Year : 
1993
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

41 - 50 of 51 Questions

# Question Ans
41.

The figure represents the graphs of y = x(2 - x) and y = (x - 1)(x - 3). What are the x-coordinates of P, Q and F respectively?

A. 1, 3, 2

B. 0, 0, 0

C. 0, 2, 3

D. 1, 2, 3

Detailed Solution

The two equations are y = x(2 - x) and y = (x - 1)(x - 3)

The root of these equations are points where the graph of the equations cuts the x axis; but at these points = 0

put y = 0; 0 = x(2 - x)

0 = (x - 1)(x - 3)

x = 0 or x = 2; x = 1 or x = 3

The values of x at P, Q, R are increasing towards the positive direction of x-axis

at P, x = 1 at Q, x = 2 at R, x = 3

P, Q, R are respectively (1, 2, 3)
42.

PQRST is a regular pentagon and PQVU is a rectangle with U and V lying on TS and SR respectively as shown in the diagram. Calculate TUP

A. 18o

B. 54o

C. 90o

D. 108o

Detailed Solution

The total angle in a regular pentagon = \((2(5) - 4) \times 90\)
= \(6 \times 90 = 540°\)
Each interior angle = \(\frac{540}{5} = 108°\)
While the interior of a quadrilateral = \(\frac{360}{4} = 90°\)
\(PTU + TPU + TUP = 180°\)
\(108° + (180° - 90°) + TUP = 180°\)
\(TUP = 180° - (108° + 18°) = 54°\)
43.

In the diagram, PQRS is a circle with O as centre and PQ/RT. If RTS = 32°. Find PSQ

A. 32o

B. 45o

C. 58o

D. 90o

Detailed Solution

< PSO = \(\frac{1}{2}\) < SOQ = \(\frac{1}{2}\)(180) = 90°

< RTS = < PQS = 32° (Alternative angle)

< PSQ = 90 - < PSQ = 90° - 32°

= 58°
44.

In the diagram, O is the centre of the circle and POQ a diameter. If POR = 96o, find the value of ORQ.

A. 84o

B. 48o

C. 45o

D. 42o

Detailed Solution

OQ = OR = radii

< ROQ = 180 - 86 = 84o

\(\bigtriangleup\)OQR = Isosceles

R = Q

R + Q + 84 = 180(angle in a \(\bigtriangleup\))

2R = 96 since R = Q

R = 48o

ORQ = 48o
45.

In the diagram, QP//ST:PQR = 34o qrs = 73o and Rs = RT. Find SRT

A. 68o

B. 102o

C. 107o

D. 141o

Detailed Solution

Construction joins R to P such that SRP = straight line

R = 180o - 107o

< p = 180o - (107o - 34o)

108 - 141o = 39o

Angle < S = 39o (corr. Ang.) But in \(\bigtriangleup\)SRT

< S = < T = 39o

SRT = 180 - (39o + 39o)

= 180o - 78o

= 102o
46.

In the figure, PT is a tangent to the circle at U and QU/RS if TUR = 35o and SRU = 50o find x

A. 95o

B. 85o

C. 50o

D. 35o

Detailed Solution

Since QRU= Xo

RSU = Xo, But RSU = 180o - (50o + 35o)

= 180o - 85o

= 95o

x = 95o
47.

In the diagram, QPS = SPR, PR = 9cm. PQ = 4cm and QS = 3cm, find SR.

A. 6\(\frac{3}{4}\)cm

B. 3\(\frac{3}{8}\)cm

C. 4\(\frac{3}{8}\)cm

D. 2\(\frac{3}{8}\)cm

Detailed Solution

Using angle bisector theorem: line PS bisects angle QPR

QS/QP = SR/PR

3/4 = SR/9

4SR = 27

SR = \(\frac{27}{4}\)

= 6\(\frac{3}{4}\)cm
48.

