JAMB UTME - Mathematics - 1988

The solution of the quadratic equation px2 + qx + b = 0 is

Answer Details

px2 + qx + b = 0
Using almighty formula
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.........(i)
Where a = p, b = q and c = b
substitute for this value in equation (i)
= $\frac{-q\pm \sqrt{q^2-4bp}}{2p}$

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

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

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 - $\frac{7}{8}$ = $\frac{1}{7}$th of his income
20% is 490, 100% is $\frac{1000}{20}$ x 490, ₦2,450.00
= $\frac{7}{8}$ of his income = ₦2,450.00

$\frac{1}{\frac{7}{8}}$ x 2450
= $\frac{8\times 2450}{7}$
= $\frac{19600}{7}$
= ₦2800.00

What is the solution of the equation x2 - x - 1 + 0?

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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.

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Let value of store = X
Ratio of Taiwo's share to kehine's is 11:9 Keyinde sells $\frac{2}{3}$ of her share to Taiwo for ₦720
$\frac{2}{3}$ of 9 = 6
∴ Sum of the ratio = 11 + 9 = 20
$\frac{6}{20}$ of x = ₦720
$\frac{6x}{20}$ = 720
∴ x = $\frac{720\times 20}{6}$
x = ₦24,000

If x varies inversely as the cube root of y and x = 1 when y = 8, find y when x = 3

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If two dice are thrown together, what is the probability of obtaining at least a score of 10?

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The total sample space when two dice are thrown together is 6 x 6 = 36
$\begin{array}{ccccccc}& 1& 2& 3& 4& 5& 6\\ 1.& 1.1& 1.2& 1.3& 1.4& 1.5& 1.6\\ 2& 2.1& 2.2& 2.3& 2.4& 2.5& 2.6\\ 3& 3.1& 3.2& 3.3& 3.4& 3.5& 3.6\\ 4& 4.1& 4.2& 4.3& 4.4& 4.5& 4.6\\ 5& 5.1& 5.2& 5.3& 5.4& 5.5& 5.6\\ 6& 6.1& 6.2& 6.3& 6.4& 6.5& 6.6\end{array}$
At least 10 means 10 and above
P(at least 10) = $\frac{6}{36}$
= $\frac{1}{6}$

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
$\frac{US}{5}$ = sin60o
US = 5 x $\frac{\sqrt{3}}{2}$
= 2.5$\sqrt{3}$
= 4.33cm

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.

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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

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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)$\pi$ = (36 - 25)$\pi$cm sq
vol. of the pipe = $\pi$ (R2 - r2)L where length (L) = 10
volume = 11$\pi$ x 10
= 110$\pi$cm3

If (IPO3)4 = 11510 find P

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1 x 43 + P x 42 + 0 x 4 + 3 = 11510
16p + 67 = 115 p = $\frac{48}{16}$
= 3

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${}^{\circ }$ = 180${}^{\circ }$

4x = 180${}^{\circ }$ + 40${}^{\circ }$

4x = 220${}^{\circ }$

x = $\frac{220}{4}$

= 55${}^{\circ }$

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($\frac{19}{24}$) = 3
-5y + 3 x $\frac{19}{24}$ = 3
-5y = $\frac{3-19}{8}$
-5 = $\frac{24-19}{8}$
= $\frac{5}{8}$
y = $\frac{5}{8\times 5}$
y = $\frac{-1}{8}$
(x, y) = ($\frac{19}{24},\frac{-1}{8}$)

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 $\frac{12}{360}$ x 150
= 5

if a = -3, b = 2, c = 4, evaluate $\frac{a^3-b^3-c^{\frac{1}{2}}}{b-a-c}$

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If (g(y)) = $\frac{y-3}{11}$ + $\frac{11}{y^2-9}$. what is g(y + 3)?

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Simplify $\frac{4a^2-49b^2}{2a^2-5ab-7b^2}$

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$\frac{4a^2-49b^2}{2a^2-5ab-7b^2}$ = $\frac{(2a{)}^2-(7b{)}^2}{(a-b)(2a+7b)}$
= $\frac{(2a+7b)(2a-7b)}{(a-b)(2a+7b)}$
= $\frac{2a-7b}{a-b}$

Simplify $\frac{1\frac{1}{2}}{\text{2 +}\frac{1}{4}\text{of 32}}$

Answer Details

$\frac{1\frac{1}{2}}{\text{2 +}\frac{1}{4}\text{of 32}}$
= $\frac{\frac{3}{2}}{2+\frac{1}{4}\times 32}$
$\frac{3}{2}$ x 4 = 6

