WAEC SSCE - General Mathematics - 1992…

Tambaya 1 Rahoto

If x varies over the set of real numbers, which of the following is illustrated in the diagram above?

-2

-3?x<2 Amsar da ta dace

-3

-3?x?2

x?2

Bayanin Amsa

Karanta bayanin darasi akan Logical Reasoning (WAEC)

Karanta bayanin darasi akan Variation (JAMB)

Logical Reasoning

Variation

Duba Amsa

Tambaya 2 Rahoto

Evaluate $\log_{10} 25 + \log_{10} 32 - \log_{10} 8$

0.2

2 Amsar da ta dace

100

409

490

Bayanin Amsa

We can use the following property of logarithms: - $\log_{a}b + \log_{a}c = \log_{a}(bc)$ - $\log_{a}b - \log_{a}c = \log_{a}\left(\dfrac{b}{c}\right)$ Using this property, we can rewrite the expression as: \begin{align*} \log_{10} 25 + \log_{10} 32 - \log_{10} 8 &= \log_{10}(25\cdot 32) - \log_{10} 8 \\ &= \log_{10}(800) - \log_{10} 8 \\ &= \log_{10}\left(\frac{800}{8}\right) \\ &= \log_{10}(100) \\ &= 2 \end{align*} Therefore, the value of the expression is 2. Thus, the answer is (2).

Karanta bayanin darasi akan Logarithms (WAEC)

Logarithms

Duba Amsa

Tambaya 3 Rahoto

Let U = {1, 2, 3, 4}, P = {2, 3} and Q = {2, 4}. What is (P∩Q)'?

(1, 2, 3)

(1, 3, 4) Amsar da ta dace

(2, 3)

(1, 3)

(1, 4)

Bayanin Amsa

In set theory, the symbol ∩ denotes the intersection of two sets, meaning the set of elements that are in both sets. Therefore, P∩Q denotes the set of elements that are in both P and Q, which is {2}. The complement of a set A, denoted by A', is the set of elements that are not in A but are in the universal set U. Therefore, (P∩Q)' denotes the set of elements that are in U but are not in P∩Q, which is {1, 3, 4}. Thus, the correct answer is: (1, 3, 4).

Karanta bayanin darasi akan Sets (WAEC)

Karanta bayanin darasi akan Sets (JAMB)

Sets

Duba Amsa

Tambaya 4 Rahoto

A ladder leans against a vertical wall at an angle
60° to the wall, if the foot of the ladder is 5 metres
away from the wall, calculate the length of the
ladder.

5√3m/3

10√3m/3

Amsar da ta dace

5√3m

10m

Bayanin Amsa

Karanta bayanin darasi akan Introductory Calculus (WAEC)

Karanta bayanin darasi akan Circles (WAEC)

Introductory Calculus

Circles

Duba Amsa

Tambaya 5 Rahoto

What is the probability of throwing a number greater than 2 with a single fair die

1/6

1/3

1/2

2/3 Amsar da ta dace

5/6

Bayanin Amsa

The total number of outcomes when rolling a fair die is 6 since there are six sides on the die. The numbers that are greater than 2 are 3, 4, 5, and 6. So, there are four possible outcomes that are greater than 2. The probability of getting a number greater than 2 with a single fair die is the number of favorable outcomes divided by the total number of possible outcomes, which is 4/6 or 2/3. Therefore, the answer is (d) 2/3.

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 6 Rahoto

Find the value of t in the diagram above

5.6cm Amsar da ta dace

6.5cm

6.6cm

6.8cm

7.0cm

Bayanin Amsa

Using the Pythagorean theorem, we can find that the length of the hypotenuse of the right triangle on the left is: $$\sqrt{5^2+12^2} = 13$$ Since the triangle on the right is similar to the triangle on the left, we can set up the following proportion: $$\frac{5}{12} = \frac{t}{13}$$ Cross-multiplying, we get: $$12t = 65$$ Therefore: $$t = \frac{65}{12} \approx 5.4$$ So the answer is closest to 5.6cm. Therefore the correct option is (A).

