JAMB UTME - Mathematics - 1987

Akọwa Nkọwa

1m = 1.1 yard, length(L)= 50m
= (50 x 1.1)yards
= 55 yards
Area(A) = length(L) y width (W)
1815 = 55y width (W)
width (w) = $\frac{1815}{55}$
= 33 years

Akọwa Nkọwa

Akọwa Nkọwa

x + y = 5.......(i)
x2 - 2y2 = 1.......(ii)
x = 5 - y.........(iii)
Subst. for x in eqn.(ii) = (5 - y)2 - 2y2 = 1
25- 10y + y2 - 2y2 = 1
25 - 1 = y2 + 10y
y2 + 10y2 - 24 = 0
(y + 12)(y - 2) = 0
Then Either y + 12 = 0 or y - 2 = 0
= (12, -2)

Akọwa Nkọwa

Z = X = Y = 180o .........(i)

2Z + Y + Y = 190 = 2Z + 2Y = 180

Z + Y = 90o........(ii)

hence x + (z + y) = 180

x + 90o = 180o

x = 180o - 90o

= 90o

Akọwa Nkọwa

A rectangle, a square and a rhombus are the only parallelogram
∴ Probability of a parallelogram = $\frac{3}{5}$

Akọwa Nkọwa

From the diagram, WYZ = 60o, XYW = 180o - 60o
= 120o
LX = 30o
XWY = 180o - 120o + 30o
XWY = 30o
WXY = XYW = 30o
Side XY = YW
YW = 8m, sin 60o = $\frac{3}{2}$
? sin 60o = $\frac{h}{YW}$, sin 60o = $\frac{h}{8}$
h = 8 x $\frac{3}{2}$
= 4$\sqrt{3}$

Akọwa Nkọwa

Sum of interior angles of any polygon is (2n - 4) right angle; n angles of the Monagon = 9
Where 3 are equal and 6 other angles = 1110o
(2 x 9 - 4)90o = (18 - 4)90o
14 x 90o = 1260o
9 angles = 12600, 6 angles = 110o
Remaining 3 angles = 1260o - 1110o = 150o
Size of one of the 3 angles $\frac{150}{3}$ = 50o

Akọwa Nkọwa

x - 1 > 4(x + 2) = x - 1 > 4x + 8
4x + 8 < x - 1 = 4x - x < -1 -8

= 3x < -9

∴ x < -3

Akọwa Nkọwa

QR is a given line and P is a given point. The construction is the perpendicular from a given point P to a given line QR

Akọwa Nkọwa

62x + 1 + 7(6x) - 5
Let 6x = y....(i)
6yx + 7y - 5 = 0
(3y + 5)(2y - 1) = 0.....(i)
Subt. for y from eqn. (i) in eqn. (ii)
= [3(6x) + 5][2(6x) - 1]

Akọwa Nkọwa

$\frac{0.021741\times 1.2047}{0.023789}$ = $\frac{0.022\times 1.2}{0.024}$ (to 216)
= $\frac{0.0264}{0.024}$
= 1.1

Akọwa Nkọwa

Re-arranging them 2, 3, 5, 7, 8, 9, 12, 14
Mean = P and median = Q
P = $\frac{2+2+5+7+8+9+12+14}{8}$
$\frac{60}{8}$ = 7.5
median (Q) = $\frac{7+8}{2}$
$\frac{15}{2}$ = 7.5
P + Q = 7.5 + 7.5 = 15

Akọwa Nkọwa

x2 + 2(m + 1)x + m + 3 from (A + B)2 = a2 + 2AB + b2
2B = $\frac{2(m+1)}{2}$ = m + 1 .......(i)
Also B2 = m + 3..........(ii)
Subst. for B in equation ..........(iii)
= (m + 1)2 = m + 3 = m2 + 2m + 1 = m + 3
= m2 + 2m - m + 1 - 3 = 0
m2 + m - 2 = 0
(m + 2)(m - 1) = 0 or m - 1 = 0
m = -2 or 1

