JAMB UTME - Mathematics - 1998

Tambaya 1 Rahoto

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

Duba Amsa

Tambaya 2 Rahoto

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}}

Bayanin Amsa

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}}

Karanta bayanin darasi akan Sequence And Series (WAEC)

Sequence And Series

Duba Amsa

Tambaya 3 Rahoto

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

Bayanin Amsa

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.

Karanta bayanin darasi akan Circles (WAEC)

Circles

Duba Amsa

Tambaya 4 Rahoto

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

Bayanin Amsa

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}

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Duba Amsa

Tambaya 5 Rahoto

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

Duba Amsa

Tambaya 6 Rahoto

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

Bayanin Amsa

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.

Karanta bayanin darasi akan Sets (WAEC)

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Sets

Surds (radicals)

Duba Amsa

Tambaya 7 Rahoto

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%

Bayanin Amsa

\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%

Karanta bayanin darasi akan Percentages (WAEC)

Percentages

Duba Amsa

Tambaya 8 Rahoto

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

Bayanin Amsa

| \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

Duba Amsa

Tambaya 9 Rahoto

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

Duba Amsa

Tambaya 10 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Areas (WAEC)

Areas

Duba Amsa

Tambaya 11 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

Tambaya 12 Rahoto

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}

Bayanin Amsa

\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}

Karanta bayanin darasi akan Algebraic Fractions (WAEC)

Algebraic Fractions

Duba Amsa

Tambaya 13 Rahoto

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}

Bayanin Amsa

| \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}

Karanta bayanin darasi akan Probability (WAEC)

Karanta bayanin darasi akan Probability (JAMB)

Probability

Duba Amsa

Tambaya 14 Rahoto

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

Bayanin Amsa

\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

Karanta bayanin darasi akan Polynomials (JAMB)

Karanta bayanin darasi akan Introductory Calculus (WAEC)

Karanta bayanin darasi akan Trigonometry (JAMB)

Polynomials

Introductory Calculus

Trigonometry

Duba Amsa

Tambaya 15 Rahoto

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

3

2

√3

√2

Bayanin Amsa

\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}

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Karanta bayanin darasi akan Graphs Of Linear And Quadratic Functions (WAEC)

Surds (radicals)

Quadratic Equations

Graphs Of Linear And Quadratic Functions

Duba Amsa

Tambaya 16 Rahoto

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

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

(r - p)(r + p0

(r - q)(r - 2p)

(2r - q)(r + p)

Bayanin Amsa

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).

Duba Amsa

Tambaya 17 Rahoto

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

6.0

2.5

\sqrt{6}

\sqrt{6}

Bayanin Amsa

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}

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Modular Arithmetic

Duba Amsa

Tambaya 18 Rahoto

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

6x2 - 5x + 5

6x2 + 5x + 5

3x2 - 5x - 5

3x2 - 5x + 3

Bayanin Amsa

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

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

Tambaya 19 Rahoto

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

\frac{3}{2}

\frac{2}{3}

\frac{4}{9}

\frac{9}{4}

Bayanin Amsa

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.

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Karanta bayanin darasi akan Areas (WAEC)

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Surds (radicals)

Areas

Quadratic Equations

Duba Amsa

Tambaya 20 Rahoto

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

3.76

1.25

0.94

0.75

Bayanin Amsa

[

\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

Karanta bayanin darasi akan Fractions, Decimals And Approximations (WAEC)

Karanta bayanin darasi akan Representation Of Data (JAMB)

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Fractions, Decimals And Approximations

Representation Of Data

Modular Arithmetic

Duba Amsa

Tambaya 21 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Differentiation (JAMB)

Differentiation

Duba Amsa

Tambaya 22 Rahoto

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

Bayanin Amsa

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.

Karanta bayanin darasi akan Areas (WAEC)

Areas

Duba Amsa

Tambaya 23 Rahoto

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

1

2

3

4

Bayanin Amsa

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

Karanta bayanin darasi akan Measures Of Dispersion (JAMB)

Measures Of Dispersion

Duba Amsa

Tambaya 24 Rahoto

Find the value of x in the diagram.

