JAMB UTME - Mathematics - 2004

Ibeere 1 Ìròyìn

Determine the locus of a point inside a square PQRS which is eqidistant from PQ and QR

The diagonal QS

the perpendicular bisector of PQ

The diagonal PR

side SR

Awọn alaye Idahun

The diagonal QS bisects the angle formed by PQ and QR
∴ [A]

Ka ìwé ẹ̀kọ́ nípa Loci (WAEC)

Ka ìwé ẹ̀kọ́ nípa Coordinate Geometry (JAMB)

Loci

Coordinate Geometry

Wo Idahun

Ibeere 2 Ìròyìn

Find the derivatives of (2 + 3x)(1 - x) with respect to x

6

-3

1 - 6x

6x - 1

Awọn alaye Idahun

To find the derivative of the given function, we need to apply the product rule of differentiation, which is: d/dx [f(x)g(x)] = f(x)g'(x) + g(x)f'(x) Here, f(x) = (2 + 3x) and g(x) = (1 - x) Taking the derivatives of the functions separately, we have: f'(x) = 3 g'(x) = -1 Substituting the values in the product rule formula, we get: d/dx [(2 + 3x)(1 - x)] = (2 + 3x)(-1) + (1 - x)(3) Simplifying the expression, we get: d/dx [(2 + 3x)(1 - x)] = -2 - 3x + 3 - 3x d/dx [(2 + 3x)(1 - x)] = -6x + 1 Therefore, the correct answer is: -6x + 1.

Ka ìwé ẹ̀kọ́ nípa Differentiation (JAMB)

Differentiation

Wo Idahun

Ibeere 3 Ìròyìn

Find P, if 4516 - P7 = 3056

627

1167

6117

1427

Awọn alaye Idahun

4516 - P7 = 3056
P7 = 4516 - 3056
P7 = 1426
convert 1426 = 1 * 62 + 4 * 61 + 2 * 60
= 36 + 24 + 2
= 62
Convert 6210 to base 7
62/7 = 8 R 6
8/7 = 1 R 1
1/7 = 0 R 1
∴P7 = 1167

Ka ìwé ẹ̀kọ́ nípa Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Wo Idahun

Ibeere 4 Ìròyìn

The sum of the interior angles of a pentagon is 6x + 6y. Find y in the terms of x

y = 90 - x

y = 150 - x

y = 60 - x

y = 120 -x

Awọn alaye Idahun

A pentagon is a five-sided polygon, and the sum of its interior angles can be found using the formula: Sum of interior angles = (n-2) * 180 degrees where n is the number of sides in the polygon. For a pentagon, n = 5, so the formula becomes: Sum of interior angles = (5-2) * 180 degrees = 3 * 180 degrees = 540 degrees Now, we are given that the sum of the interior angles of a pentagon is 6x + 6y. Therefore, we can equate this expression with 540 degrees to get: 6x + 6y = 540 Dividing both sides by 6, we get: x + y = 90 Subtracting x from both sides, we get: y = 90 - x Therefore, the value of y in terms of x is y = 90 - x. Hence, the correct answer is option (A) y = 90 - x.

Ka ìwé ẹ̀kọ́ nípa Indices (WAEC)

Indices

Wo Idahun

Ibeere 5 Ìròyìn

Find the derivatives of the function y = 2x2(2x - 1) at the point x = -1

18

16

-4

-6

Awọn alaye Idahun

y = 2x2(2x - 1)
y = 4x3 - 2x2
dy/dx = 12x2 - 4x
at x = -1
dy/dx = 12(-1)2 - 4(-1)
= 12 + 4
= 16

Ka ìwé ẹ̀kọ́ nípa Differentiation (JAMB)

Differentiation

Wo Idahun

Ibeere 6 Ìròyìn

PQRSTV is a regular polygon of side 7 cm inscribed in a circle. Find the circumference of the circle PQRSTV.

56 cm

22 cm

42 cm

44 cm

Awọn alaye Idahun

Each interior Angle of the polygon =	(n-2)	180
	_n

=	(6-2)	180
	_6

= 4 x 30
= 120

^{\circ}

i.e ∠PQR = 120

^{\circ}

∴ RQD = 120/2 = 60

^{\circ}

also
∴ Δ QOR is equilateral Δ ∴ QO = 7cm
∴ PO = QO = RO ...... = 7cm radius of the circumference of the circle = 2πr
= 2 x 22/7 x 7/1
= 44 cm

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Circles

Wo Idahun

Ibeere 7 Ìròyìn

Find the midpoint of the line joining P(-3, 5) and Q(5, -3).

