Year : 
1994
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

41 - 48 of 48 Questions

# Question Ans
41.

Find the inequality which represents the shaded portion in the diagram

A. 2x - y - 2 \(\geq\) 0

B. 2x - y - 2 \(\leq\) 0

C. 2x - y - 2 < 0

D. 2x - y - 2 > 0

Detailed Solution

2x - y - 2 \(\geq\) 0 = y \(\leq\) 2x - 2

when x = 0, y = -2, when y = 0, x = 1
42.

In the diagram, PQRS is a parallelogram. Find the value of < SQR

A. 30o

B. 50o

C. 80o

D. 100o

Detailed Solution

SQR + RQV + VQU = 18o angle on a straight line SP is parallel to QR and PV is parallel to TR

< STP = < RQV = 30o

But SQR + 30o + 50o = 180o

SQR = 180 - 80

= 100o
43.

In the diagram, O is the centre of the circle. If SOQ is a diameter and < PRS is 38o, what is the value of < PSQ

A. 148o

B. 104o

C. 80o

D. 52o

Detailed Solution

< SRP = < SQP = 38o (angle in the same segment of a circle are equal)

But < SPQ = 90o (angle in a semicircle)

also < PSQ + < SQP + < SPQ = 180o (angles in a triangle = 180o)

< PSQ + 38o + 90o = 190o

< PSQ = 128o = 180o

PSQ = 180o - 128o

PSQ = 52o
44.

In the diagram, PTS is a tangent to the circle TQR at T. Calculate < RTS

A. 120o

B. 70o

C. 60o

D. 40o

Detailed Solution

RTS = RQT (angle between a tangent and a chord at the point of contact is equal to the angle in the alternate segment) But R = Q + T = 180

RQT = 180o - (50 + 60)

= 180o - 110o

= 70o

Since RQT = RTS = 70o
45.

In the diagram. Find h

A. \(\frac{12}{7}\)cm

B. \(\frac{12}{7} \sqrt{6}\)cm

C. \(\frac{7}{12}\)cm

D. \(\frac{1}{2}\)cm

Detailed Solution

A\(\bigtriangleup\) = \(\sqrt{S(S - a) (S - b)(S - c)}\) (Hero's Formula)

S = \(\frac{a + b + c}{2}\) = \(\frac{5 + 6 + 7}{2}\)

\(\frac{18}{2} = 9\)

A\(\bigtriangleup\) \(\sqrt{9} \times 4 \times 3 \times 2\)

\(\sqrt{216} = 6 \sqrt{6}cm^3\)

A\(\bigtriangleup\) = \(\frac{1}{2} \times 6 \times h\)

6\(\sqrt{6} = \frac{1}{2} \times 7 \times h\)

h = \(\frac{12}{h} \sqrt{6}\)
46.

In the frustum of the cone, the top diagram is twice the bottom diameter. If the height of the frustum is h centimeters, find he height of the cone

A. 2h

B. 2\(\pi\)h

C. \(\pi\)h

D. \(\frac{\pi h}{2}\)

Detailed Solution

\(\frac{x}{r}\) = \(\frac{x + h}{2r}\)

2 x r = r (x + h)

Total height of cone = x + h

but x = h

total height = 2h
47.

The equation of the line in the graph is

A. 3y = 4x + 12

B. 3y = 3x + 12

C. 3y = -4x + 12

D. 3y = -4x + 9

Detailed Solution

Gradient of line = \(\frac{\text{Change in y}}{\text{Change in x}} = \frac{y_2 - y_1}{x_2 - x_1}\)

y2 = 0, y1 = 4

x2 = 3 and x1 = 0

\(\frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 4}{3 - 0} = \frac{-4}{3}\)

Equation of straight line = y = mx + c

where m = gradient and c = y

intercept = 4

y = 4x + \(\frac{4}{3}\), multiple through by 3 &
48.

The grades A1, A2, A3, C4 and F earned by students in a particular course are shown in the pie chart. What percentage of the students obtained a C4 grade?

