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If b < e < y+k and b-1 < w < k,  [#permalink]

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Difficulty:   75% (hard)

Question Stats: 46% (02:13) correct 54% (02:03) wrong based on 95 sessions

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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

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If b < e < y+k and b-1 < w < k,  [#permalink]

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GMATPrepNow wrote:
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

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

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
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Re: If b < e < y+k and b-1 < w < k,  [#permalink]

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GMATPrepNow wrote:
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

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

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If b < e < y+k and b-1 < w < k,  [#permalink]

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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.

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

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.
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Re: If b < e < y+k and b-1 < w < k,  [#permalink]

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_________________ Re: If b < e < y+k and b-1 < w < k,   [#permalink] 22 Aug 2019, 05:09
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