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Re: If b < e < y+k and b-1 < w < k,
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23 Mar 2017, 05:36
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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
If b < e < y+k and b-1 < w < k,
[#permalink]
24 Mar 2017, 03:24
1
Kudos
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.
Re: If b < e < y+k and b-1 < w < k,
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18 Oct 2020, 22:08
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