50o

60o

70o

80o

Bayanin Amsa

30 + x = 100

x = 100 - 30

= 70o

Duba Amsa

Tambaya 25 Rahoto

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}

Bayanin Amsa

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

Duba Amsa

Tambaya 26 Rahoto

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}

)

Bayanin Amsa

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)

Duba Amsa

Tambaya 27 Rahoto

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}

Bayanin Amsa

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.

Karanta bayanin darasi akan Sine, Cosine And Tangent Of An Angle (WAEC)

Karanta bayanin darasi akan Trigonometry (JAMB)

Sine, Cosine And Tangent Of An Angle

Trigonometry

Duba Amsa

Tambaya 28 Rahoto

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

\frac{2}{3}

1

k + 1

(k + 1)2

Bayanin Amsa

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%.

Karanta bayanin darasi akan Differentiation (JAMB)

Differentiation

Duba Amsa

Tambaya 29 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

Tambaya 30 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Karanta bayanin darasi akan Logarithms (WAEC)

Modular Arithmetic

Logarithms

Duba Amsa

Tambaya 31 Rahoto

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

Bayanin Amsa

\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

Karanta bayanin darasi akan Linear Inequalities (WAEC)

Karanta bayanin darasi akan Euclidean Geometry (JAMB)

Karanta bayanin darasi akan Statistics (WAEC)

Linear Inequalities

Euclidean Geometry

Statistics

Duba Amsa

Tambaya 32 Rahoto

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

Bayanin Amsa

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.

Duba Amsa

Tambaya 33 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Binary Operations (JAMB)

Binary Operations

Duba Amsa

Tambaya 34 Rahoto

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}

Bayanin Amsa

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}

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 35 Rahoto

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

Bayanin Amsa

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.

Karanta bayanin darasi akan Volumes (WAEC)

Volumes

Duba Amsa

Tambaya 36 Rahoto

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

\frac{π}{2}

π

- 2

\frac{π}{3}

π

+ 2

Bayanin Amsa

∫

_{2}^{π}

(sec2 x - tan2x)dx
∫

_{2}^{π}

dx = [X]

_{2}^{π}

=

π

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

Karanta bayanin darasi akan Integration (JAMB)

Integration

Duba Amsa

Tambaya 37 Rahoto

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}

Bayanin Amsa

< 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 $^{\circ}$

Duba Amsa

Tambaya 38 Rahoto

If 10112 + x7 = 2510, solve for X.

207

147

227

247

Bayanin Amsa

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

Karanta bayanin darasi akan Number Bases (WAEC)

Number Bases

Duba Amsa

Tambaya 39 Rahoto

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}

Bayanin Amsa

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.

Karanta bayanin darasi akan Differentiation (JAMB)

Differentiation

Duba Amsa

Tambaya 40 Rahoto

%%

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}

Bayanin Amsa

\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

Karanta bayanin darasi akan Polynomials (JAMB)

Karanta bayanin darasi akan Introductory Calculus (WAEC)

Polynomials

Introductory Calculus

Duba Amsa

Tambaya 41 Rahoto

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}

Bayanin Amsa

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}

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Karanta bayanin darasi akan Number Bases (WAEC)

Solution Of Linear Equations

Number Bases

Duba Amsa

Tambaya 42 Rahoto

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

Duba Amsa

Tambaya 43 Rahoto

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

Duba Amsa

Tambaya 44 Rahoto

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)

Bayanin Amsa

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)

Karanta bayanin darasi akan Sine, Cosine And Tangent Of An Angle (WAEC)

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Sine, Cosine And Tangent Of An Angle

Algebraic Expressions

Duba Amsa

Tambaya 45 Rahoto

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

Bayanin Amsa

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

Duba Amsa

Tambaya 46 Rahoto

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

Duba Amsa

Tambaya 47 Rahoto

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}

)

Bayanin Amsa

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).

Karanta bayanin darasi akan Coordinate Geometry (JAMB)

Coordinate Geometry

Duba Amsa

Tambaya 48 Rahoto

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