JAMB UTME - Mathematics - 1988
Tambaya 1 Rahoto
Bayanin Amsa
If there are 10 blue balls, 18 of x = 10
x = 10 x 8 = 80 balls
Green balls will be 58 x 80 = 50 and black balls = 28 x 80 = 20
Ratio = Green : black : blue
50 : 20 : 10
-10 : 10 : -
------------------
New Ratio 40 : 10 : 10
4 : 1 : 1
Tambaya 2 Rahoto
Find \(m\) such that \((m + \sqrt{3})(1 - \sqrt{3})^2 = 6 - 2\sqrt{2}\)
Bayanin Amsa
(m + √3 )(4 - 2√3 ) = 6 - 2√2
= 6 - 2√3
4m - 6 + 4 - 2m√3 = 6 - 2√3
comparing co-efficients,
4m - 6 = 6.......(i)
4 - 2m = -2.......(ii)
in both equations, m = 3
Tambaya 3 Rahoto
In the figure, PS = 7cm and RY = 9cm. IF the area of parallelogram PQRS is 56cm2. Find the area of trapezium PQTS
Bayanin Amsa
Area of parallelogram PQRS = 56cm
56 = base x height, where base = 7
7 x h = 56cm,
h = 567
= 8cm
Area of trapezium 12 (sum of two sides)x height where two sides are QT and PS but QT = QR + RY + YT = 7 +9 + 7 = 23cm
Area of trapezium PQTS = 12 (23 + 7) x 8
12 x 30 x 8 = 120cmsq
Tambaya 5 Rahoto
Bayanin Amsa
(ii) and (iii) are right
Tambaya 6 Rahoto
If (IPO3)4 = 11510 find P
Bayanin Amsa
1 x 43 + P x 42 + 0 x 4 + 3 = 11510
16p + 67 = 115 p = 4816
= 3
Tambaya 7 Rahoto
For what values of x is the curve \(y = \frac{x^2 + 3}{x + 4}\) decreasing?
Bayanin Amsa
Tambaya 8 Rahoto
In the figure, PS = RS = QS and QRS = 50o. Find QPR
Bayanin Amsa
In angle SQR, angle S = 50O
In angle QRP, 65 + 65 = 130O
Since RQP = angle RPQ = 180−1302
= 502=25o
QPR = 25O
Tambaya 9 Rahoto
A tax player is allowed \( \frac{1}{8} \)th of his income tax-free, and pays 20% on the remainder. If he pays ₦490.00 tax, what is his income?
Bayanin Amsa
He pays tax on 1 - 78
= 17
th of his income
20% is 490, 100% is 100020
x 490, ₦2,450.00
= 78
of his income = ₦2,450.00
178
x 2450
= 8×24507
= 196007
= ₦2800.00
Tambaya 10 Rahoto
Bayanin Amsa
1234561.1.11.21.31.41.51.622.12.22.32.42.52.633.13.23.33.43.53.644.14.24.34.44.54.655.15.25.35.45.55.666.16.26.36.46.56.6
At least 10 means 10 and above
P(at least 10) = 636
= 16
Tambaya 11 Rahoto
Simplify \( \frac{1\frac{1}{2}}{2+\frac{1}{4}\text{ of }32} \)
Bayanin Amsa
= 322+14×32
32 x 4 = 6
Tambaya 12 Rahoto
If \( \cos \theta = \frac{x}{y} \), find \( \cosec \theta \)
Bayanin Amsa
(hyp2) = opp2 + adj2
(hyp2) = x2 + y2
hyp = √x2+y2
Cosecθ = hyp
= x2 + y2
= 1y √x2+y2
Tambaya 13 Rahoto
Bayanin Amsa
= (2 x 5 - 4) x 90o
= 6 x 90o
= 540o
(90 - x) + (90 + x) + (110 - 2x) + (110 + 2x) + y = 540o
400o + y = 540o
y = 540 - 400o
y = 140o
Tambaya 15 Rahoto
three terms of the progression
Tambaya 16 Rahoto
Solve the following equation equation for \(x\)
\[ x^2 + \frac{2x}{r^2} + \frac{1}{r^4} = 0 \]Bayanin Amsa
(x + 1r2 ) = 0
x + 1r2 = 0
x = −1r2
Tambaya 17 Rahoto
Tambaya 18 Rahoto
Bayanin Amsa
Where actual error = 5.1 - 5.0 = 0.1
true value = 5.0g
% error = 0.15.0 x 100
= 105
= 2
Tambaya 19 Rahoto
Bayanin Amsa
log104.5 = log10 (3×3)2
log103 + log103 - log102 = 0.4771 + 0.4771 - 0.0310
= 0.6532
Tambaya 21 Rahoto
Bayanin Amsa
Tambaya 22 Rahoto
If \( g(y) = \frac{y - 3}{11} + \frac{11}{y^2 - 9} \). what is \( g(y + 3) \)?
