If b

Question

If b < e < y+k and b-1 < w < k, then which of the following MUST be true?

I. 2b-1 < w+e < y+2k
II. 1 < e-w < y
III. b < k+2

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

*kudos for all correct solutions

BrentGMATPrepNow · Accepted Answer

I designed this question to highlight two important rules about inequalities: 
#1: If the inequality symbols of two inequalities are facing in the same direction, we can ADD those inequalities. For example, if b < c and x < y, then we can say that (b + x) < (c + y)
#2: If the inequality symbols of two inequalities are facing in the same direction, we CANNOT SUBTRACT those inequalities since the resulting inequality may or may not be true. For example, if b < c and x < y, then we CANNOT conclude that (b - x) < (c - y)

I. 2b-1 < w+e < y+2k  
Given: 
b < e < y+k 
b-1 < w < k
Since the inequality symbols of two inequalities are facing in the same direction, we can ADD the inequalities.
We get: 2b-1 < w+e < y+2k
So, statement I is TRUE

II. 1 < e-w < y  
We get this inequality from SUBTRACTING b-1 < w < k from b < e < y+k
As I mentioned in rule #2 above, we cannot do this. 
If we really want to demonstrate that statement II is not necessarily true, consider the following counter-example:
b = -10, e = -5, y = 0, k = 4 and w = 0
These values satisfy both of the given inequalities (b < e < y+k and b-1 < w < k), however, when we plug these values into statement II (1 < e-w < y), we get:  1 < -5 < 0 , which is not true. 
So, statement II need NOT be true.

III. b < k+2  
Since we already know that b-1 < w < k, we can also conclude that b-1 < k
Add 1 to both sides to get: b < k + 1
Since k+1 < k+2 for ALL values of k, we can write: b < k + 1 < k + 2
This means b < k+2
So, statement III is TRUE

Answer:  D

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broall · Answer

$\begin{split}
b &< e &< y+k \
b-1 &< w &< k \
\implies 2b - 1 &< e + w &< y + 2k \
\end{split}$

Hence (I) is true.

$\begin{split}
b &< e &< y+k \
-k &< -w &< 1 -b \
\implies b - k &< e - w &< y + k + 1 -b \
\end{split}$

Hence (II) is not true

$b - 1 < w < k \implies b-1 < k \implies b < k -1 < k +2$.
Hence (III) is true.

The answer is D

Mo2men · Answer

If b < e < y+k and b-1 < w < k, then which of the following MUST be true?

I. 2b-1 < w+e < y+2k
II. 1 < e-w < y
III. b < k+2

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

To save time, I will look to most reoccurring numeral. It might help to eliminate choices. 
 
Start with to check I

As the sign are in SAME DIRECTION, we can add inequalities

b < e < y+k 
 b-1 < w < k
------------------
2b-1<e+w<y+2k................Done

Eliminate choice E

Then check III

From the inequality: 
b-1 < w < k.......then

b-1 < k........... b<k+1<k+2..........Done

Eliminate Choices A , B & C............We don not need to waste time to check II.

We only left with one choice

Answer D

Takeaway  : Always check the answer choices before rushing into the question. Checking the frequency of numeral in each choice could help us save time. If, for example, answer E were I, II &III, then I would check II. But it does not the case in the question above.

bumpbot · Answer

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If b < e < y+k and b-1 < w < k,

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