JAMB UTME - Mathematics - 1998

The shaded area represents

x

\leq

0, y

\leq

0, 2y + 3x

\leq

6

x

\geq

0, y

\geq

3, 3x + 2y

\geq

6

x

\geq

2, y

\geq

0, 3x + 2y

\leq

6

x

\geq

0, y

\geq

0, 3x + 2y

\geq

6

Akọwa Nkọwa

m =

\frac{y_{2} - y_{1}}{x_{2} - x_{2}} = \frac{3 - 0}{0 - 2} = \frac{- 3}{2}

= $\frac{y - y_{1}}{x - x_{1}}$

m = $\frac{y - 3}{x}$ $\geq$ $\frac{- 3}{2}$

2(y - 3) $\geq$ - 3x = 2y - 6 $\geq$ - 3x

= 2y + 3x $\leq$ 6 ; x $\leq$ 0, y $\leq$ 0

Lee Azịza

Ajụjụ 2 Ripọtì

A chord of a circle radius $ \sqrt{3}\,\mathrm{cm} $ subtends an angle of $60^\circ$ on the circumference of he circle. Find the length of the chord

\frac{\sqrt{3}}{2}

\frac{3}{2}

3

\sqrt{3}

cm

Circles

Lee Azịza

Ajụjụ 3 Ripọtì

For what value of $x$ does $6\sin(2x - 25)^\circ$ attain its maximum value in the range $0^\circ \le x \le 180^\circ$

12

\frac{1}{2}

32

\frac{1}{2}

57

\frac{1}{2}

147

\frac{1}{2}

Akọwa Nkọwa

The maximum value of the function y = 6sin(2x-25) occurs when the argument of the sine function is equal to 90 degrees (or pi/2 radians), because the maximum value of the sine function is 1. So we need to solve the equation 2x-25 = 90. Adding 25 to both sides gives 2x = 115, and then dividing by 2 gives x = 57.5 degrees. Therefore, the answer is x = 57.5 degrees, which is the value of x where the function 6sin(2x-25) attains its maximum value in the given range.

Gụọ akwụkwọ nkuzi na Sine, Cosine And Tangent Of An Angle (WAEC)

Gụọ akwụkwọ nkuzi na Trigonometry (JAMB)

Sine, Cosine And Tangent Of An Angle

Trigonometry

Lee Azịza

Ajụjụ 4 Ripọtì

The pie chart shows the monthly expenditure pf a public servant. The monthly expenditure on housing is twice that of school fees. How much does the worker spend on housing if his monthly income is ₦7200?

₦1000

₦2000

₦3000

₦4000

Akọwa Nkọwa

Based on the information given in the pie chart, the monthly expenditure on school fees can be represented by the fraction 1/10 (since it is one-tenth of the total expenditure) and the monthly expenditure on housing can be represented by the fraction 2/10 (since it is twice the amount of school fees). To find out how much the worker spends on housing, we need to first calculate the total amount of money he spends on all his monthly expenses. From the pie chart, we see that the total monthly expenditure is represented by the fraction 7/10. If we let x represent the amount of money the worker spends on housing each month, we can set up the following equation to solve for x: x + (1/10)*7200 = (2/10)*7200 Simplifying the equation, we get: x + 720 = 1440 Subtracting 720 from both sides, we get: x = 720 Therefore, the worker spends ₦7200/10 = ₦720 on school fees each month, and ₦720 * 2 = ₦1440 on housing each month. So the answer is ₦2000.

Lee Azịza

Ajụjụ 5 Ripọtì

The sum of the first three terms of a geometric progression is half its sum to infinity. Find the positive common ratio of the progression.

\frac{1}{4}

\sqrt{\frac{3}{2}}

\frac{1}{\sqrt{3}}

\frac{1}{\sqrt{2}}

Akọwa Nkọwa

Let the G.P be a, ar, ar2, S3 =

\frac{1}{2}

S
a + ar + ar2 =

\frac{1}{2}

(

\frac{a}{1 - r}

)
2(1 + r + r)(r - 1) = 1
= 2r3 = 3
= r3 =

\frac{3}{2}

r(

\frac{3}{2}

)

\frac{1}{3}

=

\sqrt{\frac{3}{2}}

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Sequence And Series

Lee Azịza

Ajụjụ 6 Ripọtì

Find the positive value of x if the standard deviation of the numbers 1, x + 1 is 6

1

2

3

4

Akọwa Nkọwa

mean (x) =

\frac{1 + x + 1 + 2 x + 1}{3}

=

\frac{3 x + 3}{3}

= 1 + x

\begin{matrix} X & (X - X) & (X - X)^{2} \\ 1 & - x & x^{2} \\ x + 1 & 0 & 0 \\ 2 x + 1 & x & x^{2} \\ 2 x^{2} \end{matrix}

S.D =

\sqrt{\frac{\sum (x - 7)^{2}}{\sum f}}

=

{\sqrt{(6)}}^{2}

=

\frac{2 x^{2}}{3}

= 2x2
= 18
x2 = 9
∴ x =

\pm

\sqrt{9}

=

\pm

3

Gụọ akwụkwọ nkuzi na Measures Of Dispersion (JAMB)

Measures Of Dispersion

Lee Azịza

Ajụjụ 7 Ripọtì

A market woman sells oil in cylindrical tins 10cm deep and 6cm in diameter at ₦15.00 each. If she bought a full cylindrical jug 18cm deep and 10cm in diameter for ₦50.00, how much did she make by selling all the oil?

