JAMB UTME - Mathematics - 1998

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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Algebraic Expressions

Wo Idahun

Ibeere 2 Ìròyìn

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%

Awọn alaye Idahun

\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%

Ka ìwé ẹ̀kọ́ nípa Percentages (WAEC)

Percentages

Wo Idahun

Ibeere 3 Ìròyìn

If 10112 + x7 = 2510, solve for X.

207

147

227

247

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Number Bases

Wo Idahun

Ibeere 4 Ìròyìn

In the figure, PQST is a parallelogram and TSR is a straight line. If the area of $ \triangle $QRS is 20cm², find the area of the trapezium PQRT.

35cm2

65cm2

70cm2

140cm2

Awọn alaye Idahun

A

△

=

\frac{1}{2}

x 8 x h = 20

= $\frac{1}{2}$ x 8 x h = 4h

h = $\frac{20}{4}$

= 5cm

A $△$ (PQTS) = L x H

A $△$ PQRT = A $△$ QSR + A $△$ PQTS

20 + 50 = 70cm2

ALTERNATIVE METHOD

A $△$ PQRT = $\frac{1}{2}$ x 5 x 28

= 70cm2

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Algebraic Expressions

Wo Idahun

Ibeere 5 Ìròyìn

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}

Awọn alaye Idahun

| \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}

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Ka ìwé ẹ̀kọ́ nípa Probability (JAMB)

Probability

Wo Idahun

Ibeere 6 Ìròyìn

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

Wo Idahun

Ibeere 7 Ìròyìn

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}

Awọn alaye Idahun

\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 < $\frac{1}{6}$

Wo Idahun

Ibeere 8 Ìròyìn

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}

)

Awọn alaye Idahun

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)

Wo Idahun

Ibeere 9 Ìròyìn

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

6x2 - 5x + 5

6x2 + 5x + 5

3x2 - 5x - 5

3x2 - 5x + 3

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Algebraic Expressions

Wo Idahun

Ibeere 10 Ìròyìn

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)

Awọn alaye Idahun

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)

Ka ìwé ẹ̀kọ́ nípa Sine, Cosine And Tangent Of An Angle (WAEC)

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Sine, Cosine And Tangent Of An Angle

Algebraic Expressions

Wo Idahun

Ibeere 11 Ìròyìn

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

3.76

1.25

0.94

0.75

Awọn alaye Idahun

[

\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

Ka ìwé ẹ̀kọ́ nípa Fractions, Decimals And Approximations (WAEC)

Ka ìwé ẹ̀kọ́ nípa Representation Of Data (JAMB)

Ka ìwé ẹ̀kọ́ nípa Modular Arithmetic (WAEC)

Fractions, Decimals And Approximations

Representation Of Data

Modular Arithmetic

Wo Idahun

Ibeere 12 Ìròyìn

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

\frac{π}{2}

π

- 2

\frac{π}{3}

π

+ 2

Awọn alaye Idahun

∫

_{2}^{π}

(sec2 x - tan2x)dx
∫

_{2}^{π}

dx = [X]

_{2}^{π}

=

π

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

Ka ìwé ẹ̀kọ́ nípa Integration (JAMB)

Integration

Wo Idahun

Ibeere 13 Ìròyìn

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

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

(r - p)(r + p0

(r - q)(r - 2p)

(2r - q)(r + p)

Awọn alaye Idahun

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).

Wo Idahun

Ibeere 14 Ìròyìn

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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Areas (WAEC)

Areas

Wo Idahun

Ibeere 15 Ìròyìn

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}

Awọn alaye Idahun

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}

Ka ìwé ẹ̀kọ́ nípa Solution Of Linear Equations (WAEC)

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Solution Of Linear Equations

Number Bases

Wo Idahun

Ibeere 16 Ìròyìn

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

3

2

√3

√2

Awọn alaye Idahun

\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}

Ka ìwé ẹ̀kọ́ nípa Surds (radicals) (WAEC)

Ka ìwé ẹ̀kọ́ nípa Quadratic Equations (WAEC)

Ka ìwé ẹ̀kọ́ nípa Graphs Of Linear And Quadratic Functions (WAEC)

Surds (radicals)

Quadratic Equations

Graphs Of Linear And Quadratic Functions

Wo Idahun

Ibeere 17 Ìròyìn

%%

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}

Awọn alaye Idahun

\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

Ka ìwé ẹ̀kọ́ nípa Polynomials (JAMB)

Ka ìwé ẹ̀kọ́ nípa Introductory Calculus (WAEC)

Polynomials

Introductory Calculus

Wo Idahun

Ibeere 18 Ìròyìn

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

Awọn alaye Idahun

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.

