Year : 
2002
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

41 - 48 of 48 Questions

# Question Ans
41.

In the diagram above , XZ is the diameter of the circle XZW, with center O and radius 15/2 cm. If XY = 12 cm, find the area of the triangle XYZ

A. 54 cm2

B. 45 cm2

C. 27 cm2

D. 75 cm2

Detailed Solution

o (< semi circle )
since radius = 15/2 diameter = 2 * (15/2) = 15cm
XYZ is a right angled triangle
∴ 152 = 122 + (YZ)2
225 = 144 = (YZ)2
225 - 144 = (YZ)2
81 = (YZ)2
√81 = YZ
9 = YZ
Area of ΔXYZ = 1/2bh
(1/2) * 9 * 12
9 * 6 = 54
= 54cm2
42.

The venn diagram above shows the numbers of students offering music and history in a class of 80 students. If a student is picked at random from the class, what is the probability that he offers Music only?

A. 0.38

B. 0.13

C. 0.50

D. 0.25

Detailed Solution

30 - x + x + 40 - x = 80 - 20
70 - x = 60
- x = 60 - 70
- x = - 10
∴ x = 10
Music only = 30 - x
= 30 - 10
20
P(music only) = 20/80
= 1/4
= 0.25
43.

In the diagram above, PST is a straight line, PQ = QS = RS. If ∠RST = 72o, find x

A. 36o

B. 18o

C. 72o

D. 24o

Detailed Solution

In Δ PQS, ∠PSQ = X(base ∠s of isoc Δ PQS)
In Δ QRS, ∠RQS = ∠PSQ + X(Extr ∠ = sum of two intr. opp ∠s)
∴ ∠RQS = X + X
= 2X
Also ∠QRS = 2X(base ∠s of isoc Δ QRS in Δ PRS,
72 = ∠RPS + ∠PRS(Extr, ∠ = sum of two intr. opp ∠s)
∴ 72 = x + 2x
72 = 3x
x = 72/3
x = 24o
44.

In the diagram above are two concentric circles of radii r and R respectively with center O. If r = 2/3R, express the area of the shaded portion in terms of π and R

A. 21/25πR2

B. 9/25πR2

C. 21/23πR2

D. 5/9πR2

Detailed Solution

r = \(\frac{2}{3}\)R

∴R = \(\frac{3}{3}\)R

Area of small circle = πr2

= π(\(\frac{2R}{3}\))2

Area of the big circle πr2 = π\(\frac{(3R)^2}{3}\)

Area of shaded portion = π(\(\frac{3R}{3}\))2 - π(\(\frac{2R}{3}\))2

= π[(\(\frac{3R}{3}\))2 - (\(\frac{2R}{3}\))2]

= π[(\(\frac{3R}{3}) + (\frac{2R}{3}) - (\frac{3R}{3}\)) - (\(\frac{2R}{3}\))]

= π[(\(\frac{5R}{3}\)) (\(\frac{R}{3}\))]

= π x \(\frac{5R
45.

In the diagram above, a cylinder is summounted by a hemisphere bowl. Calculate the volume of the solid.

A. 180 πcm3

B. 162 πcm3

C. 216 πcm3

D. 198 πcm3

Detailed Solution

Volume of cylinder = πr2h
= π x 32 x 20
= π x 9 x 20
= 180 πcm3
Volume of hemisphere = 2/3πr3
= 2/3 x π x 32
= 2 x π x 32
= 2 x π x 9
= 18π
= Volume of the solid = 180π x 18π
= 198πcm3
46.

The triangle PQR above is

A. an obtuse-angled triangle

B. a scalene triangle

C. an isosceles triangle

D. an equilateral triangle

Detailed Solution

∠PQR = 180 - 128 (∠s on a straight line)
= 52o
∠QPR + 76 + 52 = 180(extr ∠ = sum of intr. opp ∠s)
∠QPR + 128 = 180
∠QPR = 180 - 128
= 52o
∴ΔPQR is an isosceles triangle
47.

Table:

The distribution above shows the number of days a group of 260 students were absents from school in a particular term. How many students were absent for at least four days in the term

A. 210

B. 40

C. 120

D. 160

Detailed Solution

20 + X + 50 + 40 + 2X + 60 = 260
3X + 170 = 260
3X = 260 - 170
3x = 90
x = 30
Absent for at least 4 days

i.e 40 + 2x + 60 = 40 + (2 x 30) + 60
= 40 + 60 + 60
= 160
48.

Using the graph find the values of p and q if px + qy \(\geq\) 4

A. p = 2, q = -1

B. p = -1, q = 2

C. p = 2, q = 2

D. p = 1, q = 2

Detailed Solution

m = \(\frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 0}{0 - (4)} = \frac{2}{4} = \frac{1}{2}\)

\(\frac{y_2 - y_1}{x_2 - x_1} \geq m\)

\(\frac{y - 0}{x + 4} \geq \frac{1}{2}\)

2y \(\geq\) x + 4, -x + 2y \(\geq\) 4 = px + qy \(\geq\) 4

p = -1, q = 2
41.

