JAMB UTME - Mathematics - 1991

Question 1 Report

Musa borrows N10.00 at 2% per month simple interest and repays N8.00 after 4 months. How much does he still owe?

N10.80

N10.67

N2.80

N2.67

Answer Details

I = $\frac{P R T}{100}$

= $\frac{10 \times 2 \times 4}{100}$

= $\frac{4}{5}$

= 0.8

Total amount = N10.80

He pays N8.00

Remainder = 10.80 - 8.00

= N2.80

Read lesson note on Modular Arithmetic (WAEC)

Read lesson note on Lengths And Perimeters (WAEC)

Modular Arithmetic

Lengths And Perimeters

View Answer

Question 2 Report

A car travels from calabar to Enugu, a distance of p km with an average speed of u km per hour and continues to benin, a distance of q km, with an average speed of w km per hour. Find its average speed from Calabar to Benin

\frac{(p + q)}{p w + q u}

$\frac{u w (p + q)}{p w + q u}$

$\frac{u w (p + q)}{p w}$

$\frac{u w}{p w + q u}$

Answer Details

Average speed = $\frac{total Distance}{Total Time}$
from Calabar to Enugu in time t1, hence
t1 = $\frac{p}{u}$ also from Enugu to Benin
t2 = $\frac{q}{w}$
Average speed = $\frac{p + q}{t_{1} + t_{2}}$
= $\frac{p + q}{\frac{p}{u} + \frac{q}{w}}$
= p + q × $\frac{u w}{p w + q u}$
= $\frac{u w (p + q)}{p w + q u}$

Read lesson note on Variation (WAEC)

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

Read lesson note on Areas (WAEC)

Variation

Ratio, Proportions And Rates

Areas

View Answer

Question 3 Report

Find correct to 3 decimal places $\left(\frac{1}{0.05} + \frac{1}{5.005}\right) - (0.05 x 2.05)$

99.998

98.999

89.899

9.998

Answer Details

(

\frac{1}{0.05}

+

\frac{1}{5.005}

) - (0.05 x 2.05)
(

\frac{1}{0.05}

x 5.005) - (0.05 x 2.05)

$\frac{500.5}{0.05}$ - 1025
100.1 - 0.1025
= 99.9975
99.998 to 3 sig. fiq.

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

Fractions, Decimals And Approximations

View Answer

Question 4 Report

If 3 gallons of spirit containing 20% water are added to 5 gallons of another spirit containing 15% water, what percentage of the mixture is water?

2

\frac{4}{5}

%

16

\frac{7}{8}

%

18

\frac{1}{8}

%

18

\frac{7}{8}

%

Answer Details

% of water in the mixture
=

\frac{Total Amount of water}{Total quantity of spirit}

x

\frac{100}{1}

\frac{3 (\frac{20}{100}) + 5 (\frac{15}{100})}{3 + 5}

x

\frac{100}{1}

=

\frac{\frac{6}{10} + \frac{75}{100}}{8}

x

\frac{100}{1}

=

\frac{0.6 + 0.75}{8}

x

\frac{100}{1}

= $\frac{1.35}{8}$ x $\frac{100}{1}$
= $\frac{33.75}{2}$
= 16.875
= 16 $\frac{7}{8}$

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

Read lesson note on Modular Arithmetic (WAEC)

Fractions, Decimals And Approximations

Modular Arithmetic

View Answer

Question 5 Report

Obi borrows ₦10.00 at 2% per month simple interest and repays ₦8.00 after 4 months, how much does he still owe?

₦10.80

₦10.65

₦2.80

₦2.67

Modular Arithmetic

Areas

View Answer

Question 6 Report

Find the two values of y which satisfy the simultaneous equation 3x + y = 8, x2 + xy = 6

-1 and 5

-5 and 1

1 and 5

1 and 1

Answer Details

We can solve this system of equations by substitution or elimination method. Here, we will use substitution method. First, we solve the first equation for y in terms of x: 3x + y = 8 y = 8 - 3x Now we substitute this expression for y into the second equation and solve for x: x^2 + xy = 6 x^2 + x(8 - 3x) = 6 x^2 + 8x - 3x^2 = 6 -2x^2 + 8x - 6 = 0 x^2 - 4x + 3 = 0 (x - 3)(x - 1) = 0 x = 3 or x = 1 Now we substitute each of these values of x into the expression we found for y earlier: y = 8 - 3x When x = 3, y = -1 When x = 1, y = 5 Therefore, the two values of y that satisfy the system of equations are -1 and 5. So, the correct answer is: - -1 and 5

Read lesson note on Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

View Answer

Question 7 Report

In the diagram above, |PQ| = |QR|, |PS| = |RS|, ∠PSR = 30o and ∠PQR = 80o. Find ∠SPQ.

