JAMB UTME - Mathematics - 1988
Question 1 Report
In the figure, PS = RS = QS and QRS = 50o. Find QPR
Answer Details
In angle SQR, angle S = 50O
In angle QRP, 65 + 65 = 130O
Since RQP = angle RPQ = 180−1302
= 502=25o
QPR = 25O
Question 2 Report
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
Answer Details
volume of cone = 12π2h
= 1π3×(3)2x6=18πcm2
vol. of hemisphere = 4πr36=2πr33
= 2π3×(3)3=18πcm3
vol. of solid = 18π + 18π
= 36π cm3
Question 3 Report
Find correct to one decimal place, \(0.24633 \div 0.0306\)
Answer Details
= 246333060
= 8.05
= 8.1
Question 4 Report
Answer Details
Question 5 Report
Answer Details
PR2 = 12 + 22 - 2(1)(2) x - cos 60o
= 5 - 2(1)(2) x -12
= 5 + 2 = 7
PR = √7 cm
Question 6 Report
Simplify \( \frac{1\frac{1}{2}}{2+\frac{1}{4}\text{ of }32} \)
Answer Details
= 322+14×32
32 x 4 = 6
Question 7 Report
Answer Details
= 5
Question 8 Report
Answer Details
= (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
Question 9 Report
Answer Details
US5 = sin60o
US = 5 x √32
= 2.5√3
= 4.33cm
Question 10 Report
Answer Details
Question 11 Report
Evaluate \( \frac{8^{\frac{1}{3}} \times 5^{\frac{2}{3}}}{10^{\frac{2}{3}}} \) = \( \frac{8^{\frac{1}{3}} \times 5^{\frac{2}{3}}}{10^{\frac{2}{3}}} \)
Answer Details
= 2×5223×532
= 21 - 23
= 213
= 3√2
Question 13 Report
Answer Details
x = 2 + 3
= 5 = 10
H.C.F. of 6, 9, 15 = 3
∴ y = 3
X x y = 10 x 3
= 30
Question 14 Report
Answer Details
Question 15 Report
Simplify \( \frac{4a^2 - 49b^2}{2a^2 - 5ab - 7b^2} \)
Answer Details
= (2a+7b)(2a−7b)(a−b)(2a+7b)
= 2a−7ba−b
Question 16 Report
In the figure, PQ is a parallel to ST and QRS = \(40^\circ\). Find the value of x.
Answer Details
4x = 180∘ + 40∘
4x = 220∘
x = 2204
= 55∘
Question 17 Report
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.
Answer Details
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
Question 18 Report
Answer Details
log104.5 = log10 (3×3)2
log103 + log103 - log102 = 0.4771 + 0.4771 - 0.0310
= 0.6532
Question 19 Report
Answer Details
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
Question 20 Report
If (IPO3)4 = 11510 find P
Answer Details
1 x 43 + P x 42 + 0 x 4 + 3 = 11510
16p + 67 = 115 p = 4816
= 3
Question 21 Report
If \( \cos^2 \theta + \frac{1}{8} = \sin^2 \theta \), find \( \tan \theta \)
Answer Details
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
Question 22 Report
if a = -3, b = 2, c = 4, evaluate \( \frac{a^3-b^3-c^{\frac{1}{2}}}{b-a-c} \)
Answer Details
Question 23 Report
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 \( \frac{p}{q} \)
Answer Details
Question 24 Report
For what values of x is the curve \(y = \frac{x^2 + 3}{x + 4}\) decreasing?
Answer Details
Question 26 Report
In the figure, PQRS is a circle. If chords QR and RS are equal, calculate the value of x.
Answer Details
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∘
Question 27 Report
Find \(m\) such that \((m + \sqrt{3})(1 - \sqrt{3})^2 = 6 - 2\sqrt{2}\)
Answer Details
(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
Question 28 Report
Answer Details
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
Question 29 Report
Answer Details
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
Question 30 Report
Simplify \( \frac{x+2}{x+1} - \frac{x-2}{x+2} \)
Answer Details
= (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)
Question 31 Report
Answer Details
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
Question 32 Report
In the figure, PS = 7cm and RY = 9cm. IF the area of parallelogram PQRS is 56cm2. Find the area of trapezium PQTS
Answer Details
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
Question 33 Report
Answer Details
Where actual error = 5.1 - 5.0 = 0.1
true value = 5.0g
% error = 0.15.0 x 100
= 105
= 2
Question 34 Report
Answer Details
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
Question 35 Report
Solve the following equation equation for \(x\)
\[ x^2 + \frac{2x}{r^2} + \frac{1}{r^4} = 0 \]Answer Details
(x + 1r2 ) = 0
x + 1r2 = 0
x = −1r2
Question 36 Report
In the figure, STQ = SRP, PT = TQ = 6cm and QS = 5cm. Find SR
Answer Details
QT = 6×125=725=SR=QR−QS
= 725−5=72−255
= 475
Question 37 Report
| 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
Answer Details
Median = 2+22
= 2
= 1, 2
Question 38 Report
If \( \cos \theta = \frac{x}{y} \), find \( \cosec \theta \)
Answer Details
(hyp2) = opp2 + adj2
(hyp2) = x2 + y2
hyp = √x2+y2
Cosecθ = hyp
= x2 + y2
= 1y √x2+y2
Question 39 Report
Simplify \( \frac{x-7}{x^2-9} \, x \, \frac{x^2-3x}{x^2-49} \)
Answer Details
= x−7(x−3)(x+3) x x(x−3)(x−7)(x+7)
= x(x+3)(x+7)
Question 40 Report
Answer Details
Question 41 Report
Answer Details
-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 )
Question 42 Report
Answer Details
cos120o = -1/2
Question 43 Report
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
Answer Details
Median = ¯y = 15+162
312 = 15.5
xy = 15.615.5 = 1.0065
1.0(1 d.p)
Question 44 Report
If \( g(y) = \frac{y - 3}{11} + \frac{11}{y^2 - 9} \). what is \( g(y + 3) \)?
Answer Details
Question 45 Report
Answer Details
(ii) and (iii) are right
Question 46 Report
three terms of the progression
Answer Details
Question 47 Report
In the figure, XR and YQ are tangents to the circle YZXP if ZXR = 45o and YZX = 55o, Find ZYQ
Answer Details
< ZYQ = 90 + 45
= 135O
Question 48 Report
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?
Answer Details
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
