JAMB UTME - Mathematics - 2008

The locus of a point equidistant from two points p(6,2) and R(4,2) is a perpendicular bisector of PR passing through

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If 125x = 2010 find x

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125x = 20
1xX2 + 2xX1 + 5xX0 = 20
X2 + 2X + 5 = 20
X2 + 2X - 15 = 0
(X + 5)( X - 3) = 0
X + 5 implies X = -5
X - 3 implies X = 3
But X cannot be negative
∴X = 3

In the diagram above ∠OPQ is

Bayanin Amsa

a = a(base ∠s of Iss Δ)
∴ a+a+74 = 180
2a + 74 = 180
2a = 180-74
2a = 106
a = 53
∴∠OPQ = 53${}^{\circ }$

Find the number of ways of selecting 6 out of 10 subjects for an examination

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The result of rolling a fair die 150 times is ass summarized in the table above. What is the probability of obtaining a 5

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Total possible outcome
12+18+x+30+2x+45 = 105+3x
∴105+3x = 150
3x = 150-105
3x = 45
x = 15
P(obtaining 5) =$\frac{2x}{(105+3x)}Butx=15=\frac{2\left(15\right)}{(105+3(15\left)\right)}=\frac{30}{(105+45)}=\frac{30}{150}=\frac{1}{5}$

Differentiate sin x - x cos x

Bayanin Amsa

sin x - x cos x
dy/dx = cos x - [1.cos x + x -sin x]
= co x - [cos x - x sin x]
= cos x - cos x + x sin x
= x sin x

Find the area of the figure given

Bayanin Amsa

Area of semicircle + Area of rectangle

A = $\frac{1}{2}\pi r^2$ + LB

A = $\frac{1}{2}\times \frac{22}{77}\times (\frac{5}{2}{)}^2+(15\times 5)$

= $\frac{1}{2}\times \frac{22}{7}\times \frac{25}{4}+75$

A = $\frac{275}{28}+\frac{75}{1}$

$\frac{275+2100}{28}=\frac{2375}{28}$

A = 84.8cm2

Evaluate $\frac{(\frac{3}{8}÷\frac{1}{2}+\frac{1}{2})}{(\frac{1}{8}\times \frac{2}{3}+\frac{1}{3})}$

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Find the range of values of x which satisfy the inequalities 4x - 7 $\le$ 3x and 3x - 4 $\le$ 4x

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4X - 7 $\le$ 3X and 3X - 4 $\le$ 4X
4X - 3X $\le$ 7 and 3X - 4X $\le$ 4
X $\le$ 7 and -X $\le$ 4 = X $\ge$ -4
Range -4 $\le$ x $\le$ 7

simplify ${16}^{\frac{-1}{2}}\times 4^{\frac{-1}{2}}\times {27}^{\frac{1}{3}}$

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The bar chart shows the number of times the word a, and, in, it, he, to appear in a paragraph in a book. What is the ratio of the least frequent word?

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Express 1223456 to 3 significant figures

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Find the capacity in liters of a cylindrical well of radius 1 meter and depth 14 meters

[π = 22/7]

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V = πr2h
1m = 100cm
14cm = 1400cm
$\therefore V=\frac{22}{7}\times \frac{100x100x1400}{1000}=44,000liters$

The fifth term of an A.P is 24 and the eleventh term is 96. Find the first term.

Bayanin Amsa

U5 = 24, n = 5 and U11 = 96, n = 11
Un = a + (n-1)d
24 = a + (5-1)d imply 24 = a+4d .....eqn1
96 = a + (11-1)d imply 96 = a+10d ...eqn2
eqn1 - eqn2 -72 = -6d
d = 72/6 = 12
but 24 = a+4d
24 = a + 4(12)
24 = a + 48
a = 24-48
a = -24

Find the mean deviation of 2, 4, 5, and 9

Bayanin Amsa

X = 20/4
= 5
mean deviation = 8/4 = 2

Find the gradient of a line which is perpendicular to the line with the equation 3x + 2y + 1 = 0

Bayanin Amsa

3X + 2Y + 1 = 0
2Y = -3X - 1
$\frac{-3}{2}X-\frac{1}{2}$
Gradient of 3X + 2Y +1 = 0 is -3/2
Gradient of a line perpendicular to 3X + 2Y + 1 = 0
$=-1÷\frac{3}{2}=-1\times \frac{-2}{3}=\frac{2}{3}$

Add 11012,101112 and 1112

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11012 + 101112 + 1112 = 101011

A book seller sells Mathematics and English books. If 30 customers buy Mathematics books, 20 customers buy English books and 10 customers buy the two books, How many customers has he altogether.