In the figure, the line segment ST is tangent to two circles at S and T. O and Q are the centres of the circles wih OS = 5cm. QT = 2cm and OR = 14cm. Find ST

A. 7 \(\sqrt{3}\)cm

B. 12.9cm

C. \(\sqrt{87}\)cm

D. 7cm

Detailed Solution

SQ2 + OS2 = OQ2 + 52 = 142

SQ2 = 142 - 52

196 - 25 = 171

ST2 + TQ2 = SQ2

ST2 + 22 = 171

ST2 = 171 - 4

= 167

ST = \(\sqrt{167}\)

= 12.92 = 12.9cm
49.

In the figure, the area of the square of the square PQRS is 100cm2. If the ratio of the area of the square TUYS to the area of the area of the square XQVU is 1 : 16, Find YR

A. 6cm

B. 7cm

C. 8cm

D. 9cm

Detailed Solution

Since area of square PQRS = 100cm2

each lenght = 10cm

Also TUYS : XQVU = 1 : 16

lengths are in ratio 1 : 4, hence TU : UV = 1: 4

Let TU = x

UV = 1: 4

hence TV = x + 4x = 5x = 10cm

x = 2cm

TU = 2cm

UV = 8cm

But TU = SY and UV = YR

YR = 8cm
50.

From the figure, calculate TH in centimeters

A. \(\frac{5}{\sqrt{3} + 1}\)

B. \(\frac{5}{\sqrt{3} - 1}\)

C. \(\frac{5}{\sqrt{3}}\)

D. \(\frac{\sqrt{3}}{5}\)

Detailed Solution

TH = tan 45o, TH = QH

\(\frac{TH}{5 + QH}\) = tan 30o

TH = (b + QH) tan 30o

QH = 56 (5 + QH) \(\frac{1}{\sqrt{3}}\)

QH(1 - \(\frac{1}{\sqrt{3}}\)) = \(\frac{5}{\sqrt{3}}\)

QH = \(\frac{5\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}}\)

= \(\frac{5}{\sqrt{3} - 1}\)
41.

The figure represents the graphs of y = x(2 - x) and y = (x - 1)(x - 3). What are the x-coordinates of P, Q and F respectively?

A. 1, 3, 2

B. 0, 0, 0

C. 0, 2, 3

D. 1, 2, 3

Detailed Solution

The two equations are y = x(2 - x) and y = (x - 1)(x - 3)

The root of these equations are points where the graph of the equations cuts the x axis; but at these points = 0

put y = 0; 0 = x(2 - x)

0 = (x - 1)(x - 3)

x = 0 or x = 2; x = 1 or x = 3

The values of x at P, Q, R are increasing towards the positive direction of x-axis

at P, x = 1 at Q, x = 2 at R, x = 3

P, Q, R are respectively (1, 2, 3)
42.

PQRST is a regular pentagon and PQVU is a rectangle with U and V lying on TS and SR respectively as shown in the diagram. Calculate TUP

A. 18o

B. 54o

C. 90o

D. 108o

Detailed Solution

The total angle in a regular pentagon = \((2(5) - 4) \times 90\)
= \(6 \times 90 = 540°\)
Each interior angle = \(\frac{540}{5} = 108°\)
While the interior of a quadrilateral = \(\frac{360}{4} = 90°\)
\(PTU + TPU + TUP = 180°\)
\(108° + (180° - 90°) + TUP = 180°\)
\(TUP = 180° - (108° + 18°) = 54°\)
43.

In the diagram, PQRS is a circle with O as centre and PQ/RT. If RTS = 32°. Find PSQ

A. 32o

B. 45o

C. 58o

D. 90o

Detailed Solution

< PSO = \(\frac{1}{2}\) < SOQ = \(\frac{1}{2}\)(180) = 90°

< RTS = < PQS = 32° (Alternative angle)

< PSQ = 90 - < PSQ = 90° - 32°

= 58°
44.

