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Mathematics
UTME / JAMB 2005
Geometry and Trigonometry
Solution
Find the value of t if the standard deviation of 2t, 3t, 4t, 5t, and 6t is \sqrt2
A
1
B
2
C
3
D
4
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(mean(x) = \frac{20t}{5}\)
= S.D = \(\sqrt{\frac{((\overline{x} - x)^2}{\text{n}}}\)
\(\sqrt{2}\) = \(\sqrt{\frac{10t^2}{5}}\)
2 = \(\frac{10t^2}{5}\)
10t\(^2\) = 10
t\(^2\) = 1
t = \(\sqrt{1}\) = 1
Source: Myschool Past Questions Archive (Record #8611)
Mathematics
WAEC / WASSCE 2005
Geometry and Trigonometry
Solution
Ladi sold a car for N84,000 at a loss of 4%. How much did ladi buy the car?
A
80,500
B
80,640
C
87,360
D
87,500
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
SP = 84,000
Loss% = 4%
100% = 4% = 96% = \(\frac{96}{100}\)
Reverse \(\frac{96}{100}\) and use it to multiply SP to get the cost price.
CP = 84,000 x \(\frac{100}{96}\) = N 87500.
Source: WAEC / JAMB / NECO Classroom Archive (Record #26713)
Mathematics
WAEC / WASSCE 2005
Geometry and Trigonometry
Solution
A man bought 220 mangoes at N5x. He sold each for 3x kobo and made a gain of N8. Find the value of x
A
2
B
5
C
6
D
10
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
The cost price of the whole mangoes = N5x
The sold amount of the mangoes = 3x * 220 = N6.60x ( 660x converted to naira by dividing by hundred.)
The gain made on mangoes = N6.60x - N5x = N8.00 => N1.60x = N8 => \(x=\frac{8}{1.60}=\frac{1}{0.2}=\frac{10}{2}=5\)
Source: Myschool Past Questions Archive (Record #26675)
Mathematics
WAEC / WASSCE 2005
Geometry and Trigonometry
Solution
From a point P, R is 5km due west and 12km due south. Find the distance between P and R
A
5km
B
12km
C
13km
D
17km
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Using Pythagoras theorem
\(|PR|^2=|PO|^2+|OR|^2\\
|PR|^2=5^2+12^2\\
|PR|=\sqrt{25+144}=\sqrt{169}=13km\)
Source: Myschool Past Questions Archive (Record #26676)
Mathematics
WAEC / WASSCE 2005
Geometry and Trigonometry
Diagram
Solution
In the diagram, PR is a diameter, ∠PRQ = (3x-8)o and ∠RPQ = (2y-7)o. s x in terms of y
Attached Figure / Diagram
A
\(x=\frac{75-2y}{3}\)
B
\(x=\frac{105-3y}{2}\)
C
\(x=\frac{105-2y}{3}\)
D
\(x=\frac{75-3y}{2}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
180 = ∠RPQ + ∠PRQ + ∠PQR Since PQR = 90 (theorem: angle in a semi circle)
180 = ∠RPQ + ∠PRQ + 90 => 180o = (3x-8)o+(2y-7)o+90o; 90+8+7 = 3x+2y =>\(\frac{105-2y}{3}=x\)
Source: Myschool Past Questions Archive (Record #26679)
Mathematics
WAEC / WASSCE 2005
Geometry and Trigonometry
Solution
The lengths of the adjacent sides of a right - angled triangle are xcm, (x-1)cm. If the length of the hypotenuse is \(\sqrt{13}cm\), find the value of x
A
2
B
3
C
4
D
5
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Using Pythagoras' theorem
\(x^2+(x-1)^2 = (\sqrt{13})^2\)
x\(^2\) + x\(^2\) - 2x + 1 = 13
2x\(^2 - 2x\) - 12 = 0
dividing through by 2
x\(^2\) - x- 6 = 0; (x-3)(x-2) = 0
x = 3 or -2
Source: Myschool Past Questions Archive (Record #26684)
Mathematics
WAEC / WASSCE 2005
Geometry and Trigonometry
Diagram
Solution
In the diagram, PQRS is a parallelogram and ∠QRT = 30o. Find x
Attached Figure / Diagram
A
95o
B
100o
C
120o
D
150o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Since ∠QRS = 180o - 30o = 150o and ∠SPQ = ∠QRS (theorem opp. angles in a parallelogram are equal) ∴ x = 150o
Source: Myschool Past Questions Archive (Record #26687)
Mathematics
WAEC / WASSCE 2005
Geometry and Trigonometry
Solution
From a point R, 300m north of P, man walks eastward to a place Q which is 600m from P. Find the bearing of P from Q, correct to the nearest degreeee
A
026o
B
045o
C
210o
D
