JAMB UTME - Mathematics - 1998

Question 1 Rapport

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}

)

Détails de la réponse

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)

Voir la réponse

Question 2 Rapport

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

Détails de la réponse

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

Lisez la note de cours sur Algebraic Expressions (WAEC)

Algebraic Expressions

Voir la réponse

Question 3 Rapport

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

6x2 - 5x + 5

6x2 + 5x + 5

3x2 - 5x - 5

3x2 - 5x + 3

Détails de la réponse

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

Lisez la note de cours sur Algebraic Expressions (WAEC)

Algebraic Expressions

Voir la réponse

Question 4 Rapport

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}

Détails de la réponse

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.

Lisez la note de cours sur Sine, Cosine And Tangent Of An Angle (WAEC)

Lisez la note de cours sur Trigonometry (JAMB)

Sine, Cosine And Tangent Of An Angle

Trigonometry

Voir la réponse

Question 5 Rapport

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

\frac{π}{2}

π

- 2

\frac{π}{3}

π

+ 2

Détails de la réponse

∫

_{2}^{π}

(sec2 x - tan2x)dx
∫

_{2}^{π}

dx = [X]

_{2}^{π}

=

π

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

Lisez la note de cours sur Integration (JAMB)

Integration

Voir la réponse

Question 6 Rapport

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}

Détails de la réponse

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}

Lisez la note de cours sur Solution Of Linear Equations (WAEC)

Lisez la note de cours sur Number Bases (WAEC)

Solution Of Linear Equations

Number Bases

Voir la réponse

Question 7 Rapport

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)

Détails de la réponse

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)

Lisez la note de cours sur Sine, Cosine And Tangent Of An Angle (WAEC)

Lisez la note de cours sur Algebraic Expressions (WAEC)

Sine, Cosine And Tangent Of An Angle

Algebraic Expressions

Voir la réponse

Question 8 Rapport

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

\frac{2}{3}

1

k + 1

(k + 1)2

Détails de la réponse

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%.

Lisez la note de cours sur Differentiation (JAMB)

Differentiation

Voir la réponse

Question 9 Rapport

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

Détails de la réponse

\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

Lisez la note de cours sur Linear Inequalities (WAEC)

Lisez la note de cours sur Euclidean Geometry (JAMB)

Lisez la note de cours sur Statistics (WAEC)

Linear Inequalities

Euclidean Geometry

Statistics

Voir la réponse

Question 10 Rapport

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

Détails de la réponse

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

Lisez la note de cours sur Areas (WAEC)

Areas

Voir la réponse

Question 11 Rapport

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}}

Détails de la réponse

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}}

Lisez la note de cours sur Sequence And Series (WAEC)

Sequence And Series

Voir la réponse

Question 12 Rapport

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

3.76

1.25

0.94

0.75

Détails de la réponse

[

\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

Lisez la note de cours sur Fractions, Decimals And Approximations (WAEC)

Lisez la note de cours sur Representation Of Data (JAMB)

Lisez la note de cours sur Modular Arithmetic (WAEC)

Fractions, Decimals And Approximations

Representation Of Data

Modular Arithmetic

Voir la réponse

Question 13 Rapport

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}

Détails de la réponse

| \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}

Lisez la note de cours sur Probability (WAEC)

Lisez la note de cours sur Probability (JAMB)

Probability

Voir la réponse

Question 14 Rapport

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

6.0

2.5

\sqrt{6}

\sqrt{6}

Détails de la réponse

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}

Lisez la note de cours sur Modular Arithmetic (WAEC)

Modular Arithmetic

Voir la réponse

Question 15 Rapport

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

\frac{3}{2}

\frac{2}{3}

\frac{4}{9}

\frac{9}{4}

Détails de la réponse

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.

Lisez la note de cours sur Surds (radicals) (WAEC)

Lisez la note de cours sur Areas (WAEC)

Lisez la note de cours sur Quadratic Equations (WAEC)

Surds (radicals)

Areas

Quadratic Equations

Voir la réponse

Question 16 Rapport

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}

Détails de la réponse

\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}

Lisez la note de cours sur Algebraic Fractions (WAEC)

Algebraic Fractions

Voir la réponse

Question 17 Rapport

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

Détails de la réponse

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.

Lisez la note de cours sur Areas (WAEC)

Areas

Voir la réponse

Question 18 Rapport

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

Détails de la réponse

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

Lisez la note de cours sur Differentiation (JAMB)

Differentiation

Voir la réponse

Question 19 Rapport

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

Détails de la réponse

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.

Lisez la note de cours sur Sets (WAEC)

Lisez la note de cours sur Surds (radicals) (WAEC)

Sets

Surds (radicals)

Voir la réponse

Question 20 Rapport

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

Détails de la réponse

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

Lisez la note de cours sur Binary Operations (JAMB)

Binary Operations

Voir la réponse

Question 21 Rapport

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

Détails de la réponse

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.

