JAMB UTME - Mathematics - 1989

Question 1 Report

Solve for a positive number x such that 2(x3 - x2 - 2x) = 1

4

3

2

1

Answer Details

2(x3 - x2 - 2x) = 1
x3 - 2 - 2x = 0
x(x2 - x - 2) = 0
x2 - x - 2 = 0
(x + 1)(x - 2) = 0
x = 2

Read lesson note on Solution Of Linear Equations (WAEC)

Read lesson note on Quadratic Equations (WAEC)

Solution Of Linear Equations

Quadratic Equations

View Answer

Question 2 Report

In xyz, y = z = 30o ans XZ = 3cm. Find YZ

\frac{\sqrt{3}}{2}

cm

3

\frac{\sqrt{3}}{2}

cm

3

\sqrt{3}

cm

2

\sqrt{3}

cm

Answer Details

Y

\to

30o
X

\to

120o
Z

\to

30o

\frac{3}{sin 30}

=

\frac{Y Z}{sin 120}

YZ =

\frac{3 sin 120}{sin 30}

3 x

\frac{\sqrt{3}}{2}

x 2 = 3

\sqrt{3}

View Answer

Question 3 Report

The cost of dinner for a group of students is partly constant and partly varies directly as the number of students. If the cost is ₦74.00 when the number of is 20 and ₦96.00 when the number is 30, find the cost when there are 15 students

₦68.50

₦63.00

₦60.00

₦52.00

Answer Details

We know that the cost of dinner for the group of students is partly constant and partly varies directly as the number of students. Let the constant cost be c and the cost per student be k. Therefore, the cost of dinner for 20 students is: c + 20k = ₦74.00 ...(1) Similarly, the cost of dinner for 30 students is: c + 30k = ₦96.00 ...(2) Now, we need to find the cost of dinner when there are 15 students. We can solve for the values of c and k using simultaneous equations: Subtracting equation (1) from (2) gives: 10k = ₦22.00 Therefore, k = ₦2.20 Substituting the value of k in equation (1) gives: c + 20(2.20) = ₦74.00 Therefore, c = ₦30.00 So, the cost of dinner when there are 15 students is: c + 15k = ₦30.00 + 15(2.20) = ₦63.00 Therefore, the answer is ₦63.00.

View Answer

Question 4 Report

In the diagram, HK is parallel to QR, PH = 4cm and HQ = 3cm. What is the ratio of KR:PR?

7:3

3:7

3:4

4:3

Answer Details

Since HK is parallel to QR, angle HPQ is equal to angle KPR, and angle HQP is equal to angle KRP. Therefore, triangles HPQ and KRP are similar triangles by the angle-angle (AA) criterion. The ratio of the lengths of the corresponding sides of similar triangles is equal. Therefore: KR/PR = HQ/PQ = 3/(3+4) = 3/7 So the ratio of KR:PR is 3:7. Therefore, the correct option is 3:7.

Read lesson note on Triangles And Polygons (WAEC)

Triangles And Polygons

View Answer

Question 5 Report

A square tile has side 30cm. How many of these tiles will cover a rectangular floor of length 7.2m and width 4.2?

336

360

576

720

Answer Details

To solve this problem, we need to find the area of the rectangular floor and divide it by the area of one tile. The area of the rectangular floor is: 7.2m x 4.2m = 30.24 m² The area of one square tile is: 30cm x 30cm = 0.09 m² Now we can divide the area of the floor by the area of one tile to get the number of tiles needed to cover the floor: 30.24 m² / 0.09 m² = 336 Therefore, 336 square tiles will cover the rectangular floor. So, the answer is 336.

Read lesson note on Number Bases (WAEC)

Read lesson note on Number Bases (JAMB)

Number Bases

View Answer

Question 6 Report

If f(x) = 2x2 - 5x + 3, find f(x + 1)

2x2 - x

2x2 - x + 10

4x2 + 3x + 2

4x2 + 3x + 12

Answer Details

To find f(x + 1), we need to replace x in the given function f(x) with (x + 1): f(x + 1) = 2(x + 1)^2 - 5(x + 1) + 3 Expanding the brackets and simplifying: f(x + 1) = 2(x^2 + 2x + 1) - 5x - 5 + 3 f(x + 1) = 2x^2 + 4x + 2 - 5x - 2 f(x + 1) = 2x^2 - x Therefore, the answer is option A: 2x^2 - x.