(1, 1)

(2, 2)

(4, 4)

(4, -4)

Awọn alaye Idahun

The midpoint of a line segment is the point that is exactly halfway between the endpoints of the line segment. The formula for finding the midpoint of a line segment with endpoints (x1, y1) and (x2, y2) is: Midpoint = ((x1 + x2) / 2, (y1 + y2) / 2) In this case, we are given the endpoints of the line segment PQ as P(-3, 5) and Q(5, -3). Using the formula for the midpoint, we can find the coordinates of the midpoint as: Midpoint = ((-3 + 5) / 2, (5 - 3) / 2) = (2 / 2, 2 / 2) = (1, 1) Therefore, the midpoint of the line joining P(-3, 5) and Q(5, -3) is (1, 1). Hence, the correct answer is option (A) (1, 1).

Ka ìwé ẹ̀kọ́ nípa Coordinate Geometry (JAMB)

Coordinate Geometry

Wo Idahun

Ibeere 8 Ìròyìn

Evaluate $ \frac{\frac{1}{10}\times\frac{2}{3}+\frac{1}{4}}{\frac{\frac{1}{2}}{\frac{3}{5}}-\frac{1}{4}}} $

\frac{7}{12}

\frac{19}{35}

\frac{2}{25}

\frac{19}{60}

Awọn alaye Idahun

Explanation:

\frac{\frac{1}{10} \times \frac{2}{3} + \frac{1}{4}}{\frac{\frac{1}{2}}{\frac{3}{5}} - \frac{1}{4}} N u m e r a t o r \frac{1}{10} \times \frac{2}{3} + \frac{1}{4} = \frac{1}{5} + \frac{1}{4} = \frac{4 + 15}{60} = \frac{19}{60} d e n o m i n a t o r = \frac{\frac{1}{2}}{\frac{3}{5}} - \frac{1}{4} = \frac{1}{2} \times \frac{5}{3} - \frac{1}{4} = \frac{5}{6} - \frac{1}{4} = \frac{10 - 3}{12} = \frac{7}{12} \frac{N u m e r a t o r}{d e n o m i n a t o r} = \frac{\frac{19}{60}}{\frac{7}{12}} = \frac{19}{60} \times \frac{12}{7} = \frac{19}{35}

Wo Idahun

Ibeere 9 Ìròyìn

The nth term of two sequences are Qn = 3 . 2n - 2 and Um = 3 . 22m - 3. Find the product of Q2 and U2.

18

12

6

3

Awọn alaye Idahun

The nth term of the sequence Qn is Qn = 3*2^(n-2), and the mth term of the sequence Um is Um = 3*2^(2m-3). We are asked to find the product of Q2 and U2, which means we need to find Q2 and U2 and then multiply them together. Substituting n=2 in the formula for Qn, we get: Q2 = 3*2^(2-2) = 3*2^0 = 3*1 = 3 Substituting m=2 in the formula for Um, we get: U2 = 3*2^(2*2-3) = 3*2^1 = 3*2 = 6 So the product of Q2 and U2 is: Q2 * U2 = 3 * 6 = 18 Therefore, the answer is (A) 18.

Ka ìwé ẹ̀kọ́ nípa Sequence And Series (WAEC)

Sequence And Series

Wo Idahun

Ibeere 10 Ìròyìn

The shadow of a pole 5√3m high is 5m. Find the angle of elevation of the sun.

45o

75o

30o

60o

Awọn alaye Idahun

The height of the pole is 5√3m and its shadow on the ground is 5m. Let x be the angle of elevation of the sun. From the geometry of the situation, we can see that tan(x) = opposite/adjacent = (5√3)/5 = √3. Taking the inverse tangent of both sides, we get: x = tan^(-1)(√3) ≈ 60° Therefore, the angle of elevation of the sun is 60°. So the answer is (D) 60°.

Ka ìwé ẹ̀kọ́ nípa Trigonometry (JAMB)

Trigonometry

Wo Idahun

Ibeere 11 Ìròyìn

A farmer planted 5000 grains of maize and harvested 5000 cobs, each bearing 500 grains. What is the ratio of the number of grains sowed to the number harvested?

1 : 250 000

1 : 25 000

1 : 500

1 : 5 000

Awọn alaye Idahun

To find the ratio of the number of grains sowed to the number harvested, we need to divide the number of grains sowed by the number of grains harvested. Number of grains sowed = 5000 Number of grains harvested = 5000 cobs × 500 grains per cob = 2,500,000 So the ratio of the number of grains sowed to the number harvested is: 5000 : 2,500,000 Simplifying the ratio by dividing both sides by 5000, we get: 1 : 500 Therefore, the answer is: 1 : 500.

