(a) Copy and complete the following table for... | WAEC SSCE - General Mathematics - 2014 (Essay)

Question 1 Report

(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.

Answer

(a) Multiplication table modulo 11 (multiply, then take the remainder on division by 11).

For example \(5\otimes 5 = 25 = 2(11)+3 \equiv 3\), \(9\otimes 9 = 81 = 7(11)+4 \equiv 4\), \(10\otimes 10 = 100 = 9(11)+1 \equiv 1\).

\(\otimes\)	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) \((9\otimes 5)\otimes(10\otimes 10) = 1 \otimes 1 = \mathbf{1}\).

(ii)(1) \(10\otimes m = 2\): from the row for 10, \(10\otimes 9 = 2\), so the truth set is \(\{9\}\).

(ii)(2) \(n\otimes n = 4\): scanning the leading diagonal, only \(9\otimes 9 = 4\), so the truth set is \(\{9\}\).

(b) The fraction. In lowest terms it equals \(\tfrac{3}{4}\), so let it be \(\dfrac{3k}{4k}\).

Numerator doubled is 34 more than the denominator:

\[2(3k) = 4k + 34 \;\Rightarrow\; 6k - 4k = 34 \;\Rightarrow\; 2k = 34 \;\Rightarrow\; k = 17\] \[\text{Fraction} = \frac{3(17)}{4(17)} = \frac{51}{68}\]

Check: \(2(51) = 102 = 68 + 34\). The fraction is \(\mathbf{\tfrac{51}{68}}\).

Read lesson note on Modular Arithmetic (WAEC)

Answer Details