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Shortcut to mixed inequality and absolute DS question

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Source: — Data Sufficiency |

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by Jay@ManhattanReview » Tue Mar 14, 2017 5:02 am
Mo2men wrote:Is |3m−n|+|m−2n|>|4m−3n|?

(1) m>0
(2) 2n<m

OA: C
Hi Mo2men,

We have to see whether |3m−n|+|m−2n| > |4m−3n|?

S1: m > 0

We do not have any information about the value of n.

If m = 1 and n = 0,

|3m−n|+|m−2n| > |4m−3n| => 3 + 1 > 4. The answer is NO.

However, if m = n = 1,

|3m−n|+|m−2n| > |4m−3n| => |3−1|+|1−2| > |4−3| => 2 + 1 > 1. The answer is YES. No unique answer.

S2: 2n < m

=> If m = -2 and n= -2,

|3m−n|+|m−2n| > |4m−3n| => |-6+2|+|-2+4| > |-8+6| => 4 + 2 > 2 => 6 > 2. The answer is YES.

However, if m = 1, and n = 0,

|3m−n|+|m−2n| > |4m−3n| => 3 + 1 > 4 => 4 > 4. The answer is NO. No unique answer.

S1 and S2:

We have m > 0 and 2n < m

2n < m => m - 2n > 0

=> 3m−n > 0, 4m−3n > 0

This implies that |3m−n|+|m−2n| > |4m−3n| => 3m−n + m−2n > 4m−3n => 4m - 3n > 4m - 3n. The answer is NO. Unique answer. Sufficient.

The correct answer: C

Hope this helps!

Relevant book: Manhattan Review GMAT Data Sufficiency Guide

-Jay
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by Mo2men » Tue Mar 14, 2017 6:26 am
Jay@ManhattanReview wrote: S1 and S2:

We have m > 0 and 2n < m

2n < m => m - 2n > 0

=> 3m−n > 0, 4m−3n > 0

This implies that |3m−n|+|m−2n| > |4m−3n| => 3m−n + m−2n > 4m−3n => 4m - 3n > 4m - 3n. The answer is NO. Unique answer. Sufficient.
Hi Jay,

How can you deduce the inequality? can you elaborate please?

Thanks
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by GMATGuruNY » Tue Mar 14, 2017 6:49 am
Mo2men wrote:Is |3m−n|+|m−2n|>|4m−3n|
(1) m>0
(2) 2n<m
Let x = 3m-n and y = m-2n, with the result that x+y = 4m-3n.
Substituting x = 3m-n, y = m-2n, and x+y = 4m-3n into |3m−n|+|m−2n|>|4m−3n|, we get:
Is |x| + |y| > |x+y|?

Since an absolute value cannot be negative, both sides of the inequality in blue are NONNEGATIVE.
As a result, we can SQUARE THE INEQUALITY:
(|x| + |y|)² > (|x+y|)²
x² + y² + 2|x||y| > x² + y² + 2xy
2|x||y| > 2xy
|x||y| > xy.

The inequality in red will hold true if x and y have DIFFERENT SIGNS.
Since x = 3m-n and y = m-2n, question stem rephrased:
Do 3m-n and m-2n have different signs?

Statement 1: m>0
No way to determine the signs of 3m-n and m-2n.
INSUFFICIENT.

Statement 2: 2n<m
0 < m-2n
m-2n > 0.
No way to determine the sign of 3m-n.
INSUFFICIENT.

Statements combined:
Inequalities can be ADDED TOGETHER.
Adding together m>0 and m-2n > 0, we get:
m + (m-2n) > 0 + 0
2m - 2n > 0
m - n > 0.

Adding together m>0, m>0 and m-n>0, we get:
m + m + (m-n) > 0 + 0 + 0
3m-n > 0.

Since 3m-n > 0 and m-2n > 0, they have the SAME SIGN.
Thus, the answer to the question stem is NO.
SUFFICIENT.

The correct answer is C.
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by Jay@ManhattanReview » Wed Mar 15, 2017 12:41 am
Mo2men wrote:
Jay@ManhattanReview wrote: S1 and S2:

We have m > 0 and 2n < m

2n < m => m - 2n > 0

=> 3m−n > 0, 4m−3n > 0

This implies that |3m−n|+|m−2n| > |4m−3n| => 3m−n + m−2n > 4m−3n => 4m - 3n > 4m - 3n. The answer is NO. Unique answer. Sufficient.
Hi Jay,

How can you deduce the inequality? can you elaborate please?

Thanks
Hi Mo2men,

(3m−n) can be written as (6m − 2n)/2 = (5m + m − 2n)/2 = = [5m + (m − 2n)]/2 = [+ive number + +ive number]/2 = +ive number .

Similarly, (4m−3n) can be written as 3(8m/3−2n)/2 = (3/2)*[5m/3 + (m−2n)] = (+ive number)*(+ive number + +ive number) = +ive number

Hope this is clear.

Relevant book: Manhattan Review GMAT Math Essentials Guide

-Jay
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