JAMB UTME - Mathematics - 1987
Tambaya 1 Rahoto
If \( \tan \theta = \frac{m^2-n^2}{2mn} \) find \( \sec \theta \)
Bayanin Amsa
OppAdj 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 = m2+n22mn
Tambaya 2 Rahoto
Bayanin Amsa
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
Tambaya 3 Rahoto
Bayanin Amsa
Tambaya 4 Rahoto
Bayanin Amsa
From ratio of sides, sin -30o = -12
Cos 210o = 180o - 210o = -30o
= cos -30o = −32
But tan 30o = 13 , rationalizing this
= 13 x 33 = 33
∴ = −12 , √−32 , √33
Tambaya 5 Rahoto
An \( (n-2)^2 \) sided figure has n diagonals, find the number n of diagonals for a 25 sided figure
Bayanin Amsa
Tambaya 6 Rahoto
Simplify without using tables \( \frac{2\sqrt{14}\times3\sqrt{21}}{7\sqrt{24}\times2\sqrt{98}} \)
Bayanin Amsa
= 3√314√6
= 3√314√2×√3
= √3√2 x √2√2
= 3√228
Tambaya 7 Rahoto
Bayanin Amsa
b = 2uu + v2 + v2
= 2(1)(-1) + (-1)2 = -1
∴ 2(2a - b)(a - b∴) = [2(-2) - 1](-2 - (-1)2)
= -3 - 3
= 9
Tambaya 8 Rahoto
Bayanin Amsa
24hrs make 1 day
∴ period of darkness = 24hrs - 13hrs 30mins = 10hrs 30mins.
Angle of sector of daylight = 13×60+30×360(24×60)
= 8101440 x 360o
= 2916001440
= 202.5o
202.5o = 202.30o
darkness period = 360o - 202o 30'
= 157o30'
= 202o30' 157o 30'
Tambaya 9 Rahoto
Bayanin Amsa
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)
Tambaya 10 Rahoto
Bayanin Amsa
4x + 8 < x - 1 = 4x - x < -1 -8
= 3x < -9
∴ x < -3
Tambaya 11 Rahoto
In the diagram, PQRS is a circle center O. PQR is a diameter and ∠PRQ = 40o. Calculate ∠QSR.
Bayanin Amsa
Tambaya 12 Rahoto
Bayanin Amsa
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 25 of profit = 25 x 10,000 = ₦4,000.00
A receive 35 of profit = 35 x 10,000 = ₦6,000.00
Difference in profit received = ₦6,0000 - ₦4,000.00
= ₦2,000.00
Tambaya 13 Rahoto
Evaluate without using tables \( (0.008)^{\frac{1}{3}} \times (0.16)^{-\frac{3}{2}} \)
Bayanin Amsa
= (23)103 - 3 x (24)10 - −32
= 6258
Tambaya 14 Rahoto
Bayanin Amsa
2B = 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
Tambaya 15 Rahoto
Bayanin Amsa
From the diagram, the diagonal of one base is AC2 = 82 + 82
Ac2 = 64 + 64 = 128
AC = 8√2
but OC = 12 AC = 8√22 = 4√2 cm
OE = h = height
h2 = (4√3 )2
16 x 2 - 16 x 2
48 - 32 = 16
h = √16
= 4
Tambaya 16 Rahoto
Bayanin Amsa
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]
Tambaya 17 Rahoto
Bayanin Amsa
for the rod to be cut in exactly equal trips
Ratio 56 : 4840
564048 = 1
56 x 4840 = 240240 = 1
The least length = 240cm
Tambaya 18 Rahoto
Four boys and ten girls can cut a field in 5 hours if the boys work at \( \frac{5}{4} \) the rate at which the girls work. How many boys will be needed to cut the field in 3 hours?