₦62.50

₦35.00

₦31.00

₦25.00

Akọwa Nkọwa

The volume of the cylindrical jug is given by V = πr^2h, where r = 5cm (radius = diameter/2) and h = 18cm. Thus, V = π(5cm)^2(18cm) = 450π cm^3. The volume of each cylindrical tin is given by V = πr^2h, where r = 3cm (radius = diameter/2) and h = 10cm. Thus, V = π(3cm)^2(10cm) = 90π cm^3. Since the jug contains oil that can fill 450π/90π = 5 cylindrical tins, the market woman sells 5 tins of oil. Therefore, the amount she makes by selling all the oil is 5 × ₦15.00 = ₦75.00. Since she bought the jug for ₦50.00, her profit is ₦75.00 - ₦50.00 = ₦25.00. Therefore, the market woman made ₦25.00 by selling all the oil. Hence, the answer is option D: ₦25.00.

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Areas

Lee Azịza

Ajụjụ 8 Ripọtì

A bag contains 16 red balls and 20 blue balls only. How many white balls must be added to the bag so that the probability of randomly picking a red ball is equal to $ \frac{2}{5} $

4

20

24

40

Akọwa Nkọwa

Number of red balls = 16,
Number of blue balls = 20
Let x represent the No of white balls to be added
∴ Total number of balls = 36 + x
2(36 + x) = 80
= 2x + 80 - 72
= 8
x =

\frac{8}{2}

= 4

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Areas

Lee Azịza

Ajụjụ 9 Ripọtì

The midpoint of the segment of the line y = 4x + 3 which lies between the x-ax 1 is and the y-ax 1 is

(

\frac{3}{2}

,

\frac{3}{2}

)

(

\frac{2}{3}

,

\frac{3}{2}

)

(

\frac{3}{8}

,

\frac{3}{2}

)

(-

\frac{3}{8}

,

\frac{3}{2}

)

Akọwa Nkọwa

To find the midpoint of a line segment, we need to find the average of the endpoints. The x-intercept of the line y = 4x + 3 is found by setting y = 0 and solving for x: 0 = 4x + 3 x = -3/4 So the x-coordinate of the midpoint is the average of -3/4 and 0: x = (-3/4 + 0)/2 = -3/8 To find the y-coordinate of the midpoint, we plug in x = -3/8 to the equation of the line: y = 4(-3/8) + 3 = -3/2 + 3 = 3/2 So the midpoint is (-3/8, 3/2). Therefore, the answer is (-3/8, 3/2).

Gụọ akwụkwọ nkuzi na Coordinate Geometry (JAMB)

Coordinate Geometry

Lee Azịza

Ajụjụ 10 Ripọtì

If the distance between the points (x, 3) and (-x, 2) is 5. Find x

6.0

2.5

\sqrt{6}

\sqrt{6}

Akọwa Nkọwa

To solve the problem, we need to use the distance formula between two points in a coordinate plane. The distance formula is given by: d = sqrt[(x2 - x1)^2 + (y2 - y1)^2] where d is the distance between the two points (x1, y1) and (x2, y2). Using the given points, we have: (x2, y2) = (-x, 2) (x1, y1) = (x, 3) Substituting these values into the distance formula, we get: d = sqrt[(-x - x)^2 + (2 - 3)^2] Simplifying the expression inside the square root: d = sqrt[4x^2 + 1] We are given that d = 5, so we can substitute that into the equation and solve for x: 5 = sqrt[4x^2 + 1] 25 = 4x^2 + 1 24 = 4x^2 6 = x^2 x = sqrt(6) or x = -sqrt(6) Since we are only interested in the positive value of x, the answer is x = sqrt(6). Therefore, the correct option is:

\sqrt{6}

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Modular Arithmetic

Lee Azịza

Ajụjụ 11 Ripọtì

If $y = 243(4x + 5)^{-2}$, find $\frac{dy}{dx}$ when $x = 1$.