Read lesson note on Algebraic Expressions (WAEC)

Algebraic Expressions

View Answer

Question 7 Report

Simplify $ \frac{324 - 4x^2}{2x + 18} $

2(x - 9)

2(9 + x)

81 - x2

-2(x - 9)

Answer Details

\frac{324 - 4 x^{2}}{2 x + 18}

=

\frac{18^{2} - (2 x)^{2}}{2 x + 18}

=

\frac{(18 - 2 x) (18 + 2 x)}{(2 x + 18)}

18 - 2x = 2(9 - x)
or -2(x - 9)

Read lesson note on Algebraic Expressions (WAEC)

Algebraic Expressions

View Answer

Question 8 Report

What are K and L respectively if $ \frac{1}{2}(3y - 4x)^2 = (8x^2 + kxy + Ly^2) $

-12,

\frac{9}{2}

-6, 9

6, 9

12,

\frac{9}{2}

Answer Details

\frac{1}{2}

(3y - 4x)2 = (8x2 + kxy + Ly2)

\frac{1}{2}

(9y2 - 24xy + 16x2) = 8x2 + kxy + Ly2

\frac{9}{2}

y2 - 12xy) = kxy + Ly2
k = -12 ∴ L =

\frac{9}{2}

View Answer

Question 9 Report

if 12 = x, find x where e = 12

20

15

14

12

Number Bases

Modular Arithmetic

View Answer

Question 10 Report

a cylindrical metal pipe 1m long has an outer diameter of 7.2cm and an inner diameter of 2.8cm. Find the volume of metal used for the cylinder

440 $π$ cm3

1100 $π$ cm3

4400 $π$ cm3

1100 $π$ cm3

Answer Details

Volume of cylinder pip (V) = $π$ h(R $^{2}$ - r $^{2}$ )

= 100 $π$ (7.2 $^{2}$ - 2.8 $^{2}$ )

= 100 $π$ (51.84 - 7.84)

= 100 $π$ x 44

= 440 $π$ cm $^{3}$

Read lesson note on Volumes (WAEC)

Read lesson note on Number Bases (JAMB)

Volumes

Number Bases

View Answer

Question 11 Report

What are the values of y which satisfy the equation gy - 4 x 3y + 3 = 0?

-1 and 0

-1 and 1

1 and 3

0 and 1

Answer Details

gy - 4 x 3y + 3 = 0
Put 3y = x2 - 4x + 3 = 0
factorize (x - 3)(x - 1) = 0
x = 3 or 1
? 3y = 3, y = 1, 3y = 1, y = 0

Read lesson note on Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

View Answer

Question 12 Report

$4\sin^{2} x - 3 = 0$, find $x$ if $0 \ge x \ge 90^{o}$

30o

45o

60o

9o

Answer Details

The given equation is 4sin²x - 3 = 0. We need to find the value of x such that 0 ≤ x ≤ 90°. To solve this equation, we first need to isolate sin²x by adding 3 to both sides: 4sin²x = 3 Then, we divide both sides by 4: sin²x = 3/4 Now, we take the square root of both sides: sinx = ±√3/2 We have two possible values for sinx: √3/2 and -√3/2. To determine which of these values of sinx is valid for the given range of x, we need to examine the unit circle. The unit circle is a circle with a radius of 1 centered at the origin of the coordinate plane. It is used to visualize the values of sine, cosine, and tangent for all angles in standard position (angles whose vertex is at the origin and whose initial side lies along the positive x-axis). If we draw the unit circle and plot the points corresponding to sinx = √3/2 and sinx = -√3/2, we find that sinx = √3/2 corresponds to an angle of 60° (or π/3 radians) and sinx = -√3/2 corresponds to an angle of 300° (or 5π/3 radians). Since we are looking for a value of x such that 0 ≤ x ≤ 90°, the only valid solution is sinx = √3/2 and x = 60°. Therefore, the answer is x = 60°.