15o

25o

40o

50o

55o

Answer Details

oin PR
QRP = QPR
= 180 - 80 = 100/20 = 50o
SRP = SPR
= 180 - 30 = 150/2 = 75o
∴ SPQ = SPR - QPR
= 75 - 50 = 25o

Read lesson note on Euclidean Geometry (JAMB)

Read lesson note on Triangles And Polygons (WAEC)

Euclidean Geometry

Triangles And Polygons

View Answer

Question 8 Report

Find the total area of the surface of a solid cylinder whose base radius is 4cm and height is 5cm

56

π

cm2

72

π

cm2

96

π

cm2

192

π

cm2

Volumes

Angles

View Answer

Question 9 Report

Given that 1/2 log10 P = 1, find the value Of P

104

103

102

101

10-1

Answer Details

1/2log10P
log10P1/2 = 1
P1/2 = 10
p = 102

Read lesson note on Logarithms (WAEC)

Logarithms

View Answer

Question 10 Report

Simplify $ \frac{1}{1+\sqrt{5}} - \frac{1}{1-\sqrt{5}} $

-

\frac{1}{2} \sqrt{5}

\frac{1}{2} \sqrt{5}

--

\frac{1}{4} \sqrt{5}

5

Answer Details

\frac{1}{1 + \sqrt{5}}

-

\frac{1}{1 - \sqrt{5}}

=

\frac{3 - \sqrt{5} - 3 - \sqrt{5}}{(3 + \sqrt{5}) (3 - \sqrt{5}}

=

\frac{- 2 \sqrt{5}}{9 - 5}

=

\frac{- 2 \sqrt{5}}{4}

= -

\frac{1}{2} \sqrt{5}

Read lesson note on Number Bases (WAEC)

Read lesson note on Number Bases (JAMB)

Number Bases

View Answer

Question 11 Report

A crate of soft drinks contains 10 bottles of Coca-cola, 8 of Fanta and 6 of sprite. If one bottle is selected at random, what is the probability that it is NOT a Cocacola bottle?

$\frac{5}{12}$

$\frac{1}{3}$

$\frac{3}{4}$

$\frac{7}{24}$

Answer Details

Coca-cola = 10 bottles, Fanta = 8 bottles
Sprite = 6 bottles, Total = 24
P(cola-cola) = $\frac{10}{24}$
P(not coca-cola) = 1 - $\frac{10}{24}$
$\frac{24 - 10}{24}$ = $\frac{14}{24}$
= $\frac{7}{24}$

View Answer

Question 12 Report

Evaluate $ \frac{xy^2-x^2y}{x^2-xy^1} $, when x = -2 and y = 2

3

- $\frac{3}{5}$

$\frac{3}{5}$

-3

Answer Details

$\frac{x y^{2} - x^{2} y}{x^{2} - x y^{1}}$
= $\frac{(- 2) (3)^{2} - (- 2)^{2} (3)}{(- 2)^{2} - (- 2) (3)}$
= -3

Read lesson note on Algebraic Expressions (WAEC)

Read lesson note on Surds (radicals) (WAEC)

Read lesson note on Graphs Of Linear And Quadratic Functions (WAEC)

Algebraic Expressions

Surds (radicals)

Graphs Of Linear And Quadratic Functions

View Answer

Question 13 Report

In the figure, PQRS is a square of sides 8cm. What is the area of $ \triangle UVW $?

64 sq. cm

40 sq.cm

50 sq.cm

10 sq.cm

View Answer

Question 14 Report

Find the sum of the 20 terms in an arithmetic progression whose first term is 7 and last term 117?

