Year : 
2000
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

41 - 45 of 45 Questions

# Question Ans
41.

In the diagram above, EFGH is a circle center O. Fh is a diameter and GE is a chord which meets FH at right angle at the point N. If NH = 8cm and EG = 24cm, calculate FH.

A. 16 cm

B. 20 cm

C. 26 cm

D. 32 cm

Detailed Solution

The locus of X is the perpendicular bisector of PQ
42.

In the diagram, If < RPS = 50o, < RPQ = 30o and pq = QR, Find the value of < prs

A. 50o

B. 60o

C. 70o

D. 80o

Detailed Solution

< QPR = < PRQ = 30o (PQ = QR)

< SPQ + < QRS = 180o (supplementary)

80 + < QRS = 180o = < QRS = 180o - 80o = 100o

< QRP + < PRS = 100o = 30o + < PRS = 100o

= < PRS = 100o - 30o = 70o
43.

In the diagram, EFGH is a circle centre O. FH is a diameter and GE is a chord which meets FH at right angle at the point N. If NH = 8cm and EG = 24cm, calculate FH

A. 32cm

B. 26cm

C. 20cm

D. 16cm

Detailed Solution

\(\bigtriangleup\)OEN = \(\bigtriangleup\)OGN

OE = OG = r(radii)

(r - 8)2 + 122 = r2

r2 - 16r + 64 + 144 = r2

16r = 64 + 144 = r2

16r = 64 + 144 = 208

\(\frac{208}{16}\) = 13cm

FH = 2FO = 2OH = 2R

= 2 x 13cm

= 26cm
44.

If the diagram is the graph of y = x2, the shaded area is

A. 64 square units

B. \(\frac{126}{4}\) square units

C. 32 square units

D. \(\frac{64}{3}\) square units

Detailed Solution

A = \(\int^{x = 4}_{x = 0}\) ydx = \(\int^{4}_{0}\) x2dx = [\(\frac{x^3}{3} + c]^{4}_{0}\)

= \(\frac{64}{3}\) square units
45.

The cumulative frequency curve represents the ages of ages of students in a school. What age group do 70% of the students belongs?

A. 17.5 - 20.5

B. 16.5 - 19.5

C. 15.5 - 19.5

D. 15.5 - 18.5

A

41.

In the diagram above, EFGH is a circle center O. Fh is a diameter and GE is a chord which meets FH at right angle at the point N. If NH = 8cm and EG = 24cm, calculate FH.

A. 16 cm

B. 20 cm

C. 26 cm

D. 32 cm

Detailed Solution

The locus of X is the perpendicular bisector of PQ
42.

In the diagram, If < RPS = 50o, < RPQ = 30o and pq = QR, Find the value of < prs

A. 50o

B. 60o

C. 70o

D. 80o

Detailed Solution

< QPR = < PRQ = 30o (PQ = QR)

< SPQ + < QRS = 180o (supplementary)

80 + < QRS = 180o = < QRS = 180o - 80o = 100o

< QRP + < PRS = 100o = 30o + < PRS = 100o

= < PRS = 100o - 30o = 70o
43.

In the diagram, EFGH is a circle centre O. FH is a diameter and GE is a chord which meets FH at right angle at the point N. If NH = 8cm and EG = 24cm, calculate FH

A. 32cm

B. 26cm

C. 20cm

D. 16cm

Detailed Solution

\(\bigtriangleup\)OEN = \(\bigtriangleup\)OGN

OE = OG = r(radii)

(r - 8)2 + 122 = r2

r2 - 16r + 64 + 144 = r2

16r = 64 + 144 = r2

16r = 64 + 144 = 208

\(\frac{208}{16}\) = 13cm

FH = 2FO = 2OH = 2R

= 2 x 13cm

= 26cm
44.

If the diagram is the graph of y = x2, the shaded area is

A. 64 square units

B. \(\frac{126}{4}\) square units

C. 32 square units

D. \(\frac{64}{3}\) square units

Detailed Solution

A = \(\int^{x = 4}_{x = 0}\) ydx = \(\int^{4}_{0}\) x2dx = [\(\frac{x^3}{3} + c]^{4}_{0}\)

= \(\frac{64}{3}\) square units
45.

The cumulative frequency curve represents the ages of ages of students in a school. What age group do 70% of the students belongs?

A. 17.5 - 20.5

B. 16.5 - 19.5

C. 15.5 - 19.5

D. 15.5 - 18.5

A