Duba Amsa

Tambaya 7 Rahoto

The diagram above shows the shaded segment of a circle of radius 7cm. if the area of the triangle OXY is 12$\frac{1}{4}$cm$^2$, calculate the area of the segment
[Take ? = 22/7]

5/12cm²

7/12cm² Amsar da ta dace

11/6cm²

2 1/3cm²

6 1/6cm²

Bayanin Amsa

Karanta bayanin darasi akan Circles (WAEC)

Circles

Duba Amsa

Tambaya 8 Rahoto

Convert 77 to a number in base two

1001 101 Amsar da ta dace

111001

100110

10101

10011

Bayanin Amsa

Karanta bayanin darasi akan Number Bases (WAEC)

Number Bases

Duba Amsa

Tambaya 9 Rahoto

A fair die is rolled once. What is the probability of
obtaining 4 or 6?

1/12

1/6

1/3 Amsar da ta dace

1/2

2/3

Bayanin Amsa

When you roll a fair die, there are six possible outcomes, each with an equal chance of occurring. These outcomes are the numbers 1, 2, 3, 4, 5, and 6. To find the probability of obtaining 4 or 6, we need to count how many of the outcomes correspond to these numbers. In this case, there are two outcomes that satisfy the condition, namely 4 and 6. Therefore, the probability of obtaining 4 or 6 is 2 out of 6, or 2/6. We can simplify this fraction by dividing both the numerator and denominator by their greatest common factor, which is 2. This gives us 1/3. Therefore, the probability of obtaining 4 or 6 when rolling a fair die is 1/3.

Karanta bayanin darasi akan Probability (WAEC)

Karanta bayanin darasi akan Probability (JAMB)

Probability

Duba Amsa

Tambaya 10 Rahoto

Three balls are drawn one after the other with replacement, from a bag containing 5 red, 9white and 4 blue identical balls, What is the probability that they are one red, one white and one blue?

5/102

5/136

5/162 Amsar da ta dace

5/204

5/243

Bayanin Amsa

The probability of drawing one red ball out of 18 balls is 5/18. Similarly, the probability of drawing one white ball is 9/18 (or 1/2) and the probability of drawing one blue ball is 4/18 (or 2/9). Since we are drawing three balls, we can find the probability of getting one red, one white, and one blue by multiplying the probabilities of each event happening together. Since we are drawing with replacement, the probabilities do not change from draw to draw. Thus, the probability of getting one red, one white, and one blue is: (5/18) x (1/2) x (2/9) = 5/162 Therefore, the answer is option C: 5/162.

Karanta bayanin darasi akan Probability (WAEC)

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Probability

Surds (radicals)

Duba Amsa

Tambaya 11 Rahoto

In the diagram above, TR = TS and ?TRS = 60^o. Find the value of x

60 Amsar da ta dace

120

150

Bayanin Amsa

Since TR = TS and ?TRS = 60o, then ?TSR = ?TRS = 60o. Therefore, the angles of triangle TSP add up to 180o: ?TPS + ?TSP + ?TSR = 180o. Substituting the known values: x + 60 + 60 = 180. Simplifying: x = 60. Hence, the value of x is 60.

Duba Amsa

Tambaya 12 Rahoto

The value of tan 315°

√2/2

-1 Amsar da ta dace

-√2/2

Bayanin Amsa

To find the value of tan 315°, we can use the identity: tan (θ + 180°) = tan θ Since tan (315° + 180°) = tan 495°, and 495° is coterminal with 135°, we have: tan 315° = tan (135° + 180°) = tan 135° Using the fact that tan 45° = 1, and tan (θ + 90°) = -cot θ, we have: tan 135° = tan (45° + 90°) = -cot 45° = -1 Therefore, the value of tan 315° is -1.