A. 52.0

B. 43.2

C. 40.0

D. 12.0

D
41.

Find the inequality which represents the shaded portion in the diagram

A. 2x - y - 2 \(\geq\) 0

B. 2x - y - 2 \(\leq\) 0

C. 2x - y - 2 < 0

D. 2x - y - 2 > 0

Detailed Solution

2x - y - 2 \(\geq\) 0 = y \(\leq\) 2x - 2

when x = 0, y = -2, when y = 0, x = 1
42.

In the diagram, PQRS is a parallelogram. Find the value of < SQR

A. 30o

B. 50o

C. 80o

D. 100o

Detailed Solution

SQR + RQV + VQU = 18o angle on a straight line SP is parallel to QR and PV is parallel to TR

< STP = < RQV = 30o

But SQR + 30o + 50o = 180o

SQR = 180 - 80

= 100o
43.

In the diagram, O is the centre of the circle. If SOQ is a diameter and < PRS is 38o, what is the value of < PSQ

A. 148o

B. 104o

C. 80o

D. 52o

Detailed Solution

< SRP = < SQP = 38o (angle in the same segment of a circle are equal)

But < SPQ = 90o (angle in a semicircle)

also < PSQ + < SQP + < SPQ = 180o (angles in a triangle = 180o)

< PSQ + 38o + 90o = 190o

< PSQ = 128o = 180o

PSQ = 180o - 128o

PSQ = 52o
44.

In the diagram, PTS is a tangent to the circle TQR at T. Calculate < RTS

A. 120o

B. 70o

C. 60o

D. 40o

Detailed Solution

RTS = RQT (angle between a tangent and a chord at the point of contact is equal to the angle in the alternate segment) But R = Q + T = 180

RQT = 180o - (50 + 60)

= 180o - 110o

= 70o

Since RQT = RTS = 70o
45.

In the diagram. Find h

A. \(\frac{12}{7}\)cm

B. \(\frac{12}{7} \sqrt{6}\)cm

C. \(\frac{7}{12}\)cm

D. \(\frac{1}{2}\)cm

Detailed Solution

A\(\bigtriangleup\) = \(\sqrt{S(S - a) (S - b)(S - c)}\) (Hero's Formula)

S = \(\frac{a + b + c}{2}\) = \(\frac{5 + 6 + 7}{2}\)

\(\frac{18}{2} = 9\)

A\(\bigtriangleup\) \(\sqrt{9} \times 4 \times 3 \times 2\)

\(\sqrt{216} = 6 \sqrt{6}cm^3\)

A\(\bigtriangleup\) = \(\frac{1}{2} \times 6 \times h\)

6\(\sqrt{6} = \frac{1}{2} \times 7 \times h\)

h = \(\frac{12}{h} \sqrt{6}\)
46.

In the frustum of the cone, the top diagram is twice the bottom diameter. If the height of the frustum is h centimeters, find he height of the cone

A. 2h

B. 2\(\pi\)h

C. \(\pi\)h

D. \(\frac{\pi h}{2}\)

Detailed Solution

\(\frac{x}{r}\) = \(\frac{x + h}{2r}\)

2 x r = r (x + h)

Total height of cone = x + h

but x = h

total height = 2h
47.

The equation of the line in the graph is

A. 3y = 4x + 12

B. 3y = 3x + 12

C. 3y = -4x + 12

D. 3y = -4x + 9

Detailed Solution

Gradient of line = \(\frac{\text{Change in y}}{\text{Change in x}} = \frac{y_2 - y_1}{x_2 - x_1}\)

y2 = 0, y1 = 4

x2 = 3 and x1 = 0

\(\frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 4}{3 - 0} = \frac{-4}{3}\)

Equation of straight line = y = mx + c

where m = gradient and c = y

intercept = 4

y = 4x + \(\frac{4}{3}\), multiple through by 3 &
48.

The grades A1, A2, A3, C4 and F earned by students in a particular course are shown in the pie chart. What percentage of the students obtained a C4 grade?

A. 52.0

B. 43.2

C. 40.0

D. 12.0

D