Akọwa Nkọwa

210o = 180o - 210o = 30o
From ratio of sides, sin -30o = -$\frac{1}{2}$
Cos 210o = 180o - 210o = -30o
= cos -30o = $\frac{-3}{2}$
But tan 30o = $\frac{1}{3}$, rationalizing this
= $\frac{1}{3}$ x $\frac{3}{3}$ = $\frac{3}{3}$
∴ = $\frac{-1}{2}$, $\frac{\sqrt{-3}}{2}$, $\frac{\sqrt{3}}{3}$

Akọwa Nkọwa

Z = x2 + $\frac{1}{y^3}$
Z - x2 = $\frac{1}{y^3}$
y3 = $\frac{1}{Z{x}^2}$
y = $3\sqrt{\frac{1}{2}{x}^2}$
∴ y = $3\sqrt{Z-x^2}$

Akọwa Nkọwa

Period of day light is from 5.30 to 7.00p.m - 13hrs 30min. interval
24hrs make 1 day
∴ period of darkness = 24hrs - 13hrs 30mins = 10hrs 30mins.
Angle of sector of daylight = $\frac{13\times 60+30\times 360}{(24\times 60)}$
= $\frac{810}{1440}$ x 360o
= $\frac{291600}{1440}$
= 202.5o
202.5o = 202.30o
darkness period = 360o - 202o 30'
= 157o30'
= 202o30' 157o 30'

Akọwa Nkọwa

(0.008) -$\frac{1}{3}$ x (0.16) - $\frac{3}{2}$ = (8 x 10-3)10-3 x (16 x 10-2)-$\frac{3}{2}$
= $\frac{\left(2^3\right)}{{10}^3}$ - 3 x $\frac{\left(2^4\right)}{10}$ - $\frac{-3}{2}$
= $\frac{625}{8}$

Akọwa Nkọwa

Base of pyramid of a square of side 8cm vertex directly above the centre edge = $4\sqrt{3}$cm
From the diagram, the diagonal of one base is AC2 = 82 + 82
Ac2 = 64 + 64 = 128
AC = $8\sqrt{2}$
but OC = $\frac{1}{2}$AC = 8$\sqrt{\frac{2}{2}}$ = $4\sqrt{2}$cm
OE = h = height
h2 = ($4\sqrt{3}$)2
16 x 2 - 16 x 2
48 - 32 = 16
h = $\sqrt{16}$
= 4

Akọwa Nkọwa

a = u2 - 3y2 = (1)2 - 3(-1) = -2
b = 2uu + v2 + v2
= 2(1)(-1) + (-1)2 = -1
∴ 2(2a - b)(a - b∴) = [2(-2) - 1](-2 - (-1)2)
= -3 - 3
= 9

Akọwa Nkọwa

Akọwa Nkọwa

ABCD is the floor, by pythagoras
AC2 = 144 + 81 = $\sqrt{225}$
AC = 15cm
Height of room is 8m, diagonal of floor is 15m
∴ The cosine of the angle which a diagonal of the room makes with the floor is EC2 = 152
cosine = $\frac{\text{adj}}{\text{hyp}}$ = $\frac{15}{17}$

Akọwa Nkọwa

$\frac{2\sqrt{14}\times 3\sqrt{21}}{7\sqrt{24}\times 2\sqrt{98}}$ = $\frac{6\sqrt{14}\times 3\times \sqrt{7}\times \sqrt{3}}{7\times 2\sqrt{6}\times \sqrt{7}\times \sqrt{14}}$
= $\frac{3\sqrt{3}}{14\sqrt{6}}$
= $\frac{3\sqrt{3}}{14\sqrt{2}\times \sqrt{3}}$
= $\frac{\sqrt{3}}{\sqrt{2}}$ x $\frac{\sqrt{2}}{\sqrt{2}}$
= $\frac{3\sqrt{2}}{28}$