Karanta bayanin darasi akan Introductory Calculus (WAEC)

Karanta bayanin darasi akan Trigonometry (JAMB)

Introductory Calculus

Trigonometry

Duba Amsa

Tambaya 13 Rahoto

The data below shows the frequency distribution
of marks scored by a group of students in a class
test.

What is the modal score

4 Amsar da ta dace

Bayanin Amsa

We find the modal score by identifying the score that occurs most frequently in the data. Looking at the frequency distribution, we can see that the score 4 appears most frequently, with a frequency of 5. Therefore, the modal score is 4.

Karanta bayanin darasi akan Euclidean Geometry (JAMB)

Euclidean Geometry

Duba Amsa

Tambaya 14 Rahoto

In the diagram above PQT is a tangent to the circle QRS at Q. Angle QTR = 48° and ?QRT = 95°. Find ?QRT

48^o

45^o

37^o Amsar da ta dace

32^o

30^o

Bayanin Amsa

Karanta bayanin darasi akan Circles (WAEC)

Karanta bayanin darasi akan Triangles And Polygons (WAEC)

Circles

Triangles And Polygons

Duba Amsa

Tambaya 15 Rahoto

What must be added to the expression x$^2$ - 18x to make it a perfect square?

81 Amsar da ta dace

Bayanin Amsa

To make the expression x$^2$ - 18x a perfect square, we need to add a constant term. The constant term should be half of the coefficient of x squared, which is (18/2)$^2$ = 81. Therefore, we need to add 81 to the expression to make it a perfect square. Adding 81 to the expression gives: x$^2$ - 18x + 81 = (x - 9)$^2$ Thus, the answer is 81.

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Quadratic Equations

Duba Amsa

Tambaya 16 Rahoto

Solve the equation $\frac{m}{3} + \frac{1}{2} = \frac{3}{4} + \frac{m}{4}$

-3

-2

3 Amsar da ta dace

Bayanin Amsa

First, we simplify both sides of the equation by getting rid of the fractions. To do this, we can multiply both sides by the least common multiple (LCM) of the denominators, which is 12. So we have: \begin{align*} \frac{m}{3} + \frac{1}{2} &= \frac{3}{4} + \frac{m}{4}\\ 4 \cdot \left(\frac{m}{3} + \frac{1}{2}\right) &= 4 \cdot \left(\frac{3}{4} + \frac{m}{4}\right)\\ \frac{4m}{3} + 2 &= \frac{9}{4} + \frac{m}{4} \end{align*} Next, we can simplify further by moving all the terms involving m to one side of the equation, and all the constant terms to the other side: \begin{align*} \frac{4m}{3} - \frac{m}{4} &= \frac{9}{4} - 2\\ \frac{16m}{12} - \frac{3m}{12} &= \frac{9}{4} - \frac{8}{4}\\ \frac{13m}{12} &= \frac{1}{4}\\ 13m &= 3\\ m &= \frac{3}{13} \end{align*} Therefore, the solution to the equation is $m = \frac{3}{13}$. So the correct answer is 3.

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Karanta bayanin darasi akan Fractions, Decimals, Approximations And Percentage (JAMB)

Solution Of Linear Equations

Fractions, Decimals, Approximations And Percentage

Duba Amsa

Tambaya 17 Rahoto

The angles marked in the diagram above are measured in degrees. Find x

15^o

24^o Amsar da ta dace

30^o

36^o

48^o

Bayanin Amsa

In the given diagram, we have a triangle with two sides of equal length, which means that the angles opposite these sides are also equal. Thus, we can conclude that: 4x - 6 = 3x + 12 (using the fact that angles in a triangle add up to 180 degrees) x = 18 Therefore, the value of x is 18 degrees. Answer: (b) 24°.

Karanta bayanin darasi akan Angles And Intercepts On Parallel Lines (WAEC)

Karanta bayanin darasi akan Angles (WAEC)

Angles And Intercepts On Parallel Lines

Angles

Duba Amsa

Tambaya 18 Rahoto

The data below shows the frequency distribution
of marks scored by a group of students in a class
test.

