JAMB UTME - Mathematics - 2004

Ajụjụ 1 Ripọtì

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)

Lee Azịza

Ajụjụ 2 Ripọtì

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

Akọwa Nkọwa

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

Lee Azịza

Ajụjụ 3 Ripọtì

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

Akọwa Nkọwa

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.

Lee Azịza

Ajụjụ 4 Ripọtì

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

18

16

-4

-6

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Differentiation (JAMB)

Differentiation

Lee Azịza

Ajụjụ 5 Ripọtì

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

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Mensuration (JAMB)

Mensuration

Lee Azịza

Ajụjụ 6 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Number Bases

Algebraic Expressions

Lee Azịza

Ajụjụ 7 Ripọtì

The shaded region in the Venn diagram above is

Pc ∪ (Q ∩ R)

Pc ∩ (Q ∪ R)

P ∩ Q

Pc ∩ (Q ∩ R)

Sets

Triangles And Polygons

Lee Azịza

Ajụjụ 8 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Circles

Lee Azịza

Ajụjụ 9 Ripọtì

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

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Sequence And Series

Lee Azịza

Ajụjụ 10 Ripọtì

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

(1, 1)

(2, 2)

(4, 4)

(4, -4)

Akọwa Nkọwa

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).

Gụọ akwụkwọ nkuzi na Coordinate Geometry (JAMB)

Coordinate Geometry

Lee Azịza

Ajụjụ 11 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Trigonometry (JAMB)

Trigonometry

Lee Azịza

Ajụjụ 12 Ripọtì

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

2

10

1/5

1/10

Akọwa Nkọwa

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

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

Gụọ akwụkwọ nkuzi na Variation (JAMB)

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Sine, Cosine And Tangent Of An Angle

Variation

Modular Arithmetic

Lee Azịza

Ajụjụ 13 Ripọtì

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

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Sequence And Series

Lee Azịza

Ajụjụ 14 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Probability

Lee Azịza

Ajụjụ 15 Ripọtì

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)

Akọwa Nkọwa

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).

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 16 Ripọtì

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

1

13

2

-1

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Introductory Calculus (WAEC)

Gụọ akwụkwọ nkuzi na Differentiation (JAMB)

Gụọ akwụkwọ nkuzi na Application Of Differentiation (JAMB)

Introductory Calculus

Differentiation

Application Of Differentiation

Lee Azịza

Ajụjụ 17 Ripọtì

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

45o

75o

30o

60o

Akọwa Nkọwa

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°.

Gụọ akwụkwọ nkuzi na Trigonometry (JAMB)

Trigonometry

Lee Azịza

Ajụjụ 18 Ripọtì

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}

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Linear Inequalities (WAEC)

Linear Inequalities

Lee Azịza

Ajụjụ 19 Ripọtì

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

Akọwa Nkọwa

^{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

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Gụọ akwụkwọ nkuzi na Variation (JAMB)

Gụọ akwụkwọ nkuzi na Sets (JAMB)

Representation Of Data

Variation

Sets

Lee Azịza

Ajụjụ 20 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Loci (WAEC)

Gụọ akwụkwọ nkuzi na Coordinate Geometry (JAMB)

Loci

Coordinate Geometry

Lee Azịza

Ajụjụ 21 Ripọtì

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

45o

55o

35o

40o

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Circles

Lee Azịza

Ajụjụ 22 Ripọtì

Find the value of x in the figure above

15√6

20√6

3√6

5√6

Akọwa Nkọwa

\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}

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Surds (radicals)

Lee Azịza

Ajụjụ 23 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Circles

Triangles And Polygons

Lee Azịza

Ajụjụ 24 Ripọtì

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

-4

4

1

-1

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Solution Of Linear Equations

Modular Arithmetic

Lee Azịza

Ajụjụ 25 Ripọtì

Find P, if 4516 - P7 = 3056

627

1167

6117

1427

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Lee Azịza

Ajụjụ 26 Ripọtì

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

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Probability

Surds (radicals)

