Question Library — KalmiPrep by Kalmitech

Mathematics WAEC / WASSCE 2002 Algebraic Expressions, Polynomials & Factorisation

Factorize m(2a-b)-2n(b-2a)

A

(2a-b)(2n-m)

B

(2a+b)(m-2n)

C

(2a-b)(m+2n)

D

(2a-b)(m-2n)

medium Difficulty

Mathematics WAEC / WASSCE 2002 Fractions, Decimals, Percentages & Approximations

If q oranges are sold for t Naira, how many oranges can be bought for p naira?

A

$\frac{p}{2}t$

B

$\frac{qt}{p}$

C

$\frac{q}{pt}$

D

$\frac{pq}{t}$

medium Difficulty

Mathematics WAEC / WASSCE 2002 Indices, Logarithms & Surds

In the diagram, QRT is a straight line. If angle PTR = 90^\circ, angle PRT = 60^\circ, angle PQR = 30^\circ and |PQ| = $6\sqrt{3}cm$, calculate |RT|

Attached Figure / Diagram

A

0.3cm

B

$\frac{\sqrt{3}}{2}cm$

C

3cm

D

$3\sqrt{3}cm$

medium Difficulty

Mathematics WAEC / WASSCE 2002 Plane Geometry, Polygons & Circle Theorems

In the diagram, calculate the value of x

Attached Figure / Diagram

A

35o

B

80o

C

100o

D

115o

medium Difficulty

Mathematics WAEC / WASSCE 2002 Plane Geometry, Polygons & Circle Theorems

Diagram

In the diagram, $\bar{PS}\hspace{1mm} and \hspace{1mm}\bar{QT}$ are two altitudes of ∆PQR. Which of the following is equal to ∠RQT?

Attached Figure / Diagram

A
∠PQT

B
∠SRP

C
∠PQR

D
∠SPR
Correct

medium Difficulty

Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
$No expanded working has been attached for this question yet. The verified answer is Option D.$
Source: Myschool Past Questions Archive (Record #26560)

Mathematics WAEC / WASSCE 2002 Indices, Logarithms & Surds

Diagram Solution

In the diagram, |QR| = 10cm, PR⊥QS, angle PSR = 30^\circ and angle PQR = 45^\circ. Calculate in meters |QS|

Attached Figure / Diagram

A
$10(1+\sqrt{3})$
Correct

B
$20\sqrt{3}$

C
$10\sqrt{3}$

D
$(10+\sqrt{3})$

medium Difficulty

Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
$In $\Delta$ PQR, $\tan 45 = \frac{PR}{10} \implies PR = 10 \tan 45$ = 10m In $\Delta$ PRS, $\tan 30 = \frac{10}{RS} \implies RS = \frac{10}{\tan 30}$ = $\frac{10}{\frac{1}{\sqrt{3}}$ = $10\sqrt{3}$ PS = $10 + 10\sqrt{3}$ = $10(1 + \sqrt{3}) cm$$
Source: Myschool Past Questions Archive (Record #26561)

Mathematics WAEC / WASSCE 2002 Fractions, Decimals, Percentages & Approximations

Solution

Calculate and correct to two significant figures, the percentage error in approximating 0.375 to 0.4

A
2.0

B
2.5

C
6.6

D
6.7
Correct

medium Difficulty

Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
$Measured value = 0.375 Approximation = 0.4 Error = 0.4 - 0.375 = 0.025 Error% = $\frac{0.025}{0.375}$ x 100% = 6.67% = 6.7%$
Source: Myschool Past Questions Archive (Record #26562)

Mathematics WAEC / WASSCE 2002 Plane Geometry, Polygons & Circle Theorems

Solution

To arrive on schedule, a train is to cover a distance of 60km at 72km/hr. If it starts 10 minutes late, at what speed must it move to arrive on schedule?

A
60km/hr

B
80km/hr

C
90km/hr
Correct

D
108km/hr

medium Difficulty

Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
$speed = \frac{distance}{time}\\ 72 = \frac{60}{time}\\ t = \frac{60}{72} = \frac{5}{6}hr\) time lost = 10mis $= \frac{10}{60}hr = \frac{1}{6}$ Time required for the journey $=\frac{5}{6}-\frac{1}{6} = \frac{2}{3}\\ speed \hspace{1mm}=60 \div \frac{2}{3} = 90km/hr$
Source: Myschool Past Questions Archive (Record #26563)

Mathematics WAEC / WASSCE 2002 Plane Geometry, Polygons & Circle Theorems

Diagram Solution

In the diagram, \(QR||TP and W\hat{P}T = 88^{\circ} $. Find the value of x

Attached Figure / Diagram

A
92o

B
68o

C
67o
Correct

D
23o

medium Difficulty

Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
$Sum of the angles in a triangle = 180^\circ 3x - 180^\circ + 92^\circ + x = 180^\circ 4x - 88^\circ = 180^\circ 4x = 268^\circ x = 67^\circ$
Source: Myschool Past Questions Archive (Record #26564)

