If 0

Question

If 0 < a b + c
II. c – a > b - a
III. \(\frac{c}{a}\) < \(\frac{b}{a}\)

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

OG 2017 New Question

BrentGMATPrepNow · Accepted Answer

I. 2a > b + c Consider this scenario: a = 1, b = 2 and c = 3. This meets the given condition that 0 < a 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

sauravpaul · Answer

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.

avigutman · Answer

Video solution from Quant Reasoning: Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=u-DHNdqYkgw

KillDgmat · Answer

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.

HanoiGMATtutor · Answer

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.

JeffTargetTestPrep · Answer

We are given that 0 < a 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 - 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

meyba2ty · Answer

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

audentes · Answer

Assign Test Values Must follow rule: 0 < a 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)

divyajeswani · Answer

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

shaonkarim · Answer

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

unraveled · Answer

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 - 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.

GMATinsight · Answer

Answer: Option B

Video solution by  GMATinsight

https://www.youtube.com/watch?v=lZLbeFzzWqU

woohoo921 · Answer

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

bumpbot · Answer

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

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