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Mathematics
WAEC / WASSCE 2021
Quadratic Equations
Solution
Find The quadratic Equation Whose Roots Are -2q And 5q.
A
3x\(^2\) + 3qx - 10q\(^2\)
B
x\(^2\) + 3qx + 10q\(^2\)
C
x\(^2\) - 3qx + 10q\(^2\)
D
x\(^2\) - 3qx - 10q\(^2\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
x\(^2\) - (sum of roots)x + (products of roots) = 0
x\(^2\) - (-2q + 5q) + (-2q * 5q) = 0
x\(^2\) -(3q) + (-10q\(^2\)) = 0
x\(^2\) -3q - 10q\(^2\) = 0
Source: WAEC / JAMB / NECO Classroom Archive (Record #61232)
Mathematics
WAEC / WASSCE 2021
Quadratic Equations
Solution
(a) Copy and complete the table of values for the relation y=2x\(^2\) - x - 2 for 4 ≤ x ≤ 4.
x
-4
-3
-2
-1
0
1
2
3
4
y
19
-2
26
(b) Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of y = 2x\(^2\) - x - 2 for 4 ≤ x ≤ 4.
(c) On the same axes, draw the graph of y = 2x + 3.
(d) Use the graph to find the: (i) roots of the equation 2x-3r-5 0; (i) range of values of x for which 2x\(^2\) -x - 2<0.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
x
-4
-3
-2
-1
0
1
2
3
4
y
34
19
8
1
-2
-1
4
13
26
SEE THE GRAPH ABOVE
(c) Straight line graph of y = 2x + 3
(d) The roots of 2x\(^2\) -3x - 5 = 0 are x = -1 \pm 0.1 or 2.5 \pm 0.1
(ii) The range of values of x for which - x - 2 < 0 is (-0.7\pm 0.1) < x < (1.3 \pm 0.1)
Source: WAEC / JAMB / NECO Classroom Archive (Record #61504)
Mathematics
WAEC / WASSCE 2021
Progressions (AP & GP)
Solution
Find the 5th term of the sequence 2,5,10,17....?
A
22
B
24
C
36
D
26
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Simply add odd number starting from '3' to the next number
2
2 + 3 = 5
5 + 5 = 10
10 + 7 = 17
17 + 9 = 26
The fifth term = 26
Source: WAEC / JAMB / NECO Classroom Archive (Record #61141)
Mathematics
WAEC / WASSCE 2021
Mensuration
Diagram
Solution
In the diagram, \(\overline{PQ//RS}\) is a trapezium with QR//PS. U and T are points on \(\overline{PS}\) such that \(\overline{|PU|}\) = 5 cm,
\(\overline{|QU|}\) = 12 cm and ∠PUQ= ∠STR =90°. If the area of PQR = 20 cm\(^2\),
calculate, correct to the nearest whole number, the:
(a) perimeter; (b) area; of the trapezium
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) Area of △PQR = \(\frac{1}{2}\) * b * 12 = 20
b = \(\frac{2 x 20}{12}\) = 3.33cm
|QR| = |UT| = 3.33cm
IPQ| = \sqrt(12\(^2\) + 5\(^2\))\(^2\) = 13cm
IRT| = |QU| = 12cm
From △RTS
sin 50^\circ= \(\frac{12}{|RS|}\)
|RS| = \(\frac{12}{sin 50^\circ}\) = 15.6649 cm
|TS| = \(\frac{12}{tan 50^\circ}\) = 10.069cm
Perimeter = 13 + 3.33 + 15.6649 + 18.399
= 50.3939cm = 50cm (nearest whole number)
(b)Area= \(\frac{1}{2}\) (3.33+ 18.4004)12
6 (21.7304) = 130.3824cm\(^2\)
= 130 cm\(^2\) (nearest whole number)
Source: WAEC / JAMB / NECO Classroom Archive (Record #61507)
Mathematics
WAEC / WASSCE 2021
Lengths and Perimeters
Solution
The diagonal of a rhombus are 12cm and 5cm. calculate its perimeter
A
26cm
B
24cm
C
17cm
D
34cm
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Perimeter of a Rhombus = 4L or 2\sqrt( d1\(^2\) + d2\(^2\))
P = 2\sqrt(12\(^2\) + 5\(^2\))
P = 2\sqrt(144 + 25)
P = 2\sqrt169 = 2 * 13
P = 26cm
Source: WAEC / JAMB / NECO Classroom Archive (Record #61251)
Mathematics
WAEC / WASSCE 2021
Lengths and Perimeters
Solution
A cone has a base radius of 8cm and height 11cm. calculate , correct to 2d.p, the curved surface area
A
341.98cm\(^2\)
B
276.57cm\(^2\)
C
201.14cm\(^2\)
D
477.71cm\(^2\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Where l\(^2\) = h\(^2\) + r\(^2\)
l\(^2\) = 11\(^2\) + 8\(^2\)
l = \sqrt185
l = 13.60cm
The formula of CSA of Cone is \pirl
\(\frac{22}{7}\) * 8 * 13.60
= 341.979 or 341.98 (2d.p)
Source: WAEC / JAMB / NECO Classroom Archive (Record #61426)
Mathematics
WAEC / WASSCE 2021
Angles & Parallel Lines
Diagram
Solution
In the diagram, O is the centre of the circle PQRS, ∠PQR = 72° and OR is parallel to PS. Find ∠ OPS.