2480

1240

620

124

Answer Details

An arithmetic progression is a sequence of numbers in which the difference between any two consecutive terms is constant. Let's call this constant difference "d". We know that the first term of the sequence is 7 and the last term is 117. Let's call the number of terms in the sequence "n". Using this information, we can write two equations: 1. The nth term of an arithmetic progression formula: an = a1 + (n - 1)d, where "an" is the nth term of the sequence, "a1" is the first term of the sequence, "n" is the number of terms in the sequence, and "d" is the constant difference between any two consecutive terms in the sequence. We can use this equation to find the value of "d": 117 = 7 + (n - 1)d 110 = (n - 1)d d = 110/(n-1) 2. The sum of an arithmetic progression formula: Sn = n/2 * (a1 + an), where "Sn" is the sum of the first "n" terms of the sequence. We can use this equation to find the sum of the first 20 terms of the sequence: Sn = 20/2 * (7 + (7 + 19d)) Sn = 10 * (14 + 2090/(n-1)) Now we can substitute the value of "d" from equation (1) into equation (2) and simplify: Sn = 10 * (14 + 2090/(n-1)) Sn = 10 * (14 + 2090/(20-1)) (since there are 20 terms in the sequence) Sn = 10 * (14 + 110) Sn = 1240 Therefore, the sum of the 20 terms in the arithmetic progression is 1240

Read lesson note on Sequence And Series (WAEC)

Sequence And Series

View Answer

Question 15 Report

Simplify $ \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}} $

-2x - 2

\sqrt{x (1 + x)}

1 + 2x + 2

\sqrt{x (1 + x)}

√x(1+x)

1 + 2x - 2

\sqrt{x (1 + x)}

Positive And Negative Integers, Rational Numbers

Surds (radicals)

View Answer

Question 16 Report

Simplify $3\frac{1}{3} - 1\frac{1}{4} x \frac{2}{3} + 1\frac{2}{5}$

2

3

4

5

Answer Details

3 - $\frac{1}{3}$ - ( $\frac{5}{4}$ x $\frac{2}{3}$ ) + 1 $\frac{2}{5}$
= $\frac{10}{3}$ - $\frac{5}{6}$ + $\frac{7}{5}$
= $\frac{100 - 25 + 42}{30}$
= $\frac{117}{30}$
= 3.9
$\approx$ 4

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

Read lesson note on Number Bases (WAEC)

Read lesson note on Algebraic Expressions (WAEC)

Read lesson note on Surds (radicals) (WAEC)

Read lesson note on Logarithms (WAEC)

Fractions, Decimals And Approximations

Number Bases

Algebraic Expressions

Surds (radicals)

Logarithms

View Answer

Question 17 Report

Fifty boxes each of 50 bolts were inspected for the number which were defective. The following was the result:

No. defective per box	4	5	6	7	8	9
No. of boxes	2	7	17	10	8	6

The mean and the median of the distribution are respectively

6.7, 6

7.6, 5

5.7, 87

34, 6

Answer Details

To find the mean of the distribution, we need to calculate the sum of all the defective bolts in the boxes and divide it by the total number of boxes: Mean = (4x2 + 5x7 + 6x17 + 7x10 + 8x8 + 9x6) / 50 = 335 / 50 = 6.7 To find the median of the distribution, we need to arrange the data in order of increasing magnitude and find the middle value. In this case, we have an even number of observations (50), so the median will be the average of the two middle values, which are the 25th and 26th observations. From the table, we can see that the 25th and 26th observations are both 6, so the median is: Median = (6 + 6) / 2 = 6 Therefore, the mean and the median of the distribution are 6.7 and 6, respectively.

Read lesson note on Surds (radicals) (WAEC)

Surds (radicals)

View Answer

Question 18 Report

If g(x) = x2 + 3x + 4, find g(x + 1) - g(x)

(x + 2)

2(x + 2)

(2x + 1)

(x2 + 4)

Answer Details

The function g(x) is defined as g(x) = x^2 + 3x + 4. We are asked to find g(x + 1) - g(x). To find g(x + 1), we simply substitute (x + 1) in place of x in the function g(x): g(x + 1) = (x + 1)^2 + 3(x + 1) + 4 = x^2 + 2x + 1 + 3x + 3 + 4 = x^2 + 5x + 8 To find g(x), we substitute x in place of x in the function g(x): g(x) = x^2 + 3x + 4 Now we can find g(x + 1) - g(x) by subtracting g(x) from g(x + 1): g(x + 1) - g(x) = (x^2 + 5x + 8) - (x^2 + 3x + 4) = x^2 + 5x + 8 - x^2 - 3x - 4 = 2x + 4 Therefore, the correct option is 2(x + 2), which is equivalent to 2x + 4.