\frac{- 8}{3}

\frac{3}{8}

\frac{9}{8}

-

\frac{8}{9}

Akọwa Nkọwa

To find dy/dx, we need to differentiate y with respect to x. We can start by using the chain rule, which states that if we have a function of the form f(g(x)), the derivative of that function with respect to x is f'(g(x)) * g'(x). In this case, we have y = 243(4x + 5)-2, which can be written as y = 243/(4x + 5)^2. Using the chain rule, we have: dy/dx = d/dx(243/(4x + 5)^2) = -2 * 243 / (4x + 5)^3 * d/dx(4x + 5) = -2 * 243 / (4x + 5)^3 * 4 = -1944 / (4x + 5)^3 Now, to find the value of dy/dx when x = 1, we just need to substitute x = 1 into the expression we found above: dy/dx = -1944 / (4(1) + 5)^3 = -1944 / 729 = -8/3 Therefore, the answer is option A: -83. To summarize, we used the chain rule to differentiate y with respect to x, which gave us an expression for dy/dx in terms of x. We then substituted x = 1 to find the value of dy/dx at that point.

Gụọ akwụkwọ nkuzi na Differentiation (JAMB)

Differentiation

Lee Azịza

Ajụjụ 12 Ripọtì

If x is a positive real number, find the range of values for which $ \frac{1}{3x} + \frac{1}{2} > \frac{1}{4x} $

0 > -

\frac{1}{6}

x > 0

0 < x < 4

0 < x <

\frac{1}{6}

Akọwa Nkọwa

\frac{1}{3 x}

+

\frac{1}{2}

>

\frac{1}{4 x}

=

\frac{2 + 3 x}{6 x}

>

\frac{1}{4 x}

= 4(2 + 3x) > 6x = 12x2 - 2x = 0
= 2x(6x - 1) > 0 = x(6x - 1) > 0
Case 1 (-, -) = x < 0, 6x -1 < 0

= x < 0, x < $\frac{1}{6}$ = x < $\frac{1}{6}$ (solution)

Case 2 (+, +) = x > 0, 6x -1 > 0 = x > 0, x > $\frac{1}{6}$
Combining solutions in cases(1) and (2)
= x > 0, x < $\frac{1}{6}$ = 0 < x < $\frac{1}{6}$

Lee Azịza

Ajụjụ 13 Ripọtì

In a recent zonal championship games involving 10 teams, teams X and Y were given probabilities $ \frac{2}{5} $ and $ \frac{1}{3} $ respectively of winning the gold in the football event. What is the probability that either team will win the gold?

\frac{2}{15}

\frac{7}{15}

\frac{11}{15}

\frac{13}{15}

Akọwa Nkọwa

p(x) =

\frac{2}{5}

p(y) =

\frac{1}{3}

p(x or y) = p(x ∪ y)
= p(x) + p(y)
=

\frac{2}{5}

+

\frac{1}{3}

=

\frac{11}{5}

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Probability

Lee Azịza

Ajụjụ 14 Ripọtì

If $b^3 = a^{-2}$ and $c^{\frac{1}{3}} = a^{\frac{1}{2}}b$, express c in terms of a

a-

\frac{1}{2}

a

\frac{1}{3}

a

\frac{3}{2}

a

\frac{2}{3}

Akọwa Nkọwa

c

\frac{1}{3}

= a

\frac{1}{2}

b
= a

\frac{1}{2}

b x a-2
= a-

\frac{3}{2}

= (c

\frac{1}{3}

)3
= (a-

\frac{3}{2}

)

\frac{1}{3}

c = a-

\frac{1}{2}

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Solution Of Linear Equations

Number Bases

Lee Azịza

Ajụjụ 15 Ripọtì

When the expression pm2 + qm + 1 is divided by (m - 1), it has a remainder is 4, Find p and q respectively

2, -1

-1, 2

3, -2

-2, 3

Akọwa Nkọwa

pm2 + qm + 1 = (m - 1) Q(x) + 2
p(1)2 + q(1) + 1 = 2
p + q + 1 = 2
p + q = 1.....(i)
pm2 + qm + 1 = (m - 1)Q(x) + 4
p(-1)2 + q(-1) + 1 = 4
p - q + 1 = 4
p - q = 3....(ii)
p + q = 1, p - q = -3
2p = -2, p = -1
-1 + q = 1
q = 2

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 16 Ripọtì

Average hourly earnings(N)	5−9	10−14	15−19	20−24
No. of workers	17	32	25	24

Estimate the mode of the above frequency distribution

12.2

12.27

12.9

13.4

Akọwa Nkọwa

\begin{array}{cc} Class intervals & F & Class boundary \\ 5 - 7 & 17 & 4.5 - 9.5 \\ 10 - 14 & 32 & 9.5 - 14.5 \\ 15 - 19 & 25 & 14.5 - 19.5 \\ 20 - 24 & 24 & 19.5 - 24.5 \end{array}

mode = 9 +

\frac{D_{1}}{D_{2} + D_{1}}

x C
= 9.5 +

\frac{5 (32 - 17)}{2 (32) - 17 - 25}

= 9.5 +

\frac{75}{27}

= 12.27

\approx

2.3

Gụọ akwụkwọ nkuzi na Linear Inequalities (WAEC)

Gụọ akwụkwọ nkuzi na Euclidean Geometry (JAMB)

