Year : 
1983
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

41 - 48 of 48 Questions

# Question Ans
41.

The figure FGHK is a rhombus. What is the value of angle X?

A. 30o

B. 90o

C. 150o

D. 120o

E. 60o

Detailed Solution

< HKF = 60o, < KFG = 120o

< KFG = < KHG = x(opposite angles)

x = 120o
42.

PQRS is a desk of dimensions 2m x 0.8 which is inclined at 30o to the horizontal. Find the inclination of the diagonal PR to the horizontal

A. 30o 35'

B. 30o 32'

C. 15o 36'

D. 10o

E. 10o 42'

Detailed Solution

tan\(\theta\) = \(\frac{0.8}{2}\) = (0.4)

\(\theta\) = tan-1(0.4)

From the diagram, the inclination of the diagonal PR to the horizontal is 10o 42'
43.

If O is the centre of the circle in the figure, find the value of x

A. 50

B. 260

C. 100

D. 65

E. 130

Detailed Solution

From the diagram; The value of x = 360o - 2(130o)

= 360 - 260

= 100o
44.

PQRS isa cyclic quadrilateral which PQ = PS. PT is a tangent to the circle and PQ makes an angle of 50o with the tangent as shown in the figure. What is the size of QRS?

A. 50o

B. 40o

C. 110v

D. 80o

E. 100o

Detailed Solution

< SPQ = 80o

< SPQ + < SRQ = 180(Supplementary)

80 +
< QRS = 180o - 80o

= 100o
45.

The farm yield of four crops on a piece of land in Ondo are represented on the pie chart. what is the angle of the sector occupies by Okro in the chart?

A. 11o

B. 19\(\frac{1}{3}\)o

C. 33\(\frac{1}{3}\)o

D. 91\(\frac{1}{3}\)o

E. 91o

Detailed Solution

Adding the values of all the items together, it gives 70

Okro sector = \(\frac{145}{270}\) x \(\frac{360^o}{1}\)

= 19.33o

= 19\(\frac{1}{3}\)o
46.

In the figure, PQR is a straight line. Find the values of x and y

A. x = 22.5o, y = 33.75o

B. x = 15o, y = 52.5o

C. x = 22.5o, y = 45.0o

D. x = 56.25o, y = 11.5o

Detailed Solution

\(\frac{3}{2}\)x + 3y + 45° = 180°

3x + 6y + 90° = 360°

3x + 6y = 270.......(i) x 2,
5x + y + y = 180°

5x + 2y = 180° ... (ii) x 6

6x + 12y = 540 ... (iii)

30x + 12y = 1080 ... (iv)

eqn(iv) - eqn(iii)

24x = 540

x = 22.5° and y = 33.75°
47.

In the diagram, if PS = SR and PQ||SQ, what is the size of PQR?

A. 25o

B. 50o

C. 55o

D. 65o

E. 75o

Detailed Solution

< PQR + , PSR 180(Opposite angles in a Quadrilateral are supplementary)

< PQR + 130o = 180

< PQR = 180 - 130

= 50o
48.

In the figure PT is a tangent to the circle with centre at O. If PQT = 30o, find the value of PTO

A. 30o

B. 50o

C. 24o

D. 12o

E. 60o

Detailed Solution

FROM the diagram, PQT = 50o

PTQ = 50o(opposite angles are supplementary)
41.

The figure FGHK is a rhombus. What is the value of angle X?

A. 30o

B. 90o

C. 150o

D. 120o

E. 60o

Detailed Solution

< HKF = 60o, < KFG = 120o

< KFG = < KHG = x(opposite angles)

x = 120o
42.

PQRS is a desk of dimensions 2m x 0.8 which is inclined at 30o to the horizontal. Find the inclination of the diagonal PR to the horizontal

A. 30o 35'

B. 30o 32'

C. 15o 36'

D. 10o

E. 10o 42'

Detailed Solution

tan\(\theta\) = \(\frac{0.8}{2}\) = (0.4)

\(\theta\) = tan-1(0.4)

From the diagram, the inclination of the diagonal PR to the horizontal is 10o 42'
43.

If O is the centre of the circle in the figure, find the value of x

A. 50

B. 260

C. 100

D. 65

E. 130

Detailed Solution

From the diagram; The value of x = 360o - 2(130o)

= 360 - 260

= 100o
44.

PQRS isa cyclic quadrilateral which PQ = PS. PT is a tangent to the circle and PQ makes an angle of 50o with the tangent as shown in the figure. What is the size of QRS?

A. 50o

B. 40o

C. 110v

D. 80o

E. 100o

Detailed Solution

< SPQ = 80o

< SPQ + < SRQ = 180(Supplementary)

80 +
< QRS = 180o - 80o

= 100o
45.

The farm yield of four crops on a piece of land in Ondo are represented on the pie chart. what is the angle of the sector occupies by Okro in the chart?

A. 11o

B. 19\(\frac{1}{3}\)o

C. 33\(\frac{1}{3}\)o

D. 91\(\frac{1}{3}\)o

E. 91o

Detailed Solution

Adding the values of all the items together, it gives 70

Okro sector = \(\frac{145}{270}\) x \(\frac{360^o}{1}\)

= 19.33o

= 19\(\frac{1}{3}\)o
46.

In the figure, PQR is a straight line. Find the values of x and y

A. x = 22.5o, y = 33.75o

B. x = 15o, y = 52.5o

C. x = 22.5o, y = 45.0o

D. x = 56.25o, y = 11.5o

Detailed Solution

\(\frac{3}{2}\)x + 3y + 45° = 180°

3x + 6y + 90° = 360°

3x + 6y = 270.......(i) x 2,
5x + y + y = 180°

5x + 2y = 180° ... (ii) x 6

6x + 12y = 540 ... (iii)

30x + 12y = 1080 ... (iv)

eqn(iv) - eqn(iii)

24x = 540

x = 22.5° and y = 33.75°
47.

In the diagram, if PS = SR and PQ||SQ, what is the size of PQR?

A. 25o

B. 50o

C. 55o

D. 65o

E. 75o

Detailed Solution

< PQR + , PSR 180(Opposite angles in a Quadrilateral are supplementary)

< PQR + 130o = 180

< PQR = 180 - 130

= 50o
48.

In the figure PT is a tangent to the circle with centre at O. If PQT = 30o, find the value of PTO

A. 30o

B. 50o

C. 24o

D. 12o

E. 60o

Detailed Solution

FROM the diagram, PQT = 50o

PTQ = 50o(opposite angles are supplementary)