JAMB UTME - Mathematics - 1983

Question 1 Report

One interior angle of a convex hexagon is 170o and each of the remaining interior angles is equal to xo. Find x

120o

110o

105o

102o

100o

Answer Details

An hexagon polygon is a six sided polygon
n = 6
sum of all interior angles of hexagon polygon will be (2n - 4) x 90o
= [2 x 6 - 4] x 90
= 720o
If one angle is 170o and each of the remaining five angles is 5xo
5xo + 170o = 720o
= 5x + 550
x =

\frac{550}{5}

= 110o

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Polynomials

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Question 2 Report

In the figure, PQR is a straight line. Find the values of x and y.

^{\circ}

x = 15

^{\circ}

, y = 52.5

^{\circ}

x = 22.5

^{\circ}

, y = 45.0

^{\circ}

x = 56.25

^{\circ}

, y = 11.5

^{\circ}

Answer Details

\frac{2}{3}

+ 3y + 45

^{\circ}

= 180

^{\circ}

3x + 6y + 90 $^{\circ}$ = 360 $^{\circ}$

3x + 67 = 270.......(i) x 2, 5x + y + y = 180 $^{\circ}$

5x + 2y = 180 $^{\circ}$ .......(ii) x 6

6 x 12y = 540..........(iii)

30x + 12y = 1080.........(iv)

egn(iv) - eqn(iii)

24x = 540

x = 22.5 and y = 33.75 $^{\circ}$

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Question 3 Report

A ship H leaves a port P and sails 30 km due south. Then it sails 50km due west. What is the bearing of H from P

20o 34'

243o 26'

116o 34'

63o 26'

240o 56'

Bearings

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Question 4 Report

Solve the simultaneous equations for x in x2 + y - 8 = 0, y + 5x - 2 = 0

-28, 7

6, -28

6, -1

-1, 7

3, 2

Answer Details

x2 + y - 8 = 0, y + 5x - 2 = 0
Rearranging, x2 + y = 8.....(i)
5x + y = 2.......(ii)
Subtract eqn(ii) from eqn(i)
x2 - 5x - 6 = 0
(x - 6)(x + 1) = 0
x = 6, -1

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Solution Of Linear Equations

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Question 5 Report

y varies partly as the square of x and partly as the inverse of the square root of x. Write down the expression for y if y = 2 when x = 1 and y = 6 when x = 4.

y = $\frac{10 x^{2}}{31} + \frac{52}{31 \sqrt{x}}$

y = x2 + $\frac{1}{\sqrt{x}}$

y = x2 + $\frac{1}{x}$

y = $\frac{x^{2}}{31} + \frac{1}{31 \sqrt{x}}$

Answer Details

y = kx2 + $\frac{c}{\sqrt{x}}$
y = 2 when x = 1
2 = k + $\frac{c}{1}$
k + c = 2
y = 6 when x = 4
6 = 16k + $\frac{c}{2}$
12 = 32k + c
k + c = 2
32k + c = 12
= 31k + 10
k = $\frac{10}{31}$
c = 2 - $\frac{10}{31}$
= $\frac{62 - 10}{31}$
= $\frac{52}{31}$
y = $\frac{10 x^{2}}{31} + \frac{52}{31 \sqrt{x}}$

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Algebraic Expressions

Quadratic Equations

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Question 6 Report

PQRS is a cyclic quadrilateral. If ∠QPS = 75o, what is the size of ∠QRS?

180o

150o

105o

75o

Circles

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Question 7 Report

If x is jointly proportional to the cube of y and the fourth power of z. In what ratio is x increased or decreased when y is halved and z is doubled?

4:1 increase

2:1 increase

1:1 no change

1:4 decrease

3:4 decrease

Variation

Progression

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Question 8 Report

PQR is the diagram of a semicircle RSP with centre at Q and radius of length 3.5cm. If QPT = 60^o. Find the perimeter of the figure PTRS. $ \pi = \frac{22}{7} $

25cm

18cm

36cm

29cm

25.5cm

Answer Details

Circumference of PRS =

\frac{π}{2}

=

\frac{22}{7}

x

\frac{7}{1}

x

\frac{1}{2}

= 11cm
Side PT = 7cm, Side TR = 7cm
Perimeter(PTRS) = 11cm + 7cm + 7cm
= 25cm

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Question 9 Report

If M represents the median and D the mode of the measurements 5, 9, 3, 5, 7, 5, 8, then (M, D) is

(6, 5)

(5, 8)

(5, 7)

(5, 5)

(7, 5)

Probability

Linear Inequalities

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Question 10 Report

The figure FGHK is a rhombus. What is the value of angle X?