Find m such that (m + $\sqrt{3}$)(1 - $\sqrt{3}$)2 = 6 - 2$\sqrt{2}$

Answer Details

(m + $\sqrt{3}$)(1 - $\sqrt{3}$)2 = 6 - 2$\sqrt{2}$
(m + $\sqrt{3}$)(4 - 2$\sqrt{3}$) = 6 - 2$\sqrt{2}$
= 6 - 2$\sqrt{3}$
4m - 6 + 4 - 2m$\sqrt{3}$ = 6 - 2$\sqrt{3}$
comparing co-efficients,
4m - 6 = 6.......(i)
4 - 2m = -2.......(ii)
in both equations, m = 3

Simplify $\frac{x+2}{x+1}$ - $\frac{x-2}{x+2}$

Answer Details

$\frac{x+2}{x+1}$ - $\frac{x-2}{x+2}$ = $\frac{(x+2)(x+2)-(x-2)-(x-2)(x+1)}{(x+1)(x+2)}$
= $\frac{(x^2+4x+4)-(x^2-x-2)}{(x+1)(x+2)}$ = $\frac{x^2+4x+4-x^2+x+2}{(x+1)(x+2)}$
= $\frac{5x+6}{(x+1)(x+2)}$

In the figure, STQ = SRP, PT = TQ = 6cm and QS = 5cm. Find SR

Answer Details

From similar triangle, $\frac{QS}{QP}=\frac{TQ}{QR}=\frac{5}{12}=\frac{6}{QR}$

QT = $\frac{6\times 12}{5}=\frac{72}{5}=SR=QR-QS$

= $\frac{72}{5}-5=\frac{72-25}{5}$

= $\frac{47}{5}$

$\begin{array}{ccccccccc}Scores\left(x\right)& 0& 1& 2& 3& 4& 5& 6& 7\\ Frequency\left(f\right)& 7& 11& 6& 7& 7& 5& 3& \end{array}$

In the distribution above, the mode and median respectively are

Answer Details

From the distribution, Mode = 1 and
Median = $\frac{2+2}{2}$
= 2
= 1, 2

If cos2$\theta$ + $\frac{1}{8}$ = sin2$\theta$, find tan$\theta$

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cos2$\theta$ + $\frac{1}{8}$ = sin2$\theta$..........(i)
from trigometric ratios for an acute angle, where cos$\theta$ + sin2$\theta$ = 1 - cos$\theta$ ........(ii)
Substitute for equation (i) in (i) = cos2$\theta$ + $\frac{1}{8}$ = 1 - cos2$\theta$
= cos2$\theta$ + cos2$\theta$ = 1 - $\frac{1}{8}$
2 cos2$\theta$ = $\frac{7}{8}$
cos2$\theta$ = $\frac{7}{2\times 3}$
$\frac{7}{16}$ = cos$\theta$
$\sqrt{\frac{7}{16}}$ = $\frac{\sqrt{7}}{4}$
but cos $\theta$ = $\frac{\text{adj}}{\text{hyp}}$
opp2 = hyp2 - adj2
opp2 = 42 ($\sqrt{7}$)2
= 16 - 7
opp = $\sqrt{9}$ = 3
than $\theta$ = $\frac{\text{opp}}{\text{hyp}}$
= $\frac{3}{\sqrt{7}}$
$\frac{3}{\sqrt{7}}$ x $\frac{7}{\sqrt{7}}$ = $\frac{3\sqrt{7}}{7}$

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${}^{\circ }$ - 120${}^{\circ }$ = 60${}^{\circ }$ - (angle on a straight line)

also angle QRS = 180${}^{\circ }$ - 100${}^{\circ }$ (angle on a straight line) . In angles where QR = SR and angle SRQ = 60${}^{\circ }$

x = 100 - 60 = 40${}^{\circ }$

Solve the following equation equation for x2 + $\frac{2x}{r^2}$ + $\frac{1}{r^4}$ = 0

Answer Details

x2 + $\frac{2x}{r^2}$ + $\frac{1}{r^4}$ = 0
(x + $\frac{1}{r^2}$) = 0
x + $\frac{1}{r^2}$ = 0
x = $\frac{-1}{r^2}$

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 = $\frac{1.8\times {10}^{-2}}{800}$
= $\frac{1.8\times {10}^{-2}}{8\times {10}^2}$
= $\frac{-1.8}{8}$ x 10-4
= 0.225 x 10-4
= 2.25 x 10-5m

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 $\frac{p}{q}$

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If cos 60o = 1/2, which of the following angle has cosine of -1/2?

