JAMB UTME - Mathematics - 2008

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

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

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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
$=-1÷\frac{3}{2}=-1\times \frac{-2}{3}=\frac{2}{3}$

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?

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

Find the area of the figure given

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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

In the diagram < OPQ is

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

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

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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)

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

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X = {1,5,10}
Y = {3,4,6}
X∩Y = ∅

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

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

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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

In the diagram above ∠OPQ is

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

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

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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

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

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

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

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Express 1223456 to 3 significant figures

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Add 11012,101112 and 1112

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11012 + 101112 + 1112 = 101011

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

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 area of the figure above

[π = 22/7]

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

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

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) =$\frac{2x}{(105+3x)}Butx=15=\frac{2\left(15\right)}{(105+3(15\left)\right)}=\frac{30}{(105+45)}=\frac{30}{150}=\frac{1}{5}$

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 logx1/264 = 3, find the value of x

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

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?

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

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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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 mean deviation of 2, 4, 5, and 9

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X = 20/4
= 5
mean deviation = 8/4 = 2

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

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(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

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.

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

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

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

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

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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

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

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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

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

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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 ,

Differentiate sin x - x cos x

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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

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

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

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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
= -$\frac{1}{4}$

Evaluate ${\int }_1^2(6x^2-2x)dx$

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

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

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

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

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

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

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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
$\therefore V=\frac{22}{7}\times \frac{100x100x1400}{1000}=44,000liters$

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

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If sinθ = 3/5. Find Tanθ

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sinθ = 3/5
x2 = 52 - 32

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

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

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.

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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