30

^{\circ}

90

^{\circ}

150

^{\circ}

120

^{\circ}

60

^{\circ}

Answer Details

< HKF = 60

^{\circ}

, < KFG = 120

^{\circ}

< KFG = < KHG = x(opposite angles)

x = 120 $^{\circ}$

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Angles

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Question 11 Report

If O is the centre of the circle in the figure, find the value of x

50

260

100

65

130

Answer Details

From the diagram; The value of x = 360

^{\circ}

- 2(130

^{\circ}

)

= 360 - 260

= 100 $^{\circ}$

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Circles

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Question 12 Report

Factorize completely 81a4 - 16b4

(3a + 2b)(2a - 3b)(9a2 + 4b2)

(3a - 2b)(2a - 3b)(4a2 - 9b2)

(3a - 2b)(3a - 2b)(9a2 - 4b2)

(6a - 2b)(8a - 3b)(4a3 - 9b2)

Answer Details

81a4 - 16b4 = (9a2)2 - (4b2)2
= (9a2 + 4b2)(9a2 - 4b2)
N:B 9a2 + 4b2 = (3a - 2b)(3a - 2b)

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Algebraic Expressions

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Question 13 Report

PQRS is a desk of dimensions $2\text{m} \times 0.8$ which is inclined at $30^\circ$ to the horizontal. Find the inclination of the diagonal PR to the horizontal

30

^{\circ}

35'

30

^{\circ}

32'

15

^{\circ}

36'

10

^{\circ}

42'

Answer Details

tan

θ

=

\frac{0.8}{2}

= (0.4)

$θ$ = tan-1(0.4)

From the diagram, the inclination of the diagonal PR to the horizontal is 10 $^{\circ}$ 42'

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Measures Of Dispersion

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Question 14 Report

Simplify $ \frac{3}{2x-1} + \frac{2-x}{x-2} $

2−x

(2x−1)(x−2)

5−x

(2x−1)(x−2)

4−2x

2x−1

6−3x

(2x−1)(x−2)

Answer Details

\frac{3}{2 x - 1}

+

\frac{2 - x}{x - 2}

=

\frac{3}{2 x - 1}

-

\frac{x - 2}{x - 2}

=

\frac{3}{2 x - 1}

- 1
=

\frac{3 - (2 x - 1)}{2 x - 1}

=

\frac{3 - 2 x + 1}{2 x - 1}

=

\frac{4 - 2 x}{2 x - 1}

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Algebraic Expressions

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Question 15 Report

In a sample survey of a University Community, the following table shows the percentage distribution of the number of members per household:

No. of members per household	1	2	3	4	5	6	7	8	Total
No. of households	3	12	15	28	21	10	7	4	100

What is the median?

4

3

5

4.5

none

Answer Details

\begin{array}{cc} x & f \\ 1 & 3 \\ 2 & 13 \\ 3 & 15 \\ 4 & 28 \\ 5 & 21 \\ 6 & 10 \\ 7 & 7 \\ 8 & 4 \end{array}

Median is half of total frequency = 50th term
4 falls in the range = 4

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Question 16 Report

If $0.0000152 \times 0.042 = A \times 10^8$, where $1 \le A < 10$, find A and B

A = 9, B = 6.38

A = 6.38, B = -9

A = -6.38, B = -9

A = -9, B = -6.38

Answer Details

0.0000152 x 0.042 = A x 108

1 $\leq$ A < 10, it means values of A includes 1 - 9

0.0000152 = 1.52 x 10-5

0.00042 = 4.2 x 10-4

1.52 x 4.2 = 6.384

10-5 x 10-4

= 10-5-4

= 10-9

= 6.38 x 10-9

A = 6.38, B = -9

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Statistics

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Question 17 Report

Given that cos z = L , whrere z is an acute angle, find an expression for $ \frac{\cot z - \cosec z}{\sec z + \tan z} $

1-L

1+L

L²√3

1+L

1+L³

L²

L(L−1)