Bayanin Amsa
Tambaya 23 Rahoto
Bayanin Amsa
Tambaya 24 Rahoto
Bayanin Amsa
cos120o = -1/2
Tambaya 25 Rahoto
Bayanin Amsa
-5y + 3x = 3........{i} x 2
2y - 6x = -5.........{ii} x 5
Substituting for x in equation (i)
-5y + 3(1924 ) = 3
-5y + 3 x 1924 = 3
-5y = 3−198
-5 = 24−198
= 58
y = 58×5
y = −18
(x, y) = (1924,−18 )
Tambaya 27 Rahoto
Bayanin Amsa
x = 2 + 3
= 5 = 10
H.C.F. of 6, 9, 15 = 3
∴ y = 3
X x y = 10 x 3
= 30
Tambaya 28 Rahoto
Find correct to one decimal place, \(0.24633 \div 0.0306\)
Bayanin Amsa
= 246333060
= 8.05
= 8.1
Tambaya 29 Rahoto
Bayanin Amsa
Using almighty formula
−b±√b2−4ac2a .........(i)
Where a = p, b = q and c = b
substitute for this value in equation (i)
= −q±√q2−4bp2p
Tambaya 30 Rahoto
In the figure, XR and YQ are tangents to the circle YZXP if ZXR = 45o and YZX = 55o, Find ZYQ
Bayanin Amsa
< ZYQ = 90 + 45
= 135O
Tambaya 32 Rahoto
If \( \frac{1}{p} = \frac{a^2+2ab+b^2}{a-b} \) and \( \frac{l}{q} = \frac{a+b}{a^2-2ab+b^2} \) Find \( \frac{p}{q} \)
Bayanin Amsa
Tambaya 33 Rahoto
In the figure, PQ is a parallel to ST and QRS = \(40^\circ\). Find the value of x.
Bayanin Amsa
4x = 180∘ + 40∘
4x = 220∘
x = 2204
= 55∘
Tambaya 35 Rahoto
If \( \cos^2 \theta + \frac{1}{8} = \sin^2 \theta \), find \( \tan \theta \)
Bayanin Amsa
from trigometric ratios for an acute angle, where cosθ + sin2θ = 1 - cosθ ........(ii)
Substitute for equation (i) in (i) = cos2θ + 18 = 1 - cos2θ
= cos2θ + cos2θ = 1 - 18
2 cos2θ = 78
cos2θ = 72×3
716 = cosθ
√716 = √74
but cos θ = adjhyp
opp2 = hyp2 - adj2
opp2 = 42 (√7 )2
= 16 - 7
opp = √9 = 3
than θ = opphyp
= 3√7
3√7 x 7√7 = 3√77
Tambaya 36 Rahoto
If x and y represent the mean and the median respectively of the following set of numbers 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, find the \( \frac{x}{y} \) correct to one decimal place
Bayanin Amsa
Median = ¯y = 15+162
312 = 15.5
xy = 15.615.5 = 1.0065
1.0(1 d.p)
Tambaya 37 Rahoto
Bayanin Amsa
let outer and inner radii be R and r respectively.