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Linear Inequalities

Euclidean Geometry

Statistics

Lee Azịza

Ajụjụ 17 Ripọtì

Find the value of k if $ \frac{k}{\sqrt{3}+\sqrt{2}} = k\sqrt{3-2} $

3

2

√3

√2

Akọwa Nkọwa

\frac{k}{\sqrt{3} + \sqrt{2}}

= k

\sqrt{3 - 2}

\frac{k}{\sqrt{3} + \sqrt{2}}

x

\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}

= k

\sqrt{3 - 2}

= k(

\sqrt{3} - \sqrt{2}

)
= k

\sqrt{3 - 2}

= k

\sqrt{3}

- k

\sqrt{2}

= k

\sqrt{3 - 2}

k2 =

\sqrt{2}

k =

\frac{2}{\sqrt{2}}

=

\sqrt{2}

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Gụọ akwụkwọ nkuzi na Graphs Of Linear And Quadratic Functions (WAEC)

Surds (radicals)

Quadratic Equations

Graphs Of Linear And Quadratic Functions

Lee Azịza

Ajụjụ 18 Ripọtì

If 10112 + x7 = 2510, solve for X.

207

147

227

247

Akọwa Nkọwa

10112 + x7 = 2510 = 10112 = 1 x 23 + 0 x 22 + 1 x 21 + 1 x 2o
= 8 + 0 + 2 + 1
= 1110
x7 = 2510 - 1110
= 1410
$\begin{matrix} 7 & 14 \\ 7 & 2 R 0 \\ 0 R 2 \end{matrix}$
X = 207

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Number Bases

Lee Azịza

Ajụjụ 19 Ripọtì

Let $ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $ p = $ \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix} $ Q = $ \begin{pmatrix} u & 4 + u \\ -2v & v \end{pmatrix} $ be 2 x 2 matrices such that PQ = 1. Find (u, v)

(-

\frac{5}{2}

- 1)

(-

\frac{5}{2}

-

\frac{3}{2}

)

(-

\frac{5}{6}

- 1)

(

\frac{5}{2}

-

\frac{3}{2}

)

Akọwa Nkọwa

PQ =

(\begin{matrix} 2 & 3 \\ 4 & 5 \end{matrix})

(\begin{matrix} u & 4 + u \\ - 2 v & v \end{matrix})

=

(\begin{matrix} (2 u - 6 v & 2 (4 + u) + 3 v) \\ 4 u - 10 v & 4 (4 + u) + 5 v \end{matrix})

=

(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})

2u - 6v = 1.....(i)
4u - 10v = 0.......(ii)
2(4 + u) + 3v = 0......(iii)
4(4 + u) + 5v = 1......(iv)
2u - 6v = 1 .....(i) x 2
4u - 10v = 0......(ii) x 1

\frac{4u - 12v = 0}{-4u - 10v = 0}

-2v = 2 = v = -1
2u - 6(-1) = 1 = 2u = 5
u = -

\frac{5}{2}

∴ (U, V) = (-

\frac{5}{2}

- 1)

Lee Azịza

Ajụjụ 20 Ripọtì

Given that $ \log_{4}(y - 1) + \log_{4}\left(\frac{1}{2}x\right) = 1 $ and $ \log_{2}(y + 1) + \log_{2}x = 2 $, solve for x and y respectively

2, 3

3, 2

-2, -3

-3, -2

Akọwa Nkọwa

log4(y - 1) + log4(

\frac{1}{2}

x) = 1
log4(y - 1)(

\frac{1}{2}

x)

\to

(y - 1)(

\frac{1}{2}

x) = 4 ........(1)
log2(y + 1) + log2x = 2
log2(y + 1)x = 2

\to

(y + 1)x = 22 = 4.....(ii)
From equation (ii) x =

\frac{4}{y + 1}

........(iii)
put equation (iii) in (i) = y (y - 1)[

\frac{1}{2} (\frac{4}{y - 1}

)] = 4
= 2y - 2
= 4y + 4
2y = -6
y = -3
x =

\frac{4}{- 3 + 1}

=

\frac{4}{- 2}

X = 2
therefore x = -2, y = -3

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Modular Arithmetic

Logarithms

Lee Azịza

Ajụjụ 21 Ripọtì

If x, y can take values from the set (1, 2, 3, 4), find the probability that the product of x and y is not greater than 6

\frac{5}{8}

\frac{5}{16}

\frac{1}{2}

\frac{3}{8}

Akọwa Nkọwa

| \begin{matrix} 1 & 2 & 3 & 4 \\ 1 & 1 & 2 & 3 & 4 \\ 2 & 2 & 4 & 6 & 8 \\ 3 & 3 & 6 & 9 & 12 \\ 4 & 4 & 8 & 12 & 16 \end{matrix} |

P (product of x and y not greater than 6) =

\frac{10}{16}

=

\frac{5}{8}

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Gụọ akwụkwọ nkuzi na Probability (JAMB)