Read lesson note on Modular Arithmetic (WAEC)

Read lesson note on Fractions, Decimals, Approximations And Percentage (JAMB)

Modular Arithmetic

Fractions, Decimals, Approximations And Percentage

View Answer

Question 13 Report

In a family of 21 people, the average age is 14years. If the age of the grandfather is not counted, the average age drops to 12 years. What is the age of the grandfather?

35 years

40 years

42 years

54 years

Answer Details

Let's call the grandfather's age "G". We know that the family consists of 21 people, so if we exclude the grandfather, there are 20 people remaining. We're given that the average age of the family is 14 years, so we can write: Total age of the family = 21 x 14 We're also given that if we exclude the grandfather, the average age drops to 12 years, so we can write: Total age of the family (excluding grandfather) = 20 x 12 We can set up an equation with these two expressions and solve for G: 21 x 14 - G = 20 x 12 294 - G = 240 G = 54 Therefore, the age of the grandfather is 54 years. So the correct option is: 54 years.

Read lesson note on Surds (radicals) (WAEC)

Surds (radicals)

View Answer

Question 14 Report

Factorize completely y3 -4xy + xy3 - 4y

(y + ay)(y + 2)(y - 2)

(y - xy)(y - 2)

y(1 + x)(y + 2)(y -2)

y(1 - x)(y + 2)(y - 2)

Answer Details

y3 -4xy + xy3 - 4y = y3(1 + x) - 4y(1 + x)
(y3 - 4y)(1 + y) = (y3(1 + x) - 4y(1 + x))
∴ = y(1 + x)(y + 2)(y - 2)

Read lesson note on Algebraic Expressions (WAEC)

Algebraic Expressions

View Answer

Question 15 Report

On the curve, the points at which the gradient of the curve is equal to zero are

c, d, f. i, l

b, e, g, j, m

a, b, c, d, f, i, j, l

c, d, f, h, i, l

Answer Details

The gradient of any curve is equal to zero at the turning points. i.e. maximum or minimum points. The points in the above curve are b, e, g, j, m

Read lesson note on Differentiation (JAMB)

Read lesson note on Application Of Differentiation (JAMB)

Differentiation

Application Of Differentiation

View Answer

Question 16 Report

If a : b = 5 : 8, x : y = 25 : 16; evaluate $ \frac{a}{x} : \frac{b}{y} $

125 : 128

3 : 5

3 : 4

2 : 5

Answer Details

a : b = 5 : 8 = 2.5 : 40
x : y = 25 : 16
$\frac{a}{x}$ : $\frac{b}{y}$ = $\frac{25}{25}$ : $\frac{40}{16}$
= 1 : $\frac{40}{16}$
= 16 : 40
= 2 : 5

Read lesson note on Algebraic Fractions (WAEC)

Algebraic Fractions

View Answer

Question 17 Report

Simplify $4 - \frac{1}{2 - \sqrt{3}}$

2

\sqrt{3}

-2 -

\sqrt{3}

-2 +

\sqrt{3}

2 -

\sqrt{3}

Answer Details

4 -

\frac{1}{2 - \sqrt{3}}

= 4 -

\frac{2 + \sqrt{2}}{22 - (\sqrt{3}) 2}

= 4 -

\frac{(2 + 3)}{4 - 3}

= 4 - 2-

\sqrt{3}

= 2 -

\sqrt{3}

Read lesson note on Number Bases (WAEC)

Read lesson note on Surds (radicals) (WAEC)

Read lesson note on Number Bases (JAMB)

Number Bases

Surds (radicals)

Number Bases

View Answer

Question 18 Report

OXYZW is a pyramid with a square base such that OX = OY= OZ = OW = 5cm and XY = XW = YZ = WZ = 6cm. Find the height OT

2

\sqrt{5}

3

4

\sqrt{3}

Answer Details

xz2 = 62 + 62
36 + 36 = 72
xz =

\sqrt{72}

6

\sqrt{2}

= xT

\frac{6 \sqrt{2}}{2}

=

\frac{3}{\sqrt{2}}

OT2 = 52 + (3

\sqrt{2}

)2 = 25 + 18
OT = 4

\sqrt{3}

View Answer

Question 19 Report

If the surface area of a sphere increased by 44%, find the percentage increase in diameter

