JAMB UTME - Mathematics - 1988

Detalles de la respuesta

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) = $\frac{6}{36}$
= $\frac{1}{6}$

Detalles de la respuesta

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

Detalles de la respuesta

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)

Detalles de la respuesta

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

Detalles de la respuesta

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)
= $\frac{-q\pm \sqrt{q^2-4bp}}{2p}$

Detalles de la respuesta

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

Detalles de la respuesta

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

Detalles de la respuesta

Detalles de la respuesta

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

Detalles de la respuesta

Detalles de la respuesta

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

Detalles de la respuesta

Detalles de la respuesta

Detalles de la respuesta

Detalles de la respuesta

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

Detalles de la respuesta

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

Detalles de la respuesta

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

Detalles de la respuesta

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

< ZYQ = 90 + 45

= 135O

Detalles de la respuesta

Detalles de la respuesta

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

Detalles de la respuesta

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

Detalles de la respuesta

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

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

= $\frac{72}{5}-5=\frac{72-25}{5}$

= $\frac{47}{5}$

Detalles de la respuesta

% 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

Detalles de la respuesta

Detalles de la respuesta

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

Detalles de la respuesta

(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

Detalles de la respuesta

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

Detalles de la respuesta

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${}^{\circ }$

Detalles de la respuesta

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}$

Detalles de la respuesta

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

Detalles de la respuesta

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

Detalles de la respuesta

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

Detalles de la respuesta

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

Detalles de la respuesta

Detalles de la respuesta

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

Detalles de la respuesta

Detalles de la respuesta

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

Detalles de la respuesta

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

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

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

x = $\frac{220}{4}$

= 55${}^{\circ }$

Detalles de la respuesta

Detalles de la respuesta

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

Detalles de la respuesta

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

Detalles de la respuesta

cos60o = 1/2, cos(180o/60o) = -1/2

cos120o = -1/2

Detalles de la respuesta

Detalles de la respuesta

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

Detalles de la respuesta

Detalles de la respuesta

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

Detalles de la respuesta

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

Detalles de la respuesta

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

Detalles de la respuesta

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