Wo Idahun

Ibeere 19 Ìròyìn

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

6.0

2.5

\sqrt{6}

\sqrt{6}

Awọn alaye Idahun

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}

Ka ìwé ẹ̀kọ́ nípa Modular Arithmetic (WAEC)

Modular Arithmetic

Wo Idahun

Ibeere 20 Ìròyìn

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

Awọn alaye Idahun

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

Wo Idahun

Ibeere 21 Ìròyìn

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

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Volumes (WAEC)

Volumes

Wo Idahun

Ibeere 22 Ìròyìn

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

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Circles

Wo Idahun

Ibeere 23 Ìròyìn

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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Binary Operations (JAMB)

Binary Operations

Wo Idahun

Ibeere 24 Ìròyìn

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}

Awọn alaye Idahun

\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}

Ka ìwé ẹ̀kọ́ nípa Algebraic Fractions (WAEC)

Algebraic Fractions

Wo Idahun

Ibeere 25 Ìròyìn

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

Awọn alaye Idahun

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}

Ka ìwé ẹ̀kọ́ nípa Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Wo Idahun

Ibeere 26 Ìròyìn

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}

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Differentiation (JAMB)

Differentiation

Wo Idahun

Ibeere 27 Ìròyìn

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

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Sets (WAEC)

Ka ìwé ẹ̀kọ́ nípa Surds (radicals) (WAEC)

Sets

Surds (radicals)

Wo Idahun

Ibeere 28 Ìròyìn

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

Wo Idahun

Ibeere 29 Ìròyìn

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}

)

Awọn alaye Idahun

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).

Ka ìwé ẹ̀kọ́ nípa Coordinate Geometry (JAMB)

Coordinate Geometry

Wo Idahun

Ibeere 30 Ìròyìn

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}

Awọn alaye Idahun

< 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 $^{\circ}$

Wo Idahun

Ibeere 31 Ìròyìn

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

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Areas (WAEC)

Areas

Wo Idahun

Ibeere 32 Ìròyìn

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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Modular Arithmetic (WAEC)

Ka ìwé ẹ̀kọ́ nípa Logarithms (WAEC)

Modular Arithmetic

Logarithms

Wo Idahun

Ibeere 33 Ìròyìn

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

Awọn alaye Idahun

\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

Ka ìwé ẹ̀kọ́ nípa Polynomials (JAMB)

Ka ìwé ẹ̀kọ́ nípa Introductory Calculus (WAEC)

Ka ìwé ẹ̀kọ́ nípa Trigonometry (JAMB)

Polynomials

Introductory Calculus

Trigonometry

Wo Idahun

Ibeere 34 Ìròyìn

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}}

Awọn alaye Idahun

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}}

Ka ìwé ẹ̀kọ́ nípa Sequence And Series (WAEC)

Sequence And Series

Wo Idahun

Ibeere 35 Ìròyìn

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

Wo Idahun

Ibeere 36 Ìròyìn

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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Differentiation (JAMB)

Differentiation

Wo Idahun

Ibeere 37 Ìròyìn

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

\frac{3}{2}

\frac{2}{3}

\frac{4}{9}

\frac{9}{4}

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Surds (radicals) (WAEC)

Ka ìwé ẹ̀kọ́ nípa Areas (WAEC)

Ka ìwé ẹ̀kọ́ nípa Quadratic Equations (WAEC)

Surds (radicals)

Areas

Quadratic Equations

Wo Idahun

Ibeere 38 Ìròyìn

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

1

2

3

4

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Measures Of Dispersion (JAMB)

Measures Of Dispersion

Wo Idahun

Ibeere 39 Ìròyìn

Find the value of x in the diagram.

50o

60o

70o

80o

Awọn alaye Idahun

30 + x = 100

x = 100 - 30

= 70o

Wo Idahun

Ibeere 40 Ìròyìn

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

Wo Idahun

Ibeere 41 Ìròyìn

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

Wo Idahun

Ibeere 42 Ìròyìn

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

Wo Idahun

Ibeere 43 Ìròyìn

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

Awọn alaye Idahun

| \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

Wo Idahun

Ibeere 44 Ìròyìn

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}

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Sine, Cosine And Tangent Of An Angle (WAEC)

Ka ìwé ẹ̀kọ́ nípa Trigonometry (JAMB)

Sine, Cosine And Tangent Of An Angle

Trigonometry

Wo Idahun

Ibeere 45 Ìròyìn

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}

Awọn alaye Idahun

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}

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Probability

Wo Idahun

Ibeere 46 Ìròyìn

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

\frac{2}{3}

1

k + 1

(k + 1)2

Awọn alaye Idahun

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%.

Ka ìwé ẹ̀kọ́ nípa Differentiation (JAMB)

Differentiation

Wo Idahun

Ibeere 47 Ìròyìn