JAMB UTME - Mathematics - 1991

Swali 1 Ripoti

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

104

103

102

101

10-1

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Logarithms (WAEC)

Logarithms

Tazama Jibu

Swali 2 Ripoti

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

Maelezo ya Majibu

\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}

Soma maelezo ya somo kwenye Number Bases (WAEC)

Soma maelezo ya somo kwenye Number Bases (JAMB)

Number Bases

Tazama Jibu

Swali 3 Ripoti

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

Maelezo ya Majibu

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).

Soma maelezo ya somo kwenye Percentages (WAEC)

Percentages

Tazama Jibu

Swali 4 Ripoti

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}}

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Indices (WAEC)

Indices

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Swali 5 Ripoti

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

15o

25o

40o

50o

55o

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Euclidean Geometry (JAMB)

Soma maelezo ya somo kwenye Triangles And Polygons (WAEC)

Euclidean Geometry

Triangles And Polygons

Tazama Jibu

Swali 6 Ripoti

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

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Areas (WAEC)

Areas

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Swali 7 Ripoti

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

2, 3

3, 6

-1, 6

1, -6

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Solution Of Linear Equations (WAEC)

Soma maelezo ya somo kwenye Quadratic Equations (WAEC)

Solution Of Linear Equations

Quadratic Equations

Tazama Jibu

Swali 8 Ripoti

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

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Representation Of Data (JAMB)

Soma maelezo ya somo kwenye Trigonometry (JAMB)

Representation Of Data

Trigonometry

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Swali 9 Ripoti

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

Maelezo ya Majibu

\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}

Soma maelezo ya somo kwenye Positive And Negative Integers, Rational Numbers (WAEC)

Soma maelezo ya somo kwenye Surds (radicals) (WAEC)

Positive And Negative Integers, Rational Numbers

Surds (radicals)

Tazama Jibu

Swali 10 Ripoti

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

2

3

4

5

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Fractions, Decimals And Approximations (WAEC)

Soma maelezo ya somo kwenye Number Bases (WAEC)

Soma maelezo ya somo kwenye Algebraic Expressions (WAEC)

Soma maelezo ya somo kwenye Surds (radicals) (WAEC)

Soma maelezo ya somo kwenye Logarithms (WAEC)

Fractions, Decimals And Approximations

Number Bases

Algebraic Expressions

Surds (radicals)

Logarithms

Tazama Jibu

Swali 11 Ripoti

Simplify cos2 x (sec2x + sec2 x tan2x)

tan x

tanx secx

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

cosec2x

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Measures Of Dispersion (JAMB)

Soma maelezo ya somo kwenye Logarithms (WAEC)

Measures Of Dispersion

Logarithms

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Swali 12 Ripoti

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

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Modular Arithmetic (WAEC)

Soma maelezo ya somo kwenye Lengths And Perimeters (WAEC)

Modular Arithmetic

Lengths And Perimeters

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Swali 13 Ripoti

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

(x + 2)

2(x + 2)

(2x + 1)

(x2 + 4)

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Algebraic Expressions (WAEC)

Algebraic Expressions

Tazama Jibu

Swali 14 Ripoti

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

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Algebraic Expressions (WAEC)

Algebraic Expressions

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Swali 15 Ripoti

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

Tazama Jibu

Swali 16 Ripoti

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

Tazama Jibu

Swali 17 Ripoti

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?

$\frac{5}{12}$

$\frac{1}{3}$

$\frac{3}{4}$

$\frac{7}{24}$

Maelezo ya Majibu

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

Tazama Jibu

Swali 18 Ripoti

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

Tazama Jibu

Swali 19 Ripoti

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

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Percentages (WAEC)

Soma maelezo ya somo kwenye Variation (JAMB)

Percentages

Variation

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Swali 20 Ripoti

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

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Tazama Jibu

Swali 21 Ripoti

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)

Maelezo ya Majibu

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)

Soma maelezo ya somo kwenye Polynomials (JAMB)