Probability

Lee Azịza

Ajụjụ 22 Ripọtì

%%

In the diagram, QTR is a straight line and < PQT = $30^\circ$. find the sin of < PTR

\frac{8}{15}

\frac{2}{3}

\frac{3}{4}

\frac{15}{16}

Akọwa Nkọwa

\frac{10}{\sin 30^{o}} = \frac{15}{\sin x} = \frac{10}{0.5} = \frac{15}{\sin x}

$\frac{15}{20} = \sin x$

sin x = $\frac{15}{20} = \frac{3}{4}$

N.B x = < PRQ

Gụọ akwụkwọ nkuzi na Polynomials (JAMB)

Gụọ akwụkwọ nkuzi na Introductory Calculus (WAEC)

Polynomials

Introductory Calculus

Lee Azịza

Ajụjụ 23 Ripọtì

The locus of all points at a distance 8cm from a point N passes through points T and S. If S is equidistant from T and N, find the area of triangle STN.

4

\sqrt{3 c m^{2}}

16

\sqrt{3 c m^{2}}

32cm2

64cm2

Loci

Triangles And Polygons

Lee Azịza

Ajụjụ 24 Ripọtì

Differentiate $ \frac{x}{\cos x} $ with respect to x

1 + x sec x tan x

1 + sec2 x

cos x + x tan x

x sec x tan x + secx

Akọwa Nkọwa

let y =

\frac{x}{c o s x}

= x sec x
y = u(x) v (x0

\frac{d y}{d x}

= U

\frac{d y}{d x}

+ V

\frac{d u}{d x}

dy x [secx tanx] + secx
x = x secx tanx + secx

Gụọ akwụkwọ nkuzi na Differentiation (JAMB)

Differentiation

Lee Azịza

Ajụjụ 25 Ripọtì

The binary operation $ \oplus $ is defined by $ x \ast y = xy - y - x $ for all real values $ x $ and $ y $. If $ x \ast 3 = 2 \ast $, find $ x $

-1

4

1

5

Akọwa Nkọwa

x

*

y = xy - y - x, x

*

3 = 3x - 3 - x = 2x - 3
2

*

x = 2x - x - 2 = x - 2
∴ 2x - 3 = x - 2
x = -2 + 3
= 1

Gụọ akwụkwọ nkuzi na Binary Operations (JAMB)

Binary Operations

Lee Azịza

Ajụjụ 26 Ripọtì

Find the value of x in the diagram.

50o

60o

70o

80o

Akọwa Nkọwa

30 + x = 100

x = 100 - 30

= 70o

Lee Azịza

Ajụjụ 27 Ripọtì

The kinetic element with respect to the multiplication shown in the diagram below is

$\oplus$	$p$	$p$	$r$	$s$
$p$	$r$	$p$	$r$	$p$
$q$	$p$	$q$	$r$	$s$
$r$	$r$	$r$	$r$	$r$
$s$	$q$	$s$	$r$	$q$

p

q

r

s

Lee Azịza

Ajụjụ 28 Ripọtì

The determinant of matrix $\begin{pmatrix} x & 1 & 0 \\ 1-x & 2 & 3 \\ 1 & 1+x & 4 \end{pmatrix}$ in terms of x is

-3x2 - 17

-3x2 + 9x - 1

3x2 + 17

3x2 - 9x + 5

Akọwa Nkọwa

| \begin{matrix} x & 1 & 0 \\ 1 - x & 2 & 3 \\ 1 & 1 + x & 4 \end{matrix} |

= x

| \begin{matrix} 2 & 3 \\ 1 + x & 4 \end{matrix} |

-

| \begin{matrix} 1 - x & 3 \\ 1 & 4 \end{matrix} |

= 0
= x[8 - 3(1 + x)] - [4(1 - x)-3] - 0 = x[5 - 3x] - [1 - 4x]
= 5x - 3x2 -1 + 4x
= -3x2 + 9X - 1

Lee Azịza

Ajụjụ 29 Ripọtì

Solve for the equation $ \sqrt{x} - \sqrt{(x - 2)} - 1 = 0 $

\frac{3}{2}

\frac{2}{3}

\frac{4}{9}

\frac{9}{4}

Akọwa Nkọwa

To solve this equation, we can start by simplifying the expression inside the square root symbol by taking the common denominator. √x/(x-2) - √(x-2)/(x-2) - 1 = 0 We can simplify this further by combining the two terms inside the square root, which have a common denominator. [√x - √(x-2)]/(x-2) - 1 = 0 Now we can take the common denominator of the two terms inside the parenthesis and simplify. [√x - √(x-2) - (x-2)]/(x-2) = 0 Simplifying the numerator further, [√x - √(x-2) - x + 2]/(x-2) = 0 [√x - x + 2 - √(x-2)]/(x-2) = 0 [√x - x + 2] = √(x-2) Squaring both sides of the equation, (√x - x + 2)² = x - 2 Expanding and simplifying, x² - 2x(√x + 1) + 3 = 0 We can now use the quadratic formula to solve for x: x = [2(√x + 1) ± √(4x - 8)]/2 x = (√x + 1) ± √(x - 2) However, we need to make sure that the solution we get satisfies the original equation. We can check by substituting the value of x back into the original equation. After testing each option, we find that the only solution that satisfies the original equation is x = 9/4.

