JAMB UTME - Mathematics - 1988

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) = $\frac{6}{36}$
= $\frac{1}{6}$

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

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

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

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 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

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

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% 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 the figure, PS = RS = QS and QRS = 50o. Find QPR

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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 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

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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

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

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

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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

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$△$PUS is right angled
$\frac{US}{5}$ = sin60o
US = 5 x $\frac{\sqrt{3}}{2}$
= 2.5$\sqrt{3}$
= 4.33cm

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

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(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

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

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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 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)

In triangle PQR, PQ = 1cm, QR = 2cm and PQR = 120o Find the longest side of the triangle

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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

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

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

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

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

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For which of the following exterior angles is a regular polygon possible? i. 36o ii. 18o iii. 15o

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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 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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In the figure, PQRS is a circle. If chords QR and RS are equal, calculate the value of x.

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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${}^{\circ }$

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?

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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

Evaluate $\frac{8^{\frac{1}{3}}\times 5^{\frac{2}{3}}}{{10}^{\frac{2}{3}}}$ = $\frac{8^{\frac{1}{3}}\times 5^{\frac{2}{3}}}{{10}^{\frac{2}{3}}}$

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

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

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

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

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

= $\frac{72}{5}-5=\frac{72-25}{5}$

= $\frac{47}{5}$

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

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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}$

Four interior angles of a pentagon are 90o - xo, 90o + xo, 110o - 2xo, 110o + 2xo. Find the fifth interior angle

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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 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

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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

In the figure, XR and YQ are tangents to the circle YZXP if ZXR = 45o and YZX = 55o, Find ZYQ

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< RXZ = < ZYX = 45O(Alternate segment)

< ZYQ = 90 + 45

= 135O

If log102 = 0.3010 and log103 = 0.4771, evaluate; without using logarithm tables, log104.5

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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

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

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

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Given that 3x - 5y - 3 = 0, 2y - 6x + 5 = 0 the value of (x, y) is

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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 the figure, PQ is a parallel to ST and QRS = 40${}^{\circ }$. Find the value of x.

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From the figure, 3x + x - 40${}^{\circ }$ = 180${}^{\circ }$

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

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

x = $\frac{220}{4}$

= 55${}^{\circ }$

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

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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)
= $\frac{-q\pm \sqrt{q^2-4bp}}{2p}$

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

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?

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No of students offering Physics are $\frac{12}{360}$ x 150
= 5

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

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

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

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

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 $\frac{7}{\sqrt{7}}$ = $\frac{3\sqrt{7}}{7}$

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

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

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, 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

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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

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

In the distribution above, the mode and median respectively are

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