1997 - JAMB Mathematics Past Questions and Answers - page 3
21
Determine x + y if \(\begin{pmatrix} 2 & -3 \ -1 & 4 \end{pmatrix}\) \(\begin{pmatrix} x \ y \end{pmatrix}\) = \(\begin{pmatrix}-1 \ 8 \end{pmatrix}\)
A
3
B
4
C
7
D
12
correct option: c
Users' Answers & Comments22
Find the non-zero positive value of x which satisfies the equation \(\begin{vmatrix} x & 1 & 0 \ 1 & x & x \ 0 & 1 & x\end{vmatrix}\) = 0
A
2
B
4
C
√3
D
√2
23
Each of the base angles of a isosceles triangle is 59o and the verticles of the triangle lie on a circle. Determine rhe angle which the base of the triangle subtends at the centre of the circle.
A
128o
B
16o
C
64o
D
58o
correct option: a
Users' Answers & Comments24
A chord of a circle of a diameter 42cm subtends an angle of 60o at the centre of the circle. Find the length of the mirror arc
A
22cm
B
44cm
C
110cm
D
220cm
correct option: a
Users' Answers & Comments25
An arc of a circle subtends an angle 70o at the centre. If the radius of the circle is 6cm, calculate the area of the sector subtended by the given angle.(\(\pi\) = \(\frac{22}{7}\))
A
22cm2
B
44cm2
C
66cm2
D
88cm2
correct option: a
Users' Answers & Comments26
A cone with the sector angle of 45o is cut out of a circle of radius of the cone.
A
\(\frac{r}{16}\) cm
B
\(\frac{r}{6}\) cm
C
\(\frac{r}{8}\) cm
D
\(\frac{r}{2}\) cm
correct option: c
Users' Answers & Comments27
The angle between the positive horizontal axis and a given line is 135o. Find the equation of the line if it passes through the point (2,3)
A
x - y = 1
B
x + y = 1
C
x + y = 5
D
x - y = 5
correct option: c
m = tan 135o = -tan 45o = -1
\(\frac{y - y_1}{x - x_1}\) = m
\(\frac{y - 3}{x - 2}\) = -1
= y - 3 = -(x - 2)
= -x + 2
x + y = 5
Users' Answers & Comments\(\frac{y - y_1}{x - x_1}\) = m
\(\frac{y - 3}{x - 2}\) = -1
= y - 3 = -(x - 2)
= -x + 2
x + y = 5
28
A point P moves so that is equidistant from point L and M. If LM is 6cm, find the distance of P from LM when P is 10cm from L
A
12cm
B
10cm
C
8cm
D
6cm
29
Find the distance between the point Q (4,3) and the point common to the lines 2x - y = 4 and x + y = 2
A
3√10
B
3√5
C
√26
D
√13
correct option: d
2x - y .....(i)
x + y.....(ii)
from (i) y = 2x - 4
from (ii) y = -x + 2
2x - 4 = -x + 2
x = 2
y = -x + 2
= -2 + 2
= 0
x1 = 21
y4 = 01
x2 = 41
y2 = 3
Hence, dist. = \(\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}\)
= \(\sqrt{(3 - 0)^2}\) + (4 - 2)2
= \(\sqrt{3^2 + 2^2}\)
= \(\sqrt{13}\)
Users' Answers & Commentsx + y.....(ii)
from (i) y = 2x - 4
from (ii) y = -x + 2
2x - 4 = -x + 2
x = 2
y = -x + 2
= -2 + 2
= 0
x1 = 21
y4 = 01
x2 = 41
y2 = 3
Hence, dist. = \(\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}\)
= \(\sqrt{(3 - 0)^2}\) + (4 - 2)2
= \(\sqrt{3^2 + 2^2}\)
= \(\sqrt{13}\)
30
The angle of elevation of a building from a measuring instrument placed on the ground is 30o. If the building is 40m high, how far is the instrument from the foot of the building?
A
\(\frac{20}{√3}\)m
B
\(\frac{40}{√3}\)m
C
20√3m
D
40√3m
correct option: d
\(\frac{40}{x}\) = tan 30o
x = \(\frac{40}{tan 36}\)
= \(\frac{40}{1\sqrt{3}}\)
= 40√3m
Users' Answers & Commentsx = \(\frac{40}{tan 36}\)
= \(\frac{40}{1\sqrt{3}}\)
= 40√3m