1−L+1√1−L²

Answer Details

Given Cos z = L, z is an acute angle
$\frac{cot z - cosec z}{sec z + tan z}$ = cos z
= $\frac{cos z}{sin z}$
cosec z = $\frac{1}{sin z}$
cot z - cosec z = $\frac{cos z}{sin z}$ - $\frac{1}{sin z}$
cot z - cosec z = $\frac{L - 1}{sin z}$
sec z = $\frac{1}{cos z}$
tan z = $\frac{sin z}{cos z}$
sec z = $\frac{1}{cos z}$ + $\frac{sin z}{cos z}$
= $\frac{1}{l}$ + $\frac{sin z}{L}$
the original eqn. becomes
$\frac{cot z - cosec z}{sec z + tan z}$ = $\frac{L - \frac{1}{sin z}}{1 + s i n \frac{z}{L}}$
= $\frac{L (L - 1)}{sin z (1 + sin z)}$
= $\frac{L (L - 1)}{sin z + 1 - c o s^{2} z}$
= sin z + 1
= 1 + $\sqrt{1 - L^{2}}$
= $\frac{L (L - 1)}{1 - L + 1 \sqrt{1 - L^{2}}}$

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Matrices And Determinants

Modular Arithmetic

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Question 18 Report

PQRS is a cyclic quadrilateral which PQ = PS. PT is a tangent to the circle and PQ makes an angle of $50^\circ$ with the tangent as shown in the figure. What is the size of QRS?

50

^{\circ}

40

^{\circ}

110v

80

^{\circ}

^{\circ}

Answer Details

< SPQ = 80

^{\circ}

< SPQ + < SRQ = 180(Supplementary)

80 + < QRS = 180 $^{\circ}$ - 80 $^{\circ}$

= 100 $^{\circ}$

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Circles

Indices, Logarithms And Surds

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Question 19 Report

Find the angle of the sectors representing each item in pie chart of the following data 6, 10, 14, 16, 26

15o, 25o, 35o, 40o, 65o

60o, 100o, 140o, 160o, 260o

6o, 10o, 14o, 16o, 26o

30o, 50o, 70o, 80o, 130o

none of the above

Answer Details

6 + 10 + 14 + 16 + 26 = 72

\frac{6}{72}

x 360
= 30o
Similarly others give 30o, 50o, 70o, 80o and 130o respectively

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Polynomials

Measures Of Dispersion

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Question 20 Report

Correct each of the numbers 59.81798 and 0.0746829 to three significant figures and multiply them, giving your answer to three significant figures

4.46

4.48

4.47

4.49

4.50

Answer Details

59.81798 = 59.8 (3 s.f)
0.0746829 = 0.0747
59.8 x 0.0747 = 4.46706
= 4.48 (3s.f)

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Fractions, Decimals And Approximations

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Question 21 Report

In a class of 60 pupils, the statistical distribution of the numbers of pupils offering Biology, History, French, Geography and Additional mathematics is as shown in the pie chart. How many pupils offer Additional Mathematics?

15

10

18

12

20

Answer Details

2x - 24)

^{\circ}

+ (3x - 18)

^{\circ}

+ (2x + 12)

^{\circ}

+ (x + 12)

^{\circ}

+ x

^{\circ}

= 360

^{\circ}

9x = 360 $^{\circ}$ + 18 $^{\circ}$

x = $\frac{378}{9}$

= 42 $^{\circ}$ , if x = 42 $^{\circ}$ , then add maths = $\frac{2 x - 42}{360}$ x 60

= $\frac{2 \times 42 - 24}{360}$ x 60

= $\frac{84 - 24}{6}$

= 10

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Statistics

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Question 22 Report

If x + 2 and x - 1 are factors of the expression 1x3 + 2kx2 + 24, find the values of 1 and K

1 = -6, k = -9

1 = -2, k = 1

k = -1, 1 = -2

1 = 0, k = 1

k = 0,1 = 6

Answer Details

f(x) = Lx3 + 2kx2 + 24
f(-2) = -8L + 8k = -24
4L - 4k = 12
f(1):L + 2k = -24
L - 4k = 3
3k = -27
k = -9
1 = -6

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Algebraic Expressions

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Question 23 Report

Without using tables find the numerical value of $ \log_{7} 49 + \log_{7}\left(\frac{1}{7}\right) $

1

2

3

7

9

Answer Details

log749 + log7

\frac{1}{7}

= log77
= 1

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Positive And Negative Integers, Rational Numbers

Logarithms

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Question 24 Report

If a rod of length 250cm is measured as 255cm long in error, what is the percentage error in the measurement?