Mathematics WAEC / WASSCE 2002 Plane Geometry, Polygons & Circle Theorems

Diagram Solution

In the diagram O is the center of the circle, ∠SOR = 64^\circ and ∠PSO = 36^\circ. Calculate ∠PQR

Attached Figure / Diagram

A
100o

B
96o

C
94o

D
86o
Correct

medium Difficulty

Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
$< OSR = < ORS = $\frac{180^\circ - 64^\circ}{2}$ = 58^\circ < PSR = 36^\circ + 58^\circ = 94^\circ < PSR + < PQR = 180^\circ 94^\circ + < PQR = 180^\circ $\implies$ < PQR = 180^\circ - 94^\circ = 86^\circ$
Source: Myschool Past Questions Archive (Record #26565)

Mathematics WAEC / WASSCE 2002 Plane Geometry, Polygons & Circle Theorems

Solution

Given that $p = x-\frac{1}{x} and\hspace{1mm}q = x^2 + \frac{1}{x^2}$ express q in terms of p.

A
(p2 + 2)
Correct

B
(p - 2) 2

C
(p + 2) 2

D
(p2 - 2)

medium Difficulty

Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
$Given $p = x - \frac{1}{x}$; $q = x^2 + \frac{1}{x^2}$. $p^2 = (x - \frac{1}{x})(x - \frac{1}{x})$ $p^2 = x^2 + \frac{1}{x^2} - 2$ $p^2 = q - 2 \implies q = p^2 + 2$$
Source: Myschool Past Questions Archive (Record #26566)

Mathematics WAEC / WASSCE 2002 Permutations, Combinations & Probability

Solution

The number of goals scored by a school team in 10 netball matches are as follows: 3, 5, 7, 7, 8, 8, 8, 11, 11, 12. Find the probability that in a match, the school team will score at most 8 goals.

A
$\frac{7}{10}$
Correct

B
$\frac{2}{5}$

C
$\frac{3}{5}$

D
$\frac{1}{5}$

medium Difficulty

Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
$Number of at most 8 goals = 7 P(at most 8 goals) = $\frac{7}{10}$$
Source: Myschool Past Questions Archive (Record #26567)

Mathematics WAEC / WASSCE 2002 Indices, Logarithms & Surds

Which of the following is not a rational number?

A
-5

B
$\sqrt{4}$

C
$3\frac{3}{4}$

D
$\sqrt{90}$
Correct

medium Difficulty

Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
$No expanded working has been attached for this question yet. The verified answer is Option D.$
Source: Myschool Past Questions Archive (Record #26568)

Mathematics WAEC / WASSCE 2002 Plane Geometry, Polygons & Circle Theorems

Diagram Solution

In the diagram, LMT is a straight line. lf O is the centre of circle LMN, OMN = 20^\circ, LTN = 32^\circ and |NM| = |MT|, find LNM.

Attached Figure / Diagram

A
44o

B
46o
Correct

C
52o

D
70o

medium Difficulty

Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
$< MNT = < MTN = 32^\circ < NMT = 180^\circ - 2(32^\circ) = 116^\circ < OMN + < NMT + < LMO = 180^\circ 20^\circ + 116^\circ + < LMO = 180^\circ $\implies$ < LMO = 180^\circ - 136^\circ = 44^\circ < LMN = < LMO + < OMN = 44^\circ + 20^\circ = 64^\circ < NOM = 180^\circ - 2(20^\circ) = 140^\circ < NLM = $\frac{1}{2} \times < NOM = 70^\circ$ < LNM + < LMN + < NLM = 180^\circ < LNM + 64^\circ + 70^\circ = 180^\circ < LNM = 180^\circ - 134^\circ = 46^\circ$
Source: Myschool Past Questions Archive (Record #26569)

Mathematics NECO SSCE 2002 Sets, Logic & Venn Diagrams

Solution

Given that the universal set U={14,15,16,17,18,19,20}, J ={16,19,20} and K= {14,15,18}, find (J' ∩ K')1.

A
{17}

B
{16, 17, 19, 20}

C
{1, 4, 15, 17, 18}

D
{14, 15, 16, 18, 19, 20}
Correct

E
{14, 15, 16, 17, 18, 19, 20}

medium Difficulty

Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
$U = {14,15,16,17,18,19,20} J' = {14,15,17,18} K' = {16,17,19,20} (J' $\cap$ K') = {17} (J' $\cap$ K')' = {14,15,16,18.19,20}$
Source: Myschool Past Questions Archive (Record #41976)