2002 - JAMB Mathematics Past Questions and Answers - page 5
41
A
54 cm2
B
45 cm2
C
27 cm2
D
75 cm2
42
A
0.38
B
0.13
C
0.50
D
0.25
correct option: d
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
Users' Answers & Comments70 - 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
A
36o
B
18o
C
72o
D
24o
correct option: d
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
Users' Answers & CommentsIn 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
A
21/25πR2
B
9/25πR2
C
21/23πR2
D
5/9πR2
correct option: d
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 \(\frac{5R}{3}\) x \(\frac{R}{3}\)
= 5/9πR2
Users' Answers & Comments∴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 \(\frac{5R}{3}\) x \(\frac{R}{3}\)
= 5/9πR2
45
A
180 πcm3
B
162 πcm3
C
216 πcm3
D
198 πcm3
correct option: d
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
Users' Answers & Comments= 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
A
an obtuse-angled triangle
B
a scalene triangle
C
an isosceles triangle
D
an equilateral triangle
correct option: c
∠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
Users' Answers & Comments= 52o
∠QPR + 76 + 52 = 180(extr ∠ = sum of intr. opp ∠s)
∠QPR + 128 = 180
∠QPR = 180 - 128
= 52o
∴ΔPQR is an isosceles triangle
47
| No. of days | 1 | 2 | 3 | 4 | 5 | 6 |
| No. of students | 20 | x | 50 | 40 | 2x | 60 |
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
correct option: d
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
Users' Answers & Comments3X + 170 = 260
3X = 260 - 170
3x = 90
x = 30
Absent for at least 4 days
| 4 | 5 | 6 |
| 40 | 2x | 60 |
i.e 40 + 2x + 60 = 40 + (2 x 30) + 60
= 40 + 60 + 60
= 160
48
A
0.25
B
0.38
C
0.50
D
0.13
correct option: a
80 = (30 - x) + x + 40 - x + 20
80 = 90 - x, x = 90 - 80 = 10
N (music only) = 30 - x = 30 - 10 = 20
P (music only) = \(\frac{20}{80} = \frac{1}{4}\)
= 0.25
Users' Answers & Comments80 = 90 - x, x = 90 - 80 = 10
N (music only) = 30 - x = 30 - 10 = 20
P (music only) = \(\frac{20}{80} = \frac{1}{4}\)
= 0.25
49
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
correct option: b
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
Users' Answers & Comments\(\frac{y_2 - y_1}{x_2 - x_1} \geq m\)
\(\frac{y - 0}{x + 4} \geq \frac{1}{2}\)
2y \(\geq\) x + 4, -x + 2y \(\geq\) 4 = px + qy \(\geq\) 4
p = -1, q = 2
50
A
an isosceles triange
B
an obtuse angle
C
a scalene triangle
D
an equilateral triangle
correct option: a
< PRQ = 180o - 128o = 52o = < R
< P + < R + < Q = 180o
76o + 52o + < Q = 180o
128o + < Q = 180o
< Q = 180o - 128o
= 52o
\(\bigtriangleup\)PQR is an isosceles
Users' Answers & Comments< P + < R + < Q = 180o
76o + 52o + < Q = 180o
128o + < Q = 180o
< Q = 180o - 128o
= 52o
\(\bigtriangleup\)PQR is an isosceles