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

Ka ìwé ẹ̀kọ́ nípa Ratio, Proportions And Rates (WAEC)

Sine, Cosine And Tangent Of An Angle

Ratio, Proportions And Rates

Wo Idahun

Ibeere 12 Ìròyìn

Given that the first and forth terms of G.P are 6 and 162 respectively, find the sum of the first three terms of the progression

27

8

78

48

Awọn alaye Idahun

To find the sum of the first three terms of a G.P, we need to determine the common ratio (r) of the progression. Given that the first and fourth terms are 6 and 162 respectively, we can use the formula for the nth term of a G.P to obtain: a1 = 6, a4 = 162 a4 = a1 * r^3 162 = 6 * r^3 r^3 = 27 r = 3 Now, we can find the second and third terms of the progression using the common ratio: a2 = a1 * r = 6 * 3 = 18 a3 = a2 * r = 18 * 3 = 54 The sum of the first three terms of the G.P is: a1 + a2 + a3 = 6 + 18 + 54 = 78 Therefore, the correct answer is option (C) 78.

Ka ìwé ẹ̀kọ́ nípa Sequence And Series (WAEC)

Sequence And Series

Wo Idahun

Ibeere 13 Ìròyìn

I. Rectangular bars of equal width

II. The height of each rectangular bar is proportional to the frequency of the corresponding class interval.

III. Rectangular bars have common sides with no gaps in between

A histogram is described completely by

I, II and III

I and II

II and III

I and III

Awọn alaye Idahun

A histogram is a graphical representation of a frequency distribution of a dataset, where the data is divided into intervals or bins and the frequency of each interval is represented by the area of a rectangle or bar. To construct a histogram, we need to follow certain rules, which are described by the statements: I. Rectangular bars of equal width II. The height of each rectangular bar is proportional to the frequency of the corresponding class interval. III. Rectangular bars have common sides with no gaps in between. Therefore, a histogram is described completely by statements I, II, and III. Statement I ensures that each bar is of equal width, which is necessary to represent the intervals or bins in a consistent way. Statement II ensures that the height of each bar accurately represents the frequency of the corresponding class interval. This is important because the area of the bar represents the frequency of the interval, which is the product of the width and height of the bar. Statement III ensures that the bars are adjacent to each other without any gaps, which is necessary to accurately represent the continuity of the data. Hence, the correct answer is option (A) I, II, and III.

Ka ìwé ẹ̀kọ́ nípa Representation Of Data (JAMB)

Ka ìwé ẹ̀kọ́ nípa Measures Of Location (JAMB)

Ka ìwé ẹ̀kọ́ nípa Measures Of Dispersion (JAMB)

Representation Of Data

Measures Of Location

Measures Of Dispersion

Wo Idahun

Ibeere 14 Ìròyìn

Given that 3√42x = 16, find the value of x

4

6

3

2

Awọn alaye Idahun

3√42x = 16
this implies that (3√42x)3 = (16)3
42x = 42*3
42x = 46
∴ 2x = 6
x = 3

Ka ìwé ẹ̀kọ́ nípa Surds (radicals) (WAEC)

Surds (radicals)

Wo Idahun

Ibeere 15 Ìròyìn

The shaded region in the Venn diagram above is

P $^{\circ}$ $\cap$ (Q $\cap$ R)

P $\cap$ Q

P $^{\circ}$ $\cup$ (Q $\cap$ R)

P $^{\circ}$ $\cup$ (Q $\cup$ R)

Wo Idahun

Ibeere 16 Ìròyìn

The shaded area in the diagram above is represented by

{(x,y) : y - 3x < -6}

{(x,y) : y + 3x < 6}

{(x,y) : y + 3x < -6}

{(x,y) : y - 3x < 6}

Awọn alaye Idahun

Coordinates of the line are (x1 y1)(0,2) respectively and (x2 y2)(2,0) respectively
Gradient (m) of the line = (y2 - y1) / (x2 - x1) = (0-6) / (2-0) = -6/2
= -3
Equation of the line y-y1 = m(x-x1)
y - 6 = -3(x - 0)
y - 6 = -3x
y + 3x = 6
inequality; from the y axis, the shaded region is below the line
∴ y + 3x ∠ 6 and line is a broken line

Ka ìwé ẹ̀kọ́ nípa Linear Inequalities (WAEC)

Linear Inequalities

Wo Idahun

Ibeere 17 Ìròyìn

Find the remainder when 3x3 + 5x2 - 11x + 4 is divided by x + 3

-4

4

1

-1

Awọn alaye Idahun

x = -3
substitute x = -3 in 3x3 + 5x2 - 11x + 4
3(-3)3 + 5(-3)2 - 11(-3) + 4
-81 + 45 + 33 + 4
-81 + 82
= 1

Ka ìwé ẹ̀kọ́ nípa Solution Of Linear Equations (WAEC)

Ka ìwé ẹ̀kọ́ nípa Modular Arithmetic (WAEC)

Solution Of Linear Equations

Modular Arithmetic

Wo Idahun

Ibeere 18 Ìròyìn

In how many ways can 2 students be selected from a group of 5 students in a debating competition?