Bayanin Amsa

n(M) only = 30-10 = 20
n(E) only = 20-10 = 10
n(M∩E) = 10
∴M∪E = 20+10+10
= 40

If 2x2 - kx - 12 is divisible by x-4, Find the value of k.

Bayanin Amsa

2x2 - kx - 12 is divisible by x-4
implies x is a factor ∴ x = 4
f(4) implies 2(4)2 - k(4) - 12 = 0
32 - 4k - 12 = 0
-4k + 20 = 0
-4k = -20
k = 5

Which of the following angles is an exterior angle of a regular polygon?

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Evaluate ${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}cosxdx$

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${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}cosxdx=[sinx{]}_{\frac{-\pi }{2}}^{\frac{\pi }{2}}=sin\frac{\pi }{2}-sin\frac{-\pi }{2}$
= sin90 – sin-90
= sin90 – sin270
= 1 – (-1)
= 1+1
= 2

Find the derivative of $y=\frac{x^7-x^5}{x^4}$

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A binary operation * is defined on the set of positive integers is such x*y = 2x-3y+2 for all positive integers x and y. The binary operation is

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X * Y = 2X - 3Y + 2
2*3 = 2(2) - 3(3) + 2
=4-9+2
= -3
But -3 does not belong to positive integer

In the diagram, find the size of the angle marked ao

Bayanin Amsa

2 x s = 280o(Angle at centre = 2 x < at circum)

S = $\frac{{280}^o}{2}$

= 140

< O = 360 - 280 = 80o

60 + 80 + 140 + a = 360o

(< in a quad); 280 = a = 360

a = 360 - 280

a = 80o

If logx1/264 = 3, find the value of x

Bayanin Amsa

If logx1/264 = 3
(X 1/2)3 = 64
(X 1/2)3 = 4 3
X 1/2 = 4
X = 42
X = 16

Solve the quadratic inequalities x2 - 5x + 6 ≥0

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x2 - 5x + 6 = 0
(X-2)(X-3) = 0
X-2 = 0 implies X = 2
X-3 = 0 implies X = 3
∴ x ≤ 2, x ≥ 3

The solution of the quadratic inequality (x2 + x - 12) $\ge$ 0 is

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(x2 + x - 12) $\ge$ 0 , (x - 3)(x + 4) $\ge$ 0
For the condition to hold, each of (x - 3) and (x + 4) must be of the same sign
.i.e. x - 3 $\ge$ 0 and x + 4 $\ge$ 0
or x - 3$\le$ 0 and x + 4 $\le$ 0
when x $\ge$ 3, the condition is satisfied
when x $\ge$ -4, the condition is not satisfied.
when x $\le$ 3, the condition is not satisfied
when x $\le$ -4 , the condition is not satisfied. Thus, the solution of the inequality is x $\ge$ 3 or x $\le$ -4 ,

Factorize complete;y (4x+3y)2 - (3x-2y)2

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(4x+3y)2 - (3x-2y)2
(4x+3y+3x-2y)(4x+3y-(3x-2y))
(4x+3y+3x-2y)(4x+3y-3x+2y)
(x+5y)(7x+y)

Find the area of the figure above

[π = 22/7]

Bayanin Amsa

Area of the figure = Area of rect + area of semi circle
$=L\times h+\frac{1}{2}\pi r^{2}5\times 15+\frac{1}{2}\times \frac{22}{7}\times {\left(\frac{5}{2}\right)}^2=75+\frac{(22\times 25)}{2\times 7}=75+9\frac{23}{28}=84.8cm$

The cost of kerosine per liter increase from N60 to N85. What is the percentage rate of increase?

Bayanin Amsa

N85 - N60 = N25 increase
∴ percentage increase =$\frac{25}{60}\times \frac{100}{1}=\frac{125}{3}=41.67\%=42\%$

calculate the simple interest on N6,500 for 8 years at 5% per annum.

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If p = varies inversely as the square of q and p=8 when q=4, find when p=32

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P ∝ 1/q
P = k/q
K = q2P
= 428
∴P = 128/q
32 = 128/q
q2 = 128/32
q2 = 4
q = √4 = +/-2

Find the minimum value of X2 - 3x + 2 for all real values of x

Bayanin Amsa

y = X2 - 3x + 2, $\frac{dy}{dx}$ = 2x - 3
at turning pt, $\frac{dy}{dx}$ = 0
∴ 2x - 3 = 0
∴ x = $\frac{3}{2}$
$\frac{d^{2}y}{d{x}^2}$ = $\frac{d}{dx}$($\frac{d}{dx}$)
= 270
∴ ymin = 2$\frac{3}{2}$ - 3$\frac{3}{2}$ + 2
= $\frac{9}{4}$ - $\frac{9}{2}$ + 2
= -$\frac{1}{4}$

Find the median of 4, 1, 4, 1, 0, 4, 4, 2 and 0

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What is the mean of the data t, 2t-1, t-2, 2t-1, 4t and 2t+2?