Find the mean mark.

1.3

3.8 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Integration (JAMB)

Karanta bayanin darasi akan Statistics (WAEC)

Integration

Statistics

Duba Amsa

Tambaya 19 Rahoto

Town P is on bearing 315^o from town Q while town R is south of town P and west of town Q. lf town R is 60km away from Q, how far is R from P?

30km

42km

45km

60km Amsar da ta dace

120km

Bayanin Amsa

Karanta bayanin darasi akan Volumes (WAEC)

Karanta bayanin darasi akan Bearings (WAEC)

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Karanta bayanin darasi akan Angles Of Elevation And Depression (WAEC)

Karanta bayanin darasi akan Angles And Intercepts On Parallel Lines (WAEC)

Volumes

Bearings

Algebraic Expressions

Angles Of Elevation And Depression

Angles And Intercepts On Parallel Lines

Duba Amsa

Tambaya 20 Rahoto

Factorize the expression 2y$^2$ + xy - 3x$^2$

2y (y + x) - 3x²

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

(3x - 2y(x - y)

(2y + 3) x (y - x) Amsar da ta dace

(x – y)(2y + 3x)

Bayanin Amsa

Karanta bayanin darasi akan Polynomials (JAMB)

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Polynomials

Algebraic Expressions

Duba Amsa

Tambaya 21 Rahoto

In the diagram above, O is the center of the circle and ?POR = 126°. Find ?PQR

234^o

117^o Amsar da ta dace

72^o

63^o

54^o

Bayanin Amsa

Duba Amsa

Tambaya 22 Rahoto

A group of students measured a certain angle
(to the nearest degree) and obtained the following
results: 75^o, 76^o, 72^o, 73^o, 74^o, 79^o, 72^o, 72^o, 77^o,
72^o, 71^o, 70^o, 78^o, 73^o. Find the mode

79^o

78^o

74^o

73^o

72^o Amsar da ta dace

Bayanin Amsa

To find the mode of the given set of data, we need to identify the value that appears most frequently. From the given set of measurements, we can see that 72^o appears 4 times, which is more than any other value. Therefore, the mode of the data set is 72^o. So, the correct option is (e) 72^o.

Karanta bayanin darasi akan Angles And Intercepts On Parallel Lines (WAEC)

Karanta bayanin darasi akan Probability (JAMB)

Karanta bayanin darasi akan Angles (WAEC)

Angles And Intercepts On Parallel Lines

Probability

Angles

Duba Amsa

Tambaya 23 Rahoto

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

$\frac{x + 3}{1 - x^2}$

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

$\frac{3 - x}{1 + x^2}$

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

$\frac{3 - x}{1 - x^2}$ Amsar da ta dace

Bayanin Amsa

To add the fractions $\frac{1}{1-x}$ and $\frac{2}{1+x}$, we need to have a common denominator. We can obtain this by multiplying the first fraction by $(1+x)/(1+x)$ and the second fraction by $(1-x)/(1-x)$. This gives us: $$ \frac{1}{1-x} \cdot \frac{1+x}{1+x} + \frac{2}{1+x} \cdot \frac{1-x}{1-x} = \frac{1+x}{1-x^2} + \frac{2(1-x)}{1-x^2} $$ Combining the two fractions gives us: $$ \frac{1+x+2(1-x)}{1-x^2} = \frac{3-x}{1-x^2} $$ Therefore, the single fraction is $\boxed{\frac{3-x}{1-x^2}}$.

