JAMB UTME - Mathematics - 1988
Ajụjụ 2 Ripọtì
If x varies inversely as the cube root of y and x = 1 when y = 8, find y when x = 3
Akọwa Nkọwa
Ajụjụ 3 Ripọtì
If \( g(y) = \frac{y - 3}{11} + \frac{11}{y^2 - 9} \). what is \( g(y + 3) \)?
Akọwa Nkọwa
Ajụjụ 4 Ripọtì
Find \(m\) such that \((m + \sqrt{3})(1 - \sqrt{3})^2 = 6 - 2\sqrt{2}\)
Akọwa Nkọwa
(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
Ajụjụ 5 Ripọtì
If cos 60o = 1/2, which of the following angle has cosine of -1/2?
Akọwa Nkọwa
cos60o = 1/2, cos(180o/60o) = -1/2
cos120o = -1/2
Ajụjụ 6 Ripọtì
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
Akọwa Nkọwa
Ajụjụ 7 Ripọtì
In the figure, PS = RS = QS and QRS = 50o. Find QPR
Akọwa Nkọwa
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
Ajụjụ 8 Ripọtì
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
Akọwa Nkọwa
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
Ajụjụ 9 Ripọtì
In the figure, PS = 7cm and RY = 9cm. IF the area of parallelogram PQRS is 56cm2. Find the area of trapezium PQTS
Akọwa Nkọwa
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
Ajụjụ 10 Ripọtì
If \( \cos^2 \theta + \frac{1}{8} = \sin^2 \theta \), find \( \tan \theta \)
Akọwa Nkọwa
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
Ajụjụ 11 Ripọtì
Simplify \( \frac{x-7}{x^2-9} \, x \, \frac{x^2-3x}{x^2-49} \)
Akọwa Nkọwa
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)
Ajụjụ 12 Ripọtì
If log102 = 0.3010 and log103 = 0.4771, evaluate; without using logarithm tables, log104.5
Akọwa Nkọwa
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
Ajụjụ 14 Ripọtì
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
Akọwa Nkọwa
△
PUS is right angled
US5
= sin60o
US = 5 x √32
= 2.5√3
= 4.33cm
Ajụjụ 16 Ripọtì
Simplify \( \frac{1\frac{1}{2}}{2+\frac{1}{4}\text{ of }32} \)
Akọwa Nkọwa
1122 + 14 of 32
= 322+14×32
32
x 4 = 6
Ajụjụ 17 Ripọtì
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.
Akọwa Nkọwa
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
Ajụjụ 19 Ripọtì
Four interior angles of a pentagon are 90o - xo, 90o + xo, 110o - 2xo, 110o + 2xo. Find the fifth interior angle
Akọwa Nkọwa
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
Ajụjụ 20 Ripọtì
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
Akọwa Nkọwa
Ajụjụ 21 Ripọtì
In the figure, PQ is a parallel to ST and QRS = \(40^\circ\). Find the value of x.
Akọwa Nkọwa
From the figure, 3x + x - 40∘ = 180∘
4x = 180∘ + 40∘
4x = 220∘
x = 2204
= 55∘
Ajụjụ 23 Ripọtì
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?
Akọwa Nkọwa
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
Ajụjụ 24 Ripọtì
The solution of the quadratic equation px2 + qx + b = 0 is
Akọwa Nkọwa
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
Ajụjụ 25 Ripọtì
| 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
Akọwa Nkọwa
From the distribution, Mode = 1 and
Median = 2+22
= 2
= 1, 2
Ajụjụ 26 Ripọtì
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
Akọwa Nkọwa
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
Ajụjụ 27 Ripọtì
If \( \cos \theta = \frac{x}{y} \), find \( \cosec \theta \)
Akọwa Nkọwa
Cos θ
= xy
= adjopp
(hyp2) = opp2 + adj2
(hyp2) = x2 + y2
hyp = √x2+y2
Cosecθ
= hyp
= x2 + y2
= 1y
√x2+y2
Ajụjụ 28 Ripọtì
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
Akọwa Nkọwa
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
Ajụjụ 29 Ripọtì
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 \( \frac{p}{q} \)
Akọwa Nkọwa
Ajụjụ 31 Ripọtì
Simplify \( \frac{x+2}{x+1} - \frac{x-2}{x+2} \)
Akọwa Nkọwa
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)
Ajụjụ 32 Ripọtì
In the figure, PQRS is a circle. If chords QR and RS are equal, calculate the value of x.
