JAMB UTME - Mathematics - 2008

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

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

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

Bayanin Amsa

Express 1223456 to 3 significant figures

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

(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 ,

If 125x = 2010 find x

Bayanin Amsa

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

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 = $\frac{30}{150}$
= $\frac{1}{5}$

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

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

Bayanin Amsa

(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 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

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?

Bayanin Amsa

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

Bayanin Amsa

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

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

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

Bayanin Amsa

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

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?

Bayanin Amsa

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

Bayanin Amsa

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

Add 11012,101112 and 1112

Bayanin Amsa

11012 + 101112 + 1112 = 101011

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 = ∅

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

Bayanin Amsa

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

Bayanin Amsa

If sinθ = 3/5. Find Tanθ

Bayanin Amsa

sinθ = 3/5
x2 = 52 - 32

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

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

Bayanin Amsa

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

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 =$\frac{25}{60}\times \frac{100}{1}=\frac{125}{3}=41.67\%=42\%$

In the diagram < OPQ is

Bayanin Amsa

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

Bayanin Amsa

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

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

[π = 22/7]

Bayanin Amsa

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

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$\theta$
$\to$ $\frac{5^2}{\sqrt{41}}$ - (${\frac{4}{\sqrt{41}}}^2$) = $\frac{25}{41}$ - $\frac{16}{41}$
= ($\frac{9}{41}$)

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

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

Bayanin Amsa

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

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

Bayanin Amsa

X = 20/4
= 5
mean deviation = 8/4 = 2

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

Bayanin Amsa

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

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

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

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

Bayanin Amsa

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

Bayanin Amsa

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

Bayanin Amsa

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

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

Bayanin Amsa