2001 - JAMB Mathematics Past Questions and Answers - page 2
11
Solve the equations
m2 + n2 = 29
m + n = 7
m2 + n2 = 29
m + n = 7
A
(2, 3) and ( 3, 5)
B
(2, 5) and (5, 2)
C
(5, 2) and ( 5, 3)
D
(5, 3) and (3, 5)
correct option: b
m2 + n2 = 29 .......(1)
m + n = 7 ............(2)
From (2),
m = 7 - n
but m2 + n2 = 29, substituting;
(7-n)2 + n2 = 29
49 - 14n + n2 + n2 = 29
=> 2n2 -14n + 20 = 0
Thus n2 -7n + 10 = 0
Factorizing;
(n-5)(n-2) = 0
n - 5 = 0, => n = 5
n - 2 = 0, => n = 2.
When n = 5,
m + n = 7, => m = 2,
When n = 2,
m + n = 7, => m = 5.
Thus (m,n) = (5,2) and (2,5)
Users' Answers & Commentsm + n = 7 ............(2)
From (2),
m = 7 - n
but m2 + n2 = 29, substituting;
(7-n)2 + n2 = 29
49 - 14n + n2 + n2 = 29
=> 2n2 -14n + 20 = 0
Thus n2 -7n + 10 = 0
Factorizing;
(n-5)(n-2) = 0
n - 5 = 0, => n = 5
n - 2 = 0, => n = 2.
When n = 5,
m + n = 7, => m = 2,
When n = 2,
m + n = 7, => m = 5.
Thus (m,n) = (5,2) and (2,5)
12
An operation * is defined on the set of real numbers by a*b = a + b + 1. If the identity elements is -1, find the inverse of the element 2 under *.
A
4
B
zero
C
-2
D
-4
correct option: d
By definition a*b = a + b + 1.
Let the inverse of the element 2 be x,
Therefore 2*x = -1
i.e. 2 + x + 1 = -1
3 + x = -1
x = -1 - 3
x = -4
Users' Answers & CommentsLet the inverse of the element 2 be x,
Therefore 2*x = -1
i.e. 2 + x + 1 = -1
3 + x = -1
x = -1 - 3
x = -4
13
The sixth term of an A.P is half of its twelfth term. The first term of the A.P is equal to
A
zero
B
half of the common difference
C
double the common difference
D
the common difference
correct option: d
1st statement: U6 = 1/2(U12)
a + (n -1)d = 1/2[a + (n-1)d]
a + 5d = a + 11d
2(a + 5d) = a + 11d
2a + 10d = a + 11d
Solving, => a = d
Hence the first term is equal to the common difference
Users' Answers & Commentsa + (n -1)d = 1/2[a + (n-1)d]
a + 5d = a + 11d
2(a + 5d) = a + 11d
2a + 10d = a + 11d
Solving, => a = d
Hence the first term is equal to the common difference
14
Factorize 4x2 - 9y2 + 20x + 25
A
(2x -3y + 5)(2x - 3y - 5)
B
(2x - 3y)(2x + 3y)
C
(2x - 3y +5)(2x + 3y + 5)
D
(2x + 5)(2x - 9y +5)
correct option: c
Given: 4x2 - 9y2 + 20x + 25
Collect like terms: 4x2 + 20x + 25 - 9y2
(2x + 5)(2x + 5) - 9y2
(2x + 5)2 - (3y)2
(2x - 3y +5)(2x + 3y + 5)
Users' Answers & CommentsCollect like terms: 4x2 + 20x + 25 - 9y2
(2x + 5)(2x + 5) - 9y2
(2x + 5)2 - (3y)2
(2x - 3y +5)(2x + 3y + 5)
15
a sector of a circle of radius 7.2cm which subtends an angle of 300° at the centre is used to form a cone. What is the radius of the base of the cone?
A
8cm
B
6cm
C
9cm
D
7cm
correct option: b
(r/L) = (θ/360°)
Given θ = 300, and L = 7.2cm,
=> r = (300 x 7.2)/360
r = 6cm
Users' Answers & CommentsGiven θ = 300, and L = 7.2cm,
=> r = (300 x 7.2)/360
r = 6cm
16
A point P moves such that it is equidistant from Points Q and R. Find QR when PR = 8cm and angle PRQ = 30°
A
4√3cm
B
8cm
C
8√3cm
D
4cm
correct option: c
Hint: Make a sketch of the moving points such that the hypotenuse is 8cm and the adjacent x cm.
Cos 30 = x/8
x = 8 cos 30 = 8√3cm
Users' Answers & CommentsCos 30 = x/8
x = 8 cos 30 = 8√3cm
17
A straight line makes an angle of 30° with the positive x-axis and cuts the y-axis at y = 5. Find the equation of the straight line.
A
y = (x/10) + 5
B
y = x + 5
C
√3y = - x + 5√3
D
√3y = x + 5√3
correct option: d
Cos 30 = 5/x
x cos 30 = 5, => x = 5√3
Coordinates of P = -5, 3, 0
Coordinates of Q = 0, 5
Gradient of PQ = (y2 - y1) (X2 - X1) = (5 - 0)/(0 -5√3)
= 5/5√3 = 1/√3
Equation of PQ = y - y1 = m (x -x1)
y - 0 = 1/√3 (x -(-5√3))
Thus: √3y = x + 5√3
Users' Answers & Commentsx cos 30 = 5, => x = 5√3
Coordinates of P = -5, 3, 0
Coordinates of Q = 0, 5
Gradient of PQ = (y2 - y1) (X2 - X1) = (5 - 0)/(0 -5√3)
= 5/5√3 = 1/√3
Equation of PQ = y - y1 = m (x -x1)
y - 0 = 1/√3 (x -(-5√3))
Thus: √3y = x + 5√3
18
Given an isosceles triangle with length of 2 equal sides t units and opposite side √3t units with angle θ. Find the value of the angle θ opposite to the √3t units.
A
100°
B
120°
C
30°
D
60°
correct option: b
Cos θ° = t2 + t2 -(√3t)2 2 x t x t
= 2t2 - 3t2 2t2
= -t2/2t2
= -1/2
Thus θ = cos-1 (-0.5) = 120°
Users' Answers & Comments= 2t2 - 3t2 2t2
= -t2/2t2
= -1/2
Thus θ = cos-1 (-0.5) = 120°
19
Find the value of P if the line joining (P, 4) and (6, -2) is perpendicular to the line joining (2, P) and (-1, 3).
A
4
B
6
C
3
D
zero
correct option: d
Hint: If two lines a perpendicular, the gradient of one is equal to minus the reciprocal of the other (P, 4) and (6, -2)
Users' Answers & Comments20
Find the number of sides of a regular polygon whose interior angle is twice the exterior angle.
A
6
B
2
C
3
D
8
correct option: a
Let the ext. angle = x
Thus int. angle = 2x
But sum of int + ext = 180 (angle of a straight line).
2x + x = 180
3x = 180
x = 180/3 = 60
Each ext angle = 360/n
=> 60 = 360/n
n = 360/60 = 6
Users' Answers & CommentsThus int. angle = 2x
But sum of int + ext = 180 (angle of a straight line).
2x + x = 180
3x = 180
x = 180/3 = 60
Each ext angle = 360/n
=> 60 = 360/n
n = 360/60 = 6