55

10

5

4

2

Answer Details

% error =

\frac{Actual error}{real value}

x 100
=

\frac{5}{250}

x 100
= 2

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Percentages

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Question 25 Report

The lengths of the sides of a right angled triangle are (3x + 1)cm, (3x - 1)cm and xcm. Find x

2

6

18

12

5

Answer Details

(3x + 1)2 = (3x - 1)2 + 2 (Pythagoras's theorem)
9x2 + 6x + 1 = 9x - 6x2 + 1 + x2
x2 - 12x = 0
x(x - 12) = 0
x = 0 or 12

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Algebraic Expressions

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Question 26 Report

In the figure PT is a tangent to the circle with centre at O. If PQT = $30^\circ$, find the value of PTO

30

^{\circ}

50

^{\circ}

24

^{\circ}

12

^{\circ}

60

^{\circ}

Answer Details

FROM the diagram, PQT = 50

^{\circ}

PTQ = 50 $^{\circ}$ (opposite angles are supplementary)

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Circles

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Question 27 Report

If w varies inversely as V and U varies directly as w3, Find the relationship between u and v given that u = 1, when v = 2

u =

\frac{8}{v^{3}}

v =

\frac{8}{u^{2} v^{3}}

u = 8v3

v = 8u2

Answer Details

W

α

\frac{1}{v}

u

α

w3
w =

\frac{k_{1}}{v}

u = k2w3
u = k2(

\frac{k_{1}}{v}

)3
=

\frac{k_{2} k_{1}^{2}}{v^{3}}

k = k2k1k2
u =

\frac{k}{v^{3}}

k = uv3
= (1)(2)3
= 8
u =

\frac{8}{v^{3}}

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Variation

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Question 28 Report

On a square paper of length 2.524375cm is inscribed square diagram of length 0.524375cm. Find the area of the paper not covered by the diagram. correct to 3 significant figures.

6.00cm2

6.10cm2

6cmv

6.09cm2

4.00cm2

Answer Details

Area of the paper = area of square = L x B or S2
where s = S x k
Area of the paper = (2.524)2
area of the diagram = (0.524)2
area not covered = (2.524)2 - (0.524)2
= 6.370576 - 0.274576
= 6.096
= 6.10cm2 (2 s.f)

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Coordinate Geometry

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Question 29 Report

Simplify $ \frac{3\sqrt{27a^{-9}}}{8} $

9a²

2a³

3

2a³

2

3a²

2

Answer Details

\frac{3 \sqrt{27 a^{- 9}}}{8}

=

\frac{3 a^{- 3}}{2}

=

\frac{3}{2}

x

\frac{1}{a^{3}}

=

\frac{3}{2 a^{3}}

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Fractions, Decimals And Approximations

Surds (radicals)

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Question 30 Report

In a figure, PQR = 60o, PRS = 90o, RPS = 45o, QR = 8cm. Determine PS

2

\sqrt{3}

cm

4

\sqrt{6}

cm

2

\sqrt{6}

cm

8

\sqrt{6}

cm

8cm

Answer Details

From the diagram, sin 60o =

\frac{P R}{8}

PR = 8 sin 60 =

\frac{8 \sqrt{3}}{2}

= 4

\sqrt{3}

Cos 45o =

\frac{P R}{P S}

=

\frac{4 \sqrt{3}}{P S}

PS Cos45o = 4

\sqrt{3}

PS = 4

\sqrt{3}

x 2
= 4

\sqrt{6}

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Circles

Angles Of Elevation And Depression

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Angles

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Question 31 Report

The farm yield of four crops on a piece of land in Ondo are represented on the pie chart. what is the angle of the sector occupies by Okro in the chart?