240o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(cos\theta = \frac{adj}{hyp}=\frac{300}{600}=\frac{1}{2}\\
\theta = cos^{-1}(0.5000)=60^{\circ}\)
The bearing of P from \(Q = \theta + 180 = 60 + 180 = 240^{\circ}\)
Source: Myschool Past Questions Archive (Record #26688)
Mathematics
WAEC / WASSCE 2005
Geometry and Trigonometry
Solution
What is the diameter of a circle of area 77cm2 [Take \(\pi = \frac{22}{7}\)]
A
\(\frac{\sqrt{2}}{7}cm\)
B
\(3\frac{1}{2}cm\)
C
7cm
D
\(7\sqrt{2}cm\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(A=\pi r^2 \Rightarrow 77 = \frac{22}{7} \times \frac{r^2}{1}\Rightarrow r^2 = \frac{77\times 7}{22} \Rightarrow r^2 = \frac{49}{2}\\
\Rightarrow r = \sqrt{\frac{49}{2}}=\frac{7}{\sqrt{2}}. Since D = 2r ∴ D = 2 \times \frac{7}{\sqrt{2}} = \frac{14}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = 7\sqrt{2}cm\)
Source: Myschool Past Questions Archive (Record #26689)
Mathematics
WAEC / WASSCE 2005
Geometry and Trigonometry
Solution
An arc of a circle of radius 14cm subtends angle 300o at the center. Find the perimeter of the sector formed by the arc
A
14.67cm
B
42.67cm
C
101.33cm
D
543.33cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
The perimeter of the sector \(=2r+\frac{\theta}{360}\times 2\pi r \\
\Rightarrow 28 + \frac{300}{360} \times \frac{2}{1} \times \frac{22}{7}\times \frac{14}{1} = \frac{220}{3}+28\\
73.133+28=101.33cm\)
Source: Myschool Past Questions Archive (Record #26697)
Mathematics
WAEC / WASSCE 2005
Geometry and Trigonometry
which of the following statements describes the locus of a point R which moves in a plane such that its equidistant from two intersecting lines?
A
the bisector of the angle formed by the lines
B
the point of intersection of the two lines
C
A cone, with two intersecting lines as slant height
D
A circle, with the point of intersection of the two lines as the center
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 #26698)
Mathematics
WAEC / WASSCE 2005
Geometry and Trigonometry
Solution
A train moving at a uniform speed covers 36km in 21 minutes. How long does it take to cover 60km?
A
35 mins
B
40 mins
C
45 mins
D
90 mins
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Since 36km in 21 minutes
1km in \(\frac{21}{36}\)minutes
60km in \(\left(\frac{21}{36}\times 60\right)\)minutes = \(\frac{70}{2}\)min
= 35 mins
Source: Myschool Past Questions Archive (Record #26700)
Mathematics
WAEC / WASSCE 2005
Geometry and Trigonometry
Solution
A regular polygon has 9 sides. What is the size of one of its exterior angles?
A
20o
B
40o
C
90o
D
140o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Exterior angle \(=\frac{360}{n}\) where n = no of sides
\(=\frac{360}{9}=40^{\circ}\)
Source: Myschool Past Questions Archive (Record #26702)
Mathematics
WAEC / WASSCE 2005
Geometry and Trigonometry
Diagram
Solution
In the diagram PQRS is a circle, |PT| = |QT| and ∠QPT = 70o what is the size of ∠PRS?
Attached Figure / Diagram
A
40o
B
70o
C
80o
D
140o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
∠QPT = ∠QTP = 70o. Since ∠QTR = 140o
∠PRS = TSR = 1/2(180-40) = 70o
Source: Myschool Past Questions Archive (Record #26708)
Mathematics
WAEC / WASSCE 2005
Geometry and Trigonometry
Solution
XOY is a real sector of a circle center O of radius 3.5cm which subtends an angle of 144o at the center. Calculate, in term of \(\pi\), the area of the sector
A
1.4\(\pi\)cm2
B
2.8\(\pi\)cm2
C
4.9\(\pi\)cm2
D
9.8\(\pi\)cm2
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Area of a sector \(=\frac{\theta}{360}\times \pi r^2\\
\left(\frac{144}{360}\times \pi \times \frac{3.5}{1} \times 3.5\right)cm^2 = 4.9\pi cm^2\)
Source: Myschool Past Questions Archive (Record #26710)