Read lesson note on Algebraic Expressions (WAEC)

Algebraic Expressions

View Answer

Question 19 Report

A flagstaff stands on the top of a vertical tower. A man standing 60 m away from the tower observes that the angles of elevation of the top and bottom of the flagstaff are 64o and 62o respectively. Find the length of the flagstaff.

60 (tan 62o - tan 64o)

60 (cot 64o - cot 62o)

60 (cot 62o - cot 64o)

60 (tan 64o - tan 62o)

Answer Details

\frac{B C}{60}

=

\frac{t a n 62}{1}

BC = 60 tan 62

\frac{A C}{60}

=

\frac{t a n 62}{1}

AC = 60 tan 64
AB = AC - BC
= 60(tan 64o - tan 62o)

View Answer

Question 20 Report

Simplify cos2 x (sec2x + sec2 x tan2x)

tan x

tanx secx

\frac{1 + t a n^{2} x}{s e c^{2} x}

cosec2x

Answer Details

To simplify cos2 x (sec2x + sec2 x tan2x), we can use the trigonometric identity: 1 + tan2 x = sec2 x. First, we can simplify sec2 x + sec2 x tan2 x: sec2 x + sec2 x tan2 x = sec2 x (1 + tan2 x) Next, we can substitute sec2 x (1 + tan2 x) into the original equation: cos2 x (sec2 x + sec2 x tan2 x) = cos2 x (sec2 x (1 + tan2 x)) Using the identity 1 + tan2 x = sec2 x, we can simplify: cos2 x (sec2 x (1 + tan2 x)) = cos2 x (sec2 x)(sec2 x) Finally, using the identity sec2 x = 1/cos2 x, we can simplify: cos2 x (sec2 x)(sec2 x) = cos2 x (1/cos2 x)(1/cos2 x) = 1/cos2 x = sec2 x Therefore, the answer is 1+tan2xsec2x.

Read lesson note on Measures Of Dispersion (JAMB)

Read lesson note on Logarithms (WAEC)

Measures Of Dispersion

Logarithms

View Answer

Question 21 Report

PMN and PQR are two secants of the circle MQTRN and PT is a tangent. If PM = 5cm, PN = 12cm and PQ = 4.8cm, calculate the respective lengths of PR and PT in centimeters

7.3, 5.9

7.7, 12.5

12.5, 7.7

5.9, 7.3

Answer Details

In this problem, we are given a circle MQTRN and two secants PMN and PQR intersecting at point T and a tangent PT. First, we can use the theorem that states that the product of the lengths of the two segments of each secant is equal. So we have: PM * PN = PT * PR Substituting the given values, we get: 5cm * 12cm = PT * PR 60cm² = PT * PR Next, we need to find the respective lengths of PR and PT. To do this, we can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides. In triangle PRT, PT is the hypotenuse, so we have: PT² = PR² + RT² We can also use the fact that RT is perpendicular to PT, which means that triangle PRT is a right triangle. To find RT, we can use the theorem that states that the product of the segments of a secant and its external part is equal. So we have: PT * TP = QT * TR Substituting the given values, we get: PT * PT = 4.8cm * TR PT² = 4.8cm * TR TR = PT² / 4.8cm Now we can substitute RT in the Pythagorean theorem and simplify: PT² = PR² + (PT² / 4.8cm)² PT⁴ / (4.8cm)² = PR² PR = sqrt(PT⁴ / (4.8cm)²) = PT² / 4.8cm Finally, we can substitute the value of PR in the first equation we obtained and solve for PT: 60cm² = PT * (PT² / 4.8cm) PT³ = 288cm³ PT = cube root of 288cm³ ≈ 6.67cm Now we can substitute the value of PT in the equation we obtained for PR: PR = PT² / 4.8cm ≈ 7.7cm Therefore, the respective lengths of PR and PT are approximately 7.7cm and 6.67cm, respectively. So the answer is 7.7, 12.5.

Read lesson note on Circles (WAEC)

Read lesson note on Coordinate Geometry (JAMB)

Circles

Coordinate Geometry

View Answer

Question 22 Report

Find the volume of the figure.

\frac{a π^{2}}{3} (x + 3 y)

a

π^{2}

y

\frac{a π^{2}}{3}

(y + x)

(

\frac{1}{3} a π^{2} x + y

)

View Answer

Question 23 Report

Find the positive number n such that thrice its square is equal to 12 times the number.