Polynomials

Tazama Jibu

Swali 22 Ripoti

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)

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Algebraic Expressions (WAEC)

Algebraic Expressions

Tazama Jibu

Swali 23 Ripoti

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

136

^{\circ}

124

^{\circ}

12

^{\circ}

68

^{\circ}

Maelezo ya Majibu

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 $^{\circ}$

Soma maelezo ya somo kwenye Angles And Intercepts On Parallel Lines (WAEC)

Soma maelezo ya somo kwenye Angles (WAEC)

Angles And Intercepts On Parallel Lines

Angles

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Swali 24 Ripoti

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

2480

1240

620

124

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Sequence And Series (WAEC)

Sequence And Series

Tazama Jibu

Swali 25 Ripoti

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}

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

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

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

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Variation (WAEC)

Soma maelezo ya somo kwenye Ratio, Proportions And Rates (WAEC)

Soma maelezo ya somo kwenye Areas (WAEC)

Variation

Ratio, Proportions And Rates

Areas

Tazama Jibu

Swali 26 Ripoti

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)

Maelezo ya Majibu

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)

Soma maelezo ya somo kwenye Sequence And Series (WAEC)

Sequence And Series

Tazama Jibu

Swali 27 Ripoti

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

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Solution Of Linear Equations (WAEC)

Soma maelezo ya somo kwenye Sequence And Series (WAEC)

Solution Of Linear Equations

Sequence And Series

Tazama Jibu

Swali 28 Ripoti

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

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Bearings (WAEC)

Bearings

Tazama Jibu

Swali 29 Ripoti

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

2

-4

-2

4

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Graphs Of Linear And Quadratic Functions (WAEC)

Graphs Of Linear And Quadratic Functions

Tazama Jibu

Swali 30 Ripoti

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}

Maelezo ya Majibu

\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

Soma maelezo ya somo kwenye Number Bases (WAEC)

Soma maelezo ya somo kwenye Number Bases (JAMB)

Number Bases

Tazama Jibu

Swali 31 Ripoti

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

Maelezo ya Majibu

(

\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.

Soma maelezo ya somo kwenye Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

Tazama Jibu

Swali 32 Ripoti

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

Maelezo ya Majibu

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}

Tazama Jibu

Swali 33 Ripoti

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}}

Maelezo ya Majibu

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}}

Soma maelezo ya somo kwenye Surds (radicals) (WAEC)

Soma maelezo ya somo kwenye Logarithms (WAEC)

Surds (radicals)

Logarithms

Tazama Jibu

Swali 34 Ripoti

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

)

Tazama Jibu

Swali 35 Ripoti

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)

Maelezo ya Majibu

\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)

Tazama Jibu

Swali 36 Ripoti

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

5

3

1

\frac{1}{25}

Maelezo ya Majibu

\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

Soma maelezo ya somo kwenye Fractions, Decimals And Approximations (WAEC)

Soma maelezo ya somo kwenye Number Bases (WAEC)

Soma maelezo ya somo kwenye Modular Arithmetic (WAEC)

Soma maelezo ya somo kwenye Logarithms (WAEC)

Fractions, Decimals And Approximations

Number Bases

Modular Arithmetic

Logarithms

Tazama Jibu

Swali 37 Ripoti

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

Tazama Jibu

Swali 38 Ripoti

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)

Tazama Jibu

Swali 39 Ripoti

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

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Surds (radicals) (WAEC)

Surds (radicals)

Tazama Jibu

Swali 40 Ripoti

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

1

-1

-4

4

Tazama Jibu

Swali 41 Ripoti

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

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Circles (WAEC)

Soma maelezo ya somo kwenye Coordinate Geometry (JAMB)

Circles

Coordinate Geometry

Tazama Jibu

Swali 42 Ripoti

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

1

2

3

4

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Number Bases (WAEC)

Soma maelezo ya somo kwenye Modular Arithmetic (WAEC)

Number Bases

Modular Arithmetic

Tazama Jibu

Swali 43 Ripoti

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