25 ways

10 ways

15 ways

20 ways

16 ways

Permutation And Combination

Wo Idahun

Ibeere 19 Ìròyìn

In the diagram above, PQ = 4 cm and TS = 6 cm. If the area of parallelogram PQTU is 32 cm2, find the area of the trapezium PQRU

60 cm2

72 cm2

48 cm2

24 cm2

Awọn alaye Idahun

To find the area of the trapezium PQRU, we need to know the length of its height or perpendicular distance between the parallel sides PQ and RU. We can find this height by using the area of the parallelogram PQTU, which has the same base PQ and height as the trapezium, and a known area of 32 cm². Area of parallelogram PQTU = base × height 32 cm² = 4 cm × height Height = 8 cm Now we can use the formula for the area of a trapezium: Area of trapezium PQRU = (sum of parallel sides × height) ÷ 2 = (PQ + RU) × height ÷ 2 = (4 cm + 10 cm) × 8 cm ÷ 2 = 56 cm² Therefore, the area of trapezium PQRU is 56 cm². Answer option (B) is correct.

Ka ìwé ẹ̀kọ́ nípa Mensuration (JAMB)

Mensuration

Wo Idahun

Ibeere 20 Ìròyìn

An unbiased die is rolled 100 times and the outcome is tabulated above.

What is the probability of obtaining a 5?

1/5

1/2

1/6

1/4

Awọn alaye Idahun

Number of times 5 was obtained = 20
P(obtaining 5) = 20/100
= 1/5

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Probability

Wo Idahun

Ibeere 21 Ìròyìn

Factorize completely ac - 2bc - a + 4b2

(a - 2b)(c - a - 2b)

(a - 2b)(c + a +2b)

(a - 2b)(c - a + 2b)

(a - 2b)(c + a - 2b)

Awọn alaye Idahun

To factorize the expression, we can group the terms as follows: ac - 2bc - a + 4b^2 = a(c - 1) - 2b(c - 2b) Now we can factor out the common factors: a(c - 1) - 2b(c - 2b) = (c - 1)(a - 2b) So the completely factored form of the expression is: ac - 2bc - a + 4b^2 = (c - 1)(a - 2b) Therefore, the correct answer is option A: (a - 2b)(c - a - 2b).

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Algebraic Expressions

Wo Idahun

Ibeere 22 Ìròyìn

In the diagram above, KLMN is a cyclic quadrilateral. /KL/ = /KN/, ∠NKM = 55o and ∠KML = 40o. Find ∠LKM.

35o

40o

45o

50o

55o

Wo Idahun

Ibeere 23 Ìròyìn

Find the value of α2 + β2 if α + β = 2 and the distance between points (1, α) and (β, 1)is 3 units

14

3

5

11

Awọn alaye Idahun

PQ=√(β−1)2+(1−α)23=√(β2−2β2+1+1−2α+α2)3=√(α2+β2−2α+2β+2)3=√(α2+β2−2(α+β)+2)3=√(α2+β2−2∗2+2)3=√(α2+β2−2)9=(α2+β2−2)α2+β2=9+2α2+β2=11

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Number Bases

Algebraic Expressions

Wo Idahun

Ibeere 24 Ìròyìn

A container has 30 gold medals, 22 silver medals and 18 bronze medals. If one medals is selected at the random from the container, what is the probability that it is not a gold medal?

9/35

11/35

4/7

3/7

Awọn alaye Idahun

There are a total of 30 + 22 + 18 = 70 medals in the container. The probability of selecting a gold medal is 30/70. The probability of not selecting a gold medal is equal to the probability of selecting either a silver or a bronze medal. Therefore, the probability of not selecting a gold medal is (22 + 18)/70 = 40/70. This fraction can be simplified to 4/7, which is the answer.

