Mathematical Reasoning (Paper 2) Difficulty 4

TMUA practice: Necessary Sufficient, problem 1

A TMUA-calibre necessary sufficient problem at difficulty 4 of 5, with the full worked solution.

The question

Let

x, y, z, w

be positive real numbers satisfying: (i) if

x > y

then

z > w

; (ii) if

x > z

then

y < w

. Which of the following is a valid conclusion?

\text{if } x<y \text{ then } z<w

\text{if } x<z \text{ then } y>w

\text{if } x>y+z \text{ then } z<y

\text{if } x>y+z \text{ then } z>y

The correct answer is highlighted. D

Worked solution

Suppose $x > y + z$ . Since $y$ and $z$ are positive, $y + z > y$ and $y + z > z$ , so

$x > y \qquad \text{and} \qquad x > z.$

From $x > y$ , premise (i) gives $z > w$ .
From $x > z$ , premise (ii) gives $y < w$ .

Chaining these: $y < w < z$ , hence $z > y$ .

So $x > y + z \implies z > y$ — exactly statement D.

(The other options fail: (i) and (ii) are one-directional implications, so the converse-style claims in A and B do not follow, and C is the negation of what we just proved.)

Answer: D.