(a) Copy and complete the following table for multiplication modulo 11.
| \(\otimes\) | 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 :(1) \(10 \otimes m = 2\); (2) \(n \otimes n = 4\)
(b) When a fraction is reduced to its lowest term, it is equal to \(\frac{3}{4}\). The numerator of the fraction when doubled would be 34 greater than the denominator. Find the fraction.
(a)
| \(\otimes\) | 1 | 5 | 9 | 10 |
| 1 | 1 | 5 | 9 | 10 |
| 5 | 5 | 3 | 1 | 6 |
| 9 | 9 | 1 | 4 | 2 |
| 10 | 10 | 6 | 2 | 1 |
(i) \((9 \otimes 5) \otimes (10 \otimes 10) = 1 \otimes 1 = 1\)
(ii) (1) \(10 \otimes m = 2\)
By comparison, \(10 \otimes 9 = 2\).
(2) \(n \otimes n = 4\)
From the table, \(9 \otimes 9 = 4\)
Hence, n = 9.
(b) Let the fraction = \(\frac{m}{n}\)
\(\frac{m}{n} = \frac{3}{4}..... (1)\)
\(\implies n = \frac{4m}{3} ....... (2)\)
\(2m = n + 34 ...... (3)\)
Put (2) in (3),
\(2m = \frac{4m}{3} + 34\)
\(2m - \frac{4m}{3} = 34 \implies \frac{2m}{3} = 34\)
\(m = \frac{34 \times 3}{2} = 51\)
\(n = \frac{4m}{3} = \frac{4 \times 51}{3} = 68\)
Hence, the fraction = \(\frac{51}{68}\).
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