44

30

22

20

Answer Details

Surface Area of Sphere A = 4

π r^{2}

∴ A = 4

π

\frac{(D)^{2}}{2}

=

\frac{(D)^{2}}{2}

=

π

D2
When increased by 44% A =

\frac{144 π D^{2}}{100}

π

\frac{(12 D)^{2}}{10}

=

π

\frac{(6 D)^{2}}{5}

Increase in diameter =

\frac{6 D}{5}

- D =

\frac{1}{5}

D
Percentage increase =

\frac{1}{5}

x

\frac{1}{100}

%
= 20%

View Answer

Question 20 Report

Evaluate 2700 000 x 0.03 + 18 000

4.5 x 10o

4.5 x 101

4.5 x 102

4.5 x 103

Representation Of Data

View Answer

Question 21 Report

What is the different between 0.007685 correct to three significant figures and 0.007685 correct to four places of decimal?

10-5

7 x 10-4

8 x 10-5

10-6

Answer Details

When we round 0.007685 to three significant figures, we get 0.00768. The last digit (5) is rounded down because it is less than 5. Therefore, the difference between 0.007685 correct to three significant figures and 0.007685 correct to four places of decimal is: 0.007685 - 0.0077 = -0.000015 So the answer is -1.5 x 10^-5. Therefore, the correct option is: - 10-5

Read lesson note on Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

View Answer

Question 22 Report

Make R the subject of the fomula $S = \sqrt{\frac{2R + T}{2RT}}$

R =

\frac{T}{(T S^{2} + 1)}

R =

\frac{T}{(T S^{2} + 1)}

R =

\frac{T}{2 (T S^{2} + 1)}

R =

\frac{R}{2 (T S^{2} + 1)}

Answer Details

S =

\sqrt{\frac{2 R + T}{2 R T}}

squaring both sides = S2 =

\frac{2 R + T}{2 R T}

S2(2RT) = 2R + T
2S2RT - 2R = T
R(2S2T - 2) = T
R = R =

\frac{T}{2 (T S^{2} - 2)}

Read lesson note on Change Of Subject Of A Formula/Relation (WAEC)

Read lesson note on Algebraic Expressions (WAEC)

Change Of Subject Of A Formula/Relation

Algebraic Expressions

View Answer

Question 23 Report

If PQR is a straight line with QS = QR, calculate TPQ, If QT\\SR and TQS = 3yo

62o

56

\frac{3}{2}

o

20

\frac{3}{2}

o

18o

Answer Details

Since QS = QR

then, angle SQR = angle SRQ

2 SQR = 180 - 56, SQR =

\frac{124}{2}

= 62o

QTP = 62o

QTP = 62o, corresponding angle

3y + 56 + 62 = 180 = 3y = 180 - 118

3y = 62 = 180

3y = 180 - 118

3y = 62

y =

\frac{62}{3}

= 20

\frac{3}{2}

View Answer

Question 24 Report

If PST is a straight line and PQ = QS = SR in the diagram, find y.

24

^{\circ}

48

^{\circ}

72

^{\circ}

84

^{\circ}

Answer Details

< PSQ = < SQR = < SRQ = 24

^{\circ}

< QSR = 180 $^{\circ}$ - 48 $^{\circ}$ = 132 $^{\circ}$

< PSQ + < QSR + y + 180 (angle on a straight lines)

24 + 132 + y = 180 $^{\circ}$ = 156 $^{\circ}$ + y = 180

y = 180 $^{\circ}$ - 156 $^{\circ}$

= 24 $^{\circ}$

Read lesson note on Loci (JAMB)

Read lesson note on Trigonometry (JAMB)

Read lesson note on Triangles And Polygons (WAEC)

Loci

Trigonometry

Triangles And Polygons

View Answer

Question 25 Report

Simplify $ \frac{x(x+1)^{\frac{1}{2}}-(x+1)^{\frac{1}{2}}}{(x+1)^{\frac{1}{2}}} $

