JAMB UTME - Mathematics - 1986

If x varies directly as y3 and x = 2 when y = 1, find x when y = 5

Answer Details

x $\alpha$ y3
x = ky3
k = $\frac{x}{y^3}$
when x = 2, y = 1
k = 2
Thus x = 2y3 - equation of variation
= 2(5)3
= 250

The table below gives the scores of a group of students in a Mathematical test.

$\begin{array}{ccccccccc}Scores& 1& 2& 3& 4& 5& 6& 7& 8\\ Frequency& 2& 4& 7& 14& 12& 6& 4& 1\end{array}$

If the mode in m and the number of students who scored 4 or less is s. What is (s, m)?

Answer Details

M = mode = the number having the highest frequency = 4
S = Number of students with 4 or less marks
= 14 + 7 + 4 + 2
= 27
∴ (M,S) = (27, 4)

Three boys shared some oranges. The first received 1/3 of the oranges and the second received 2/3 of the remaining. If the third boy received the remaining 12 oranges, how many oranges did they share

Answer Details

Let x = the number of oranges
The 1st received 1/3 of x = 1/3x
∴Remainder = x - 1/3x = 2x/3
The 2nd received 2/3 of 2x/3 = 2/3 * 2x/3 = 4x/3
The 3rd received 12 oranges
∴1/3x + 4x/9 + 12 = x
(3x + 4x + 108)/9 = x
3x + 4x + 108 = 9x
7x + 108 = 9x
9x - 7x = 108
2x = 108
x = 54 oranges

In 1984, Ike was 24 yrs old and his father was 45 yrs old. In what year was Ike exactly half his father's age?

Answer Details

Let the no. of years be y
24 - y = $\frac{1}{2}$(45 - y)
45 - y = 2(24 - y)
45 - y = 48 - 2y
2y - y = 48 - 45
∴ y = 3
The exact year = 1984
1984 - 3 = 1981

A man kept 6 black, 5 brown and 7 purple shirts in a drawer. What is the probability of his picking a purple shirt with his eyes closed?

Answer Details

Prob. of purple shirt is $\frac{7}{18}$

Simplify $\frac{9^{\frac{1}{3}}\times {27}^{-\frac{1}{3}}}{3^{-\frac{1}{6}}\times 3^{\frac{2}{3}}}$

Answer Details

Find the probability that a number selected at random from 40 to 50 is a prime

Answer Details

From 40 to 50 = 11 & number are prime i.e. 41, 43, 47
prob. of selecting a prime No. is $\frac{3}{11}$

Find the reciprocal of $\frac{\frac{2}{3}}{\frac{1}{2}+\frac{1}{3}}$

Answer Details

$\frac{2}{3}$ = $\frac{2}{3}$
= $\frac{2}{3}$ x $\frac{6}{5}$
= $\frac{4}{5}$
reciprocal of $\frac{4}{5}$ = $\frac{1}{\frac{4}{5}}$
= $\frac{5}{4}$

Udoh deposited ₦150.00 in the bank. At the end of 5 years the simple interest on the principal was ₦55.00. At what rate per annum was the interest paid?

Answer Details

.I = $\frac{PTR}{100}$
R = $\frac{100}{PT}$
$\frac{100\times 50}{150\times 6}$ = $\frac{22}{3}$
= 7$\frac{1}{3}$%

The figure is a solid with the trapezium PQRS as its uniform cross-section. Find its volume

Answer Details

Volume of solid = cross section x H

Since the cross section is a trapezium

= $\frac{1}{2}(6+11)\times 12\times 8$

= 6 x 17 x 8 = 816m3

The ages of Tosan and Isa differ by 6 and the product of their ages is 187. Write their ages in the form (x, y), where x > y.

Answer Details

x - y = 6.......(i)
xy = 187.......(ii)
From equation (i), x(6 + y)
sub. for x in equation (ii) = y(6 + y)
= 187
y2 + 6y = 187
y2 + 6y - 187 = 0
(y + 17)(y - 11) = 0
y = -17 or y = 11
y cannot be negative, y = 11
Sub. for y in equation(i) = x - 11
= 16
x = 6 + 11
= 17
∴(x, y) = (17, 11)

Find n if log24 + log27 - log2n = 1

Answer Details

log24 + log27 - log2n = 1
= log2(4 x 7) - log2n = 1
= log228 - log2n
= log$\frac{28}{2n}$
$\frac{28}{2n}$ = 21
= 2
2n = 28
∴ n = 14

Two chords QR and NP of a circle intersect inside the circle at x. If RQP = 37o, RQN = 49o and QPN = 35o, find PRQ

