JAMB UTME - Mathematics - 1988
Question 1 Report
Find \(m\) such that \((m + \sqrt{3})(1 - \sqrt{3})^2 = 6 - 2\sqrt{2}\)
Answer Details
(m + √3
)(1 - √3
)2 = 6 - 2√2
(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 2 Report
If \( \cos^2 \theta + \frac{1}{8} = \sin^2 \theta \), find \( \tan \theta \)
Answer Details
cos2θ
+ 18
= sin2θ
..........(i)
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 3 Report
Tope bought X oranges at N5.00 each and some mangoes at N4.00 each. if she bought twice as many mangoes as oranges and spent at least N65.00 and at most N130.00, find the range of values of X.
Answer Details
Question 4 Report
Given that 3x - 5y - 3 = 0, 2y - 6x + 5 = 0 the value of (x, y) is
Answer Details
3x - 5y = 3, 2y - 6x = -5
-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 5 Report
In the figure, STQ = SRP, PT = TQ = 6cm and QS = 5cm. Find SR
Answer Details
From similar triangle, QSQP=TQQR=512=6QR
QT = 6×125=725=SR=QR−QS
= 725−5=72−255
= 475
Question 6 Report
If 7 and 189 are the first and fourth terms of a geometric progression respectively find the sum for the first
three terms of the progression
Answer Details
Question 7 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
The volume of the solid = vol. of cone + vol. of hemisphere
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 8 Report
Simplify \( \frac{x+2}{x+1} - \frac{x-2}{x+2} \)
Answer Details
x+2x+1
- x−2x+2
= (x+2)(x+2)−(x−2)−(x−2)(x+1)(x+1)(x+2)
= (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 9 Report
If a metal pipe 10cm long has an external diameter of 12cm and a thickness of 1cm find the volume of the metal used in making the pipe
Answer Details
The volume of the pipe is equal to the area of the cross section and length.
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 10 Report
A 5.0g of salt was weighted by Tunde as 5.1g. What is the percentage error?
Answer Details
% error = actual errortrue value
x 100
Where actual error = 5.1 - 5.0 = 0.1
true value = 5.0g
% error = 0.15.0
x 100
= 105
= 2
Question 11 Report
In the figure, PQRS is a circle. If chords QR and RS are equal, calculate the value of x.
Answer Details
SRT is a straight line, where QRT = 120
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 12 Report
In the figure, PQ is a parallel to ST and QRS = \(40^\circ\). Find the value of x.
Answer Details
From the figure, 3x + x - 40∘ = 180∘
4x = 180∘ + 40∘
4x = 220∘
x = 2204
= 55∘
Question 13 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
Let value of store = X
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 14 Report
If \( \cos \theta = \frac{x}{y} \), find \( \cosec \theta \)
Answer Details
Cos θ
= xy
= adjopp
(hyp2) = opp2 + adj2
(hyp2) = x2 + y2
hyp = √x2+y2
Cosecθ
= hyp
= x2 + y2
= 1y
√x2+y2
Question 15 Report
In the figure, XR and YQ are tangents to the circle YZXP if ZXR = 45o and YZX = 55o, Find ZYQ
Answer Details
< RXZ = < ZYX = 45O(Alternate segment) < ZYQ = 90 + 45 = 135O
Question 16 Report
In a class of 150 students, the sector in a pie chart representing the students offering Physics has angle 12o. How many students are offering Physics?
Answer Details
No of students offering Physics are 12360
x 150
= 5
Question 17 Report
If x varies inversely as the cube root of y and x = 1 when y = 8, find y when x = 3
Answer Details
Question 18 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
Mean ¯x
= 15610
= 15.6
Median = ¯y
= 15+162
312
= 15.5
xy
= 15.615.5
= 1.0065
1.0(1 d.p)
Question 19 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 20 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
From the figure, PS = QR = YT = 7cm
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 21 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
Question 22 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 23 Report
For what values of x is the curve \(y = \frac{x^2 + 3}{x + 4}\) decreasing?