Lisez la note de cours sur Circles (WAEC)

Circles

Voir la réponse

Question 22 Rapport

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}

Détails de la réponse

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

Voir la réponse

Question 23 Rapport

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

Détails de la réponse

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}

Lisez la note de cours sur Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Voir la réponse

Question 24 Rapport

If 10112 + x7 = 2510, solve for X.

207

147

227

247

Détails de la réponse

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

Lisez la note de cours sur Number Bases (WAEC)

Number Bases

Voir la réponse

Question 25 Rapport

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}

Détails de la réponse

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}

Lisez la note de cours sur Probability (WAEC)

Probability

Voir la réponse

Question 26 Rapport

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

Voir la réponse

Question 27 Rapport

Find the value of x in the diagram.

50o

60o

70o

80o

Détails de la réponse

30 + x = 100

x = 100 - 30

= 70o

Voir la réponse

Question 28 Rapport

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%

Détails de la réponse

\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%

Lisez la note de cours sur Percentages (WAEC)

Percentages

Voir la réponse

Question 29 Rapport

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

Détails de la réponse

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

Lisez la note de cours sur Modular Arithmetic (WAEC)

Lisez la note de cours sur Logarithms (WAEC)

Modular Arithmetic

Logarithms

Voir la réponse

Question 30 Rapport

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

Voir la réponse

Question 31 Rapport

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

Détails de la réponse

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

Voir la réponse

Question 32 Rapport

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

Voir la réponse

Question 33 Rapport

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}

Détails de la réponse

\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}

Lisez la note de cours sur Change Of Subject Of A Formula/Relation (WAEC)

Lisez la note de cours sur Simple Operations On Algebraic Expressions (WAEC)

Change Of Subject Of A Formula/Relation

Simple Operations On Algebraic Expressions

Voir la réponse

Question 34 Rapport

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

Détails de la réponse

| \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

Voir la réponse

Question 35 Rapport

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

Détails de la réponse

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

Lisez la note de cours sur Algebraic Expressions (WAEC)

Algebraic Expressions

Voir la réponse

Question 36 Rapport

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

Détails de la réponse

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.

Lisez la note de cours sur Volumes (WAEC)

Volumes

Voir la réponse

Question 37 Rapport

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

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

(r - p)(r + p0

(r - q)(r - 2p)

(2r - q)(r + p)

Détails de la réponse

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).

Voir la réponse

Question 38 Rapport

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

Voir la réponse

Question 39 Rapport

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

Détails de la réponse

\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

Lisez la note de cours sur Polynomials (JAMB)

Lisez la note de cours sur Introductory Calculus (WAEC)

Lisez la note de cours sur Trigonometry (JAMB)

Polynomials

Introductory Calculus

Trigonometry

Voir la réponse

Question 40 Rapport

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}

)

Détails de la réponse

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).

Lisez la note de cours sur Coordinate Geometry (JAMB)

Coordinate Geometry

Voir la réponse

Question 41 Rapport

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

1

2

3

4

Détails de la réponse

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

Lisez la note de cours sur Measures Of Dispersion (JAMB)

Measures Of Dispersion

Voir la réponse

Question 42 Rapport

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}

Détails de la réponse

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.

Lisez la note de cours sur Differentiation (JAMB)

Differentiation

Voir la réponse

Question 43 Rapport

%%

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}

Détails de la réponse

\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

Lisez la note de cours sur Polynomials (JAMB)

Lisez la note de cours sur Introductory Calculus (WAEC)

Polynomials

Introductory Calculus

Voir la réponse

Question 44 Rapport

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

3

2

√3

√2

Détails de la réponse

\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}

Lisez la note de cours sur Surds (radicals) (WAEC)

Lisez la note de cours sur Quadratic Equations (WAEC)

Lisez la note de cours sur Graphs Of Linear And Quadratic Functions (WAEC)

Surds (radicals)

Quadratic Equations

Graphs Of Linear And Quadratic Functions

Voir la réponse

Question 45 Rapport

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

Détails de la réponse

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.

Voir la réponse

Question 46 Rapport

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

Voir la réponse

Question 47 Rapport

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}

Détails de la réponse

< 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 $^{\circ}$

Voir la réponse

Question 48 Rapport

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

Voir la réponse

\(\oplus\)	\(p\)	\(p\)	\(r\)	\(s\)
\(p\)	\(r\)	\(p\)	\(r\)	\(p\)
\(q\)	\(p\)	\(q\)	\(r\)	\(s\)
\(r\)	\(r\)	\(r\)	\(r\)	\(r\)
\(s\)	\(q\)	\(s\)	\(r\)	\(q\)






Appuyez et maintenez pour déplacer
Cliquez ici pour fermer