Area of the cross section = (R2 - r2)
where R = 6 and r = 6 - 1
= 5cm
Area of the cross section = (62 - 52)π = (36 - 25)π cm sq
vol. of the pipe = π (R2 - r2)L where length (L) = 10
volume = 11π x 10
= 110π cm3
Tambaya 38 Rahoto
Simplify \( \frac{4a^2 - 49b^2}{2a^2 - 5ab - 7b^2} \)
Bayanin Amsa
= (2a+7b)(2a−7b)(a−b)(2a+7b)
= 2a−7ba−b
Tambaya 39 Rahoto
Bayanin Amsa
Tambaya 40 Rahoto
In the figure, a solid consists of a hemisphere surmounted by a right circular cone, with radius 3.0cm and height 6.0cm. Find the volume of the solid
Bayanin Amsa
volume of cone = 12π2h
= 1π3×(3)2x6=18πcm2
vol. of hemisphere = 4πr36=2πr33
= 2π3×(3)3=18πcm3
vol. of solid = 18π + 18π
= 36π cm3
Tambaya 41 Rahoto
Evaluate \( \frac{8^{\frac{1}{3}} \times 5^{\frac{2}{3}}}{10^{\frac{2}{3}}} \) = \( \frac{8^{\frac{1}{3}} \times 5^{\frac{2}{3}}}{10^{\frac{2}{3}}} \)
Bayanin Amsa
= 2×5223×532
= 21 - 23
= 213
= 3√2
Tambaya 42 Rahoto
Simplify \( \frac{x+2}{x+1} - \frac{x-2}{x+2} \)
Bayanin Amsa
= (x2+4x+4)−(x2−x−2)(x+1)(x+2) = x2+4x+4−x2+x+2(x+1)(x+2)
= 5x+6(x+1)(x+2)
Tambaya 43 Rahoto
Tambaya 44 Rahoto
| Scores(\(x\)) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| Frequency(\(f\)) | 7 | 11 | 6 | 7 | 7 | 5 | 3 |
In the distribution above, the mode and median respectively are
Bayanin Amsa
Median = 2+22
= 2
= 1, 2
Tambaya 45 Rahoto
In the figure, PQRS is a circle. If chords QR and RS are equal, calculate the value of x.
Bayanin Amsa
SRQ = 180∘ - 120∘ = 60∘ - (angle on a straight line)
also angle QRS = 180∘ - 100∘ (angle on a straight line) . In angles where QR = SR and angle SRQ = 60∘
x = 100 - 60 = 40∘
Tambaya 46 Rahoto
Simplify \( \frac{x-7}{x^2-9} \, x \, \frac{x^2-3x}{x^2-49} \)
Bayanin Amsa
= x−7(x−3)(x+3) x x(x−3)(x−7)(x+7)
= x(x+3)(x+7)
Tambaya 47 Rahoto
Bayanin Amsa
18 x 103m = 1.8 x 10-2m
One leaf = 1.8×10−2800
= 1.8×10−28×102
= −1.88 x 10-4
= 0.225 x 10-4
= 2.25 x 10-5m
Tambaya 48 Rahoto
Two sisters, Taiwo and Keyinde, own a store. The ratio of Taiwo's share to Kehinde's is 11:9. Later, Keyinde sells \( \frac{2}{3} \) of her share to Taiwo for ₦720.00. Find the value of the store.
Bayanin Amsa
Ratio of Taiwo's share to kehine's is 11:9 Keyinde sells 23 of her share to Taiwo for ₦720
23 of 9 = 6
∴ Sum of the ratio = 11 + 9 = 20
620 of x = ₦720
6x20 = 720
∴ x = 720×206
x = ₦24,000
Tambaya 49 Rahoto
Bayanin Amsa
US5 = sin60o
US = 5 x √32
= 2.5√3
= 4.33cm
