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Question 99 of 104 in Unit 6

Unit 6

4 marks · Normal

5. Let y be inversely proportional to x, and y=8 when x=3.

(a)

Write down the formula connecting y and x.

[1 mark]
(b)

Find the value of y when x = 12.

[2 marks]
(c)

Set A = { x | y ≥ 1, x > 0 }. Using set notation, describe A.

[1 mark]

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(a) y = 24/x

(b) y = 2

(c) A = { x | 0 < x ≤ 24 }

Common mistakes: Incorrect proportional constant; Wrong set notation

Show mark scheme and worked solution

Marking points

M1 (a) · 1 mark · States formula y=k/x using the given values.

M2 (b) · 1 mark · Evaluates with correct k. (follow-through)

A1 (b) · 1 mark · Finds correct value (2). (follow-through)

B1 (c) · 1 mark · Correctly uses set notation to describe all x with y≥1, x>0.

Worked solution

a) y ∝ 1/x ⇒ y = k/x. When y=8, x=3, so 8= k/3 ⇒ k=24. So y=24/x.

b) y = 24/12 = 2.

c) For y ≥ 1: 24/x ≥ 1 ⇒ x ≤ 24. Also x > 0. So A = { x | 0 < x ≤ 24 }.

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