option B

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Is that a rule that we can "subtract ANY VALUE (such as a) from both sides, the inequality remains valid"? Many thanks in advance