Answer Details

In PNO, ONP
= 180 - (35 + 86)
= 180 - 121
= 59
PRQ = QNP = 59(angles in the same segment of a circle are equal)

Simplify $\frac{0.0324\times 0.00064}{0.48\times 0.012}$

Answer Details

$\frac{0.0324\times 0.00064}{0.48\times 0.012}$ = $\frac{324\times {10}^{-4}\times {10}^{-5}}{48x{10}^{-2}x12x{10}^{-3}}$
$\frac{324\times 64\times 10^{-9}}{48\times 12\times {10}^{-5}}$ = 36 x 10-4
= 3.6 x 10-3

Evaluate (212)3 - (121)3 + (222)3

Answer Details

If U and V are two distinct fixed points and W is a variable points such that UWV is a right angle, what is the locus of W?

Answer Details

Solve the equation 3x2 + 6x - 2 = 0

Answer Details

3x2 + 6x - 2 = 0
Using almighty formula i.e. x = $\frac{b\pm \sqrt{b^2-4ac}}{2a}$
a = 3, b = 6, c = -2
x = $\frac{-6\pm \sqrt{6^2-4\left(3\right)(-2)}}{2\left(3\right)}$
x = $\frac{6\pm \sqrt{36-24}}{6}$
x = $\frac{6\pm \sqrt{60}}{6}$
x = $\frac{-6\pm \sqrt{4\times 15}}{6}$
x = $\frac{-1\pm \sqrt{15}}{3}$

An open rectangular box externally measures 4m x 3m x 4m. Find the total cost of painting the box externally if it costs ₦2.00 to paint one square meter

Answer Details

Total surface area(s) = 2(4 x 3) + 2(4 x 4)
= 2(12) + 2(16)
= 24 + 32
= 56cm2
1m2 costs ₦2.00
∴ 56m∴ will cost 56 x ₦2.00
= ₦112.00

Find the median of the numbers 89, 141, 130, 161, 120, 131, 131, 100, 108 and 119

Answer Details

Arrange in ascending order
89, 100, 108, 119, |120, 130|, 131, 131, 141, 161
Median = $\frac{120+130}{2}$
= 125

Divide the L.C.M of 48, 64, and 80 by their H.C.F

Answer Details

48 = 24 x 3, 64 = 26, 80 = 24 x 5
L.C.M = 26 x 3 x 5
H.C.F = 24
$\frac{2^6\times 3\times 5}{2^4}$
= 22 x 3 x 5
= 4 x 3 x 5
= 12 x 5
= 60

Simplify ($\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}$ x $\frac{1}{\sqrt{3}}$)

Answer Details

($\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}$ x $\frac{1}{\sqrt{3}}$)
$\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}$ = $\frac{\sqrt{5}-\sqrt{3}-(\sqrt{5}+\sqrt{3})}{(\sqrt{5}+\sqrt{3})\left(5\sqrt{3}\right)}$
= $\frac{-2\sqrt{3}}{5-3}$
= $\frac{-2\sqrt{3}}{7}$

Find all real numbers x which satisfy the inequality $\frac{1}{3}$(x + 1) - 1 > $\frac{1}{5}$(x + 4)

Answer Details

$\frac{1}{3}$(x + 1) - 1 > $\frac{1}{5}$(x + 4)
= $\frac{x+1}{3}$ - 1 > $\frac{x+4}{5}$
$\frac{x+1}{3}$ - $\frac{x+4}{5}$ - 1 > 0
= $\frac{5x+5-3x-12}{15}$
= 2x - 7 > 15
= 2x > 12
= x > 11

An arc of a circle of radius 6cm is 8cm long. Find the area of the sector

Answer Details

Radius of the circle r = 6cm, Length of the arc = 8cm
Area of sector = $\frac{\theta }{360}$ x 2$\pi$r2........(i)
Length of arc = $\frac{\theta }{360}$ x 2$\pi$r........(ii)
from eqn. (ii) $\theta$ = $\frac{240}{\pi }$, subt. for $\theta$ in eqn (i)
Area x $\frac{240}{1}$ x $\frac{1}{360}$ x $\frac{\pi 6}{1}$
= 24cm2

A number of pencils were shared out among Bisi, Sola and Tunde in the ratio of 2 : 3 : 5 respectively. If Bisi got 5, how many were share out?