Akọwa Nkọwa
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∘
Ajụjụ 33 Ripọtì
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
Ajụjụ 34 Ripọtì
If (IPO3)4 = 11510 find P
Akọwa Nkọwa
1 x 43 + P x 42 + 0 x 4 + 3 = 11510
16p + 67 = 115 p = 4816
= 3
Ajụjụ 35 Ripọtì
Solve the following equation equation for \(x\)
\[ x^2 + \frac{2x}{r^2} + \frac{1}{r^4} = 0 \]Akọwa Nkọwa
x2 + 2xr2
+ 1r4
= 0
(x + 1r2
) = 0
x + 1r2
= 0
x = −1r2
Ajụjụ 36 Ripọtì
If two dice are thrown together, what is the probability of obtaining at least a score of 10?
Akọwa Nkọwa
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
Ajụjụ 37 Ripọtì
Find correct to one decimal place, \(0.24633 \div 0.0306\)
Akọwa Nkọwa
0.246330.03060
multiplying throughout by 100,000
= 246333060
= 8.05
= 8.1
Ajụjụ 38 Ripọtì
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
Akọwa Nkọwa
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
Ajụjụ 39 Ripọtì
A 5.0g of salt was weighted by Tunde as 5.1g. What is the percentage error?
Akọwa Nkọwa
% 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
Ajụjụ 40 Ripọtì
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
Akọwa Nkọwa
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
Ajụjụ 41 Ripọtì
For which of the following exterior angles is a regular polygon possible? i. 36o ii. 18o iii. 15o
Akọwa Nkọwa
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
Ajụjụ 42 Ripọtì
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.
Akọwa Nkọwa
Ajụjụ 43 Ripọtì
In the figure, XR and YQ are tangents to the circle YZXP if ZXR = 45o and YZX = 55o, Find ZYQ
Akọwa Nkọwa
< RXZ = < ZYX = 45O(Alternate segment) < ZYQ = 90 + 45 = 135O
Ajụjụ 44 Ripọtì
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
Akọwa Nkọwa
Mean ¯x
= 15610
= 15.6
Median = ¯y
= 15+162
312
= 15.5
xy
= 15.615.5
= 1.0065
1.0(1 d.p)
Ajụjụ 45 Ripọtì
For what values of x is the curve \(y = \frac{x^2 + 3}{x + 4}\) decreasing?
Akọwa Nkọwa
Ajụjụ 46 Ripọtì
Simplify \( \frac{4a^2 - 49b^2}{2a^2 - 5ab - 7b^2} \)
Akọwa Nkọwa
4a2−49b22a2−5ab−7b2
= (2a)2−(7b)2(a−b)(2a+7b)
= (2a+7b)(2a−7b)(a−b)(2a+7b)
= 2a−7ba−b
Ajụjụ 48 Ripọtì
Evaluate \( \frac{8^{\frac{1}{3}} \times 5^{\frac{2}{3}}}{10^{\frac{2}{3}}} \) = \( \frac{8^{\frac{1}{3}} \times 5^{\frac{2}{3}}}{10^{\frac{2}{3}}} \)
Akọwa Nkọwa
813×5322103
= (23)13×532(2×5)23
= 2×5223×532
= 21 - 23
= 213
= 3√2
Ajụjụ 49 Ripọtì
Given that 3x - 5y - 3 = 0, 2y - 6x + 5 = 0 the value of (x, y) is
Akọwa Nkọwa
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
)