1

2

3

4

Answer Details

The problem can be translated into an equation using algebra. Let's start by translating the words into math symbols. Let n be the positive number we are looking for. The phrase "thrice its square" means 3 times n^2 or 3n^2. The phrase "12 times the number" means 12 times n or 12n. So we can translate the sentence "thrice its square is equal to 12 times the number" into the equation: 3n^2 = 12n Now we can solve for n. First, we can simplify the equation by dividing both sides by 3: n^2 = 4n Next, we can rearrange the equation by subtracting 4n from both sides: n^2 - 4n = 0 Now we can factor out an n: n(n - 4) = 0 So the solutions to this equation are n = 0 and n - 4 = 0, which gives n = 4. However, we are looking for a positive value of n, so we can discard the solution n = 0. Therefore, the positive number n such that thrice its square is equal to 12 times the number is n = 4. Therefore, the answer is 4.

Read lesson note on Number Bases (WAEC)

Read lesson note on Modular Arithmetic (WAEC)

Number Bases

Modular Arithmetic

View Answer

Question 24 Report

Simplify $ \frac{3}{x} + \frac{15}{2y} + \frac{6}{xy} $

\frac{2 y - 5 x}{4}

\frac{9 (2 x - 5 x)}{x^{2} y^{2}}

\frac{5 x - 2 y}{2}

\frac{c^{2} y^{2}}{18 y - 45 x}

\frac{4}{2 y - 5 x}

Algebraic Expressions

Quadratic Equations

View Answer

Question 25 Report

Which of the following is a factor of rs + tr - pt - ps?

(p - s)

(s - p)

(r - p)

(1 + p)

Solution Of Linear Equations

Algebraic Expressions

View Answer

Question 26 Report

Find the gradient of the line passing through the points (-2, 0) and (0, -4).

2

-4

-2

4

Answer Details

To find the gradient of a line passing through two given points, we need to use the formula: Gradient = (change in y-coordinate) / (change in x-coordinate) Let's take the two given points: (-2, 0) and (0, -4) The change in y-coordinate is -4 - 0 = -4 (final y-coordinate - initial y-coordinate) The change in x-coordinate is 0 - (-2) = 2 (final x-coordinate - initial x-coordinate) Now, we can substitute the values in the formula: Gradient = -4 / 2 = -2 Therefore, the gradient of the line passing through the points (-2, 0) and (0, -4) is -2. Note: The gradient (also known as slope) of a line represents how steep the line is. A positive gradient means the line is sloping upwards, while a negative gradient means the line is sloping downwards. The larger the magnitude of the gradient, the steeper the line.

Read lesson note on Graphs Of Linear And Quadratic Functions (WAEC)

Graphs Of Linear And Quadratic Functions

View Answer

Question 27 Report

3% of a family's income is spent on electricity, 58% on food, 20% on transport, 11% on education and 7% on extended family. The angles subtended at the centre of the pie chart under education and food are respectively

76.8o and 25.2o

10.8o and 224.6o

112.4o and 72.0o

39.6o and 212.4o

Answer Details

To find the angles subtended at the center of the pie chart under education and food, we need to first calculate the percentage of the chart that each category occupies. The family's income is divided into five categories: electricity, food, transport, education, and extended family. The percentages of income spent on each category are given as follows: - Electricity: 3% - Food: 58% - Transport: 20% - Education: 11% - Extended family: 7% To calculate the percentage of the chart that each category occupies, we can multiply each percentage by 360 (the total number of degrees in a circle) and divide by 100. This gives us: - Electricity: (3/100) x 360 = 10.8 degrees - Food: (58/100) x 360 = 208.8 degrees - Transport: (20/100) x 360 = 72 degrees - Education: (11/100) x 360 = 39.6 degrees - Extended family: (7/100) x 360 = 25.2 degrees So, the angles subtended at the center of the pie chart under education and food are 39.6 degrees and 208.8 degrees, respectively. However, the options given in the question are in terms of rounded values, so we need to find the closest options to our calculated values. The closest option to 39.6 degrees is 39.6 degrees itself. The closest option to 208.8 degrees is 212.4 degrees. Therefore, the answer is (D) 39.6 degrees and 212.4 degrees.