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Surds (radicals)

Areas

Quadratic Equations

Lee Azịza

Ajụjụ 30 Ripọtì

TQ is tangent to circle XYTR, $ \angle YXT = 32^\circ $, $ \angle RTQ = 40^\circ $. Find $ \angle YTR $

108

^{\circ}

121

^{\circ}

140

^{\circ}

148

^{\circ}

Akọwa Nkọwa

< TWR = < QTR = 40

^{\circ}

(alternate segment)

< TWR = < TXR = 40 $^{\circ}$ (Angles in the same segments)

< YXR = 40 $^{\circ}$ + 32 $^{\circ}$ = 72 $^{\circ}$

< YXR + < YTR = 180 $^{\circ}$ (Supplementary)

72 $^{\circ}$ + < YTR = 180 $^{\circ}$

< YTR = 180 $^{\circ}$ - 72 $^{\circ}$

= 108 $^{\circ}$

Lee Azịza

Ajụjụ 31 Ripọtì

a cylindrical drum of diameter 56 cm contains 123.2 litres of oil when full. Find the height of the drum in centimeters

12.5

25.0

45.0

50.00

Akọwa Nkọwa

To solve the problem, we need to use the formula for the volume of a cylinder, which is V = πr²h, where V is the volume, r is the radius, and h is the height of the cylinder. We are given the diameter of the drum, which is 56 cm. To find the radius, we need to divide the diameter by 2: radius (r) = diameter / 2 = 56 cm / 2 = 28 cm We are also given that the drum contains 123.2 litres of oil when full. To convert litres to cubic centimeters, we need to multiply by 1000: 123.2 litres * 1000 = 123200 cubic centimeters Now we can use the formula for the volume of a cylinder to find the height (h): V = πr²h h = V / (πr²) h = 123200 / (π28²) h ≈ 123200 / 2463.47 h ≈ 50.00 cm Therefore, the height of the drum is approximately 50.00 cm.

Gụọ akwụkwọ nkuzi na Volumes (WAEC)

Volumes

Lee Azịza

Ajụjụ 32 Ripọtì

The bar chart shows the distribution of marks scored by 60 pupils in a test in which the maximum score was 10. If the pass mark was 5, what percentage of the pupils failed the test?

59.4%

50.0%

41.7%

25.0%

Akọwa Nkọwa

\begin{array}{cc} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ f & 1 & 9 & 4 & 7 & 10 & 8 & 7 & 9 & 8 & 2 & 1 \end{array}

no pupils who failed the test = 1 + 3 + 4 + 7 + 10

= 25

5 of pupils who fail = $\frac{25}{60}$ x 100%

= 41.70%

Gụọ akwụkwọ nkuzi na Percentages (WAEC)

Percentages

Lee Azịza

Ajụjụ 33 Ripọtì

Evaluate $ \int_{2}^{\pi} (\sec^2 x - \tan^2 x)\,dx $

\frac{π}{2}

π

- 2

\frac{π}{3}

π

+ 2

Akọwa Nkọwa

∫

_{2}^{π}

(sec2 x - tan2x)dx
∫

_{2}^{π}

dx = [X]

_{2}^{π}

=

π

- 2 + c
when c is an arbitrary constant of integration

Gụọ akwụkwọ nkuzi na Integration (JAMB)

Integration

Lee Azịza

Ajụjụ 34 Ripọtì

In the diagram above; O is the centre of the circle and |BD| = |DC|. If ∠DCB = 35o, find ∠BAO.

20o

25o

30o

35o

40o

Circles

Lee Azịza

Ajụjụ 35 Ripọtì

find the equation of the curve which passes through by 6x - 5

6x2 - 5x + 5

6x2 + 5x + 5

3x2 - 5x - 5

3x2 - 5x + 3

Akọwa Nkọwa

m =

\frac{d y}{d v}

= 6x - 5
∫dy = ∫(6x - 5)dx
y = 3x2 - 5x + C
when x = 2, y = 5
∴ 5 = 3(2)2 - 5(2) +C
C = 3
∴ y = 3x2 - 5x + 3

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 36 Ripọtì

Make $ \frac{a}{x} $ the subject of formula $ \frac{x+1}{x-a} $

\frac{m - 1}{m + 1}

\frac{m + 1}{1 - m}

\frac{m - 1}{1 + m}

\frac{m + 1}{m - 1}

Akọwa Nkọwa

\frac{x + a}{x - a}

= m
x + a = mx - ma
a + ma = mx - x
a(m + 1) = x(m - 1)

