JAMB UTME - Mathematics - 1986

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

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$\theta$

Bayanin Amsa

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 + $\frac{b^2-a^2}{a^2}$
= $\frac{a^2+b^2-a^2}{a^2}$
= $\frac{b^2}{a^2}$

Factorize completely 8a + 125ax3

Bayanin Amsa

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

Bayanin Amsa

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${}^{\circ }$

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?

Bayanin Amsa

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

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)?

Bayanin Amsa

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)

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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?

Bayanin Amsa

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

Solve the equation 3x2 + 6x - 2 = 0

Bayanin Amsa

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 = $\frac{-1\pm \sqrt{15}}{3}$

Find n if log24 + log27 - log2n = 1

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

$\frac{1}{3}$(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

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?

Bayanin Amsa

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

Bayanin Amsa

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${}^{\circ }$

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?

Bayanin Amsa

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

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

Bayanin Amsa

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

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?

Bayanin Amsa

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

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

Bayanin Amsa

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

Factorize (4a + 3)2 - (3a - 2)2

Bayanin Amsa

(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)

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.

Bayanin Amsa

PT x QT = TR x TS

24 x 8 = TR x 12

TR = $\frac{24\times 8}{12}$

= = 16cm

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

Bayanin Amsa

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

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

Bayanin Amsa

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

$\frac{a}{c}$ = $\frac{c}{d}$ = 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 = $\frac{3b^2{k}^2-b{k}^{2}d+d{k}^2}{3b^2-bd+d^2}$

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.

Bayanin Amsa

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)

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

Bayanin Amsa

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

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?

Bayanin Amsa

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

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

Bayanin Amsa

cos z = $\frac{y^2+x^2-z^2}{2yx}$

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

= $\frac{29}{36}$

Factorize x2 + 2a + ax + 2x

Bayanin Amsa

Evaluate (212)3 - (121)3 + (222)3

Bayanin Amsa

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

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