\frac{- 1}{x + 1}

\frac{1}{x + 1}

\frac{1}{x}

\frac{1}{x - 1}

Answer Details

\frac{x}{(x + 1)}

-

\frac{\sqrt{(x + 1)}}{\sqrt{x + 1}}

=

\frac{x}{x + 1}

- 1

\frac{x - x - 1}{x + 1}

=

\frac{- 1}{x + 1}

Read lesson note on Solution Of Linear Equations (WAEC)

Read lesson note on Algebraic Expressions (WAEC)

Solution Of Linear Equations

Algebraic Expressions

View Answer

Question 26 Report

The prime factors of 2520 are

2, 9, 5

2, 9, 7

2, 3, 5, 7

2, 3, 7, 9

Answer Details

To find the prime factors of 2520, we need to divide it by its prime factors until we are left with prime numbers only. We start by dividing 2520 by 2, which gives 1260. 1260 is still an even number, so we divide it by 2 again to get 630. 630 is divisible by 2 and 3, so we divide it by 2 to get 315, and then by 3 to get 105. 105 is divisible by 3 and 5, so we divide it by 3 to get 35, and then by 5 to get 7. 7 is a prime number, so we can't divide it any further. Therefore, the prime factors of 2520 are 2, 2, 2, 3, 3, 5, and 7. So the answer is 2, 3, 5, 7.

Read lesson note on Number Bases (WAEC)

Number Bases

View Answer

Question 27 Report

Express $ \frac{1}{x+1} - \frac{1}{x-2} $ as a single algebraic fraction

\frac{- 3}{(x + 1) (2 - x)}

\frac{3}{(x + 1) (2 - x)}

\frac{- 1}{(x + 1)}

\frac{1}{(x + 1) (x - 2)}

Answer Details

To express the given expression as a single algebraic fraction, we first need to find a common denominator for all the terms. The common denominator is (x+1)(2-x). Then, we can simplify each term by multiplying the numerator and denominator by the missing factor in the denominator to obtain: 1(x+1)(2-x) / (x+1) + 1(2-x) / (2-x)(x+1) - 1x(x+1) / (2-x)(x+1) Simplifying further by combining like terms, we get: [(x+1)(2-x) + 1(1-x)(x+1) - x(x+1)] / [(2-x)(x+1)] Simplifying the numerator by distributing, we get: [-x^2 + 3x - 1] / [(2-x)(x+1)] Therefore, the expression 1x+1 / (x+1) + 1 / (2-x) - 1x?2 / (x+1)(2-x) can be simplified to -[-x^2 + 3x - 1] / [(2-x)(x+1)]. So, the answer is (a) -3(x+1)(2-x) / [-x^2 + 3x - 1].

Read lesson note on Algebraic Expressions (WAEC)

Read lesson note on Algebraic Fractions (WAEC)

Algebraic Expressions

Algebraic Fractions

View Answer

Question 28 Report

The sum of the first two terms of a geometric progression is x and sum of the last terms is y. If there are n terms in all, then the common ratio is

\frac{x}{y}

\frac{y}{x}

(

\frac{x}{y}

)

\frac{1}{n - 2}

D) (

\frac{y}{x}

)

\frac{1}{n - 2}

Answer Details

Sum of nth term of a G.P = Sn =

\frac{a r^{n} - 1}{r - 1}

sum of the first two terms =

\frac{a r^{2} - 1}{r - 1}

x = a(r + 1)
sum of the last two terms = Sn - Sn - 2
=

\frac{a r^{n} - 1}{r - 1}

-

\frac{(a r^{n - 1})}{r - 1}

=

\frac{a (r^{n} - 1 - r^{n - 2} + 1)}{r - 1}

(r2 - 1)
∴

\frac{a r^{n - 2} (r + 1) (r - 1)}{1}

= arn - 2(r + 1) = y
= a(r + 1)r^n - 2
y = xrn - 2
= yrn - 2

\frac{y}{x}

= r = (

\frac{y}{x}

)

\frac{1}{n - 2}

Read lesson note on Sequence And Series (WAEC)

Sequence And Series

View Answer

Question 29 Report

Which of the following is in descending order?