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A binary operation on the real set of numbers excluding -1 is such that for all m, n ∈ R, mΔn = m+n+mn. Find the identity element of the operation.

Bayanin Amsa

mΔn = m+n+mn
Let e be the identity element
∴mΔe = eΔm = m
m+e+me = m
e+me = m-m
e+me = 0
e(1+m) = 0
e = 0 / (1+m)
e = 0

The bar chart above shows the number of times the word a, and , in, it, the , to appear in a paragraph in a book. What is the ratio of the least frequent word?

Bayanin Amsa

Ratio of least to most = 3:12
= 3/12
= 1/4

Find the minimum value of the function y = x(1+x)

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The probability of picking a letter T fr4om the word OBSTRUCTION is

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OBSTRUCTION
Total possible outcome = 11
Number of chance of getting T = 2
P(picking T) = 2/11

If x - 3 is directly proportional to the square of y and x = 5 when y =2, find x when y = 6.

Bayanin Amsa

(x – 3) ∝ y2
X-3 = Ky2
K = X-3 / y2
= 5-2/22
= 2/4
= 1/2
∴X-3 = 1/2y2
X-3 = 1/2(6)2
X-3 = 1/2 x 30/1
X-3 = 18
X = 21

The result of rolling a fair die 150 times is as summarized in the table given.

$\begin{array}{ccccccc}\text{Number}& 1& 2& 3& 4& 5& 6\\ \text{Frequency}& 12& 18& x& 30& 2x& 45\end{array}$

What is the probability of obtaining a 5?

Bayanin Amsa

$\begin{array}{ccccccc}Number& 1& 2& 3& 4& 5& 6\\ Frequency& 12& 18& x& 30& 2x& 45\end{array}$
12 + 18 + x + 30 + 2x + 45 = 150
3x + 105 = 150
3x = 150 - 105
3x = 45
x = $\frac{45}{3}$
x = 15
probability of 5 = $\frac{30}{150}$
= $\frac{1}{5}$

If tan $\theta$ = $\frac{5}{4}$ find sin2$\theta$ - cos2$\theta$

Bayanin Amsa

(tan $\theta$ = $\frac{opp}{adj}$)
|AB|2 = 52 + 42 $\to$ |AB|2 = 41
$\to$ AB = $\sqrt{41}$ sin2$\theta$ - cos2$\theta$
$\to$ $\frac{5^2}{\sqrt{41}}$ - (${\frac{4}{\sqrt{41}}}^2$) = $\frac{25}{41}$ - $\frac{16}{41}$
= ($\frac{9}{41}$)

If sinθ = 3/5. Find Tanθ

Bayanin Amsa

sinθ = 3/5
x2 = 52 - 32

Make Q the subject of formula when $L=\frac{4}{3}M\sqrt{PQ}$

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If X = {n2 + 1:n = 0,2,3} and Y = {n+1:n=2,3,5}, find X∩Y.

Bayanin Amsa

X = {1,5,10}
Y = {3,4,6}
X∩Y = ∅

In how many ways can the letters of the word ACCEPTANCE be arranged?

Bayanin Amsa

ACCEPTANCE = 10 Letters
A = 2 letters
C = 3 letters
E = 2 letters
Can be arranged in 10! / (2!3!2!) ways

If x > 0, find the range of number x-3, 3x+2,x-1, 4x, 2x-1, x-2, 2x-2, 3x and 3x+1

Bayanin Amsa

x-3, 3x+2,x-1, 4x, 2x-1, x-2, 2x-2, 3x, 3x+1
Range = 3x+2 - (x-3)
= 3x+2 - x - 3
= 2x + 5

If $\frac{1+\sqrt{2}}{1-\sqrt{2}}$ is expressed in the form of x+y√2 find the values of x and y

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In the diagram < OPQ is

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Evaluate ${\int }_1^2(6x^2-2x)dx$

Bayanin Amsa

${\int }_1^2(6x^2-2x)dx=[\frac{6x^3}{3}-\frac{2x^2}{2}{]}_1^2=[2x^3-x^2{]}_1^2$
= [2(2)3 - (2)2] – [2(1)3 - (1)2]
= [16-4] – [2-1]
= 12 – 1
= 11