Question Library
Browse, filter, and study verified past examination questions with step-by-step mathematical reasoning.
Searching 18,000+ Questions...
Sifting topics, KaTeX formulas & verified solutions
Showing 2716–2730 of 4910 verified questions
Sort by:
Mathematics
UTME / JAMB 2003
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram above, PQ is parallel to RS. What is the value of α + β + γ?
Attached Figure / Diagram
A
360o
B
200o
C
180o
D
90o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
γ + S = 180(Allied angles)
S = 180 - γ
α + P = 180(Allied angles)
p = 180 - α
β = P + S
β = 180 - α + 180 - γ
β = 360 - α - γ
β + α + γ = 360
Source: Myschool Past Questions Archive (Record #11106)
Mathematics
UTME / JAMB 2003
Inequalities & Linear Programming
Diagram
Solution
XYZ is a circle with center O and a radius 7cm. Find the area of the shaded region.
Attached Figure / Diagram
A
84cm2
B
77cm2
C
38cm2
D
14cm2
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\theta = 45 x 2 (∠ at center twice ∠ at \theta)
\theta = 90^\circ
Area of sector = \(\frac{\theta}{360}\) x \pir\(^2\)
= \(\frac{90}{360}\) x \(\frac{22}{7}\) x 7 x 7
= \(\frac{77}{2}\)cm\(^2\)
Area of triangle = \(\frac{1}{2}\) ab sin \theta
= \(\frac{1}{2}\) x 7 x 7 sin 90
= \(\frac{1}{2}\) x 7 x 7 x 1
= \(\frac{49}{2}\) cm\(^{2}\)
Area of shaded region = \(\frac{77}{2}\) - \(\frac{49}{2}\)
= \(\frac{28}{2}\)
= 14cm\(^2\)
Source: Myschool Past Questions Archive (Record #11107)
Mathematics
UTME / JAMB 2003
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram above, PQR is a straight line and PS is a tangent to the circle QRS with /PS/ = /SR/ and ∠SPR = 40^\circ. Find ∠PSQ
Attached Figure / Diagram
A
40o
B
30o
C
20o
D
10o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
R = P (base ∠s of isc \Delta, ps = RS)
∴R = 40^\circ
∠PSQ = R (∠s in alternate segment)
∴∠PSQ = 40^\circ
Source: Myschool Past Questions Archive (Record #11108)
Mathematics
UTME / JAMB 2003
Coordinate Geometry
Solution
A triangle has vertices P(-1,6), Q(-3, -4) and R(1, -4).Find the midpoint of PQ and QR respectively.
A
(0,-2) and (-1, -4)
B
(-1,0) and (-1,-1)
C
(-2,1) and (0,1)
D
(-2,1) and (-1,-4)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Midpoint of PQ \(\frac{x_2 + x_1}{2};\frac{y_2 + y_1}{2}\)
=\(\frac{-1+(-3)}{2};\frac{6-4}{2}\)
=\(\frac{-4}{2};\frac{2}{2}\)
= -2, 1
Midpoint of QR = \(\frac{x_2 + x_1}{2};\frac{y_2 + y_1}{2}\)
\(\frac{-3+1}{2};\frac{-4-4}{2}\)
\(\frac{-2}{2};\frac{-8}{2}\)
-1, -4
Midpoint of PQ and QR = (-2,1) and (-1,-4)
Source: Myschool Past Questions Archive (Record #11109)
Mathematics
UTME / JAMB 2003
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
The mean of the numbers 3, 6, 4, x, and 7 is 5. Find the standard deviation.
A
\(\sqrt{2}\)
B
\(\sqrt{3}\)
C
2
D
3
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Standard deviation = \(\frac{\sum(x - \overline{x})^2}{\text{n}}\)
\(\overline{x}\) = \(\frac{ 3 + 6 + 4 + x + 7}{5}\) = 5
= 20 + x = 25
x = 25 - 20 = 5.
x
x - \(\overline{x}\)
(x - \(\overline{x}\))\(^2\)
3
3 - 5
-2 = 4
4
4 - 5
-1 = 1
5
5 - 5
0 = 0
6
6 - 5
1 = 1
7
7 - 5
2 = 4
\(\sum\) (x - \(\overline{x}\))\(^2\) = 10
Standard deviation = \(\frac{\sum(x - \overline{x})^2}{\text{n}}\) = \(\frac{\sqrt{10}}{5}\) = \(\sqrt{2}\)
Source: Myschool Past Questions Archive (Record #11110)
Mathematics
WAEC / WASSCE 2003
Plane Geometry, Polygons & Circle Theorems
Solution
Simplify \(\left(\frac{3}{4} - \frac{1}{3}\right)\times 4\frac{1}{3}\div 3\frac{1}{4}\)
A
\(\frac{5}{9}\)
B
\(\frac{5}{16}\)
C
\(\frac{8}{13}\)
D
\(\frac{16}{5}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\((\frac{3}{4} - \frac{1}{3}) \times 4\frac{1}{3} \div 3\frac{1}{4}\)
= \((\frac{9 - 4}{12}) \times \frac{13}{3} \div \frac{13}{4}\)
= \(\frac{5}{12} \times \frac{13}{3} \times \frac{4}{13}\)
= \(\frac{5}{9}\)
Source: Myschool Past Questions Archive (Record #26570)
Mathematics
WAEC / WASSCE 2003
Statistics (Data Representation, Central Tendency & Dispersion)
Diagram
The table below gives the distribution of marks obtained by a number of pupils in a class test.