9	4	3	17
10	5	4	10

4	9	3	17
5	10	4	20

9	17	4	3
10	10	5	4

4	9	17	3
5	10	10	4

Answer Details

$\frac{9}{10} \frac{4}{5} \frac{3}{4} \frac{17}{10}$ = $\frac{18, 16, 15, 34}{20}$
$\frac{4}{5} \frac{9}{10} \frac{3}{4} \frac{17}{20}$ = $\frac{16, 18, 15, 17}{20}$
$\frac{9}{10} \frac{17}{10} \frac{4}{5} \frac{3}{4}$ = $\frac{18, 17, 16, 15}{20}$
∴ $\frac{4}{5} \frac{9}{10} \frac{17}{10} \frac{3}{4}$ is in descending order

Read lesson note on Areas (WAEC)

Areas

View Answer

Question 30 Report

Oke deposited ₦800.00 in the bank at the rate of $12\frac{1}{2}\%$ simple interest. After some time the total amount was one and half times the principal. For how many years was the money left in the bank?

2

\frac{2}{3}

4

5

\frac{1}{3}

8

Answer Details

This problem is asking us to find the number of years that Oke's ₦800.00 deposit was left in the bank at a certain interest rate, given that the total amount he received was one and a half times the original deposit amount. We can start by using the simple interest formula, which is: Simple Interest = Principal * Rate * Time Here, we know the principal is ₦800.00 and the rate is 12.12%. We don't know the time, which is what we're trying to find. Let's call it "t". We also know that the total amount Oke received after some time was one and a half times the original deposit amount, or: Total Amount = 1.5 * Principal Substituting the values we know, we get: Total Amount = 1.5 * ₦800.00 = ₦1200.00 We can use this equation to solve for "t" by first finding the simple interest: Simple Interest = Total Amount - Principal Simple Interest = ₦1200.00 - ₦800.00 = ₦400.00 Then, we can rearrange the simple interest formula to solve for "t": Time = Simple Interest / (Principal * Rate) Substituting the values we know, we get: Time = ₦400.00 / (₦800.00 * 0.1212) = 4 years (rounded to the nearest whole number) Therefore, Oke's money was left in the bank for 4 years.

Read lesson note on Representation Of Data (JAMB)

Read lesson note on Modular Arithmetic (WAEC)

Representation Of Data

Modular Arithmetic

View Answer

Question 31 Report

MN is tangent to the given circle at M, MR and MQ are two chords. IF QNM is 60o and MNQ is 40o. Find RMQ

120o

110o

60o

20o

Answer Details

QMN = 60o

MRQ = 60o(angle in the alternate segment are equal)

MQN = 80o(angle sum of a $△$ = 180o)

60 = x = 80o(exterior angle = sum of opposite interior angles)

x = 80o - 60o = 20o

RMQ = 20o

Read lesson note on Circles (WAEC)

Circles

View Answer

Question 32 Report

If n is the median and m is the mode of the following set of numbers, 2.4, 2.1, 1.6, 2.6, 2.6, 3.7, 2.1, 2.6, then (n, m) is

(2.6, 2.6)

(2.5, 2.6)

(2.6, 2.5)

(2.5, 2.1)

Answer Details

Arrange the numbers in order, 1.6, 2.1, 2.1| 2.4, 2.6| 2.6, 2.6, 3.7
n = median =

\frac{2.4 + 2.6}{2}

= 2.5
m = mode = 2.6
∴ (n, m) = (2.5, 2.6)

Read lesson note on Probability (WAEC)

Read lesson note on Sets (JAMB)

Probability

Sets

View Answer

Question 33 Report

Solve the pair of pair of equation for x and y respectively 2x-t - 3y-1 = 4, 4x-1 + y-1 = 1

-1, 2

1, 2

2, 1

2, -1

Answer Details

2x-t - 3y-1 = 4, 4x-1 + y-1 = 1
Let x - 1 = a and y - 1= b
2a - 3b = 4 .......(i)
4a + b = 1 .........(ii)
(i) x 3 = 12a + 3b = 3........(iii)
2a - 3b = 4 ...........(i)
(i) + (iii) = 14a
= 7
∴ a =

\frac{7}{14}

=

\frac{1}{2}

From (i), 3b = 2a - 4
3b = 1 - 4
3b = -3
∴ x = -1
From substituting, 2-1 = x - 1
∴ x = 2
y-1 = -1, y = -1