Karanta bayanin darasi akan Algebraic Fractions (WAEC)

Algebraic Fractions

Duba Amsa

Tambaya 24 Rahoto

Sin 60° has the same value as I. Sin 120° II. cos 240° III. -sin 150° IV. cos 210° V. sin 240°

I only Amsar da ta dace

II only

IV only

III only

IV and V only

Bayanin Amsa

Karanta bayanin darasi akan Sine, Cosine And Tangent Of An Angle (WAEC)

Karanta bayanin darasi akan Measures Of Dispersion (JAMB)

Sine, Cosine And Tangent Of An Angle

Measures Of Dispersion

Duba Amsa

Tambaya 25 Rahoto

The value of sin 210° is

$-\frac{1}{2}$ Amsar da ta dace

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

$\frac{1}{2}$

$\frac{\sqrt{2}}{2}$

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

Bayanin Amsa

Karanta bayanin darasi akan Sine, Cosine And Tangent Of An Angle (WAEC)

Karanta bayanin darasi akan Trigonometry (JAMB)

Sine, Cosine And Tangent Of An Angle

Trigonometry

Duba Amsa

Tambaya 26 Rahoto

The nth term of a sequence is given by 3.2$^{n-2}$. Write down the first three terms of the sequence.

2/3, 0, 6

3/2, 3, 6, Amsar da ta dace

2/3, 3, 8/3

2/3, ³/₄, 6

2/3, 3, 1/3

Bayanin Amsa

Karanta bayanin darasi akan Sequence And Series (WAEC)

Sequence And Series

Duba Amsa

Tambaya 27 Rahoto

A cliff on the bank of a river is 300 metres
high. if the angle of depression of a point on the
opposite side of the river is 60° find the width of
the river.

100m

75√3m

100√3m Amsar da ta dace

200√3m

300m

Bayanin Amsa

Karanta bayanin darasi akan Sine, Cosine And Tangent Of An Angle (WAEC)

Karanta bayanin darasi akan Change Of Subject Of A Formula/Relation (WAEC)

Sine, Cosine And Tangent Of An Angle

Change Of Subject Of A Formula/Relation

Duba Amsa

Tambaya 28 Rahoto

The volume of a cone of height 9cm is 1848cm$^3$. Find its radius. [Take π = 22/7]

7cm

14cm Amsar da ta dace

28cm

98cm

196cm

Bayanin Amsa

Karanta bayanin darasi akan Volumes (WAEC)

Volumes

Duba Amsa

Tambaya 29 Rahoto

In the diagram above, PQT is an isosceles triangle.|PQ| = |QT|, ?SRQ = 75°, ?QPT = 25° and PQR is straight line. Find ?RST

20^o

50^o

55^o Amsar da ta dace

70^o

75^o

Bayanin Amsa

Duba Amsa

Tambaya 30 Rahoto

A bricklayer measured the length of a wall and obtained 4.10m. If the actual length of the wall is 4.25m, find his percentage error.

3 9/17% Amsar da ta dace

3 27/41%

15%

35 5/17%

36 24/41%

Bayanin Amsa

To find the percentage error, we need to find the absolute difference between the measured value and actual value, divide it by the actual value, and then multiply by 100 to get a percentage. Absolute difference = |actual value - measured value| = |4.25 - 4.10| = 0.15 Percentage error = (|actual value - measured value| / actual value) x 100% Percentage error = (0.15 / 4.25) x 100% = 3.529% (rounded to 3 decimal places) Therefore, the percentage error is approximately 3.529%, which corresponds to 3 9/17%.

Karanta bayanin darasi akan Percentages (WAEC)

Karanta bayanin darasi akan Fractions, Decimals, Approximations And Percentage (JAMB)

Percentages

Fractions, Decimals, Approximations And Percentage

Duba Amsa

Tambaya 31 Rahoto

The data above shows the frequency distribution
of marks scored by a group of students in a class
test.

How many students took the test?

