Unit 7

5 marks · Normal

A maths club has 36 members. Every member is given a membership number which is a recurring decimal less than 1. For a club quiz, these numbers are put into two lists:

List A: Members whose membership numbers are 0.3\dot{7} or 0.2\dot{5}.

List B: Members whose membership numbers are 0.2\dot{5} or 0.6\dot{2}.

Let S = {members of the club}, A = {members with numbers 0.3\dot{7} or 0.2\dot{5}}, B = {members with numbers 0.2\dot{5} or 0.6\dot{2}}.

(a)

Write each recurring decimal above as a fraction in its simplest form.

[2 marks]

(b)

In the club, 6 members are in both lists A and B. 11 members are only in list A. 14 members are only in list B. The rest are in neither list.

A member is chosen at random.

Calculate the probability that the member is in exactly one of the lists. Give your answer as a fraction in its simplest form.

[3 marks]

Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.

Show answer

(a) · must show Fraction form for each decimal

(b) · must show Add members only in A to those only in B, divide by 36, simplest fraction

Common mistakes: Not simplifying the fractions; Incorrectly counting members in one set only

Show mark scheme and worked solution

Marking points

M1 (a) · 1 mark · For correctly setting up or starting to convert one recurring decimal to a correct fraction expression (e.g., forming equations for x = 0.3\dot{7} etc.).

A1 (a) · 1 mark · For all three recurring decimals written as correct, fully simplified fractions.

M2 (b) · 1 mark · For identifying the total members in exactly one list (11 + 14 = 25).

M3 (b) · 1 mark · For dividing the number of members in exactly one list by the total to get a probability (). (follow-through)

A2 (b) · 1 mark · For writing the final answer as a simplified fraction.

Worked solution

a) Let . Then:

Multiply both sides by 100 (since 2 digits repeat):

Subtract x from 100x:

Let :

Multiply both sides by 10 (one digit repeats):

Subtract y from 10y:

Let :

10z = 6.222...; 10z - z = 9z = 5.6; z = 5.6/9 = 56/90 = 28/45 (simplified).

So:

b) Members in exactly one list are those only in A or only in B. That is 11 (only in A) + 14 (only in B) = 25.

Total number of members: 36