Year : 
2005
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

21 - 30 of 47 Questions

# Question Ans
21.

The sum of the interior angle of a regular polygon is 1800o. Calculate the size of one exterior angle of the polygon

A. 30o

B. 24o

C. 18o

D. 12o

Detailed Solution

Sum of interior ∠s = 1800o
∴(n - 2) 180o = 1800o
180n -360o = 1800o
180n = 1800o + 360o
180n = 2160o
n = 2160o/180o
n = 12 sides
Each exterior ∠ = 360o/n
= 360o/12
= 30o
22.

In the diagram above, O is the center of the circle,

∠UOT = 70o and ∠RST = 100o. Calculate ∠RUO.

A. 20o

B. 25o

C. 50o

D. 80o

Detailed Solution

∠OUT = ∠OTU = a (Base ∠s of 180 Δ)
∴ a + a + 70 = 180o (sum of ∠s of a Δ)
2a = 180o - 70o
2a = 110o
a = 55o
But ∠RUT + ∠RBT = 180o (opposite ∠s of a Cyclic quad)
∴ x + a = 100 = 180
x + 55 + 100 = 180
x = 180 - 155
x = 25
so ∠RUO = x = 25o
23.

A sector of a circle has an area of 55 cm2. If the radius of the circle is 10 cm, calculate the angle of the sector
[π = 22/7]

A. 45o

B. 63o

C. 75o

D. 90o

Detailed Solution

Area of a sector = \(\frac{\theta}{360}\times \pi r^2\\55=\frac{\theta}{360}\times \frac{22}{7}\times \frac{10\times 10}{1}\\\theta=\frac{360 \times 55 \times 7}{22 \times 10 \times 10}\\\theta = 63^{\circ}\)
24.

Find the curved surface area of a cone with circular base diameter 10 cm and height 12 cm

A. 25 πcm2

B. 65 πcm2

C. 120 πcm2

D. 156 πcm2

Detailed Solution

L2 = 122 + 52
= 144 + 25
= 169
L = √169
= 13
Curved surface Area = πrL
= 5/1 * 13/1 * π
= 65πcm2
25.

Two lines PQ and ST intersect at 75°. The locus of points equidistant from PQ and ST lies on the

A. perpendicular bisector of PQ

B. perpendicular bisector of ST

C. bisector of the angles between lines PQ and ST

D. bisector of the angles between lines PT and QS

C
26.

Find the equation of the perpendicular at the point (4,3) to the line y + 2x = 5

A. 2y - x = 4

B. y + 2x = 3

C. y + 2x = 5

D. 2y - x = 2

Detailed Solution

Gradient of the line y + 2x = 5 = -2
Gradient of the line perpendicular to
y + 2x + 5 = 1/2
equation of a line perpendicular to
y + 2x = 5 at the point (4, 3)
y - y1 = m(x - x1)
y - 3 = 1/2(x - 4)
2y - 6 = x -4
2y - x = 2
27.

A chord of a circle subtends an angle of 60o at the length of a circle of radius 14 cm. Find the length of the chord

A. 7 cm

B. 14 cm

C. 21 cm

D. 28 cm

Detailed Solution

Chord Length (L) = \(2 \times Radius \times (sin \frac{angle}{2})\)
= \(2 \times 14 \times (sin \frac{60}{2})\)
= \( 28 \times (sin 30)\)
= \(28 \times \frac{1}{2}\)
= \(14cm\)
28.

If sin θ = -1/2 for 0 < θ < 360o, the value of θ is

A. 30o and 150o

B. 150o and 210o

C. 210o and 330o

D. 150o and 330o

Detailed Solution

sin θ = -1/2
= -0.5
θ = sin-1 (0.5)
θ = 30o
Since θ is negative
θ = 180 + 30 = 210o
θ = 360 - 30 = 330o
∴θ = 210o and 330o
29.

If y = (1 - 2x)\(^3\), find the value of dy/dx at x = -1

A. 57

B. 27

C. -6

D. -54

Detailed Solution

y = (1 - 2x)\(^3\)
let u = 1-2x
du/dx = -2
∴y = u\(^3\)
dy/du = 3u\(^2\)
But dy/dx = du/dx * dy/du
= -2 * 3u\(^2\)
= -6u\(^2\)
= -6(1-2x)\(^2\)
At x = -1: dy/dx = -6(1 - 2(-1))\(^2\)
dy/dx = -6(1 + 2)\(^2\)
= -6 * 3\(^2\)
= -6 * 9
= -54
30.

Find the derivative of y = sin(2x\(^3\) + 3x - 4)

A. cos (2x3 + 3x - 4)

B. -cos (2x3 + 3x - 4)

C. (6x2 + 3) cos (2x3 + 3x - 4)

D. -(6x2 + 3) cos (2x3 + 3x - 4)

Detailed Solution

y = sin (2x\(^3\) + 3x - 4)
let u = 2x\(^3\) + 3x - 4
∴du/dx = 6x\(^2\)
y = sin u
dy/du = cos u
dy/dx = du/dx * dy/du
∴dy/dx = (6x\(^2\) + 3) cos u
= (6x\(^2\) + 3)cos(2x\(^3\) + 3x - 4)
21.