15 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Representation Of Data (JAMB)

Karanta bayanin darasi akan Integration (JAMB)

Karanta bayanin darasi akan Statistics (WAEC)

Representation Of Data

Integration

Statistics

Duba Amsa

Tambaya 32 Rahoto

A regular polygon has each of its exterior angles as 18^o. How many sides has the polygon

20 Amsar da ta dace

Bayanin Amsa

The exterior angles of a polygon add up to 360 degrees. If each exterior angle of a regular polygon is 18 degrees, then the polygon has 360/18 = 20 exterior angles. Since a regular polygon has the same number of sides as exterior angles, the polygon has 20 sides. Therefore, the answer is (d) 20.

Karanta bayanin darasi akan Sets (JAMB)

Karanta bayanin darasi akan Triangles And Polygons (WAEC)

Sets

Triangles And Polygons

Duba Amsa

Tambaya 33 Rahoto

Points X and Y are respectively 12m Nonh and
5m of point Z. Calculate XY

12m

13m Amsar da ta dace

17m

18m

Bayanin Amsa

To find the distance between points X and Y, we can use the Pythagorean theorem, which states that for a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, we can consider points X, Y, and Z as the vertices of a right triangle, where the right angle is at point Z. Then, XY is the hypotenuse, and we know that XZ = 12m and YZ = 5m. Therefore, we have: XY^2 = XZ^2 + YZ^2 XY^2 = 12^2 + 5^2 XY^2 = 144 + 25 XY^2 = 169 XY = √169 XY = 13m Therefore, the distance between points X and Y is 13m. So the answer is option (C) 13m.

Karanta bayanin darasi akan Inequalities (JAMB)

Karanta bayanin darasi akan Number Bases (JAMB)

Karanta bayanin darasi akan Angles (WAEC)

Inequalities

Number Bases

Angles

Duba Amsa

Tambaya 34 Rahoto

Construct a quadratic equation whose roots are $-\frac{1}{2}$ and 2.

3x²-3x+2=0

3x²+3x-2=0

2x²+3x-2=0

2x²-3x+2=0

2x²-3x-2=0 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Quadratic Equations

Duba Amsa

Tambaya 35 Rahoto

The data below shows the number of worker?
employed in the various sections of a construction
company in Lagos.

Carpenters 24 Labourers 27
Plumbers 12 Plasterers 15
Painters 9 Messengers 3
Bricklayers 18

If one of the workers is absent on a certain day,
what is the probability that he is a bricklayer?

1/12

1/9

2/9

1/6 Amsar da ta dace

1/4

Bayanin Amsa

The total number of workers in the company is 24+27+12+15+9+3+18 = 108. The probability of an absent worker being a bricklayer is the ratio of the number of bricklayers to the total number of workers, which is 18/108 = 1/6. Therefore, the probability that the absent worker is a bricklayer is 1/6. Answer: 1/6.

Karanta bayanin darasi akan Statistics (WAEC)

Statistics

Duba Amsa

Tambaya 36 Rahoto

In the diagram above, O is the center of the circle QOR is a diameter and ?PSR is 37°. Find ?PRQ

37^o

53^o Amsar da ta dace

65^o

127^o

147^o

Bayanin Amsa

Since QOR is a diameter, it divides the circle into two equal halves, which means that ?QOR is a right angle (90°). Thus, we have: ?PQR = 180° - ?QOR (sum of angles in a triangle) ?PQR = 180° - 90° (since ?QOR is a right angle) ?PQR = 90° Now, we have a right triangle PQR with right angle at Q. Since we know that ?PSR is 37°, we can use the fact that the sum of angles in a triangle is 180° to find ?PRQ: ?PQR + ?QPR + ?PRQ = 180° 90° + ?QPR + 37° = 180° ?QPR = 180° - 90° - 37° ?QPR = 53° Therefore, the answer is 53°.

