JAMB UTME - Mathematics - 1988

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

Awọn alaye Idahun

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

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.

Awọn alaye Idahun

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

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

Awọn alaye Idahun

% 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

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

Awọn alaye Idahun

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) = ($\frac{19}{24},\frac{-1}{8}$)

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?

Awọn alaye Idahun

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

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

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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 $\frac{1}{p}$ = $\frac{a^2+2ab+b^2}{a-b}$ and $\frac{l}{q}$ = $\frac{a+b}{a^2-2ab+b^2}$ Find $\frac{p}{q}$

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

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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

Awọn alaye Idahun

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 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

Awọn alaye Idahun

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

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

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

In the distribution above, the mode and median respectively are

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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

< ZYQ = 90 + 45

= 135O

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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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)
= $\frac{-q\pm \sqrt{q^2-4bp}}{2p}$

If (IPO3)4 = 11510 find P

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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

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

Awọn alaye Idahun

(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

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

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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

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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

Awọn alaye Idahun

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 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

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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

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

= $\frac{72}{5}-5=\frac{72-25}{5}$

= $\frac{47}{5}$

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?

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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

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

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

x = $\frac{220}{4}$

= 55${}^{\circ }$

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

Awọn alaye Idahun

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

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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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

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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}$

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

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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

Awọn alaye Idahun

cos60o = 1/2, cos(180o/60o) = -1/2

cos120o = -1/2

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

Awọn alaye Idahun

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

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

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

Awọn alaye Idahun

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

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

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

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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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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 = $\frac{-1}{r^2}$

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

Awọn alaye Idahun

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

Awọn alaye Idahun

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) = $\frac{6}{36}$
= $\frac{1}{6}$

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

Awọn alaye Idahun

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 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

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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