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Ka ìwé ẹ̀kọ́ nípa Surds (radicals) (WAEC)

Probability

Surds (radicals)

Wo Idahun

Ibeere 25 Ìròyìn

The pie chart above shows the distribution of the crops harvested from a farmland in a year. If 3000 tonnes of millet is harvested, what amount of beans is harvested

6000 tonnes

1500 tonnes

1200 tonnes

9000 tonnes

Awọn alaye Idahun

Sectorial angle of beans = 360 - (150 + 90 + 60)
= 360 -300
= 60

^{\circ}

if 150

^{\circ}

represents 3000 tonnes
1

^{\circ}

3000/150
60

^{\circ}

will represents (3000/150) * (60/1)
= 1200 tonnes

Ka ìwé ẹ̀kọ́ nípa Sets (JAMB)

Sets

Wo Idahun

Ibeere 26 Ìròyìn

The length L of a simple pendulum varies directly as the square of its period T. If a pendulum with period 4 sec. is 64 cm long, find the length of pendulum whose period is 9 sec

96 cm

324 cm

36 cm

144 cm

Awọn alaye Idahun

The problem describes a direct variation between the length of a simple pendulum and the square of its period, which means that we can write the relationship as: L = kT^2 where L is the length of the pendulum, T is its period, and k is a constant of proportionality. To find the value of k, we can use the information given in the problem that a pendulum with a period of 4 seconds has a length of 64 cm. Substituting these values into the equation above, we get: 64 = k(4^2) 64 = 16k k = 4 Now that we know the value of k, we can use the same equation to find the length of a pendulum with a period of 9 seconds: L = 4(9^2) L = 324 Therefore, the length of the pendulum whose period is 9 seconds is 324 cm.

Ka ìwé ẹ̀kọ́ nípa Variation (WAEC)

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Variation

Probability

Algebraic Expressions

Wo Idahun

Ibeere 27 Ìròyìn

The mean age of a group of students is 15 years. When the age of a teacher, 45 years old, is added to the age of the students, the mean of their ages becomes 18 years. Find the number of the students in the group

9

7

42

15

Awọn alaye Idahun

Let's start by defining some variables: - n: the number of students in the group - sum_age_students: the sum of the ages of the students We know that the mean age of the group of students is 15, so we can write: mean_age_students = sum_age_students / n = 15 From the problem, we also know that when the age of the teacher is added to the age of the students, the mean of their ages becomes 18. So we can write another equation: (mean_age_students + 45) = (sum_age_students + 45) / (n + 1) = 18 Now we have two equations with two variables (n and sum_age_students), and we can solve for them simultaneously. Let's start by simplifying the second equation: sum_age_students + 45 = 18(n + 1) sum_age_students = 18n + 27 Now we can substitute this expression for sum_age_students in the first equation: sum_age_students / n = 15 (18n + 27) / n = 15 18n + 27 = 15n 3n = 27 n = 9 Therefore, there are 9 students in the group.

Ka ìwé ẹ̀kọ́ nípa Integration (JAMB)

Ka ìwé ẹ̀kọ́ nípa Euclidean Geometry (JAMB)

Integration

Euclidean Geometry

Wo Idahun

Ibeere 28 Ìròyìn

Find the sum to infinity of the series 1/2 , 1/6, 1/18, .....

2/3

1/3

3/4

1

Awọn alaye Idahun

The given series is a geometric series with the first term a = 1/2 and the common ratio r = 1/3. To find the sum to infinity of the series, we can use the formula: S = a/(1 - r) where S is the sum to infinity. Substituting the values of a and r, we get: S = (1/2)/(1 - 1/3) = (1/2)/(2/3) = 1/2 * 3/2 = 3/4 Therefore, the sum to infinity of the series 1/2, 1/6, 1/18, ... is 3/4. So the answer is (C) 3/4.

Ka ìwé ẹ̀kọ́ nípa Sequence And Series (WAEC)

Sequence And Series

Wo Idahun

Ibeere 29 Ìròyìn

A committee of six is to be formed by a state governor from nine state commissioners and three members of the state house of assembly. In how many ways can the members of the committee be chosen so as to include one member of the house of assembly

378 ways

462 ways

840 ways

924 ways

Awọn alaye Idahun

^{3} C_{1} \times^{9} C_{5} = \frac{3!}{(3 - 1)! 1!} \times \frac{9!}{(9 - 5)! 5!} = \frac{3!}{2! 1!} \times \frac{9!}{4! 5!} = \frac{3 \times 2!}{2! \times 1} \times \frac{9 \times 8 \times 7 \times 6 \times 5!}{4 \times 3 \times 2 \times 1 \times 5!} = 3 \times 9 \times 2 \times 7 = 378 w a y s

Ka ìwé ẹ̀kọ́ nípa Representation Of Data (JAMB)

Ka ìwé ẹ̀kọ́ nípa Variation (JAMB)

Ka ìwé ẹ̀kọ́ nípa Sets (JAMB)

Representation Of Data

Variation

Sets

Wo Idahun

Ibeere 30 Ìròyìn

y = is inversely proportional to x and y = 4 when x = 1/2. Find x when y = 10.