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cos60o = 1/2, cos(180o/60o) = -1/2

cos120o = -1/2

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

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In the figure, PS = 7cm and RY = 9cm. IF the area of parallelogram PQRS is 56cm2. Find the area of trapezium PQTS

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From the figure, PS = QR = YT = 7cm

Area of parallelogram PQRS = 56cm

56 = base x height, where base = 7

7 x h = 56cm,

h = $\frac{56}{7}$

= 8cm

Area of trapezium $\frac{1}{2}$ (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 = $\frac{1}{2}$(23 + 7) x 8

$\frac{1}{2}$ x 30 x 8 = 120cmsq

If cos $?$ = $\frac{x}{y}$, find cosec$?$

Answer Details

Cos $\theta$ = $\frac{x}{y}$ = $\frac{\text{adj}}{\text{opp}}$
(hyp2) = opp2 + adj2
(hyp2) = x2 + y2
hyp = $\sqrt{x^2+y^2}$
Cosec$\theta$ = hyp
= x2 + y2
= $\frac{1}{y}$$\sqrt{x^2+y^2}$

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 = $\frac{1}{2}{\pi }^{2}h$

= $\frac{1\pi }{3}\times (3{)}^{2}x6=18\pi c{m}^2$

vol. of hemisphere = $\frac{4\pi r^3}{6}=\frac{2\pi r^3}{3}$

= $\frac{2\pi }{3}\times (3{)}^3=18\pi c{m}^3$

vol. of solid = 18$\pi$ + 18$\pi$

= 36$\pi$cm3

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. $\frac{360}{18}$ = 20 sides and $\frac{360}{15}$ = 24 sides
(ii) and (iii) are right

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

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

$\frac{8^{\frac{1}{3}}\times 5^{\frac{3}{2}}}{\frac{2}{{10}^3}}$ = $\frac{(2^3{)}^{\frac{1}{3}}\times 5^{\frac{3}{2}}}{(2\times 5{)}^{\frac{2}{3}}}$
= $\frac{2\times 5}{2^{\frac{2}{3}}\times 5^{\frac{3}{2}}}$
= 21 - $\frac{2}{3}$
= 2$\frac{1}{3}$
= 3$\sqrt{2}$

Simplify $\frac{x?y}{x^{\frac{1}{3}}?x^{\frac{1}{3}}}$

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For what values of x is the curve y = $\frac{x^2+3}{x+4}$ decreasing?

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Find correct to one decimal place, 0.24633 $÷$ 0.0306

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$\frac{0.24633}{0.03060}$ multiplying throughout by 100,000
= $\frac{24633}{3060}$
= 8.05
= 8.1

Factorize completely (x2 + x)2 - (2x + 2)2

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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

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Mean $\overline{x}$ = $\frac{156}{10}$ = 15.6
Median = $\overline{y}$ = $\frac{15+16}{2}$
$\frac{31}{2}$ = 15.5
$\frac{x}{y}$ = $\frac{15.6}{15.5}$ = 1.0065
1.0(1 d.p)

A 5.0g of salt was weighted by Tunde as 5.1g. What is the percentage error?

Answer Details

% error = $\frac{\text{actual error}}{\text{true value}}$ x 100
Where actual error = 5.1 - 5.0 = 0.1
true value = 5.0g
% error = $\frac{0.1}{5.0}$ x 100
= $\frac{10}{5}$
= 2

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 -$\frac{1}{2}$
= 5 + 2 = 7
PR = $\sqrt{7}$cm

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 = $\frac{180-130}{2}$

= $\frac{50}{2}={25}^o$

QPR = 25O

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 $\frac{(3\times 3)}{2}$
log103 + log103 - log102 = 0.4771 + 0.4771 - 0.0310
= 0.6532

Simplify $\frac{x-7}{x^2-9}$ x $\frac{x^2-3x}{x^2-49}$

Answer Details

$\frac{x-7}{x^2-9}$ x $\frac{x^2-3x}{x^2-49}$
= $\frac{x-7}{(x-3)(x+3)}$ x $\frac{x(x-3)}{(x-7)(x+7)}$
= $\frac{x}{(x+3)(x+7)}$

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, $\frac{1}{8}$ of x = 10
x = 10 x 8 = 80 balls
Green balls will be $\frac{5}{8}$ x 80 = 50 and black balls = $\frac{2}{8}$ x 80 = 20
Ratio = Green : black : blue
50 : 20 : 10
-10 : 10 : -
------------------
New Ratio 40 : 10 : 10
4 : 1 : 1

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

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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

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