Karanta bayanin darasi akan Circles (WAEC)

Circles

Duba Amsa

Tambaya 37 Rahoto

The curved surface area of a cylindrical tin is 704cm$^2$. Calculate the height when the radius is 8cm. [Take π = 22/7]

3.5cm

7cm

14cm Amsar da ta dace

28cm

32cm

Bayanin Amsa

Karanta bayanin darasi akan Inequalities (JAMB)

Inequalities

Duba Amsa

Tambaya 38 Rahoto

Simplify: $(\frac{16}{81})^{\frac{1}{4}}$

8/27

1/3

4/9

2/3 Amsar da ta dace

-4/3

Bayanin Amsa

To simplify $(\frac{16}{81})^{\frac{1}{4}}$, we need to take the fourth root of $\frac{16}{81}$. The fourth root of a number is the number that, when multiplied by itself four times, gives that number. So, to find the fourth root of $\frac{16}{81}$, we need to find a number that, when multiplied by itself four times, gives $\frac{16}{81}$. We can rewrite $\frac{16}{81}$ as $(\frac{2}{3})^4$, since $2^4=16$ and $3^4=81$. Therefore, $(\frac{16}{81})^{\frac{1}{4}} = ((\frac{2}{3})^4)^{\frac{1}{4}} = \frac{2}{3}$, which is.

Karanta bayanin darasi akan Fractions, Decimals And Approximations (WAEC)

Karanta bayanin darasi akan Number Bases (WAEC)

Fractions, Decimals And Approximations

Number Bases

Duba Amsa

Tambaya 39 Rahoto

In the diagram above, O is the center of the circle PQRS and ?QPS = 36°. Find ?QOS

36^o

72^o Amsar da ta dace

108^o

144^o

288^o

Bayanin Amsa

Karanta bayanin darasi akan Circles (WAEC)

Circles

Duba Amsa

Tambaya 40 Rahoto

Simplify (³/₄ + ¹/₃) x 4¹/₃ + 3¹/₄

1/2

13/12

10/9

17/12

13/9 Amsar da ta dace

Bayanin Amsa

To simplify this expression, we need to follow the order of operations, which is Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction (PEMDAS). First, we can simplify the expression inside the parentheses: (³/₄ + ¹/₃) x 4¹/₃ + 3¹/₄ = (⁹/₁₂ + ⁴/₁₂) x 4¹/₃ + 3¹/₄ = ¹³/₁₂ x 4¹/₃ + 3¹/₄ Next, we can simplify the exponent 4¹/₃. To do this, we can convert the mixed number to an improper fraction: 4¹/₃ = ^{(4 x 3) + 1}/₃ = ¹³/₃ Now we can substitute the value of 4¹/₃ into the expression: ¹³/₁₂ x 4¹/₃ + 3¹/₄ = ¹³/₁₂ x ¹³/₃ + ¹³/₄ To simplify this, we can first simplify the multiplication of fractions: ¹³/₁₂ x ¹³/₃ = ¹⁶⁹/₃₆ Now we can add the two fractions: ¹⁶⁹/₃₆ + ¹³/₄ = ¹⁶⁹/₃₆ + ³⁹/₃₆ = ²⁰⁸/₃₆ = ^13}{9 Therefore, the simplified expression is ^13}{9, which corresponds to option (E).

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Tambaya 41 Rahoto

Marks	1	2	3	4	5	6	7	8	9	10
No of students	1	3	2	0	1	6	1	0	1	0

The table above shows the scores of 15 students in a Physics test.

How many students scored at least 5?

9 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

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Tambaya 42 Rahoto

In the diagram above, ML//PQ and NP//QR, if ?LMN = 40° and ?MNP = 55°. Find ?QPR

15^o Amsar da ta dace

25^o

35^o

40^o

55^o

Bayanin Amsa

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Tambaya 43 Rahoto

Find the median of the following set of scores:
65, 72, 55, 48, 78

63.5

65 Amsar da ta dace

Bayanin Amsa

To find the median of a set of scores, we need to arrange the scores in order of magnitude from smallest to largest. The scores in this set are: 48, 55, 65, 72, 78 There are five scores in this set, so the median will be the middle score. Since the scores are already in order, we can simply find the middle score, which is 65. Therefore, the median of this set of scores is 65.