Answer Details
Question 25 Report
The solutions of x2 - 2x - 1 = 0 are the points of intersection of two graphs. if one of the graphs is y = 2 + x - x2, find the second graph
Answer Details
Question 26 Report
if x is the addition of the prime numbers between 1 and 6; and y the H.C.F. of 6, 9, 15. Find the product of x and y
Answer Details
Prime numbers between 1 and 6 are 2, 3 and 5
x = 2 + 3
= 5 = 10
H.C.F. of 6, 9, 15 = 3
∴ y = 3
X x y = 10 x 3
= 30
Question 27 Report
If the sum of the 8th and 9th terms of an arithmetic progression is 72 and the 4th term is -6, find the common difference
Answer Details
Question 28 Report
PQR is a triangle in which PQ = 10cm and QPR = 60oS is a point equidistant from P and Q. Also S is a point equidistant from PQ and PR. If U is the foot of the perpendicular from S on PR, find the length SU in cm to one decimal place
Answer Details
△
PUS is right angled
US5
= sin60o
US = 5 x √32
= 2.5√3
= 4.33cm
Question 29 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
From the distribution, Mode = 1 and
Median = 2+22
= 2
= 1, 2
Question 30 Report
if a = -3, b = 2, c = 4, evaluate \( \frac{a^3-b^3-c^{\frac{1}{2}}}{b-a-c} \)
Answer Details
Question 31 Report
For which of the following exterior angles is a regular polygon possible? i. 36o ii. 18o iii. 15o
Answer Details
for a regular polygon to be possible, it must have all sides angles equal. 36018
= 20 sides and 36015
= 24 sides
(ii) and (iii) are right
Question 32 Report
Find correct to one decimal place, \(0.24633 \div 0.0306\)
Answer Details
0.246330.03060
multiplying throughout by 100,000
= 246333060
= 8.05
= 8.1
Question 33 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
813×5322103
= (23)13×532(2×5)23
= 2×5223×532
= 21 - 23
= 213
= 3√2
Question 34 Report
The solution of the quadratic equation px2 + qx + b = 0 is
Answer Details
px2 + qx + b = 0
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 35 Report
Solve the following equation equation for \(x\)
\[ x^2 + \frac{2x}{r^2} + \frac{1}{r^4} = 0 \]Answer Details
x2 + 2xr2
+ 1r4
= 0
(x + 1r2
) = 0
x + 1r2
= 0
x = −1r2
Question 36 Report
If \( g(y) = \frac{y - 3}{11} + \frac{11}{y^2 - 9} \). what is \( g(y + 3) \)?
Answer Details
Question 37 Report
The thickness of an 800 pages of book is 18mm. Calculate the thickness of one leaf of the book giving your answer in meters and in standard form
Answer Details
Thickness of an 800 pages book = 18mm to meter
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 39 Report
If two dice are thrown together, what is the probability of obtaining at least a score of 10?
Answer Details
The total sample space when two dice are thrown together is 6 x 6 = 36
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 40 Report
A basket contain green, black and blue balls in the ratio 5 : 2 : 1. If there are 10 blue balls. Find the corresponding new ratio when 10 green and 10 black balls are removed from the basket
Answer Details
Let x represent total number of balls in the basket.
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 41 Report
Simplify \( \frac{4a^2 - 49b^2}{2a^2 - 5ab - 7b^2} \)
Answer Details
4a2−49b22a2−5ab−7b2
= (2a)2−(7b)2(a−b)(2a+7b)
= (2a+7b)(2a−7b)(a−b)(2a+7b)
= 2a−7ba−b
Question 42 Report
If cos 60o = 1/2, which of the following angle has cosine of -1/2?
Answer Details
cos60o = 1/2, cos(180o/60o) = -1/2
cos120o = -1/2
Question 43 Report
Four interior angles of a pentagon are 90o - xo, 90o + xo, 110o - 2xo, 110o + 2xo. Find the fifth interior angle
Answer Details
Let the fifth interior angle be y: sum of interior angle of a pentagon
= (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 44 Report
If log102 = 0.3010 and log103 = 0.4771, evaluate; without using logarithm tables, log104.5
Answer Details
If log102 = 0.3010 and log103 = 0.4771,
log104.5 = log10 (3×3)2
log103 + log103 - log102 = 0.4771 + 0.4771 - 0.0310
= 0.6532
Question 45 Report
In the figure, PS = RS = QS and QRS = 50o. Find QPR
Answer Details
In the figure PS = RS = QS, they will have equal base QR = RP
In angle SQR, angle S = 50O
In angle QRP, 65 + 65 = 130O
Since RQP = angle RPQ = 180−1302
= 502=25o
QPR = 25O
Question 46 Report
Simplify \( \frac{1\frac{1}{2}}{2+\frac{1}{4}\text{ of }32} \)
Answer Details
1122 + 14 of 32
= 322+14×32
32
x 4 = 6
Question 48 Report
In triangle PQR, PQ = 1cm, QR = 2cm and PQR = 120o Find the longest side of the triangle
Answer Details
PR2 = PQ2 + QR2 - 2(QR)(PQ) COS 120o
PR2 = 12 + 22 - 2(1)(2) x - cos 60o
= 5 - 2(1)(2) x -12
= 5 + 2 = 7
PR = √7
cm
Question 49 Report
Simplify \( \frac{x-7}{x^2-9} \, x \, \frac{x^2-3x}{x^2-49} \)
Answer Details
x−7x2−9
x x2−3xx2−49
= x−7(x−3)(x+3)
x x(x−3)(x−7)(x+7)
= x(x+3)(x+7)
