JAMB UTME - Mathematics - 1988

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

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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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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

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

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

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 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) = $\frac{6}{36}$
= $\frac{1}{6}$

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${}^{\circ }$

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

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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

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

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

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

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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)
= $\frac{-q\pm \sqrt{q^2-4bp}}{2p}$

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 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 (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

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

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}$

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

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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

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

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

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

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 = $\frac{24-19}{8}$
= $\frac{5}{8}$