Lee Azịza

Ajụjụ 27 Ripọtì

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

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Gụọ akwụkwọ nkuzi na Measures Of Location (JAMB)

Gụọ akwụkwọ nkuzi na Measures Of Dispersion (JAMB)

Representation Of Data

Measures Of Location

Measures Of Dispersion

Lee Azịza

Ajụjụ 28 Ripọtì

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

2/3

1/3

3/4

1

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Sequence And Series

Lee Azịza

Ajụjụ 29 Ripọtì

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

Akọwa Nkọwa

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,

Gụọ akwụkwọ nkuzi na Linear Inequalities (WAEC)

Gụọ akwụkwọ nkuzi na Inequalities (JAMB)

Linear Inequalities

Inequalities

Lee Azịza

Ajụjụ 30 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Circles

Lee Azịza

Ajụjụ 31 Ripọtì

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

2 -√3

-2 - √3

2 + √3

-2 + √3

Akọwa Nkọwa

\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}

Gụọ akwụkwọ nkuzi na Positive And Negative Integers, Rational Numbers (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Positive And Negative Integers, Rational Numbers

Surds (radicals)

Lee Azịza

Ajụjụ 32 Ripọtì

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

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Integration (JAMB)

Gụọ akwụkwọ nkuzi na Differentiation (JAMB)

Integration

Differentiation

Lee Azịza

Ajụjụ 33 Ripọtì

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

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Probability

Lee Azịza

Ajụjụ 34 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Sets (JAMB)

Sets

Lee Azịza

Ajụjụ 35 Ripọtì

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

2.5

2.0

1.0

1.5

1.2

Measures Of Dispersion

Statistics

Lee Azịza

Ajụjụ 36 Ripọtì

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

8/3

4/3

3/4

3/8

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Positive And Negative Integers, Rational Numbers (WAEC)

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Positive And Negative Integers, Rational Numbers

Logarithms

Lee Azịza

Ajụjụ 37 Ripọtì

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

6

-3

1 - 6x

6x - 1

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Differentiation (JAMB)

Differentiation

Lee Azịza

Ajụjụ 38 Ripọtì

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

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Indices (WAEC)

Indices

Lee Azịza

Ajụjụ 39 Ripọtì

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

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Gụọ akwụkwọ nkuzi na Measures Of Dispersion (JAMB)

Surds (radicals)

Measures Of Dispersion

Lee Azịza

Ajụjụ 40 Ripọtì

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

Akọwa Nkọwa

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.

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

Gụọ akwụkwọ nkuzi na Ratio, Proportions And Rates (WAEC)

Sine, Cosine And Tangent Of An Angle

Ratio, Proportions And Rates

Lee Azịza

Ajụjụ 41 Ripọtì

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

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Variation (WAEC)

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Variation

Probability

Algebraic Expressions

Lee Azịza

Ajụjụ 42 Ripọtì

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}

Akọwa Nkọwa

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}

Lee Azịza

Ajụjụ 43 Ripọtì

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

\frac{2}{3}

- \frac{2}{3}

- 6 \frac{2}{3}

6 \frac{2}{3}

Akọwa Nkọwa

\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}

Gụọ akwụkwọ nkuzi na Integration (JAMB)

Gụọ akwụkwọ nkuzi na Trigonometry (JAMB)

Integration

Trigonometry

Lee Azịza

Ajụjụ 44 Ripọtì

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

4

6

3

2

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Surds (radicals)

Lee Azịza

Ajụjụ 45 Ripọtì

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

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Integration (JAMB)

Gụọ akwụkwọ nkuzi na Euclidean Geometry (JAMB)

Integration

Euclidean Geometry

Lee Azịza

Ajụjụ 46 Ripọtì

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

35o

40o

45o

50o

55o

Lee Azịza

Ajụjụ 47 Ripọtì

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

Lee Azịza

Ajụjụ 48 Ripọtì

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%

Akọwa Nkọwa

\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%

Gụọ akwụkwọ nkuzi na Percentages (WAEC)

Percentages

Lee Azịza






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