Answer Details

Let x r3epresent total number of pencils shared

B : S : T = 2 + 3 + 5 = 10

2 : 3 : 5

= 210 $\frac{2}{10}$ x y

= 5

2y =5

2y = 50

∴ y = 502 $\frac{50}{2}$

= 25

Simplify $\frac{1}{x-2}$ + $\frac{1}{x+2}$ + $\frac{2x}{x^2-4}$

Answer Details

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

Factorize x2 + 2a + ax + 2x

Answer Details

If P = 18, Q = 21, R = -6 and S = -4, Calculate $\frac{(P-Q{)}^3+S^2}{R^3}$ + S2

Answer Details

$\frac{(P-Q{)}^3+S^2}{R^3}$ = $\frac{(18-21{)}^3+(-4{)}^2}{(-6{)}^3}$
= $\frac{-27+16}{R^3}$
= $\frac{-11}{-216}$
= $\frac{11}{216}$

If $\frac{a}{c}$ = $\frac{c}{d}$ = k, find the value of $\frac{3a^2?ac+c^2}{3b^2?bd+d^2}$

Answer Details

$\frac{a}{c}$ = $\frac{c}{d}$ = k
∴ $\frac{a}{b}$ = bk
$\frac{c}{d}$ = k
∴ c = dk
= $\frac{3a^2-ac+c^2}{3b^2-bd+d^2}$
= $\frac{3(bk{)}^2-(bk\left)\right(dk)+d{k}^2}{3b^2-bd+a^2}$
= $\frac{3b^2{k}^2-b{k}^{2}d+d{k}^2}{3b^2-bd+d^2}$
k = $\frac{3b^2{k}^2-b{k}^{2}d+d{k}^2}{3b^2-bd+d^2}$

make U the subject of the formula S = $\sqrt{\frac{6}{u}-\frac{w}{2}}$

Answer Details

S = $\sqrt{\frac{6}{u}-\frac{w}{2}}$
S = $\frac{12-uw}{2u}$
2us2 = 12 - uw
u(2s2 + w) = 12
u = $\frac{12}{2s^2+w}$

If 5(x + 2y) = 5 and 4(x + 3y) = 16, find 3(x + y)

Answer Details

5(x + 2y) = 5
∴ x + 2y = 1.....(i)
4(x + 3y) = 16 = 42
x + 3y = 2 .....(ii)
x + 2y = 1.....(i)
x + 3y = 2......(ii)
y = 1
Substitute y = 1 into equation (i) = x + 2y = 1
∴ x + 2(1) = 1
x + 2 = 1
∴ x = 1
∴ 3x + y = 3-1 + 1
= 3 = 1

Of the nine hundred students admitted in a university in 1979, the following was the distribution by state: Anambrs 185, Imo 135, Kaduna 90, Kwara 110, Ondo 155, Oyo 225. In a pie chart drawn to represent this distribution, The angle subtended at the centre by anambra is

Answer Details

Anambra = $\frac{185}{900}$ x $\frac{360}{1}$
= 74o

Find the total surface area of solid cone of radius 2$\sqrt{3}$cm and slanting side 4$\sqrt{3}$

Answer Details

Total surface area of a solid cone
r = 2$\sqrt{3}$
= $\pi r^2$ + $\pi$rH
H = 4$\sqrt{3}$, $\pi$r(r + H)
∴ Area = $\pi$2$\sqrt{3}$ [2$\sqrt{3}$ + 4$\sqrt{3}$]
= $\pi$2$\sqrt{3}$(6$\sqrt{3}$)
= 12$\pi$ x 3
= 36$\pi$cm2

If Musa scored 75 in biology instead of 57, his average mark in four subjects would have been 60. What was his total mark?

Answer Details

Let x represent Musa's total mark when he scores 57 in biology and Let Y represent Musa's total mark when he now scored 75 in biology, if he scored 75 in biology his new total mark will be $\frac{Y}{4}$ = 60, y = 4 x 60 = 240
To get his total mark when he scored 57, subtract 57 from 75 to give 18, then subtract this 18 from the new total mark(ie. 240)
= 240 - 18
= 222

If cos$\theta$ = $\frac{a}{b}$, find 1 + tan2$\theta$

Answer Details

cos$\theta$ = $\frac{a}{b}$, Sin$\theta$ = $\sqrt{\frac{b^2-a^2}{a}}$
Tan$\theta$ = $\sqrt{\frac{b^2-a^2}{a^2}}$, Tan 2 = $\sqrt{\frac{b^2-a^2}{a^2}}$
1 + tan2$\theta$ = 1 + $\frac{b^2-a^2}{a^2}$
= $\frac{a^2+b^2-a^2}{a^2}$
= $\frac{b^2}{a^2}$

If y = $\frac{x}{x-3}$ + $\frac{x}{x+4}$ find y when x = -2

Answer Details

y = $\frac{x}{x-3}$ + $\frac{x}{x+4}$ when x = -2
y = $\frac{-2}{-5}$ + $\frac{(-2)}{-2+4}$
= $\frac{-2}{5}$ + $\frac{-2}{2}$
= $\frac{4+-10}{10}$
= $\frac{-14}{10}$
= -$\frac{7}{5}$

A girl walks 45 meters in the direction 050o from a point Q to a point X. She then walks 24 meters in the direction 140o from X to a point Y. How far is she then from Q?

