WAEC Mathematics Past Questions & Answers - Page 9

Reflex ∠MOR = 2 × 124° = 248° (angle at the centre is twice the angle at the circumference)
∠MOR = 360° - 248° = 112° (sum of angle at a point is 360°)
∠OMN = 360° - (124°+ 48° + 112°) (sum of angles in a quadrilateral is 360°)
= 360° - 284°
∴ ∠OMN = 76°

Let the two numbers be x and y 

$\frac{1}{3}( x + y) = 12$

then x + y = 36 ..........i

2( x - y) = 12

x - y = 6 ............ii

add equations i and ii 

2x = 42

x = 21, put x = 21 into equation i

x + y = 36

21 + y = 36 

y = 36 - 21 = 15

therefore the numbers are 21 and 15

From the diagram above; Tan$\theta = \frac{opp}{adj}$

tan50° = $\frac{124}{d}$

d = $\frac{124}{tan50}$

therefore, d = 104.05m

Let Mr Manu be x years old and Adu be y years old.

But Mr Manu is four times as old as Adu then, x = 4y.

7 years ago, the sum of their ages was 76.

( x - 7) + ( y - 7) = 76

x + y - 14 = 76 

x + y = 76 + 14 = 90

But x = 4y 

Therefore, 4y + y = 90

5y = 90 

y = $\frac{90}{5}$

y = 18

Therefore Adu is 18 years old.

Using the venn diagram above

μ = 30
n(W) = 15
n(M) = 13
$n(W ∪ M)^1 = 6$
Let x = number of students that study both woodwork and metalwork
i.e. n(W ∩ M) = x
Number of students that study only woodwork,$n(W ∩ M^1)$ = $15 - x$
Number of students that study only metalwork, $n(W^1 ∩ M)$ = $13 - x$
Bringing all together,
$n(W ∩ M^1)$ +$n(W^1 ∩ M)$ + $n(W ∩ M)$ + $n(W ∪ M)^1$ = $μ$
∴ (15 - x) + (13 - x) + x + 6 = 30
⇒ 34 - x = 30
⇒ 34 - 30 = x
∴ x = 4
$n(W ∩ M^1)$ = $15 - 4 = 11$
∴ The number of students that study woodwork but not metalwork is 11.