$$If\ \ \ w>x>y>z>0\ ,\ is\ z<4?$$ $$(1)\ \ \ \frac{1}{w}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1.$$ $$(2)\ \ \frac{1}{w}>\frac{1}{4}.$$ [spoiler]OA=D[/spoiler].
I'd appreciate an explanation here. I couldn't solve this DS question. <i class="em em-frowning"></i>
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If w > x > y > z > 0, is z < 4?
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Statement 1
Let's first recognize that if w > x > y > z, then 1/z > 1/y > 1/x > 1/w. For instance, if w = 4, x = 3, y = 2, and x = 1, then 4 > 3 > 2 > 1 and 1/1 > 1/2 > 1/3 > 1/4.
If w = x = y = z, then all four variables would equal 4, to give 1/4 + 1/4 + 1/4 + 1/4 = 1. However, we know that all four of these fractions must have different values, meaning that some must be less than 1/4 and some must be greater than 1/4 in order for them to add up to 1. Since 1/z must be the largest of these fractions, we know that 1/z must be greater than 1/4, making z less than 4. Sufficient.
Statement 2
If 1/w > 1/4, then w < 4 (e.g. if w = 2, then 1/2 > 1/4). If w < 4, and z < w, then z must also be less than 4. Sufficient.
Let's first recognize that if w > x > y > z, then 1/z > 1/y > 1/x > 1/w. For instance, if w = 4, x = 3, y = 2, and x = 1, then 4 > 3 > 2 > 1 and 1/1 > 1/2 > 1/3 > 1/4.
If w = x = y = z, then all four variables would equal 4, to give 1/4 + 1/4 + 1/4 + 1/4 = 1. However, we know that all four of these fractions must have different values, meaning that some must be less than 1/4 and some must be greater than 1/4 in order for them to add up to 1. Since 1/z must be the largest of these fractions, we know that 1/z must be greater than 1/4, making z less than 4. Sufficient.
Statement 2
If 1/w > 1/4, then w < 4 (e.g. if w = 2, then 1/2 > 1/4). If w < 4, and z < w, then z must also be less than 4. Sufficient.
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