Read lesson note on Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

View Answer

Question 34 Report

A rectangular polygon of (2k + 1) sides has 140o as the size of each interior angle. Find k

4

\frac{1}{2}

8

\frac{1}{2}

Answer Details

A rectangular has all sides and all angles equal. If each interior angle is 140o each exterior angle must be
180o - 140o = 40o
The number of sides must be

\frac{360^{o}}{40^{o}}

= 9 sides
hence 2k + 1 = 9
2k = 9 - 1
8 = 2k
k =

\frac{8}{2}

= 4

Read lesson note on Polynomials (JAMB)

Read lesson note on Number Bases (WAEC)

Read lesson note on Angles And Intercepts On Parallel Lines (WAEC)

Polynomials

Number Bases

Angles And Intercepts On Parallel Lines

View Answer

Question 35 Report

If -8, m, n, 19 are in arithmetic progression, find (m, n)

1, 10

2, 10

3, 13

4, 16

Answer Details

To solve this problem, we need to use the arithmetic progression formula: a_n = a_1 + (n-1)d where a_n is the nth term of the arithmetic progression, a_1 is the first term, n is the number of terms, and d is the common difference. We are given that -8, m, n, and 19 are in arithmetic progression. So we can set up the following equations: m = -8 + d n = -8 + 2d 19 = -8 + 3d We can solve for d by subtracting the first equation from the second equation: n - m = 2d - d n - m = d We can substitute this expression for d into the third equation: 19 = -8 + 3(n - m) Simplifying this equation gives: 27 = 3(n - m) 9 = n - m We can substitute this expression for n - m into the equation we derived earlier: n - m = d So we have: d = 9 Substituting this value of d into any of the earlier equations will allow us to solve for m and n. For example, using the equation: m = -8 + d gives: m = -8 + 9 m = 1 And using the equation: n = -8 + 2d gives: n = -8 + 2(9) n = 10 Therefore, the solution is (m, n) = (1, 10)

Read lesson note on Sequence And Series (WAEC)

Sequence And Series

View Answer

Question 36 Report

find the range of values of values of r which satisfies the following inequality, where a, b and c are positive $ \frac{r}{a} + \frac{r}{b} + \frac{r}{c} > 1 $

r >

\frac{a b c}{b c + a c + a b}

r < abc

r >

\frac{1}{a}

+

\frac{1}{b}

+

\frac{1}{c}

r <

\frac{1}{a b c}

Answer Details

\frac{r}{a}

+

\frac{r}{b}

+

\frac{r}{c}

> 1 =

\frac{b c r + a c r + a b r}{a b c}

> 1
r(bc + ac + ba > abc) = r >

\frac{a b c}{b c + a c + a b}

Read lesson note on Linear Inequalities (WAEC)

Read lesson note on Number Bases (JAMB)

Read lesson note on Statistics (WAEC)

Linear Inequalities

Number Bases

Statistics

View Answer

Question 37 Report

What value of Q will make the expression 4x2 + 5x + Q a complete square?

\frac{25}{16}

\frac{25}{16}

\frac{25}{16}

\frac{25}{16}

Answer Details

4x2 + 5x + Q
To make a complete square, the coefficient of x2 must be 1
= x2 +

\frac{5 x}{4}

+

\frac{Q}{4}

Then (half the coefficient of x2) should be added
i.e. x2 +

\frac{5 x}{4}

+

\frac{25}{64}

?

\frac{Q}{4}

=

\frac{25}{64}

Q =

\frac{4 \times 25}{64}

=

\frac{25}{16}

Read lesson note on Quadratic Equations (WAEC)

Quadratic Equations

View Answer

Question 38 Report

Simplify $ \sqrt[3]{(64r^6)^{\frac{1}{2}}} $

r

2

2r

2

r

64r

Answer Details

\sqrt[3]{(64 r^{6})^{\frac{1}{2}}}

= (64 r^{6})^{\frac{1}{2} \cdot \frac{1}{3}}

= 64 r^{6 \cdot \frac{1}{6}}

= 64 r

Read lesson note on Surds (radicals) (WAEC)

Surds (radicals)

View Answer

Question 39 Report

In a class of 30 students, the marks scored in an examination are displayed in the histogram. What percentage of the student scored more than 40%?