\frac{a}{x}

=

\frac{m - 1}{m + a}

Gụọ akwụkwọ nkuzi na Change Of Subject Of A Formula/Relation (WAEC)

Gụọ akwụkwọ nkuzi na Simple Operations On Algebraic Expressions (WAEC)

Change Of Subject Of A Formula/Relation

Simple Operations On Algebraic Expressions

Lee Azịza

Ajụjụ 37 Ripọtì

Factorize r2 - r(2p + q) + 2pq

(r - 2q)(2r - p)

(r - p)(r + p0

(r - q)(r - 2p)

(2r - q)(r + p)

Akọwa Nkọwa

To factorize the expression r² - r(2p + q) + 2pq, we need to look for two numbers whose product is 2pq and whose sum is -r(2p + q). Let's try to break up -r(2p + q) into two parts such that their product is 2pq. We can write -r(2p + q) as -2rpq - rq². Then, we can rewrite the expression as: r² - 2rpq - rq² + 2pq Now, we can group the first two terms and the last two terms together and factor them separately: r(r - 2pq) - q(r - 2pq) We can see that r - 2pq is a common factor, so we can factor it out: (r - 2pq)(r - q) Therefore, the factorization of r² - r(2p + q) + 2pq is (r - 2pq)(r - q). So the correct option is (c) (r - q)(r - 2p).

Lee Azịza

Ajụjụ 38 Ripọtì

If p + 1, 2P - 10, 1 - 4p2are three consecutive terms of an arithmetic progression, find the possible values of p

-4, 2

-3,

\frac{4}{11}

-

\frac{4}{11}

, 2

5, -3

Akọwa Nkọwa

2p - 10 =

\frac{p + 1 + 1 - 4 P^{2}}{2}

(Arithmetic mean)
= 2(2p - 100 = p + 2 - 4P2)
= 4p - 20 = p + 2 - 4p2
= 4p2 + 3p - 22 = 0
= (p - 2)(4p + 11) = 0
∴ p = 2 or -

\frac{4}{11}

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Lee Azịza

Ajụjụ 39 Ripọtì

2x2 + 11x2 + 17x + 6 by 2x + 1

x2 + 5x + 6

2x2 + 5x - 6

2x2 + 5x + 6

x2 - 5x + 6

Solution Of Linear Equations

Algebraic Expressions

Graphs Of Linear And Quadratic Functions

Lee Azịza

Ajụjụ 40 Ripọtì

Express in partial fractions $ \frac{11+2}{6x^2-x-1} $

\frac{1}{3 x - 1}

+

\frac{3}{2 x + 1}

\frac{3}{3 x + 1}

-

\frac{1}{2 x - 1}

\frac{3}{3 x + 1}

-

\frac{1}{2 x - 1}

\frac{1}{3 x + 1}

+

\frac{3}{2 x - 1}

Akọwa Nkọwa

\frac{11 + 2}{6 x^{2} - x - 1}

=

\frac{11 + 2}{3 x + 1}

=

\frac{A}{3 x + 1}

+

\frac{B}{2 x - 1}

11x = 2 = A(2x - 1) + B(3x + 1)
put x =

\frac{1}{2}

= -

\frac{- 5}{3}

= -

\frac{- 5}{3}

A

\to

A = 1
∴

\frac{11 x + 2}{6 x^{2} - x - 1}

=

\frac{1}{3 x + 1}

+

\frac{3}{2 x - 1}

Gụọ akwụkwọ nkuzi na Algebraic Fractions (WAEC)

Algebraic Fractions

Lee Azịza

Ajụjụ 41 Ripọtì

If m and n are the mean and median respectively of the set of numbers 2, 3, 9, 7, 6, 7, 8, 5, find m + 2n to the nearest whole number

19

18

13

12

Akọwa Nkọwa

To find the mean (m), you need to add up all the numbers in the set and then divide by the total number of numbers. In this case, the sum is 47 and the total number of numbers is 8, so the mean is 47/8 = 5.875. To find the median (n), you need to arrange the numbers in order from smallest to largest and then find the middle number. In this case, the numbers in order are: 2, 3, 5, 6, 7, 7, 8, 9. The middle number is 7, so the median is 7. To find m + 2n, you just need to substitute the values of m and n into the expression and solve. m + 2n = 5.875 + 2(7) = 19.875. To the nearest whole number, m + 2n is 20. Therefore, the answer is: 19.