Read lesson note on Percentages (WAEC)

Read lesson note on Variation (JAMB)

Percentages

Variation

View Answer

Question 28 Report

One angle of a rhombus is 60o. The shorter of the two diagonals is 8cm long. Find the length of the longer one.

8

\sqrt{3}

\frac{16}{\sqrt{3}}

\sqrt{3}

\frac{10}{\sqrt{3}}

Answer Details

In a rhombus, all sides are congruent, and opposite angles are equal. If one angle of the rhombus is 60 degrees, then the opposite angle is also 60 degrees. Let the shorter diagonal of the rhombus be divided into two equal halves. Let the half of the shorter diagonal be `a`. Using the properties of a 30-60-90 degree triangle, we know that the length of the longer diagonal of the rhombus is `2a√3`. Since the shorter diagonal is 8cm, `a = 4cm`. Therefore, the length of the longer diagonal is: `2a√3 = 2(4cm)√3 = 8√3 cm` Therefore, the length of the longer diagonal of the rhombus is `8√3 cm`. So the correct answer is: 8√3.

Read lesson note on Indices (WAEC)

Indices

View Answer

Question 29 Report

Rationalize $ \frac{2\sqrt{3}+3\sqrt{2}}{3\sqrt{2}-2\sqrt{3}} $

5 - 2

\sqrt{6}

5 + 2

\sqrt{6}

5

\sqrt{6}

5

Answer Details

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

=

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

x

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

\frac{4 (3) + 9 (2) + 2 (6) \sqrt{6}}{9 (2) - 4 (3)}

\frac{12 + 18 + 12 \sqrt{6}}{1 ‘ 8 - 12}

=

\frac{30 + 12 \sqrt{6}}{6}

= 5 + 2

\sqrt{6}

Read lesson note on Positive And Negative Integers, Rational Numbers (WAEC)

Read lesson note on Surds (radicals) (WAEC)

Positive And Negative Integers, Rational Numbers

Surds (radicals)

View Answer

Question 30 Report

The area of a square is 144 sq cm. Find the length of its diagonal

11

\sqrt{3}

cm

12cm

12

\sqrt{2}

cm

13cm

Answer Details

The area of a square is given as 144 sq cm. To find the length of its diagonal, we need to use the Pythagorean theorem, which states that in a right-angled triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the lengths of the other two sides. In the case of a square, we know that all sides are equal in length, so we can use this fact to simplify the equation. Let's call the length of one side of the square "s". Then, we know that the area of the square is s^2 = 144 sq cm. Solving for "s", we get s = sqrt(144) = 12 cm. Now, let's draw a diagonal line across the square, splitting it into two right-angled triangles. We can label the hypotenuse of each triangle as "d" (which is the length of the diagonal we want to find), and the other two sides as "s". Using the Pythagorean theorem, we get: d^2 = s^2 + s^2 d^2 = 2s^2 d^2 = 2(12^2) d^2 = 288 d = sqrt(288) = 12sqrt(2) So the length of the diagonal is 12sqrt(2) cm.

Read lesson note on Areas (WAEC)

Areas

View Answer

Question 31 Report

Multiply (x2 - 3x + 1) by (x - a)

x3 - (3 + a) x2 + (1 + 3a)x - a

x3 - (3 - a)x2 + 3ax - a

x3 - (3 - a)x2 - (1 = 3a) - a

x3 + (3 - a)x2 + (1 + 3a) - a

Answer Details

To multiply the expression (x^2 - 3x + 1) by (x - a), we need to use the distributive property of multiplication over addition. We start by multiplying each term in the first expression (x^2 - 3x + 1) by (x - a) and then add the resulting terms. (x^2 - 3x + 1) * (x - a) = x^3 - ax^2 - 3x^2 + 3ax + x - a = x^3 - (a+3)x^2 + (3a+1)x - a Therefore, the correct option is: - x^3 - (a+3)x^2 + (3a+1)x - a

Read lesson note on Algebraic Expressions (WAEC)

Algebraic Expressions

View Answer

Question 32 Report

If 2257 is the result of subtracting 4577 from 7056 in base n, find n

8

\frac{17}{30}

9

\frac{9}{10}

10

\frac{1}{10}

11

\frac{11}{36}

Answer Details

\begin{array}{cc} 7056 \\ - 4577 \\ 2257 \end{array}

By trial and error method
Let the base to 8
i.e. Let n = 8 and it is easily verified that the subtraction holds.
The subtraction does not hold when other values of n are tried
n = 8

Read lesson note on Number Bases (WAEC)

Read lesson note on Number Bases (JAMB)

Number Bases

View Answer

Question 33 Report

Fifty boxes each of 50 bolts were inspected for the number which were defective. The following was the result:

No. defective per box	4	5	6	7	8	9
No. of boxes	2	7	17	10	8	6

Find the percentage of boxes containing at least 5 defective bolts each.

