2010 - JAMB Mathematics Past Questions and Answers - page 5
41
If \(\left[\begin{array}{cc}
x & 3\\ 2 & 7\end{array}\right]\)= 15 find the value of x
x & 3\\ 2 & 7\end{array}\right]\)= 15 find the value of x
A
3
B
4
C
5
D
2
correct option: a
\(\left[\begin{array}{cc}
x & 3\\ 2 & 7\end{array}\right]\)= 15
x x 7 - 2 x 3 = 15
7x - 6 = 15
7x = 15+6
7x = 21
x = 21/7
x = 3
Users' Answers & Commentsx & 3\\ 2 & 7\end{array}\right]\)= 15
x x 7 - 2 x 3 = 15
7x - 6 = 15
7x = 15+6
7x = 21
x = 21/7
x = 3
42
From the diagram above, find x
A
65o
B
50o
C
75o
D
55o
correct option: a
∠RTS = 90o (∠ in semi circle)
∠UTS = 25o (∠ in alternate segment)
X + 90 + 25 = 180(sum of ∠s on a str line)
X + 115 = 180
X = 180 - 115
X = 65o
Users' Answers & Comments∠UTS = 25o (∠ in alternate segment)
X + 90 + 25 = 180(sum of ∠s on a str line)
X + 115 = 180
X = 180 - 115
X = 65o
43
From the cyclic quadrilateral TUVW above, find the value of x
A
26o
B
24o
C
20o
D
23o
correct option: b
TUVW is a cyclic quad
3χ + 20 + 88 = 180 (opp ∠ s of a cyclic quad are supplementary)
3χ + 108 = 180
3χ = 180 - 108
3χ = 72
χ = 72/3
χ = 24o
Users' Answers & Comments3χ + 20 + 88 = 180 (opp ∠ s of a cyclic quad are supplementary)
3χ + 108 = 180
3χ = 180 - 108
3χ = 72
χ = 72/3
χ = 24o
44
If two smaller sides of a right angled triangle are 4cm and 5cm, find its area
A
10 cm2
B
6 cm2
C
24 cm2
D
8 cm2
correct option: a
Area of a triangle = 1/2 base x height
= 1/2 x 4 x 5
= 10 cm2
Users' Answers & Comments= 1/2 x 4 x 5
= 10 cm2
45
An arc subtends an angle of 50o at the center of circle of radius 6cm. Calculate the area of the sector formed
A
100/7
B
80/7
C
110/7
D
90/7
correct option: c
| Area of a sector = | θ | x πr2 |
| 360 |
= 50/360 x 22/7 x 36/1
= 110/7
46
What is the locus of point that is equidistant from points P(1,3) and Q(3,5)?
A
y = -χ + 6
B
y = -χ - 6
C
y = χ + 6
D
y = χ - 6
correct option: a
Locus of a point equidistant from two fixed points is the perpendicular bisector of the straight line joining the points Gradients of PQ = (5-3 / 3-1) = 2/2 = 1
Gradient bisector = -1
Equation of perpendicular bisector
y - y1 = m(χ - χ1
y - 4 = -χ + 2
y = -χ + 6
Users' Answers & CommentsGradient bisector = -1
Equation of perpendicular bisector
y - y1 = m(χ - χ1
y - 4 = -χ + 2
y = -χ + 6
47
If the area of ΔPQR above is 12√3 cm2, find the value of q?
A
6 cm
B
7 cm
C
8 cm
D
5 cm
correct option: a
Area of a triangle = 1/2 ab Sinθ
12√3 = 1/2 x 8 x q sin 60
12√3 = 4 x q x √3/2
12√3 = 2q√3
q = 6
Users' Answers & Comments12√3 = 1/2 x 8 x q sin 60
12√3 = 4 x q x √3/2
12√3 = 2q√3
| q = | 12√3 |
| 2√3 |
q = 6
48
If y = (2x + 1)3 find dy/dx
A
3(2x+1)2
B
6(2x+1)
C
3(2x+1)
D
6(2x+1)2
correct option: d
y = (2x + 1)3
dy/dx = 3(2x + 1)3-1 x 2
= 3(2x + 1)2 x 2
= 6(2x + 1)2
Users' Answers & Commentsdy/dx = 3(2x + 1)3-1 x 2
= 3(2x + 1)2 x 2
= 6(2x + 1)2
49
Find ∫(sin x + 2)dx
A
cos x + x2 + K
B
cos x 2x + K
C
-cos x + x2 + K
D
-cos x + 2x + K
50
| Marks | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| No. of students | 3 | 1 | 5 | 2 | 4 | 2 | 3 |
from the table above, if the pass mark is 5, how many students failed the test?
A
6
B
2
C
7
D
9