2017 - JAMB Mathematics Past Questions and Answers - page 4
31
In how many ways can the word MATHEMATICS be arranged?
A
\(\frac{11!}{9!2!}\)
B
\(\frac{11!}{9!2!2!}\)
C
\(\frac{11!}{2!2!2!}\)
D
\(\frac{11!}{2!2!}\)
correct option: c
MATHEMATICS is an eleven letter word = 11!
There are 2Ms and 2As and 2Es
Divide the number of repeating letters
= \(\frac{11!}{2!2!2!}\)
Users' Answers & CommentsThere are 2Ms and 2As and 2Es
Divide the number of repeating letters
= \(\frac{11!}{2!2!2!}\)
32
In how many ways can the word MACICITA be arranged?
A
\(\frac{8!}{2!}\)
B
\(\frac{8!}{3! 2!}\)
C
\(\frac{8!}{2! 2! 2!}\)
D
8!
correct option: c
MACICITA is an eight letter word = 8!
Since we have repeating letters, we have to divide to remove duplicates accordingly. There are 2A, 2C, 2I
∴ \(\frac{8!}{2! 2! 2!}\)
Users' Answers & CommentsSince we have repeating letters, we have to divide to remove duplicates accordingly. There are 2A, 2C, 2I
∴ \(\frac{8!}{2! 2! 2!}\)
33
y is inversely proportional to x and y and 6 when x = 7. Find the constant of the variation
A
47
B
42
C
54
D
46
correct option: b
Y ∝ \(\frac{1}{2}\)
Y = 6, X = 7
Y = \(\frac{k}{x}\) where k is constant
6 = \(\frac{k}{7}\)
k = 42
Users' Answers & CommentsY = 6, X = 7
Y = \(\frac{k}{x}\) where k is constant
6 = \(\frac{k}{7}\)
k = 42
34
In the diagram MN, PQ and RS are parallel lines. What is the value of the angle marked X?
A
123o
B
170o
C
117o
D
137o
correct option: c
MN || PQ || RS
MN = PQ = RS (parallel lines)
Label the angle in the lines
a = i (corresponding angles are equal)
b = x (corresponding angles are equal)
If |MN| = |RS|
If a = i
and a = 63 = i
a + b = 180 (Adjacent interior angles are supplementary i.e add to 180)
∴ i + x = 180
63 + x = 180
x = 180 - 63
x = 1170
Users' Answers & CommentsMN = PQ = RS (parallel lines)
Label the angle in the lines
a = i (corresponding angles are equal)
b = x (corresponding angles are equal)
If |MN| = |RS|
If a = i
and a = 63 = i
a + b = 180 (Adjacent interior angles are supplementary i.e add to 180)
∴ i + x = 180
63 + x = 180
x = 180 - 63
x = 1170
35
Find the equation of the locus of a point p (x, y) such that pv = pw, where v= (1, 1) and w = (3, 5)
A
2x + 2y = 9
B
2x + 3y = 8
C
2x + y = 9
D
x + 2y = 8
correct option: d
The locus of a point p(x, y) such that pv = pw where v = (1, 1)
and w = (3, 5). This means that the point p moves so that its distance from v and w are equidistance
\(\sqrt{(x − x_1)^2 + (y − y_1)^2}\) = \(\sqrt{(x − x_2)^2 + (y − y_2)^2}\)
\(\sqrt{(x -1)^2 + (y - 1)^2}\) = \(\sqrt{(x - 3)^2 + (y - 5)^2}\)
square both sides
(x - 1)2 + (y - 1)2 = (x - 3)2 + (y - 5)2
x2 - 2x + 1 + y2 - 2y + 1 = x2 - 6x + 9 + y2 - 10y + 25
x2 + y2 -2x -2y + 2 = x2 + y2 - 6x - 10y + 34
Collecting like terms
x2 - x2 + y2 - y2 - 2x + 6x -2y + 10y = 34 - 2
4x + 8y = 32
Divide through by 4
x + 2y = 8
Users' Answers & Commentsand w = (3, 5). This means that the point p moves so that its distance from v and w are equidistance
\(\sqrt{(x − x_1)^2 + (y − y_1)^2}\) = \(\sqrt{(x − x_2)^2 + (y − y_2)^2}\)