2

10

1/5

1/10

Awọn alaye Idahun

y ∝ 1/x
y = K/x
K = xy
K = (1/2) * 4 ( when y = 4 and x = 1/2)
K = 2
∴y = 2/x
10 = 2/x (when y = 10)
10x = 2
x = 2/10
x = 1/5

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

Ka ìwé ẹ̀kọ́ nípa Variation (JAMB)

Ka ìwé ẹ̀kọ́ nípa Modular Arithmetic (WAEC)

Sine, Cosine And Tangent Of An Angle

Variation

Modular Arithmetic

Wo Idahun

Ibeere 31 Ìròyìn

If the shadow of a pole 7m high is 1/2 its length what is the angle of elevation of the sun, correct to the nearest degree?

90o

63o

60o

26o

0o

Awọn alaye Idahun

tan θ - 7/3.5 = 2
tan θ= 2.0000
θ = 63o

Ka ìwé ẹ̀kọ́ nípa Trigonometry (JAMB)

Trigonometry

Wo Idahun

Ibeere 32 Ìròyìn

Three teachers shared a packet of chalk. The first teacher got 2/5 of the chalk and the second teacher received 2/15 of the remainder. What fraction the the third teacher receive?

8/15

11/25

12/25

13/25

Awọn alaye Idahun

Let's assume the packet of chalk has a total of 15 units (the lowest common multiple of 5 and 15). The first teacher received 2/5 of this, which is 6 units of chalk. The remainder is 9 units of chalk (15 - 6). The second teacher received 2/15 of this remainder, which is (2/15) x 9 = 2/5 units of chalk. So the third teacher received the remaining chalk, which is 9 - 2/5 = 35/5 - 2/5 = 33/5 units of chalk. Therefore, the fraction of chalk that the third teacher received is 33/5, which can be simplified to 6 3/5 or 13/5. So the answer is option D: 13/25.

Wo Idahun

Ibeere 33 Ìròyìn

Some white balls put in a basket containing twelve red balls and sixteen black balls. If the probability of picking a white balls from the baskets is 3/7, how many white balls were introduced?

12

21

28

32

Awọn alaye Idahun

Let's say the number of white balls introduced is x. The total number of balls in the basket before introducing the white balls was 12 + 16 = 28. After introducing the x white balls, the total number of balls becomes 12 + 16 + x. Now, the probability of picking a white ball is given as 3/7. We can express this probability as: Number of white balls / Total number of balls = 3/7 Substituting the values, we get: x / (12 + 16 + x) = 3/7 Solving for x: 7x = 3(28 + x) 7x = 84 + 3x 4x = 84 x = 21 Therefore, the number of white balls introduced is 21. So, option (B) is the correct answer.

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Probability

Wo Idahun

Ibeere 34 Ìròyìn

P, R and S lie on a circle center as shown above while Q lies outside the circle. Find ∠PSO

45o

55o

35o

40o

Awọn alaye Idahun

xo = 35 + 29 (Exterior angle = sum of two interior opposite angles)
x = 55o (∠ at the center twice ∠ at circumference) y = 110o
∠PSO = ∠SPO (base ∠S of 1sc Δ b/c PO = SO)
∴ ∠PSO = (180 - 110)/2
= 35o

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Circles

Wo Idahun

Ibeere 35 Ìròyìn

The graph above shows the cumulative frequency curve of the distribution of marks in a class test. What percentage of the students scored more than 20 marks?

68%

28%

17%

8%

Awọn alaye Idahun

\begin{matrix} class-marks(x) & (freq. (f) & cum. freq. \\ 9 & 3.....3 \\ 14 & 6....3 + 6 = 9 \\ 19 & 9.....9 + 9 = 18 \\ 24 & 3.....18 + 3 = 21 \\ 29 & 2.....21 + 2 = 23 \\ 34 & 2.....23 + 2 = 25 \\ 25 \end{matrix}

% of students scoring more than 20 marks

= $\frac{7}{25}$ x 100% = 28%

Ka ìwé ẹ̀kọ́ nípa Percentages (WAEC)

Percentages

Wo Idahun

Ibeere 36 Ìròyìn

What is the rate of change of the volume v of a hemisphere with respect to its radius r when r = 2?

8π

16π

2π

4π

Awọn alaye Idahun

To find the rate of change of the volume of a hemisphere with respect to its radius, we need to differentiate the formula for the volume of a hemisphere with respect to its radius. The formula for the volume of a hemisphere is V = (2/3)πr^3. Taking the derivative of this formula with respect to r gives us dV/dr = 2πr^2. Therefore, the rate of change of the volume of the hemisphere with respect to its radius is 2πr^2. When r = 2, the rate of change is 2π(2^2) = 2π(4) = 8π. Thus, the answer is 8π.