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

CF (6) = 5 + 6 + 8 + 10 + 6 + 3
= 38

Akọwa Nkọwa

Total amount invested by A and B = ₦5,000
farm yield was sold for ₦15,000.00
profit = 15,000.00 - 5,000.00
= ₦10,000.00
Profit was shared in ratio 2 : 3
2 + 3 = 5
A received $\frac{2}{5}$ of profit = $\frac{2}{5}$ x 10,000 = ₦4,000.00
A receive $\frac{3}{5}$ of profit = $\frac{3}{5}$ x 10,000 = ₦6,000.00
Difference in profit received = ₦6,0000 - ₦4,000.00
= ₦2,000.00

Akọwa Nkọwa

1/2log10P
log10P1/2 = 1
P1/2 = 10
p = 102

Akọwa Nkọwa

Circumference of Latitude 0oS where R is the radius of the earth is = 2$\pi$ R cos $\theta$

Akọwa Nkọwa

Let x represents number of boys that can work at $\frac{5}{4}$ the rate at which the 10 girls work
For 1hr. x boys will work for $\frac{\frac{1}{5}}{4}$ x 10
x = $\frac{4}{5}$ x 10 = 8 boys
8 boys will do the work of ten girls at the same rate 4 + 8 = 12 boys cut the field in 5 hrs
For 3 hrs, $\frac{12\times 5}{3}$ boys will be needed = 20 boys

Akọwa Nkọwa

Akọwa Nkọwa

x2−y22x2+xy−y2 $\frac{x^2-y^2}{2x^2+xy-y^2}$

2x2+xy−y2=2x2−xy+2xy−y2 $2x^2+xy-y^2=2x^2-xy+2xy-y^2$

= x(2x−y)+y(2x−y) $x(2x-y)+y(2x-y)$

= (x+y)(2x−y) $(x+y)(2x-y)$

x2−y22x2+xy−y2=(x+y)(x−y)(x+y)(2x−y) $\frac{x^2-y^2}{2x^2+xy-y^2}=\frac{(x+y)(x-y)}{(x+y)(2x-y)}$

= x−y2x−y

Akọwa Nkọwa

profit 8% of ₦1.35 = $\frac{8}{100}$ x ₦1.35 = ₦0.08
Cost price = ₦1.35 - ₦0.10 = ₦1.25
If he sells the 20 oranges for ₦1.10 now
%loss = $\frac{\text{actual loss}}{\text{Cost price}}$ x 100
$\frac{125-1.10}{1.25}$ x 100
= $\frac{0.15\times 100}{1.25}$
= $\frac{15}{1.25}$
= 12%

Akọwa Nkọwa

Let x rep. number of goals
$\begin{array}{ccc}"f& "& frequency\\ x& f& fx\\ 0& 4& 0\\ 1& 3& 3\\ 2& 15& 30\\ 3& 16& 48\\ 4& 1& 4\\ 5& 0& 0\\ 6& 1& 6\end{array}$
$\sum fx$ = 91
Total number of goals scored is $\sum fx$= 91

Akọwa Nkọwa

Q = 1.5 + 0.5n gives the cost 1(in Naira) of feeding n people for a week. Extra cost of feeding one additional person = n = 1
Subt. for n in the formula Q = 1.5 + 0.5(1)
= 15 + 0.5(2)
Q = ₦2
= 200k

Akọwa Nkọwa

a = 4, r = $\frac{4}{2}$
$\frac{8}{4}$ = 2
n = 11
Tn = arn - 1
T11 = 4(2)11 - 1
4(2)10 = 212
since 4 = 22
= 212

Akọwa Nkọwa

The least length is $\frac{40}{48}$ = $\frac{5}{8}$
for the rod to be cut in exactly equal trips
Ratio $\frac{5}{6}$ : $\frac{48}{40}$
$\frac{\frac{5}{6}}{\frac{40}{48}}$ = 1
$\frac{5}{6}$ x $\frac{48}{40}$ = $\frac{240}{240}$ = 1
The least length = 240cm

