JAMB UTME - Mathematics - 1983

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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

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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

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

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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$\sqrt{6}$

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

< KFG = < KHG = x(opposite angles)

x = 120${}^{?}$

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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}$${}^{?}$

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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

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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

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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

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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

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

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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

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

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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

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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${}^{\circ }$

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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

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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

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

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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

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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${}^{\circ }$

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

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

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

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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

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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$\frac{1}{2}$

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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)

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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

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

= 360 - 260

= 100${}^{\circ }$

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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 + $\sqrt{1-L^2}$
= $\frac{L(L-1)}{1-L+1\sqrt{1-L^2}}$

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

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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

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x = -2 or x = $\frac{1}{4}$

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

y = 2 - 7x - 4x2

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

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

< SPQ + < SRQ = 180(Supplementary)

80 + < QRS = 180${}^{?}$ - 80${}^{?}$

= 100${}^{?}$

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

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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)

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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 }$

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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

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

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

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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

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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'

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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 = $\frac{8}{v^3}$

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