Ka ìwé ẹ̀kọ́ nípa Integration (JAMB)

Ka ìwé ẹ̀kọ́ nípa Differentiation (JAMB)

Integration

Differentiation

Wo Idahun

Ibeere 37 Ìròyìn

Find the mean deviation of 1, 2, 3 and 4.

2.5

2.0

1.0

1.5

1.2

Measures Of Dispersion

Statistics

Wo Idahun

Ibeere 38 Ìròyìn

Evaluate $ \int_{1}^{3}(X^2 - 1)\,dx $

\frac{2}{3}

- \frac{2}{3}

- 6 \frac{2}{3}

6 \frac{2}{3}

Awọn alaye Idahun

\int_{1}^{3} (x^{2} - 1) d x = {[\frac{1}{3} x^{2} - x]}_{1}^{2} = (9 - 3) - (\frac{1}{3} - 1) = 6 - (- \frac{2}{3}) = 6 + \frac{2}{3} = 6 \frac{2}{3}

Ka ìwé ẹ̀kọ́ nípa Integration (JAMB)

Ka ìwé ẹ̀kọ́ nípa Trigonometry (JAMB)

Integration

Trigonometry

Wo Idahun

Ibeere 39 Ìròyìn

If the operation * on the set of integers is defined by $p * Q = \sqrt{pq}$, find the value of $4 * (8 * 32)$.

8

3

16

4

Awọn alaye Idahun

P×Q=√PQ4×(8×32)=4×√8×32=4×√256=4×16=√4×16=√64=8

Wo Idahun

Ibeere 40 Ìròyìn

Simplify $ \frac{1}{\sqrt{3}+2} $ in the form $ a+b\sqrt{3} $

2 -√3

-2 - √3

2 + √3

-2 + √3

Awọn alaye Idahun

\frac{1}{\sqrt{3} + 2} = \frac{1}{\sqrt{3} + 2} \times \frac{\sqrt{3} - 2}{\sqrt{3} - 2} = \frac{\sqrt{3} - 2}{(\sqrt{3})^{2} - 2^{2}} = \frac{\sqrt{3} - 2}{3 - 4} = \frac{\sqrt{3} - 2}{1} = - \sqrt{3} + 2 = 2 - \sqrt{3}

Ka ìwé ẹ̀kọ́ nípa Positive And Negative Integers, Rational Numbers (WAEC)

Ka ìwé ẹ̀kọ́ nípa Surds (radicals) (WAEC)

Positive And Negative Integers, Rational Numbers

Surds (radicals)

Wo Idahun

Ibeere 41 Ìròyìn

If y = 3 cos(x/3), find dy/dx when x = (3π/2)

1

13

2

-1

Awọn alaye Idahun

y = 3cos(x/3)
dy/dx = 3x(1/3)x - sin (x/3)
= - sin x/3
But x = 3π/2 ∴ -sin(x/3) = -sin(3π/6)
= -sin (3 * 180)/6
= - sin 90
= -1

Ka ìwé ẹ̀kọ́ nípa Introductory Calculus (WAEC)

Ka ìwé ẹ̀kọ́ nípa Differentiation (JAMB)

Ka ìwé ẹ̀kọ́ nípa Application Of Differentiation (JAMB)

Introductory Calculus

Differentiation

Application Of Differentiation

Wo Idahun

Ibeere 42 Ìròyìn

The locus of a point which is 5 cm from the line LM is a

line distance 10 cm from LM and parallel to LM

pair of line on opposite sides of LM and parallel to it, each distance 5 cm from LM

line parallel to LM and 5 cm from LM

pair of parallel lines on one side of LM and parallel to LM

Awọn alaye Idahun

The locus of a point which is 5cn from the line LM is a pair of lines on opposite sides of LM and parallel to it, each distance 5cm from LM

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Ka ìwé ẹ̀kọ́ nípa Triangles And Polygons (WAEC)

Circles

Triangles And Polygons

Wo Idahun

Ibeere 43 Ìròyìn

Find the value of x in the figure above

15√6

20√6

3√6

5√6

Awọn alaye Idahun

\frac{X}{s i n 45} = \frac{15}{s i n 60} X = \frac{15 s i n 45}{s i n 60} X = \frac{15 \times (\frac{1}{\sqrt{2}})}{\sqrt{\frac{3}{2}}} X = \frac{15}{\sqrt{2}} \times \frac{2}{\sqrt{3}} X = \frac{30}{\sqrt{6}} X = \frac{30}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{30 \sqrt{6}}{6} = 5 \sqrt{6}

Ka ìwé ẹ̀kọ́ nípa Surds (radicals) (WAEC)

Surds (radicals)

Wo Idahun

Ibeere 44 Ìròyìn

If 6logx2 - 3logx3 = 3log50.2, find x.