\(\sqrt{(x -1)^2 + (y - 1)^2}\) = \(\sqrt{(x - 3)^2 + (y - 5)^2}\)
square both sides
(x - 1)2 + (y - 1)2 = (x - 3)2 + (y - 5)2
x2 - 2x + 1 + y2 - 2y + 1 = x2 - 6x + 9 + y2 - 10y + 25
x2 + y2 -2x -2y + 2 = x2 + y2 - 6x - 10y + 34
Collecting like terms
x2 - x2 + y2 - y2 - 2x + 6x -2y + 10y = 34 - 2
4x + 8y = 32
Divide through by 4
x + 2y = 8
36
Find ∫(x2 + 3x − 5)dx
A
\(\frac{x_3}{3}\) - \(\frac{3x_2}{2}\) - 5x + k
B
\(\frac{x_3}{3}\) - \(\frac{3x_2}{2}\) + 5x + k
C
\(\frac{x_3}{3}\) + \(\frac{3x_2}{2}\) - 5x + k
D
\(\frac{x_3}{3}\) + \(\frac{3x_2}{2}\) + 5x + k
correct option: c
∫xndx = \(\frac{x_{n + 1}}{n + 1}\)
∫dx = x + k
where k is constant
∫(x2 + 3x − 5)dx
∫x2 dx + ∫3xdx − ∫5dx
\(\frac{2_{2 + 1}}{2 + 1}\) + \(\frac{3x^{1 + 1}}{1 + 1}\) − 5x + k
\(\frac{x_3}{3}\) + \(\frac{3x_2}{2}\) − 5x + k
Users' Answers & Comments∫dx = x + k
where k is constant
∫(x2 + 3x − 5)dx
∫x2 dx + ∫3xdx − ∫5dx
\(\frac{2_{2 + 1}}{2 + 1}\) + \(\frac{3x^{1 + 1}}{1 + 1}\) − 5x + k
\(\frac{x_3}{3}\) + \(\frac{3x_2}{2}\) − 5x + k
37
In the diagram below MN is a chord of a circle KMN centre O and radius 10cm. If
A
10cm
B
18cm
C
17cm
D
12cm
correct option: a
Find the diagram
Sin 70o
x = 10 Sin 70o
= 9.3969
Hence, length of chord MN = 2x
= 2 × 9.3969
= 18.79
= 19cm (nearest cm)
Users' Answers & CommentsSin 70o
x = 10 Sin 70o
= 9.3969
Hence, length of chord MN = 2x
= 2 × 9.3969
= 18.79
= 19cm (nearest cm)
38
If m * n = [mn − nm] for m, n belong to R, evaluate − 3 * 4
A
3
B
4
C
5
D
6
correct option: c
m * n = \(\frac{m}{n}\) - \(\frac{m}{n}\)
m = − 3
n = 4
∴ − 3 × 4 = \(\frac{-3}{4}\) - \(\frac{-4}{-3}\)
= \(\frac{3(−3) − (− 4 × 4)}{12}\)
= \(\frac{− 9 + 16}{12}\)
= \(\frac{7}{12}\)
Users' Answers & Commentsm = − 3
n = 4
∴ − 3 × 4 = \(\frac{-3}{4}\) - \(\frac{-4}{-3}\)
= \(\frac{3(−3) − (− 4 × 4)}{12}\)
= \(\frac{− 9 + 16}{12}\)
= \(\frac{7}{12}\)
39
Factorize completely x2 + 12xy + y + 3x + 3y - 18
A
(x + y + 6)(x + y -3)
B
(x - y - 6)(x - y + 3)
C
(x - y + 6)(x - y - 3)
D
(x + y - 6)(x + y + 3)
correct option: a
x + 2xy + y + 3x + 3y - 18
x + 2xy + 3x + y + 3y -18
x + 2xy - 3x + 6x + y -3y + 6y -18
x + 2xy -3x + y -3y + 6x + 6y -18
x + xy -3x + xy + y - 3y + 6x + 6y -18
x(x + y - 3) + y(x + y - 3) + 6(x + y - 3)
= (x + y - 3)(x + y + 6)
= (x + y + 6)(x + y -3)
Users' Answers & Commentsx + 2xy + 3x + y + 3y -18
x + 2xy - 3x + 6x + y -3y + 6y -18
x + 2xy -3x + y -3y + 6x + 6y -18
x + xy -3x + xy + y - 3y + 6x + 6y -18
x(x + y - 3) + y(x + y - 3) + 6(x + y - 3)
= (x + y - 3)(x + y + 6)
= (x + y + 6)(x + y -3)
40
Make S the subject of the relation
p = s + \(\frac{sm^2}{nr}\)
p = s + \(\frac{sm^2}{nr}\)
A
s = \(\frac{nrp}{nr + m^2}\)
B
s = nr + \(\frac{m^2}{mrp}\)
C
s = \(\frac{nrp}{mr}\) + m2
D
s = \(\frac{nrp}{nr}\) + m2
correct option: d
p = s + \(\frac{sm^2}{nr}\)
p = s + ( 1 + \(\frac{m^2}{nr}\))
p = s (1 + \(\frac{nr + m^2}{nr}\))
nr × p = s (nr + m2)
s = \(\frac{nrp}{nr + m^2}\)
Users' Answers & Commentsp = s + ( 1 + \(\frac{m^2}{nr}\))
p = s (1 + \(\frac{nr + m^2}{nr}\))
nr × p = s (nr + m2)
s = \(\frac{nrp}{nr + m^2}\)