Bayanin Amsa
For 1hr. x boys will work for 154 x 10
x = 45 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, 12×53 boys will be needed = 20 boys
Tambaya 19 Rahoto
Bayanin Amsa
AC2 = 144 + 81 = √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 = adjhyp = 1517
Tambaya 20 Rahoto
Bayanin Amsa
= (50 x 1.1)yards
= 55 yards
Area(A) = length(L) y width (W)
1815 = 55y width (W)
width (w) = 181555
= 33 years
Tambaya 21 Rahoto
Bayanin Amsa
84 = 2
n = 11
Tn = arn - 1
T11 = 4(2)11 - 1
4(2)10 = 212
since 4 = 22
= 212
Tambaya 22 Rahoto
Convert 241 in base 5 to base 8,
Bayanin Amsa
2415 = 2 x 52 + 4 x 5 + 1 x 5 + 1 x 5o
50 + 20 + 1 = 7110
Convert 7110 to base 8
87188R781R080R1
= 1078
Tambaya 23 Rahoto
Bayanin Amsa
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 1503 = 50o
Tambaya 24 Rahoto
Simplify \( \frac{x^2-y^2}{2x^2+xy-y^2} \)
Bayanin Amsa
x2−y22x2+xy−y2
2x2+xy−y2=2x2−xy+2xy−y2
= x(2x−y)+y(2x−y)
= (x+y)(2x−y)
x2−y22x2+xy−y2=(x+y)(x−y)(x+y)(2x−y)
= x−y2x−y
Tambaya 25 Rahoto
Bayanin Amsa
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 = Area of rectangleArea of square
120×30100×100 = 96100
= 24 : 25
Tambaya 26 Rahoto
Bayanin Amsa
= 180 - (35 + 86)
= 180 - 121
= 59
PRQ = QNP = 59(angles in the same segment of a circle are equal)
Tambaya 27 Rahoto
The figure is an example of the construction of a
Bayanin Amsa
Tambaya 28 Rahoto
Bayanin Amsa
Subt. for n in the formula Q = 1.5 + 0.5(1)
= 15 + 0.5(2)
Q = ₦2
= 200k
Tambaya 29 Rahoto
Bayanin Amsa
Mean = P and median = Q
P = 2+2+5+7+8+9+12+148
608 = 7.5
median (Q) = 7+82
152 = 7.5
P + Q = 7.5 + 7.5 = 15
Tambaya 30 Rahoto
In the diagram above, QR // TS, QR : TS = 2.3, Find the ratio of the area of triangle PQR to the area of the trapezium QRST.
Bayanin Amsa
A△PQRATP(QRST)=12×2×h112×h2(2+3)
25×h1h2=25×2
= 45
Tambaya 31 Rahoto
Bayanin Amsa
and v KR2 .......(ii)
(where k is constant)
Subst. for v in equation (i) = p = 12KR .....(ii)
when r = 7, p = 2
2 = k72
k = 2 x 49
= 98
Subt. foe k in ....(iii)
P = 98R2
PR2 = 98
Tambaya 33 Rahoto
In the figure, find the value of x
Bayanin Amsa
2Z + Y + Y = 190 = 2Z + 2Y = 180
Z + Y = 90o........(ii)
hence x + (z + y) = 180
x + 90o = 180o
x = 180o - 90o
= 90o
Tambaya 34 Rahoto
Bayanin Amsa
Tambaya 35 Rahoto
Make \(y\) the subject of the formula \(Z = x^2 + \frac{1}{y^3}\)
Bayanin Amsa
Z - x2 = 1y3
y3 = 1Zx2
y = 3√12x2
∴ y = 3√Z−x2
Tambaya 36 Rahoto
Bayanin Amsa
log10P1/2 = 1
P1/2 = 10
p = 102
Tambaya 37 Rahoto
If \( P = \frac{2}{3}\left(\frac{1-r^2}{n^2}\right) \), find n when r = \( \frac{1}{3} \) and p = 1
Bayanin Amsa
p = 2(1−r2)3n2 when r = 13 and p = 1
1 = 23 (1−(13)2)n2
n2 = 2(3−1)3×3
n2 = 2×23×3
= 49
n = 49
= 23
Tambaya 40 Rahoto
Bayanin Amsa
= 120o
LX = 30o
XWY = 180o - 120o + 30o
XWY = 30o
WXY = XYW = 30o
Side XY = YW
YW = 8m, sin 60o = 32
? sin 60o = hYW , sin 60o = h8
h = 8 x 32
= 4?3
Tambaya 41 Rahoto
Bayanin Amsa
Cost price = ₦1.35 - ₦0.10 = ₦1.25
If he sells the 20 oranges for ₦1.10 now
%loss = actual lossCost price x 100
125−1.101.25 x 100
= 0.15×1001.25
= 151.25
= 12%
Tambaya 42 Rahoto
Reduce each number to two significant figures and then evaluate \( \frac{0.021741 \times 1.2047}{0.023789} \)
Bayanin Amsa
= 0.02640.024
= 1.1
Tambaya 43 Rahoto
Bayanin Amsa
Average speed from O to P = 45km/h
Average for the entire journey = 90 + 45
1352 = 67.50 km/h
Tambaya 44 Rahoto
Simplify without using tables \( \frac{\log_2 6}{\log_2 8} - \frac{\log_2 3}{\log_2 \frac{1}{2}} \)
Bayanin Amsa
log28 - 212 log28 - log212 log28 - log214
log22log232=log225log22
= 15
N.B. log22 = 1
Tambaya 45 Rahoto
The minimum value of \(y\) in the equation \(y=x^2-6x+8\)
Tambaya 46 Rahoto
The goals scored by 40 football teams from three league divisions are recorded below:
| Number of goals | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| frequency | 4 | 3 | 15 | 16 | 1 | 0 | 1 |
What is the total number of goals scored by all the terms?
Bayanin Amsa
"f"frequencyxffx0401332153031648414500616
∑fx = 91
Total number of goals scored is ∑fx = 91