Attached Figure / Diagram
A
18°
B
108°
C
54°
D
36°
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Angle at the centre O is twice the angle at the circumference.
i.e O = 72^\circ * 2 = 144^\circ
∠OPS = 180 - 144 = 36^\circ
Source: WAEC / JAMB / NECO Classroom Archive (Record #61348)
Mathematics
WAEC / WASSCE 2021
Angles & Parallel Lines
Solution
The equation of a line is given as 3 x - 5y = 7. Find its gradient (slope)
A
\(\frac{5}{3}\).
B
\(\frac{3}{5}\).
C
\(\frac{-3}{5}\).
D
\(\frac{-5}{3}\).
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
the form y=mx+c
where m is the gradient and c is the y-intercept.
the equation to gives 5y=3x - 7.
divide both side by 5 the coefficient of y → \(\frac{3}{5}\)x - \(\frac{7}{5}\)
comparing this with the general equation y=mx+c,
you can see that m= the gradient= \(\frac{3}{5}\).
Source: WAEC / JAMB / NECO Classroom Archive (Record #61465)
Mathematics
WAEC / WASSCE 2021
Triangles and Polygons
Solution
The sum of the interior angles of a regular polygon with k sides is (3k-10) right angles. Find the size of the exterior angle?
A
60°
B
40°
C
90°
D
120°
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
In a polygon with n sides, the sum of the angles =
(n - 2)180, where n = the number of sides.
In our problem, n = k.
So, we have:
(3k - 10)90 = (k - 2)180
270k - 900 = 180k - 360
Simplifying:
90k = 540
k = 6. So, we have a regular hexagon.
Now, each exterior angle = 360/n
We have: 360/6 = 60
Conclusion: Each exterior angle is 60 degrees.
Source: WAEC / JAMB / NECO Classroom Archive (Record #61197)
Mathematics
WAEC / WASSCE 2021
Triangles and Polygons
In △LMN, |LM| = 6cm, ∠LNM = x and sin x = sin x = \(\frac{3}{5}\).
Find the area of △LMN
A
60cm\(^2\)
B
48cm\(^2\)
C
24cm\(^2\)
D
30cm\(^2\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: WAEC / JAMB / NECO Classroom Archive (Record #61208)
Mathematics
WAEC / WASSCE 2021
Triangles and Polygons
Which of the following is not an exterior angle of a regular polygon?
A
66°
B
72°
C
24°
D
15°
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: WAEC / JAMB / NECO Classroom Archive (Record #61308)
Mathematics
WAEC / WASSCE 2021
Coordinate Geometry
Solution
\(\overline{XY}\) is a line segments with the coordinates X (- 8,- 12) and Y(p,q). if the midpoint of \(\overline{XY}\) is (-4,-2) find the coordinates of Y.
A
(-6,-2)
B
(0,8)
C
(4,10)
D
(0,4)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
The formula for midpoint = \(\frac{x_1 + x_2}{2}\), \(\frac{y_1 + y_2}{2}\)
(-4,-2) = (x,y)
x = \(\frac{x_1 + x_2}{2}\)
-4 = \(\frac{-8 + p}{2}\)
-4 * 2 = -8 + p
-8 + 8 = p
: p = 0
y = \(\frac{y_1 + y_2}{2}\)
-2 = \(\frac{-12 + q}{2}\)
-2 * 2 = -12 + q
-4 + 12 = q
: q = 8
Source: WAEC / JAMB / NECO Classroom Archive (Record #61431)
Mathematics
WAEC / WASSCE 2021
Angles of Elevation & Depression
Solution
(a) A cottage is on a bearing of 200° and 110° from Dogbe's and Manu's farms respectively. If Dogbe walked 5 km and Manu 3 km from the cottage to their farms, find, correct to: (i) two significant figures, the distance between the two farms, (ii) the nearest degree, the bearing of Manu's farm from Dogbe's.
(b) A ladder 10 m long leaned against a vertical wall xm high. The distance between the wall and the foot of the ladder is 2 m longer than the height of the wall.
Calculate the value of x
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
diagram above(i)
(i) Distance between Manu and Dogbe's farm =\sqrt(3\(^2\)+5\(^2\))
= 5.83km or 5.8 km
(i) tanØ = 0.6
Ø= tan\(^{-1}\) (O.6)
= 30.96^\circ
Bearing = 30.96^\circ+200^\circ =230.96
= 231^\circ (nearest degree)
(b) x\(^2\) + (x+2)\(^2\) = 10\(^2\)
2x\(^2\) + 4x - 96 = 0
(x-6) (x+8) = 0
x = 6,x = -8
: x = 6cm
Source: WAEC / JAMB / NECO Classroom Archive (Record #61561)
Mathematics
WAEC / WASSCE 2021
Bearings and Distances
Solution
The circumference of a circular track is 9km. A cyclist rides round it a number of times and stops after covering a distance of 302km. How far is the cyclist from the starting point?
A
5km
B
6km
C
7km
D
3km
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Circumference of the circular track = 9km
Distance covered = 302km
Number of complete circles or revolutions from the starting point = 302/9 =33 circles and additional 5km.
So, the distance of the cyclist from the starting point would be 5km
Source: WAEC / JAMB / NECO Classroom Archive (Record #61155)
Mathematics
WAEC / WASSCE 2021
Bearings and Distances
From a point T, a man moves 12km due west and then moves 12km due south to another point Q. Calculate the bearing of T from Q.
A
225°
B
315°
C
045°
D
135°
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: WAEC / JAMB / NECO Classroom Archive (Record #61312)