8/3

4/3

3/4

3/8

Awọn alaye Idahun

6logx2 - 3logx3 = 3log50.2
= logx26 - 3logx33 = log5(0.2)3
= logx(64/27) = log5(1/5)3
logx(64/27) = log5(1/125)
let logx(64/27) = y
∴xy = 64/27
and log5(1/125) = y
∴ 5y = 1/125
5y = 125-1
5y = 5-3
∴ y = -3
substitute y = -3 in xy = 64/27
implies x-3 = 64/27
1/x3 = 64/27
64x3 = 27
x3 = 27/64
x3 = 3√27/64
x = 3/4

Ka ìwé ẹ̀kọ́ nípa Positive And Negative Integers, Rational Numbers (WAEC)

Ka ìwé ẹ̀kọ́ nípa Logarithms (WAEC)

Positive And Negative Integers, Rational Numbers

Logarithms

Wo Idahun

Ibeere 45 Ìròyìn

What are the integer values of x which satisfy the inequality $ -1 < 3 - 2x \le 5 $?

-1. 0, 1, 2

-2, 1, 0 , -1

0, 1, 2

-1, 0, 1

Awọn alaye Idahun

The inequality given is: -1 < 3 - 2x ≤ 5. We need to find the integer values of x that satisfy this inequality. First, let's solve for the left side of the inequality: -1 < 3 - 2x. Add 2x to both sides: 2x - 1 < 3 Add 1 to both sides: 2x < 4 Divide both sides by 2: x < 2 Now, let's solve for the right side of the inequality: 3 - 2x ≤ 5. Subtract 3 from both sides: -2x ≤ 2 Divide both sides by -2, remembering to flip the inequality: x ≥ -1 So, the integer values of x that satisfy the inequality are -1, 0, 1, and 2. Therefore, the correct option is: - -1, 0, 1,

Ka ìwé ẹ̀kọ́ nípa Linear Inequalities (WAEC)

Ka ìwé ẹ̀kọ́ nípa Inequalities (JAMB)

Linear Inequalities

Inequalities

Wo Idahun

Ibeere 46 Ìròyìn

The shaded region in the Venn diagram above is

Pc ∪ (Q ∩ R)

Pc ∩ (Q ∪ R)

P ∩ Q

Pc ∩ (Q ∩ R)

Sets

Triangles And Polygons

Wo Idahun

Ibeere 47 Ìròyìn

The weight of 10 pupils in a class are 15 kg, 16 kg, 17 kg, 18 kg, 16 kg, 17 kg, 17 kg, 17 kg, 18 kg, and 16 kg. What is the range of this distribution?

4

3

2

1

Awọn alaye Idahun

The range of a distribution is the difference between the maximum and minimum values. In this case, the maximum weight is 18 kg and the minimum weight is 15 kg, so the range is 18 kg - 15 kg = 3 kg. Therefore, the answer is option (B) 3.

Ka ìwé ẹ̀kọ́ nípa Surds (radicals) (WAEC)

Ka ìwé ẹ̀kọ́ nípa Measures Of Dispersion (JAMB)

Surds (radicals)

Measures Of Dispersion

Wo Idahun

Ibeere 48 Ìròyìn

An arc of length 22cm subtends an angle of θ at the center of the circle. What is the value of θ if the radius of the circle is 15cm?[Take π = 22/7]

70o

84o

96o

156o

168o

Awọn alaye Idahun

In a circle, the ratio of the arc length to the circumference of the circle is equal to the ratio of the angle subtended by the arc at the center of the circle to the angle of one full revolution (360 degrees). Using this property, we can find the value of the angle θ in the following way: The circumference of the circle is given by: C = 2πr = 2 x (22/7) x 15 = 94.29 cm The given arc length is 22 cm. Therefore, the ratio of the arc length to the circumference of the circle is: 22/94.29 = 0.2332 Let the angle of one full revolution be 360 degrees. Then, the ratio of the angle subtended by the arc at the center of the circle to the angle of one full revolution is also 0.2332. Thus, we can write: θ/360 = 0.2332 Multiplying both sides by 360, we get: θ = 360 x 0.2332 = 83.952 degrees (approximately) Therefore, the value of θ is approximately 84 degrees. The correct option is: - 84o

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Circles

Wo Idahun






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