JAMB UTME - Mathematics - 1991

Ajụjụ 1 Ripọtì

Express $62 + 3$ as decimal correct to 3 significant figures.

20.6

20.667

20.67

20.7

Fractions, Decimals And Approximations

Fractions, Decimals, Approximations And Percentage

Lee Azịza

Ajụjụ 2 Ripọtì

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)

Akọwa Nkọwa

m3 - 2m2 - m + 2
Let f(m) = m3 - 2m2 - m2 + 2
= f(1)
= 1 - 2 - 2 + 2 = 0
∴ m - 1 is factor

\frac{m^{3} - 2 m^{2} - m^{2} + 2}{m - 1}

= m2 - m - 2
= (m - 1)m2 - m - 2
= (m - 1)(m + 1)(m - 2)

Gụọ akwụkwọ nkuzi na Polynomials (JAMB)

Polynomials

Lee Azịza

Ajụjụ 3 Ripọtì

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}

Akọwa Nkọwa

\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

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Number Bases (JAMB)

Number Bases

Lee Azịza

Ajụjụ 4 Ripọtì

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

2480

1240

620

124

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Sequence And Series

Lee Azịza

Ajụjụ 5 Ripọtì

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)

Lee Azịza

Ajụjụ 6 Ripọtì

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

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Areas

Lee Azịza

Ajụjụ 7 Ripọtì

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

136

^{\circ}

124

^{\circ}

12

^{\circ}

68

^{\circ}

Akọwa Nkọwa

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 $^{\circ}$

Gụọ akwụkwọ nkuzi na Angles And Intercepts On Parallel Lines (WAEC)

Gụọ akwụkwọ nkuzi na Angles (WAEC)

Angles And Intercepts On Parallel Lines

Angles

Lee Azịza

Ajụjụ 8 Ripọtì

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}

%

Akọwa Nkọwa

% 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 $\frac{7}{8}$

Gụọ akwụkwọ nkuzi na Fractions, Decimals And Approximations (WAEC)

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Fractions, Decimals And Approximations

Modular Arithmetic

Lee Azịza

Ajụjụ 9 Ripọtì

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

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Bearings (WAEC)

Bearings

Lee Azịza

Ajụjụ 10 Ripọtì

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}}

Akọwa Nkọwa

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}}

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Surds (radicals)

Logarithms

Lee Azịza

Ajụjụ 11 Ripọtì

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

Akọwa Nkọwa

\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}

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 12 Ripọtì

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}

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

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

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Variation (WAEC)

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

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Variation

Ratio, Proportions And Rates

Areas

Lee Azịza

Ajụjụ 13 Ripọtì

Simplify cos2 x (sec2x + sec2 x tan2x)

tan x

tanx secx

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

cosec2x

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Measures Of Dispersion (JAMB)

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Measures Of Dispersion

Logarithms

Lee Azịza

Ajụjụ 14 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Solution Of Linear Equations

Sequence And Series

Lee Azịza

Ajụjụ 15 Ripọtì

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

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Gụọ akwụkwọ nkuzi na Coordinate Geometry (JAMB)

Circles

Coordinate Geometry

Lee Azịza

Ajụjụ 16 Ripọtì

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}}

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Indices (WAEC)

Indices

Lee Azịza

Ajụjụ 17 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Gụọ akwụkwọ nkuzi na Lengths And Perimeters (WAEC)

Modular Arithmetic

Lengths And Perimeters

Lee Azịza

Ajụjụ 18 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 19 Ripọtì

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

2

-4

-2

4

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Graphs Of Linear And Quadratic Functions (WAEC)

Graphs Of Linear And Quadratic Functions

Lee Azịza

Ajụjụ 20 Ripọtì

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

104

103

102

101

10-1

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Logarithms

Lee Azịza

Ajụjụ 21 Ripọtì

Solve the equation (x - 2) (x - 3) = 12

2, 3

3, 6

-1, 6

1, -6

Akọwa Nkọwa

(x - 2) (x - 3) = 12
x2 - 3x - 2x + 6 = 12
x2 - 5x - 6 = 0
(x - 1)(x - 6) = 0
x = -1 or 6

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Solution Of Linear Equations

Quadratic Equations

Lee Azịza

Ajụjụ 22 Ripọtì

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

(x + 2)

2(x + 2)

(2x + 1)

(x2 + 4)

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 23 Ripọtì

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

Lee Azịza

Ajụjụ 24 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Sets (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Sets

Surds (radicals)

Logarithms

Quadratic Equations

Lee Azịza

Ajụjụ 25 Ripọtì

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

Akọwa Nkọwa

(

\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.