The sum of the interior angle of a regular polygon is 1800o. Calculate the size of one exterior angle of the polygon

A. 30o

B. 24o

C. 18o

D. 12o

Detailed Solution

Sum of interior ∠s = 1800o
∴(n - 2) 180o = 1800o
180n -360o = 1800o
180n = 1800o + 360o
180n = 2160o
n = 2160o/180o
n = 12 sides
Each exterior ∠ = 360o/n
= 360o/12
= 30o
22.

In the diagram above, O is the center of the circle,

∠UOT = 70o and ∠RST = 100o. Calculate ∠RUO.

A. 20o

B. 25o

C. 50o

D. 80o

Detailed Solution

∠OUT = ∠OTU = a (Base ∠s of 180 Δ)
∴ a + a + 70 = 180o (sum of ∠s of a Δ)
2a = 180o - 70o
2a = 110o
a = 55o
But ∠RUT + ∠RBT = 180o (opposite ∠s of a Cyclic quad)
∴ x + a = 100 = 180
x + 55 + 100 = 180
x = 180 - 155
x = 25
so ∠RUO = x = 25o
23.

A sector of a circle has an area of 55 cm2. If the radius of the circle is 10 cm, calculate the angle of the sector
[π = 22/7]

A. 45o

B. 63o

C. 75o

D. 90o

Detailed Solution

Area of a sector = \(\frac{\theta}{360}\times \pi r^2\\55=\frac{\theta}{360}\times \frac{22}{7}\times \frac{10\times 10}{1}\\\theta=\frac{360 \times 55 \times 7}{22 \times 10 \times 10}\\\theta = 63^{\circ}\)
24.

Find the curved surface area of a cone with circular base diameter 10 cm and height 12 cm

A. 25 πcm2

B. 65 πcm2

C. 120 πcm2

D. 156 πcm2

Detailed Solution

L2 = 122 + 52
= 144 + 25
= 169
L = √169
= 13
Curved surface Area = πrL
= 5/1 * 13/1 * π
= 65πcm2
25.

Two lines PQ and ST intersect at 75°. The locus of points equidistant from PQ and ST lies on the

A. perpendicular bisector of PQ

B. perpendicular bisector of ST

C. bisector of the angles between lines PQ and ST

D. bisector of the angles between lines PT and QS

C
26.

Find the equation of the perpendicular at the point (4,3) to the line y + 2x = 5

A. 2y - x = 4

B. y + 2x = 3

C. y + 2x = 5

D. 2y - x = 2

Detailed Solution

Gradient of the line y + 2x = 5 = -2
Gradient of the line perpendicular to
y + 2x + 5 = 1/2
equation of a line perpendicular to
y + 2x = 5 at the point (4, 3)
y - y1 = m(x - x1)
y - 3 = 1/2(x - 4)
2y - 6 = x -4
2y - x = 2
27.

A chord of a circle subtends an angle of 60o at the length of a circle of radius 14 cm. Find the length of the chord

A. 7 cm

B. 14 cm

C. 21 cm

D. 28 cm

Detailed Solution

Chord Length (L) = \(2 \times Radius \times (sin \frac{angle}{2})\)
= \(2 \times 14 \times (sin \frac{60}{2})\)
= \( 28 \times (sin 30)\)
= \(28 \times \frac{1}{2}\)
= \(14cm\)
28.

If sin θ = -1/2 for 0 < θ < 360o, the value of θ is

A. 30o and 150o

B. 150o and 210o

C. 210o and 330o

D. 150o and 330o

Detailed Solution

sin θ = -1/2
= -0.5
θ = sin-1 (0.5)
θ = 30o
Since θ is negative
θ = 180 + 30 = 210o
θ = 360 - 30 = 330o
∴θ = 210o and 330o
29.

If y = (1 - 2x)\(^3\), find the value of dy/dx at x = -1

A. 57

B. 27

C. -6

D. -54

Detailed Solution

y = (1 - 2x)\(^3\)
let u = 1-2x
du/dx = -2
∴y = u\(^3\)
dy/du = 3u\(^2\)
But dy/dx = du/dx * dy/du
= -2 * 3u\(^2\)
= -6u\(^2\)
= -6(1-2x)\(^2\)
At x = -1: dy/dx = -6(1 - 2(-1))\(^2\)
dy/dx = -6(1 + 2)\(^2\)
= -6 * 3\(^2\)
= -6 * 9
= -54
30.

Find the derivative of y = sin(2x\(^3\) + 3x - 4)

A. cos (2x3 + 3x - 4)

B. -cos (2x3 + 3x - 4)

C. (6x2 + 3) cos (2x3 + 3x - 4)

D. -(6x2 + 3) cos (2x3 + 3x - 4)

Detailed Solution

y = sin (2x\(^3\) + 3x - 4)
let u = 2x\(^3\) + 3x - 4
∴du/dx = 6x\(^2\)
y = sin u
dy/du = cos u
dy/dx = du/dx * dy/du
∴dy/dx = (6x\(^2\) + 3) cos u
= (6x\(^2\) + 3)cos(2x\(^3\) + 3x - 4)