96

94

92

90

Answer Details

To find the percentage of boxes containing at least 5 defective bolts each, we need to first add up the number of boxes that have 5, 6, 7, 8, or 9 defective bolts. These are the boxes that contain at least 5 defective bolts. From the table, we can see that there are: - 7 boxes with 5 defective bolts - 17 boxes with 6 defective bolts - 10 boxes with 7 defective bolts - 8 boxes with 8 defective bolts - 6 boxes with 9 defective bolts Adding these up, we get a total of 7+17+10+8+6 = 48 boxes. Therefore, out of the 50 boxes that were inspected, 48 of them contain at least 5 defective bolts each. To find the percentage, we can use the formula: percentage = (part/whole) x 100 In this case, the "part" is the number of boxes containing at least 5 defective bolts (48), and the "whole" is the total number of boxes inspected (50). Substituting in the values, we get: percentage = (48/50) x 100 = 96% Therefore, the answer is 96%, option (A).

Read lesson note on Percentages (WAEC)

Percentages

View Answer

Question 34 Report

If w varies inversely as $ \frac{ur}{u+r} $ and is equal to 8 when u = 2 and r = 6, find a relationship between u, v, w.

urw = 16(u + v)

16ur = 3w(u + v)

urw = 12(u + v)

12urw = u + v

Answer Details

W

α

\frac{\frac{1}{u v}}{u + v}

∴ w =

\frac{\frac{k}{u v}}{u + v}

=

\frac{k (u + v)}{u v}

w =

\frac{k (u + v)}{u v}

w = 8, u = 2 and v = 6
8 =

\frac{k (2 + 6)}{2 (6)}

=

\frac{k (8)}{12}

k = 12

Read lesson note on Solution Of Linear Equations (WAEC)

Read lesson note on Sequence And Series (WAEC)

Solution Of Linear Equations

Sequence And Series

View Answer

Question 35 Report

Simplify $2\log \frac{2}{5} - \log \frac{72}{125} + \log 9$

1 - 4 log3

-1 + 2 log 3

-1 + 5 log2

1 - 2log 2

Answer Details

2log

\frac{2}{5}

- log

\frac{72}{125}

+ log 9
[

\frac{2}{5}

)2 x 9] = log

\frac{4}{25}

x

\frac{9}{1}

x

\frac{125}{72}

= log

\frac{72}{125}

= log

\frac{5}{2}

= log

\frac{10}{4}

= log 10 - log 4
= log10 10 - log10 22
= 1 - 2 log2

Read lesson note on Sets (WAEC)

Read lesson note on Surds (radicals) (WAEC)

Read lesson note on Logarithms (WAEC)

Read lesson note on Quadratic Equations (WAEC)

Sets

Surds (radicals)

Logarithms

Quadratic Equations

View Answer

Question 36 Report

Factory P produces 20,000 bags of cement per day while factory Q produces 15,000 bags per day. If P reduces production by 5% and Q increases production by 5%, determine the effective loss in the number of bags produced per day by the two factories

250

750

1000

1250

Answer Details

Factory P produces 20,000 bags of cement per day and reduces production by 5%, which means it produces: 20,000 - 5% of 20,000 = 20,000 - 1,000 = 19,000 bags per day. Factory Q produces 15,000 bags of cement per day and increases production by 5%, which means it produces: 15,000 + 5% of 15,000 = 15,000 + 750 = 15,750 bags per day. The total number of bags produced per day by both factories before the changes were: 20,000 + 15,000 = 35,000 bags per day. The total number of bags produced per day by both factories after the changes are: 19,000 + 15,750 = 34,750 bags per day. The effective loss in the number of bags produced per day by the two factories is the difference between the original number of bags produced and the new number of bags produced: 35,000 - 34,750 = 250 bags per day. Therefore, the effective loss in the number of bags produced per day by the two factories is 250

Read lesson note on Representation Of Data (JAMB)

Read lesson note on Trigonometry (JAMB)

Representation Of Data

Trigonometry

View Answer

Question 37 Report

From a point Z, 60 m north of X, a man walks 60√3m eastwards to another point Y. Find the bearing of Y from X.