In the diagram above , XZ is the diameter of the circle XZW, with center O and radius 15/2 cm. If XY = 12 cm, find the area of the triangle XYZ

A. 54 cm2

B. 45 cm2

C. 27 cm2

D. 75 cm2

Detailed Solution

o (< semi circle )
since radius = 15/2 diameter = 2 * (15/2) = 15cm
XYZ is a right angled triangle
∴ 152 = 122 + (YZ)2
225 = 144 = (YZ)2
225 - 144 = (YZ)2
81 = (YZ)2
√81 = YZ
9 = YZ
Area of ΔXYZ = 1/2bh
(1/2) * 9 * 12
9 * 6 = 54
= 54cm2
42.

The venn diagram above shows the numbers of students offering music and history in a class of 80 students. If a student is picked at random from the class, what is the probability that he offers Music only?

A. 0.38

B. 0.13

C. 0.50

D. 0.25

Detailed Solution

30 - x + x + 40 - x = 80 - 20
70 - x = 60
- x = 60 - 70
- x = - 10
∴ x = 10
Music only = 30 - x
= 30 - 10
20
P(music only) = 20/80
= 1/4
= 0.25
43.

In the diagram above, PST is a straight line, PQ = QS = RS. If ∠RST = 72o, find x

A. 36o

B. 18o

C. 72o

D. 24o

Detailed Solution

In Δ PQS, ∠PSQ = X(base ∠s of isoc Δ PQS)
In Δ QRS, ∠RQS = ∠PSQ + X(Extr ∠ = sum of two intr. opp ∠s)
∴ ∠RQS = X + X
= 2X
Also ∠QRS = 2X(base ∠s of isoc Δ QRS in Δ PRS,
72 = ∠RPS + ∠PRS(Extr, ∠ = sum of two intr. opp ∠s)
∴ 72 = x + 2x
72 = 3x
x = 72/3
x = 24o
44.

In the diagram above are two concentric circles of radii r and R respectively with center O. If r = 2/3R, express the area of the shaded portion in terms of π and R

A. 21/25πR2

B. 9/25πR2

C. 21/23πR2

D. 5/9πR2

Detailed Solution

r = \(\frac{2}{3}\)R

∴R = \(\frac{3}{3}\)R

Area of small circle = πr2

= π(\(\frac{2R}{3}\))2

Area of the big circle πr2 = π\(\frac{(3R)^2}{3}\)

Area of shaded portion = π(\(\frac{3R}{3}\))2 - π(\(\frac{2R}{3}\))2

= π[(\(\frac{3R}{3}\))2 - (\(\frac{2R}{3}\))2]

= π[(\(\frac{3R}{3}) + (\frac{2R}{3}) - (\frac{3R}{3}\)) - (\(\frac{2R}{3}\))]

= π[(\(\frac{5R}{3}\)) (\(\frac{R}{3}\))]

= π x \(\frac{5R
45.

In the diagram above, a cylinder is summounted by a hemisphere bowl. Calculate the volume of the solid.

A. 180 πcm3

B. 162 πcm3

C. 216 πcm3

D. 198 πcm3

Detailed Solution

Volume of cylinder = πr2h
= π x 32 x 20
= π x 9 x 20
= 180 πcm3
Volume of hemisphere = 2/3πr3
= 2/3 x π x 32
= 2 x π x 32
= 2 x π x 9
= 18π
= Volume of the solid = 180π x 18π
= 198πcm3
46.

The triangle PQR above is

A. an obtuse-angled triangle

B. a scalene triangle

C. an isosceles triangle

D. an equilateral triangle

Detailed Solution

∠PQR = 180 - 128 (∠s on a straight line)
= 52o
∠QPR + 76 + 52 = 180(extr ∠ = sum of intr. opp ∠s)
∠QPR + 128 = 180
∠QPR = 180 - 128
= 52o
∴ΔPQR is an isosceles triangle
47.

Table:

The distribution above shows the number of days a group of 260 students were absents from school in a particular term. How many students were absent for at least four days in the term

A. 210

B. 40

C. 120

D. 160

Detailed Solution

20 + X + 50 + 40 + 2X + 60 = 260
3X + 170 = 260
3X = 260 - 170
3x = 90
x = 30
Absent for at least 4 days

i.e 40 + 2x + 60 = 40 + (2 x 30) + 60
= 40 + 60 + 60
= 160
48.

Using the graph find the values of p and q if px + qy \(\geq\) 4

A. p = 2, q = -1

B. p = -1, q = 2

C. p = 2, q = 2

D. p = 1, q = 2

Detailed Solution

m = \(\frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 0}{0 - (4)} = \frac{2}{4} = \frac{1}{2}\)

\(\frac{y_2 - y_1}{x_2 - x_1} \geq m\)

\(\frac{y - 0}{x + 4} \geq \frac{1}{2}\)

2y \(\geq\) x + 4, -x + 2y \(\geq\) 4 = px + qy \(\geq\) 4

p = -1, q = 2