Gụọ akwụkwọ nkuzi na Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

Lee Azịza

Ajụjụ 26 Ripọtì

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

2

3

4

5

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Fractions, Decimals And Approximations (WAEC)

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Fractions, Decimals And Approximations

Number Bases

Algebraic Expressions

Surds (radicals)

Logarithms

Lee Azịza

Ajụjụ 27 Ripọtì

If the exterior angles of a pentagon are xo, (x + 5)o, (x + 10), (x + 15)o and (x + 20)o, find x

118o

72o

62o

36o

Algebraic Expressions

Lee Azịza

Ajụjụ 28 Ripọtì

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

Lee Azịza

Ajụjụ 29 Ripọtì

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

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Surds (radicals)

Lee Azịza

Ajụjụ 30 Ripọtì

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

Akọwa Nkọwa

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).

Gụọ akwụkwọ nkuzi na Percentages (WAEC)

Percentages

Lee Azịza

Ajụjụ 31 Ripọtì

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

)

Lee Azịza

Ajụjụ 32 Ripọtì

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

15o

25o

40o

50o

55o

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Euclidean Geometry (JAMB)

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Euclidean Geometry

Triangles And Polygons

Lee Azịza

Ajụjụ 33 Ripọtì

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

Akọwa Nkọwa

\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}

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Number Bases (JAMB)

Number Bases

Lee Azịza

Ajụjụ 34 Ripọtì

At what value of x is the function $y = x^2 - 2x -3$ minimum?

1

-1

-4

4

Lee Azịza

Ajụjụ 35 Ripọtì

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)

Akọwa Nkọwa

\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)

Lee Azịza

Ajụjụ 36 Ripọtì

What is the product of $ \frac{27}{5^1}(3)^{-3} $ and $ \frac{(1)^{-1}}{5} $?

5

3

1

\frac{1}{25}

Akọwa Nkọwa

\frac{27}{5^{1}}

(3)-3 x

\frac{(1)^{- 1}}{5}

=

\frac{27}{5}

x

\frac{1}{3^{3}}

x

\frac{1}{\frac{1}{5}}

=

\frac{27}{5}

x

\frac{1}{27}

x

\frac{5}{1}

= 1

Gụọ akwụkwọ nkuzi na Fractions, Decimals And Approximations (WAEC)

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Fractions, Decimals And Approximations

Number Bases

Modular Arithmetic

Logarithms

Lee Azịza

Ajụjụ 37 Ripọtì

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

Lee Azịza

Ajụjụ 38 Ripọtì

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

1

2

3

4

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Number Bases

Modular Arithmetic

Lee Azịza

Ajụjụ 39 Ripọtì

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?

$\frac{5}{12}$

$\frac{1}{3}$

$\frac{3}{4}$

$\frac{7}{24}$

Akọwa Nkọwa

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

Lee Azịza

Ajụjụ 40 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Lee Azịza

Ajụjụ 41 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Gụọ akwụkwọ nkuzi na Trigonometry (JAMB)

Representation Of Data

Trigonometry

Lee Azịza

Ajụjụ 42 Ripọtì

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

Akọwa Nkọwa

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}

Lee Azịza

Ajụjụ 43 Ripọtì

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)

Akọwa Nkọwa

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)

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Sequence And Series

Lee Azịza

Ajụjụ 44 Ripọtì

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

Lee Azịza

Ajụjụ 45 Ripọtì

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

Lee Azịza

Ajụjụ 46 Ripọtì

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

3

- $\frac{3}{5}$

$\frac{3}{5}$

-3

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Gụọ akwụkwọ nkuzi na Graphs Of Linear And Quadratic Functions (WAEC)

Algebraic Expressions

Surds (radicals)

Graphs Of Linear And Quadratic Functions

Lee Azịza

Ajụjụ 47 Ripọtì

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

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Percentages (WAEC)

Gụọ akwụkwọ nkuzi na Variation (JAMB)

Percentages

Variation

Lee Azịza

Ajụjụ 48 Ripọtì

Factorize 1 - (a - b)2

(1 - a - b)(1 - a + b)

(1 + a - b)(1 - a + b)

(1 - a + b)(1 - a + b)

(1 + a + b)(1 + a + b)

Akọwa Nkọwa

1 - (a - b)2 = [1 + (a - b)][1 - a + b]
= (1 + a - b)(1 - a + b)

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza






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