0.30o

045o

060o

090o

Answer Details

To find the bearing of Y from X, we need to determine the angle between the direction of due north and the direction of XY. First, we can draw a diagram to represent the problem. Let X be the origin, and let Y be the point that is 60√3 m east and some distance north of X. Let Z be the point that is 60 m due north of X, and let W be the point that is due west of Y and due south of Z. [Diagram not included as plain text] We can see that △XYZ is a right-angled triangle, with XY as the hypotenuse, XZ as the adjacent side, and YZ as the opposite side. We know that XZ has a length of 60 m, and YZ has a length of 60√3 m. Therefore, we can use the tangent ratio to find the angle between XZ and XY: tan θ = YZ/XZ tan θ = (60√3)/60 tan θ = √3/1 θ = tan⁻¹(√3) θ = 60° Therefore, the bearing of Y from X is 060°. Option (C) is the correct answer.

Read lesson note on Bearings (WAEC)

Bearings

View Answer

Question 38 Report

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

(x2 - 1)-

\frac{1}{2}

(x2 - 1)1

(x2 - 1)

(x2 - 1)-1

Answer Details

x2(x2 - 1)-

\frac{1}{2}

- (x2 - 1)

\frac{1}{2}

=

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

-

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

=

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

=

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

= (x2 - 1)-

\frac{1}{2}

View Answer

Question 39 Report

In this figure, $PQ = PR = PS$ and $\angle SRT = 68^\circ$. Find $\angle QPS$.

136

^{\circ}

124

^{\circ}

12

^{\circ}

68

^{\circ}

Answer Details

Since PQRS is quadrilateral 2y + 2x = QPS = 360

^{\circ}

i.e. 2(y + x) + QPS = 360 $^{\circ}$

QPS = 360 $^{\circ}$ - 2(y + x)

But x + y + 68 $^{\circ}$ = 180 $^{\circ}$

x + y = 180 $^{\circ}$ - 68 $^{\circ}$ = 180 $^{\circ}$

x + y = 180 $^{\circ}$ - 68 $^{\circ}$

= 112 $^{\circ}$

QPS = 360 $^{\circ}$ - 2(112 $^{\circ}$ )

360 $^{\circ}$ - 224 $^{\circ}$ = 136 $^{\circ}$

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

Read lesson note on Angles (WAEC)

Angles And Intercepts On Parallel Lines

Angles

View Answer

Question 40 Report

If $ \cos x = \sqrt{\frac{a}{b}} $, find coses x

\frac{b}{\sqrt{b - a}}

\sqrt{\frac{b}{a}}

\sqrt{\frac{b}{b - a}}

\sqrt{\frac{b - a}{a}}

Answer Details

cosx =

\sqrt{\frac{a}{b}}

y2 +

\sqrt{(a)^{2}}

=

\sqrt{(b)^{2}}

by pythagoras
y2 = b - a
∴ y = b - a
cosec x =

\frac{1}{s i n x}

=

\frac{1}{y}

\frac{b}{y}

=

\frac{\sqrt{b}}{\sqrt{b - a}}

=

\sqrt{\frac{b}{b - a}}

Read lesson note on Surds (radicals) (WAEC)

Read lesson note on Logarithms (WAEC)

Surds (radicals)

Logarithms

View Answer

Question 41 Report

What is the nth term of the progression 27, 9, 3,......?

27

\frac{1}{3}

n - 1

3n + 2

27 + 18(n - 1)

27 + 6(n - 1)

Answer Details

The given progression is decreasing by a factor of 3 in each term, starting from 27. Therefore, the nth term can be calculated by multiplying 27 by 1/3 raised to the power of (n-1). This is because in the nth term, the factor of 3 has been applied (n-1) times to the first term 27. Therefore, the answer is: 27 x (1/3)^(n-1)

Read lesson note on Sequence And Series (WAEC)

Sequence And Series

View Answer

Question 42 Report

Factorize m3 - 2m2 - m + 2

(m2 + 1)(m - 2)

(m - 1)(m + 1)(m + 2)

(m - 2)(m + 1)(m - 1)

(m2 + 2)(m - 1)

Answer Details