Answer Details

QY = 452 + 242 = 2025 + 576
= 2601
QY = $\sqrt{2601}$
= 51

In the diagram, PQ and RS are chords of a circle centre O which meet at T outside the circle. If TP = 24cm. TQ = 8cm and TS = 12cm, find TR.

Answer Details

PT x QT = TR x TS

24 x 8 = TR x 12

TR = $\frac{24\times 8}{12}$

= = 16cm

Find the smallest number by which 252 can be multiplied to obtain a perfect square

Answer Details

Let the smallest number be x and the perfect square be y 252x = y.
By trial and error method, 252 x 9 = 1764
Check if y = 1764
y2 = 42
x = 7

Factorize completely 8a + 125ax3

Answer Details

8a + 125ax3 = 23a + 53ax3
= a(23 + 53x3)
∴a[23 + (5x)3]
a3 + b3 = (a + b)(a2 - ab + b2)
∴ a(23 + (5x)3)
= a(2 + 5x)(4 - 10x + 25x2)

In $△$XYZ, determine the cosine of angle Z.

Answer Details

cos z = $\frac{y^2+x^2-z^2}{2yx}$

= $\frac{9+36-16}{2\left(3\right)\left(6\right)}$

= $\frac{29}{36}$

Find the values of x which satisfy the equation 16x - 5 x 4x + 4 = 0

Answer Details

16x - 5 x 4x + 4 = 0
(4x)2 - 5(4x) + 4 = 0
let 4x = y
y2 - 5y + 4 = 0
(y - 4)(y - 1) = 0
y = 4 or 1
4x = 4
x = 1
4x = 1
i.e. 4x = 4o, x = 0
∴ x = 1 or 0

In the figure, $△$PQT is isosceles. PQ = QT, SRQ = 35${}^{\circ }$, TPQ = 20${}^{\circ }$ and PQR is a straight line.Calculate TSR

Answer Details

Given $△$ isosceles PQ = QT, SRQ = 35${}^{\circ }$

TPQ = 20${}^{\circ }$

PQR = is a straight line

Since PQ = QT, angle P = angle T = 20${}^{\circ }$

Angle PQR = 180${}^{\circ }$ - (20 + 20) = 140${}^{\circ }$

TQR = 180${}^{\circ }$ - 140${}^{\circ }$ = 40${}^{\circ }$ < on a straight line

QSR = 180${}^{\circ }$ - (40 + 35)${}^{\circ }$ = 105${}^{\circ }$

TSR = 180${}^{\circ }$ - 105${}^{\circ }$

= 75${}^{\circ }$

PQ and PR are tangents from P to a circle centre O as shown in the figure. If QRP = 34${}^{\circ }$, find the angle marked x

Answer Details

From the circle centre 0, if PQ & PR are tangents from P and QRP = 34${}^{\circ }$

Then the angle marked x i.e. QOP

34${}^{\circ }$ x 2 = 68${}^{\circ }$

Factorize (4a + 3)2 - (3a - 2)2

Answer Details

(4a + 3)2 - (3a - 2)2 = a2 - b2
= (a + b)(a - b)
= [(4a + 3) + (3a - 2)][(4a + 3) + (3a - 2)]
= [(4a + 3 + 3a - 2)][(4a + 3 - 3a + 2)]
= (7a + 1)(a + 5)
∴ (a + 5)(7a + 1)

At what points does the straight line y = 2x + 1 intersect the curve y = 2x2 + 5x - 1?

(-2, -3) and($\frac{1}{2}$, 0)

Answer Details

Simplify $\frac{1}{5x+5}$ + $\frac{1}{7x+7}$

Answer Details

$\frac{1}{5x+5}$ + $\frac{1}{7x+7}$ = $\frac{1}{5(x+1)}$ + $\frac{1}{7(x+1)}$
= $\frac{7+5}{35(x+1)}$
= $\frac{12}{35(x+1)}$

The people in a city with a population of 0.9 million were grouped according to their ages. Use the diagram to determine the number of people in the 15 - 29 years group

Answer Details

15 - 29 years is represented by 104${}^{\circ }$

Number of people in the group is $\frac{104}{360}$ x 0.9m

= 260000 = 26 x 104

A regular polygon of n sides has 160o as the size of each interior angle. Find n

Answer Details

Interior + exterior = 360
160 + exterior = 360
Exterior = 360 - 160
Exterior = 20
n = 360/exterior
n = 360/20
n = 18