JAMB UTME - Mathematics - 1983

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?

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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 }$

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

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

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$\frac{3}{2x-1}$ + $\frac{2-x}{x-2}$
= $\frac{3}{2x-1}$ - $\frac{x-2}{x-2}$
= $\frac{3}{2x-1}$ - 1
= $\frac{3-(2x-1)}{2x-1}$
= $\frac{3-2x+1}{2x-1}$
= $\frac{4-2x}{2x-1}$

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

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

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

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< SPQ = 80${}^{?}$

< SPQ + < SRQ = 180(Supplementary)

80 + < QRS = 180${}^{?}$ - 80${}^{?}$

= 100${}^{?}$

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

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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

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

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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

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

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

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

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From the diagram; The value of x = 360${}^{\circ }$ - 2(130${}^{\circ }$)

= 360 - 260

= 100${}^{\circ }$

Find the mean of the following 24.57, 25.63, 24.32, 26.01, 25.77

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$\frac{24.57+25.63+24.32+26.01+25.77}{5}$
mean = $\frac{126.3}{5}$
= 25.26

In the figure, find PRQ

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Angle subtended at any part of the circumference of the circle $\frac{{125}^o}{2}$ at centre = 360${}^{?}$ - 235${}^{?}$ = 125${}^{?}$

$\overline{PQR}$ = $\frac{125}{2}$

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

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

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($\frac{2}{3}$)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

Which of the following equations represents the graph above?

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x = -2 or x = $\frac{1}{4}$

(x + 2)(4x - 1) = 0

y = 2 - 7x - 4x2

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

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(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

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.

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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$\frac{1}{2}$

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.

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

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

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W $\alpha$ $\frac{1}{v}$u $\alpha$ 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 = $\frac{8}{v^3}$

Simplify log10 a$\frac{1}{3}$ + $\frac{1}{4}$log10 a - $\frac{1}{12}$log10a7

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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

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.

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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)

Find the missing value in the following table:

$\begin{array}{ccccccc}x& -2& -1& 0& 1& 2& 3\\ y=x^3-x+3& & 3& 3& 3& 9& 27\end{array}$

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

Factorize completely 81a4 - 16b4

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

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

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$\frac{2}{3}$ + 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${}^{\circ }$

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

$\begin{array}{cccccccccc}\text{No. of members per household}& 1& 2& 3& 4& 5& 6& 7& 8& \text{Total}\\ \text{No. of households}& 3& 12& 15& 28& 21& 10& 7& 4& 100\end{array}$

What is the median?

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$\begin{array}{cc}x& f\\ 1& 3\\ 2& 13\\ 3& 15\\ 4& 28\\ 5& 21\\ 6& 10\\ 7& 7\\ 8& 4\end{array}$
Median is half of total frequency = 50th term
4 falls in the range = 4

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

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Vowels of letters are 6 in numbers
prob. of vowel = $\frac{6}{13}$

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

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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

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?

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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 = $\frac{2x-42}{360}$ x 60

= $\frac{2\times 42-24}{360}$ x 60

= $\frac{84-24}{6}$

= 10

If 0.0000152 x 0.042 = A x 108, where 1 ≤ $\le$ A < 10, find A and B

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0.0000152 x 0.042 = A x 108

1 ≤ $\le$ 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

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

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

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

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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

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

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By using cosine formula, p2 = Q2 + R2 - 2QR cos p
Cos P = $\frac{Q^2+R^2-p^2}{2QR}$
= $\frac{(3{)}^2+2(\sqrt{3}{)}^2-3^2}{2\sqrt{3}}$
= $\frac{3+12-9}{12}$
= $\frac{6}{12}$
= $\frac{1}{2}$
= 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${}^{\circ }$

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

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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?

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Speed = $\frac{distance}{time}$
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

Find x if (xbase4)2 = 100100base2

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(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 = $\begin{array}{cc}4& 6\\ 1r2& \end{array}$
= 124

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

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6 + 10 + 14 + 16 + 26 = 72
$\frac{6}{72}$ x 360
= 30o
Similarly others give 30o, 50o, 70o, 80o and 130o respectively

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

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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

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?

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Without using tables find the numerical value of log749 + log7($\frac{1}{7}$)

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log749 + log7$\frac{1}{7}$
= log77
= 1

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

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< HKF = 60${}^{?}$, < KFG = 120${}^{?}$

< KFG = < KHG = x(opposite angles)

x = 120${}^{?}$

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?

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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?

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

Simplify $\frac{3\sqrt{27a^{-9}}}{8}$

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$\frac{3\sqrt{27a^{-9}}}{8}$ = $\frac{3a^{-3}}{2}$
= $\frac{3}{2}$ x $\frac{1}{a^3}$
= $\frac{3}{2a^3}$

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

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% error = $\frac{\text{Actual error}}{\text{real value}}$ x 100
= $\frac{5}{250}$ x 100
= 2

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

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FROM the diagram, PQT = 50${}^{\circ }$

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

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

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Circumference of PRS = $\frac{\pi }{2}$ = $\frac{22}{7}$ x $\frac{7}{1}$ x $\frac{1}{2}$
= 11cm
Side PT = 7cm, Side TR = 7cm
Perimeter(PTRS) = 11cm + 7cm + 7cm
= 25cm

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

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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

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

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From the diagram, sin 60o = $\frac{PR}{8}$
PR = 8 sin 60 = $\frac{8\sqrt{3}}{2}$
= 4$\sqrt{3}$
Cos 45o = $\frac{PR}{PS}$ = $\frac{4\sqrt{3}}{PS}$
PS Cos45o = 4$\sqrt{3}$
PS = 4$\sqrt{3}$ x 2
= 4$\sqrt{6}$

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

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tan$\theta$ = $\frac{0.8}{2}$ = (0.4)

$\theta$ = tan-1(0.4)

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