Find the median of the distribution
Attached Figure / Diagram
A
1
B
2
C
3
D
4
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: Myschool Past Questions Archive (Record #26572)
Mathematics
WAEC / WASSCE 2003
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
The table below gives the distribution of marks obtained by a number of pupils in a class test.
How many pupils scored at least 2 marks?
Attached Figure / Diagram
A
12
B
23
C
37
D
49
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
= 12 + 18 + 11 + 8
= 49 students
Source: Myschool Past Questions Archive (Record #26573)
Mathematics
WAEC / WASSCE 2003
Inequalities & Linear Programming
Diagram
Interprete the inequality represented on the number line
Attached Figure / Diagram
A
-4< X ≤ 5
B
-4 ≤ x< 5
C
-4
D
-4≤x≤5
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: Myschool Past Questions Archive (Record #26574)
Mathematics
WAEC / WASSCE 2003
Fractions, Decimals, Percentages & Approximations
Solution
A student bought 3 notebooks and 1 pen for N35. After misplacing these items, she again bought 2 notebooks and 2 pens, all of the same type for
N30. What is the cost of a pen?
A
N5.00
B
N7.50
C
N10.00
D
N15.00
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Represent notebook with x
Represent pen with y
∴ 3x+y=35 ---- (i)
2x+2y=30 ---- (ii)
Multiply eqn (i) by 2 and eqn (ii) by 1
6x+2y=70 ---- (iii)
2x+2y=30 ---- (iv)
Subtract eqn (iv) from eqn (iii)
4x = 40
X = 10
Substitute x = 10 in eqn (i)
3x+y=35
3(10)+y=35
Y = 5
∴ The cost of pen is N5.00
Source: Myschool Past Questions Archive (Record #26575)
Mathematics
WAEC / WASSCE 2003
Indices, Logarithms & Surds
Solution
Simplify \(\frac{4\sqrt{18}}{\sqrt{8}}\)
A
2
B
3
C
6
D
12
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{4\sqrt{18}}{\sqrt{8}}\\
= \frac{4\sqrt{9\times 2}}{\sqrt{4 \times 2}} = \frac{12\sqrt{2}}{2\sqrt{2}}=\frac{6\sqrt{2}}{\sqrt{2}}\\
Rationalising \frac{6\sqrt{2}}{\sqrt{2}}\\
=\frac{6\sqrt{2}}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = 6\)
Source: Myschool Past Questions Archive (Record #26577)
Mathematics
WAEC / WASSCE 2003
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, PQ/RT /RS/ =/QS/, ∠QST= 70o and ∠PQW = 55o
Find the size of reflex ∠WQS
Attached Figure / Diagram
A
235o
B
290o
C
305o
D
325o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Reflex ∠WQS = 360o - (55o + 70o)
=360o - 125o = 235o
Source: Myschool Past Questions Archive (Record #26578)
Mathematics
WAEC / WASSCE 2003
Fractions, Decimals, Percentages & Approximations
Solution
Three men, Bedu, Bakre and Kofi shared' N500 in the ratio 3:2: x respectively. If Bedu’s share is N150, find the value of x.
A
1
B
4
C
5
D
6
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(5+x\rightarrow 500\)
3 \(\rightarrow 150\)
∴ 3 \(\times 500 = 150 \times\) (5+x)
1500 = 750 + 150x
x = \(\frac{750}{150}\) = 5
Source: Myschool Past Questions Archive (Record #26579)
Mathematics
WAEC / WASSCE 2003
Algebraic Expressions, Polynomials & Factorisation
Solution
Simplify \(\left(1\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2\)
A
\(2\frac{1}{3}\)
B
\(1\frac{1}{3}\)
C
1
D
\(\frac{3}{7}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\left(1\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2=\left(\frac{5}{3}\right)^2 - \left(\frac{2}{3}\right)^2\)
Difference of two squares
\(\left(\frac{5}{3}-\frac{2}{3}\right)\left(\frac{5}{3}+\frac{2}{3}\right)=\left(\frac{3}{3}\right)\left(\frac{7}{3}\right)\\
\frac{7}{3}=2\frac{1}{3}\)
Source: Myschool Past Questions Archive (Record #26580)
Mathematics
WAEC / WASSCE 2003
Plane Geometry, Polygons & Circle Theorems
Solution
Find the value of x which satisfies the equation
5(x-7)=7-2x
A
x =2
B
x=4
C
x=6
D
x = 14
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
5(x - 7) = 7 - 2x
5x - 35 = 7 - 2x
5x + 2x = 7 + 35
7x = 42
x = \(\frac{42}{7}\)
= 6
Source: Myschool Past Questions Archive (Record #26581)