WAEC 2015 · Paper 2 · Q1

(a)
Without using mathematical tables or a calculator, simplify $3\frac49 \div \left(5\frac13 - 2\frac34\right) + 5\frac{9}{10}$ .
(b)
A number is selected at random from each of the sets $\{2, 3, 4\}$ and $\{1, 3, 5\}$ . Find the probability that the sum of the two numbers is greater than 3 and less than 7.

Worked solution (try it first)

(a)

Brackets first.
Change to improper fractions: $5\frac13 - 2\frac34 = \frac{16}{3} - \frac{11}{4}$
$= \frac{64 - 33}{12}$
$= \frac{31}{12}$ .
Then divide: $3\frac49 \div \frac{31}{12} = \frac{31}{9} \times \frac{12}{31}$
$= \frac{12}{9}$
$= \frac43$ .
Finally add: $\frac43 + 5\frac{9}{10} = \frac{40}{30} + \frac{177}{30}$
$= \frac{217}{30}$
$= 7\frac{7}{30}$ .

(b)

List the sample space: each of 2, 3, 4 with each of 1, 3, 5 gives $3 \times 3 = 9$ equally likely pairs.
Their sums:
1 3 5

2 3 5 7

3 4 6 8

4 5 7 9
"Greater than 3 and less than 7" means a sum of 4, 5 or 6.
There are 4 such pairs: $(2, 3)$ , $(3, 1)$ , $(3, 3)$ and $(4, 1)$ .
So the probability is $\frac49$ .

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