Brent@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
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
Cheers,
Brent