11

^{\circ}

19

\frac{1}{3}

^{\circ}

33

\frac{1}{3}

^{\circ}

91

\frac{1}{3}

^{\circ}

91

^{\circ}

Answer Details

Adding the values of all the items together, it gives 70

Okro sector = $\frac{145}{270}$ x $\frac{360^{o}}{1}$

= 19.33 $^{\circ}$

= 19 $\frac{1}{3}$ $^{\circ}$

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Angles

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Question 32 Report

In a triangle PQT, QR = $ \sqrt{3}\text{ cm} $, PR = 3cm, PQ = $ 2\sqrt{3} $cm and PQR = 30^o. Find angles P and R

P = 60o and R = 90o

P = 30o and R = 120o

P = 90o and R = 60o

P = 60o and R 60o

P = 45o and R = 105o

Answer Details

By using cosine formula, p2 = Q2 + R2 - 2QR cos p
Cos P =

\frac{Q^{2} + R^{2} - p^{2}}{2 Q R}

=

\frac{(3)^{2} + 2 (\sqrt{3})^{2} - 3^{2}}{2 \sqrt{3}}

=

\frac{3 + 12 - 9}{12}

=

\frac{6}{12}

=

\frac{1}{2}

= 0.5
Cos P = 0.5
p = cos-1 0.5 = 60

^{\circ}

= < P = 60

^{\circ}

If < P = 60

^{\circ}

and < Q = 30

< R = 180 $^{\circ}$ - 90 $^{\circ}$
angle P = 60 $^{\circ}$ and angle R is 90 $^{\circ}$

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Question 33 Report

Find the missing value in the following table:

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$y=x^3-x+3$		$3$	$3$	$3$	$9$	$27$

-3

3

-9

13

9

Answer Details

When x = -2, y = x3 - x + 3
= -23 - (-2) + 3
= -8 + 2 + 3
= -3

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Question 34 Report

Find x if (xbase4)2 = 100100base2

6

12

100

210

110

Answer Details

(x2)4 to base10 gives (1 x 25) + (1 x 25) + (1 x 2)
32 + 4 = 36
x2 = 36, x = 6
610 to base 4 =

\begin{array}{cc} 4 & 6 \\ 1 r 2 \end{array}

= 124

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Number Bases

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Question 35 Report

Solve the following equations 4x - 3 = 3x + y = x - y = 3, 3x + y = 2y + 5x - 12

x = 5, y = 2

x = 2, y = 5

x = 5, y = -2

x = -2, y = -5

x = -5, y = -2

Answer Details

4x - 3 = 3x + y = x - y = 3.......(i)
3x + y = 2y + 5x - 12.........(ii)
eqn(ii) + eqn(i)
3x = 15
x = 5
substitute for x in equation (i)
5 - y = 3
y = 2

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Solution Of Linear Equations

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Question 36 Report

In the figure, find PRQ

66

\frac{1}{2}

^{\circ}

^{\circ}

125

^{\circ}

105

^{\circ}

65

^{\circ}

Answer Details

Angle subtended at any part of the circumference of the circle

\frac{125^{o}}{2}

at centre = 360

^{\circ}

- 235

^{\circ}

= 125

^{\circ}

$\bar{P Q R}$ = $\frac{125}{2}$

= 62 $\frac{1}{2}$ $^{\circ}$

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Representation Of Data

Triangles And Polygons

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Question 37 Report

The value of $ (0.303)^3 - (0.02)^3 $ is

0.019

0.0019

0.00019

0.000019

0.0278

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Question 38 Report

Make T the subject of the equation $ \frac{av}{1-v} = \sqrt{\frac{2v+T}{a+2T}} $

T = $\frac{3 a v}{1 - v}$

T = $\frac{1 + v}{2 a^{2} v^{3}}$

T = $\frac{2 v (1 - v)^{3} - a^{4} v^{3}}{2 a^{3} v^{3} + (1 - v)^{2}}$

\frac{2 v (1 - v)^{3} - a^{4} v^{3}}{2 a^{3} v^{3} - (1 - v)^{3}}

Answer Details

$\frac{a v}{1 - v}$ = $\sqrt{\frac{2 v + T}{a + 2 T}}$
$\frac{(a v)^{3}}{(1 - v)^{3}}$ = $\frac{2 v + T}{a + 2 T}$
$\frac{a^{3} v^{3}}{(1^{3} - v)^{3}}$ = $\frac{2 v + T}{a + 2 T}$
= $\frac{2 v (1 - v)^{3} - a^{4} v^{3}}{2 a^{3} v^{3} - (1 - v)^{3}}$