14%

40%

40

\frac{3}{4}

%

53

\frac{1}{3}

%

Answer Details

This histogram is transferred into this frequency table

\begin{array}{cc} M a r k s & 20 & 40 & 60 & 80 & 100 \\ s t u d e n t s & 9 & 7 & 6 & 6 & 2 \end{array}

Students who scored more than 40 = 6 + 6 + 2 = 14

i.e. $\frac{14}{30}$ x 100% = 46 $\frac{3}{4}$ %

View Answer

Question 40 Report

Factorize 4a2 - 12ab - C2 + 9b2

4a(a - 3b) + (3b - c)2

(2a + 3b - c)(2a + 3b + c)

(2a - 3b + c)(2a - 3b - c)

4a(a - 3b) + (3b + c)

Answer Details

4a2 - 12ab - C2 + 9b2
rearranges: (4a2 - 12ab + 9b2) - c2
(2a - 3b)(2a - 3b) - c2 = (2a - 3b)2 - c2
= (2a - 3b + c)(2a - 3b - c)

Read lesson note on Algebraic Expressions (WAEC)

Algebraic Expressions

View Answer

Question 41 Report

In preparing rice cutlets, a cook used 75g of rice, 40g of margarine, 105g of meat and 20g of bread crumbs. Find the angle of the sector which represents meat in pie chart

30o

60o

112.5o

157.5o

Answer Details

To find the angle of the sector which represents meat in the pie chart, we first need to find the total weight of the ingredients used: Total weight = 75g + 40g + 105g + 20g = 240g Now, we can find the percentage of meat used: Percentage of meat = (Weight of meat / Total weight) x 100% = (105g / 240g) x 100% = 43.75% To find the angle of the sector in the pie chart, we use the formula: Angle of sector = Percentage of sector / 100% x 360 degrees Substituting the percentage of meat, we get: Angle of sector = 43.75 / 100 x 360 degrees = 157.5 degrees Therefore, the answer is 157.5 degrees.

View Answer

Question 42 Report

PQRS is a rhombus. If PR2 + QS2 = KPQ2, determine k

1

2

3

4

Answer Details

PR2 + QS2 = KPQ2
SQ2 = SR2 + RQ2
PR2 + SQ2 = PQ2 + SR2 + 2RQ2
= 2PQ2 + 2RQ2
= 4PQ2
∴ K = 4

Read lesson note on Polynomials (JAMB)

Read lesson note on Triangles And Polygons (WAEC)

Polynomials

Triangles And Polygons

View Answer

Question 43 Report

Find P in terms of q if log3P + 3log3q = 3

(

\frac{3}{q}

)3

(

\frac{q}{3}

)

\frac{1}{3}

(

\frac{q}{3}

)3

(

\frac{3}{q}

)

\frac{1}{3}]

Answer Details

log3P + 3log3q = 3
log3(Pq3) = 3
Pq3 = 33
P = (

\frac{3}{P}

)3

Read lesson note on Logarithms (WAEC)

Logarithms

View Answer

Question 44 Report

if x : y = 5 : 12 and z = 52cm, find the perimeter of the triangle

68cm

84cm

100cm

120cm

Answer Details

13 = 52
1 =

\frac{52}{13}

= 4
5 + 12 + 13 = 30
Total perimeter = 30 x 4
= 120cm

Read lesson note on Lengths And Perimeters (WAEC)

Read lesson note on Ratio, Proportions And Rates (WAEC)

Lengths And Perimeters

Ratio, Proportions And Rates

View Answer

Question 45 Report

The pilot of an aeroplane, flying 10km above the ground in the direction of a landmark, views the landmark to have angles of depression of 35o and 55o. Find the distance between the two points of observation

10(sin 35o - sin 55o)

10 cos35o - cos 55o

10(tan35o - tan55o)

10(cot35o - cot55o)

Answer Details

x = 10 cot35o - 10 cot55o
= 10(cot35o - cot55o)

Read lesson note on Trigonometry (JAMB)

Trigonometry

View Answer






Press & Hold to Drag Around
Click Here to Close