WAEC 2014 · Paper 2 · Q8

(a)
Copy and complete the table for multiplication $\otimes$ modulo 11 on the set $\{1, 5, 9, 10\}$ .

$\otimes$ 1 5 9 10

1 1 5 9 10

5 5

9 9

10 10

Use the table to: (i) evaluate $(9 \otimes 5) \otimes (10 \otimes 10)$ ; (ii) find the truth set of $10 \otimes m = 2$ ; (iii) find the truth set of $n \otimes n = 4$ .
Model answer

$\otimes$ 1 5 9 10

1 1 5 9 10

5 5 3 1 6

9 9 1 4 2

10 10 6 2 1

Multiply, then take the remainder on dividing by 11: for example $9 \otimes 10 = 90 = 8 \times 11 + 2$ , so the entry is 2. From the table, (i) $1 \otimes 1 = 1$ ; (ii) $10 \otimes 9 = 2$ , so $\{9\}$ ; (iii) $9 \otimes 9 = 4$ , so $\{9\}$ .
(b)
When a fraction is reduced to its lowest terms, it is equal to $\frac34$ . The numerator of the fraction when doubled would be 34 greater than the denominator. Find the fraction.
Show the answer
$\frac{51}{68}$

$\otimes$	1	5	9	10
1	1	5	9	10
5	5
9	9
10	10

$\otimes$	1	5	9	10
1	1	5	9	10
5	5	3	1	6
9	9	1	4	2
10	10	6	2	1

Worked solution (try it first)

$\otimes$	1	5	9	10
1	1	5	9	10
5	5	3	1	6
9	9	1	4	2
10	10	6	2	1

From the table, $9 \otimes 5 = 1$ and $10 \otimes 10 = 1$ , so $(9 \otimes 5) \otimes (10 \otimes 10) = 1 \otimes 1$
$= 1$ .

In the row of 10, the entry 2 is in the column of 9: the truth set of $10 \otimes m = 2$ is $\{9\}$ .

On the diagonal, $n \otimes n = 4$ only for $n = 9$ : the truth set is $\{9\}$ .

Let the fraction be $\frac{x}{y}$ .
It equals $\frac34$ , so $4x = 3y$ .
Doubling the numerator gives 34 more than the denominator: $2x = y + 34$ , so $y = 2x - 34$ .
Then $4x = 3(2x - 34) = 6x - 102$ , so $2x = 102$ , $x = 51$ and $y = 68$ .
The fraction is $\frac{51}{68}$ .