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Variation

Change Of Subject Of A Formula/Relation

Solution Of Linear Equations

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Question 39 Report

The letters of the word MATRICULATION are cut and put into a box. One letter is drawn at random from the box. Find the probability of drawing a vowel

7

13

5

13

6

13

8

13

4

13

Answer Details

Vowels of letters are 6 in numbers
prob. of vowel =

\frac{6}{13}

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Probability

Surds (radicals)

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Question 40 Report

Simplify $ \log_{10} a^{\frac{1}{3}} + \frac{1}{4}\log_{10} a - \frac{1}{12}\log_{10} a^{7} $

1

\frac{7}{6}

log10 a

zero

10

Answer Details

log10 a

\frac{1}{3}

+

\frac{1}{4}

log10 a -

\frac{1}{12}

log10a7 = log10 a

\frac{1}{3}

+ log10

\frac{1}{4}

- log10 a

\frac{7}{12}

= log10 a

\frac{7}{12}

- log10 a

\frac{7}{12}

= log10 1 = 0

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Logarithms

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Question 41 Report

A man drove for 4 hours at a certain speed, he then doubled his speed and drove for another 3 hours. Although he covered 600 kilometers. At what speed did he drive for the last 3 hours?

120km/hr

60km/hr

670km/hr

40km/hr

Answer Details

Speed =

\frac{d i s t a n c e}{t i m e}

let x represent the speed, d represent distance
x =

\frac{d}{4}

d = 4x
2x =

\frac{600 - d}{3}

6x = 600 - d
6x = 600 - 4x
10x = 600
x =

\frac{600}{10}

= 60km/hr

View Answer

Question 42 Report

Given a regular hexagon, calculate each interior angle of the hexagon

60o

30o

120o

45o

135o

Answer Details

Sum of interior angles of polygon = (2n - 4)rt < s

sum of interior angles of an hexagon
(2 x 6 - 4) x 90o = (12 - 4) x 90o
= 8 x 90o
= 720o
each interior angle will have $\frac{720^{o}}{6}$
= 120o

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Angles

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Question 43 Report

If $ \left(\frac{2}{3}\right)^m \left(\frac{3}{4}\right)^n = \frac{252}{729} $, find the values of m and n

m = 4, n = 2

m = -4, n = -2

m = -4, n =2

m = 4, n = -2

m = -2, n = 4

Answer Details

(

\frac{2}{3}

)m (

\frac{3}{4}

)n =

\frac{252}{729}

\frac{2^{m}}{4^{n}}

x

\frac{3^{n}}{3^{m}}

= 283-6
2m - 2n x 3n - m = 28 x 3-6
m - 2n = 8........(i)
-m + n = -6........(ii)
Solving the equations simultaneously
m = 4, n = -2

Read lesson note on Number Bases (WAEC)

Number Bases

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Question 44 Report

The scores of set of final year students in the first semester examination in a paper are 41, 29, 55, 21, 47, 70, 70, 40, 43, 56, 73, 23, 50, 50. Find the median of the scores.

47

50

48

\frac{1}{2}

48

49

Answer Details

By re-arranging 21, 23, 29, 40, 41, 43| 47, 50| 50, 55, 56, 70, 70, 73
The median =

\frac{47 + 50}{2}

\frac{97}{2}

= 48

\frac{1}{2}

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Question 45 Report

The scores of 16 students in a mathematics test are 65, 65, 55, 60, 60, 65, 60, 70, 75, 70, 65, 70, 60, 65, 65,
70. What is the sum of the median and modal scores?

125

130

140

150

Measures Of Dispersion

Statistics

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Question 46 Report

A construction company is owned by two partners X and Y and it is agreed that their profit will be divided in the ratio 4:5. At the end of the year, Y received ₦5,000 more than X. What is the total profit of the company for the year?

₦20, 000

₦25,000

₦30,000

₦15,000

₦45, 000

Answer Details

Total sharing ratio is 9
X has 4, Y has 4 + 1
If 1 is ₦5000
Total profit = 5000 x 9
= ₦45,000

Read lesson note on Sine, Cosine And Tangent Of An Angle (WAEC)

Read lesson note on Ratio, Proportions And Rates (WAEC)

Sine, Cosine And Tangent Of An Angle

Ratio, Proportions And Rates

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