Gụọ akwụkwọ nkuzi na Sets (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Sets

Surds (radicals)

Lee Azịza

Ajụjụ 42 Ripọtì

Evaluate \[ \frac{1}{0.03} \div \frac{1}{0.024} \]⁻¹ correct to 2 decimal places

3.76

1.25

0.94

0.75

Akọwa Nkọwa

[

\frac{1}{0.03}

+

\frac{1}{0.024}

]
= [

\frac{1}{0.03 \times 0.024}

]-1
= [

\frac{0.024}{0.003}

]-1
=

\frac{0.03}{0.024}

=

\frac{30}{24}

= 1.25

Gụọ akwụkwọ nkuzi na Fractions, Decimals And Approximations (WAEC)

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Fractions, Decimals And Approximations

Representation Of Data

Modular Arithmetic

Lee Azịza

Ajụjụ 43 Ripọtì

In the venn diagram, the shaded region is

(P

\cap

Q)

\cup

R

(P

\cap

Q)

\cap

R

(P

\cap

Q1)

\cap

R

(P

\cap

Q1)

\cup

R

Matrices And Determinants

Sets

Lee Azịza

Ajụjụ 44 Ripọtì

Two chords PQ and RS of a circle intersected at right angles at a point inside the circle. If ∠QPR = 35o,find ∠PQS

35o

45o

55o

70o

90o

Akọwa Nkọwa

Since PQ and RS intersect inside the circle at right angles, then the line joining the point of intersection to the center of the circle will bisect both chords. Let O be the center of the circle, and let T be the point of intersection of the two chords. Then, angle QTR = 90 degrees and the angle subtended by chord PQ at the center O is twice angle QPR. Therefore, angle POQ = 2 * angle QPR = 70 degrees (since angle QPR = 35 degrees). Similarly, angle ROS = 70 degrees. Since PQ and RS are chords of a circle, then angle POQ = angle PTS and angle ROS = angle TQS. Thus, angle PTS + angle TQS = 140 degrees. Also, angle PTS + angle PTQ + angle QTS = 180 degrees (because they form a straight line). Therefore, angle TQS = 180 - 140 - 90 = 50 degrees. Since angle PQT = angle RQT (because they are opposite angles), then angle PQS = angle RQS = (180 - angle QTS)/2 = (180 - 50)/2 = 65 degrees. Therefore, the answer is 55 degrees.

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Circles

Lee Azịza

Ajụjụ 45 Ripọtì

Find the variance of the numbers k, k+1, k+2.

\frac{2}{3}

1

k + 1

(k + 1)2

Akọwa Nkọwa

To find the variance of the numbers k, k+1, k+2, we can use the formula for variance which is the average of the squared differences from the mean. First, we need to find the mean of the three numbers. Mean = (k + k + 1 + k + 2) / 3 = (3k + 3) / 3 = k + 1 So, the mean is k + 1. Next, we find the squared differences from the mean for each number: For k, the difference from the mean is k - (k+1) = -1. The squared difference is (-1)^2 = 1. For k+1, the difference from the mean is (k+1) - (k+1) = 0. The squared difference is 0^2 = 0. For k+2, the difference from the mean is (k+2) - (k+1) = 1. The squared difference is 1^2 = 1. Now we can find the variance: Variance = [(1^2 + 0^2 + 1^2) / 3] = 2/3 = 0.67 (rounded to two decimal places) Therefore, the answer is option (A) 2/3 or as a percentage approximately 66.7%.

Gụọ akwụkwọ nkuzi na Differentiation (JAMB)

Differentiation

Lee Azịza

Ajụjụ 46 Ripọtì

From the top of a vertical mast 150m high., two huts on the same ground level are observed. One due east and the other due west of the mast. Their angles of depression are 60o and 45o respectively. Find the distance between the huts

150(1 +

\sqrt{3}

)m

50(

\sqrt{3}

-

\sqrt{3}

)m

150

\sqrt{3}

m

\frac{50}{\sqrt{3}}

m

Akọwa Nkọwa

\frac{150}{Z}

= tan 60o,
Z =

\frac{150}{t a n 60^{o}}

=

\frac{150}{3}

= 50

\sqrt{3}

cm

\frac{150}{X x Z}

= tan45o = 1
X + Z = 150
X = 150 - Z
= 150 - 50

\sqrt{3}

= 50(

\sqrt{3}

-

\sqrt{3}

)m

Gụọ akwụkwọ nkuzi na Polynomials (JAMB)

Gụọ akwụkwọ nkuzi na Introductory Calculus (WAEC)

Gụọ akwụkwọ nkuzi na Trigonometry (JAMB)

Polynomials

Introductory Calculus

Trigonometry

Lee Azịza

Ajụjụ 47 Ripọtì

A man is paid r naira per hour for normal work and double rate for overtime. if he does a 35-hour week which includes q hours of overtime, what is his weekly earning in naira?

r(35 + q)

q(35r - q)

q(35 + r)

r(35 + 2q)

Akọwa Nkọwa

The cost of normal work = 35r
The cost of overtime = q x 2r = 2qr
The man's total weekly earning = 35r + 2qr
= r(35 + 2q)

Gụọ akwụkwọ nkuzi na Sine, Cosine And Tangent Of An Angle (WAEC)

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Sine, Cosine And Tangent Of An Angle

Algebraic Expressions

Lee Azịza

Ajụjụ 48 Ripọtì