Akọwa Nkọwa

Average speed from P to Q = 90km\h
Average speed from O to P = 45km/h
Average for the entire journey = 90 + 45
$\frac{135}{2}$ = 67.50 km/h

Akọwa Nkọwa

Tan $\theta$ = $\frac{m^2-n^2}{2mn}$
$\frac{\text{Opp}}{\text{Adj}}$ by pathagoras theorem
= Hyp2 = Opp2 + Adj2
Hyp2 = (m2 - n2) + (2mn)2
Hyp2 = m4 - 2m2n4 - 4m2 - n2
Hyp2 = m4 + 2m2 + n2n
Hyp2 = (m2 - n2)2
Hyp2 = $\frac{m^2+n^2}{2mn}$

Akọwa Nkọwa

2415 = 2 x 52 + 4 x 5 + 1 x 5 + 1 x 5o
50 + 20 + 1 = 7110
Convert 7110 to base 8
$\begin{array}{cc}8& 71\\ 8& 8R7\\ 8& 1R0\\ 8& 0R1\end{array}$
= 1078

Akọwa Nkọwa

In PNO, ONP
= 180 - (35 + 86)
= 180 - 121
= 59
PRQ = QNP = 59(angles in the same segment of a circle are equal)

Akọwa Nkọwa

Length and width of a square is 100%
Length increased by 20% and
Width decreased by 20% to form a rectangle
Length of rectangle = 120% to form a rectangle
Length of rectangle = 120% and
Width of rectangle = 80%
Area of rectangle = L x W
Area of square = W
Ratio of the area of the rectangle to the area of the square
A = $\frac{\text{Area of rectangle}}{\text{Area of square}}$
$\frac{120\times 30}{100\times 100}$ = $\frac{96}{100}$
= 24 : 25

Akọwa Nkọwa

h1 = $\frac{h_1+h_2}{1.5}$ = $\frac{h_1}{h_2}=2$

$\frac{A_{△PQR}}{A_{TP\left(QRST\right)}}=\frac{\frac{1}{2}\times 2\times h_1}{\frac{1}{2}\times h_2(2+3)}$

$\frac{2}{5}\times \frac{h_1}{h_2}=\frac{2}{5}\times 2$

= $\frac{4}{5}$

Akọwa Nkọwa

If P = $\frac{2}{3}$ ($\frac{1-r^2}{n^2}$), find n when r = $\frac{1}{3}$ and p = 1
p = $\frac{2(1-r^2)}{3n^2}$ when r = $\frac{1}{3}$ and p = 1
1 = $\frac{2}{3}$ $\frac{(1-(\frac{1}{3}{)}^2)}{n^2}$
n2 = $\frac{2(3-1)}{3\times 3}$
n2 = $\frac{2\times 2}{3\times 3}$
= $\frac{4}{9}$
n = $\frac{4}{9}$
= $\frac{2}{3}$

Akọwa Nkọwa

From the diagram, PQRS is a rectangle

Area of shaded part = 72sq.cm

But 72 = 3h - 4 + 6h - 4 + 4h

= 72 - 16h - 8

= 16h - 72 + 8

=16h = 80

h = 8016 $\frac{80}{16}$

= 5cm

Akọwa Nkọwa

log26 - log23 = log2 ($\frac{6}{3}$) = log22
log28 - 2$\frac{1}{2}$ log28 - log2$\frac{1}{2}$ log28 - log2$\frac{1}{4}$
$\frac{lo{g}_{2}2}{lo{g}_{2}32}=\frac{lo{g}_{2}2}{5lo{g}_{2}2}$
= $\frac{1}{5}$
N.B. log22 = 1

Akọwa Nkọwa

x2 + 33 = 52
x2 + 9 = 25
x2 = 25 – 9
x2 = 16

x = √ 16 
= 4 cm
∴ The locus is a circle of radius 4 cm with the center 0