In the diagram, O is the centre of the circle and POQ a diameter. If POR = 96o, find the value of ORQ.

A. 84o

B. 48o

C. 45o

D. 42o

Detailed Solution

OQ = OR = radii

< ROQ = 180 - 86 = 84o

\(\bigtriangleup\)OQR = Isosceles

R = Q

R + Q + 84 = 180(angle in a \(\bigtriangleup\))

2R = 96 since R = Q

R = 48o

ORQ = 48o
45.

In the diagram, QP//ST:PQR = 34o qrs = 73o and Rs = RT. Find SRT

A. 68o

B. 102o

C. 107o

D. 141o

Detailed Solution

Construction joins R to P such that SRP = straight line

R = 180o - 107o

< p = 180o - (107o - 34o)

108 - 141o = 39o

Angle < S = 39o (corr. Ang.) But in \(\bigtriangleup\)SRT

< S = < T = 39o

SRT = 180 - (39o + 39o)

= 180o - 78o

= 102o
46.

In the figure, PT is a tangent to the circle at U and QU/RS if TUR = 35o and SRU = 50o find x

A. 95o

B. 85o

C. 50o

D. 35o

Detailed Solution

Since QRU= Xo

RSU = Xo, But RSU = 180o - (50o + 35o)

= 180o - 85o

= 95o

x = 95o
47.

In the diagram, QPS = SPR, PR = 9cm. PQ = 4cm and QS = 3cm, find SR.

A. 6\(\frac{3}{4}\)cm

B. 3\(\frac{3}{8}\)cm

C. 4\(\frac{3}{8}\)cm

D. 2\(\frac{3}{8}\)cm

Detailed Solution

Using angle bisector theorem: line PS bisects angle QPR

QS/QP = SR/PR

3/4 = SR/9

4SR = 27

SR = \(\frac{27}{4}\)

= 6\(\frac{3}{4}\)cm
48.

In the figure, the line segment ST is tangent to two circles at S and T. O and Q are the centres of the circles wih OS = 5cm. QT = 2cm and OR = 14cm. Find ST

A. 7 \(\sqrt{3}\)cm

B. 12.9cm

C. \(\sqrt{87}\)cm

D. 7cm

Detailed Solution

SQ2 + OS2 = OQ2 + 52 = 142

SQ2 = 142 - 52

196 - 25 = 171

ST2 + TQ2 = SQ2

ST2 + 22 = 171

ST2 = 171 - 4

= 167

ST = \(\sqrt{167}\)

= 12.92 = 12.9cm
49.

In the figure, the area of the square of the square PQRS is 100cm2. If the ratio of the area of the square TUYS to the area of the area of the square XQVU is 1 : 16, Find YR

A. 6cm

B. 7cm

C. 8cm

D. 9cm

Detailed Solution

Since area of square PQRS = 100cm2

each lenght = 10cm

Also TUYS : XQVU = 1 : 16

lengths are in ratio 1 : 4, hence TU : UV = 1: 4

Let TU = x

UV = 1: 4

hence TV = x + 4x = 5x = 10cm

x = 2cm

TU = 2cm

UV = 8cm

But TU = SY and UV = YR

YR = 8cm
50.

From the figure, calculate TH in centimeters

A. \(\frac{5}{\sqrt{3} + 1}\)

B. \(\frac{5}{\sqrt{3} - 1}\)

C. \(\frac{5}{\sqrt{3}}\)

D. \(\frac{\sqrt{3}}{5}\)

Detailed Solution

TH = tan 45o, TH = QH

\(\frac{TH}{5 + QH}\) = tan 30o

TH = (b + QH) tan 30o

QH = 56 (5 + QH) \(\frac{1}{\sqrt{3}}\)

QH(1 - \(\frac{1}{\sqrt{3}}\)) = \(\frac{5}{\sqrt{3}}\)

QH = \(\frac{5\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}}\)

= \(\frac{5}{\sqrt{3} - 1}\)