Karanta bayanin darasi akan Representation Of Data (JAMB)

Karanta bayanin darasi akan Integration (JAMB)

Karanta bayanin darasi akan Probability (WAEC)

Representation Of Data

Integration

Probability

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Tambaya 44 Rahoto

A number is chosen at random from the set (1 ,2,3
....,9, 10). What is the probability that the number
is greater than or equal to 7?

1/10

3/10

2/5 Amsar da ta dace

3/5

7/10

Bayanin Amsa

There are 10 numbers in the set (1, 2, 3, ..., 9, 10). Out of these, the numbers that are greater than or equal to 7 are 7, 8, 9, and 10. So, there are 4 numbers that satisfy the given condition. Therefore, the probability of choosing a number that is greater than or equal to 7 is the number of numbers that satisfy the condition divided by the total number of numbers in the set, which is: 4/10 = 2/5 Hence, the probability that the number chosen at random from the set (1, 2, 3, ..., 9, 10) is greater than or equal to 7 is 2/5.

Karanta bayanin darasi akan Number Bases (WAEC)

Number Bases

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Tambaya 45 Rahoto

The angle of a sector of a circle of radius 8cm is 240°. This sector is bent to form a cone. Find the radius of the base of the cone.

16/3cm Amsar da ta dace

15/3cm

16/5cm

8/3cm

16/10cm

Bayanin Amsa

The sector of the circle with an angle of 240° forms a cone. To find the radius of the base of the cone, we need to use the formula relating the arc length of the sector and the circumference of the base of the cone. The arc length of the sector is (240/360) * 2π * 8 = (4/3) * π * 8 = (32/3)π cm. The circumference of the base of the cone is equal to the arc length of the sector, so it is (32/3)π cm. We know that the circumference of a circle is equal to 2πr, where r is the radius of the circle. Therefore, we can write: (32/3)π = 2πr Simplifying the equation, we get: r = (16/3) cm So, the radius of the base of the cone is 16/3 cm. Therefore, the correct option is: 16/3cm

Karanta bayanin darasi akan Circles (WAEC)

Karanta bayanin darasi akan Coordinate Geometry (WAEC)

Circles

Coordinate Geometry

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Tambaya 46 Rahoto

The data below shows the number of worker?
employed in the various sections of a construction
company in Lagos.

Carpenters 24 Labourers 27
Plumbers 12 Plasterers 15
Painters 9 Messengers 3
Bricklayers 18

If a worker is retrenched, what is the probability that he is a plumber or plasterer.

1/27

1/9

5/36

1/4 Amsar da ta dace

3/4

Bayanin Amsa

To calculate the probability that a retrenched worker is a plumber or plasterer, we need to first find the total number of workers in the company, and then find the number of workers who are plumbers or plasterers. The total number of workers in the company is: 24 (Carpenters) + 27 (Labourers) + 12 (Plumbers) + 15 (Plasterers) + 9 (Painters) + 3 (Messengers) + 18 (Bricklayers) = 108 The number of workers who are plumbers or plasterers is: 12 (Plumbers) + 15 (Plasterers) = 27 Therefore, the probability that a retrenched worker is a plumber or plasterer is: 27 (number of plumbers and plasterers) / 108 (total number of workers) = 1/4 So the correct answer is option D: 1/4.

Karanta bayanin darasi akan Probability (WAEC)

Karanta bayanin darasi akan Sets (JAMB)

Probability

Sets

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Tambaya 47 Rahoto

The area of a parallelogram is 513cm$^2$ and the height is 19cm. Calculate the base.

13.5cm

25cm

27cm Amsar da ta dace

54cm

108cm

Bayanin Amsa

The area of a parallelogram is given by the formula: Area = base × height In this problem, the area and height of the parallelogram are given as 513cm$^2$ and 19cm respectively. We can rearrange the formula to solve for the base: base = Area ÷ height Substituting the given values, we get: base = 513cm$^2$ ÷ 19cm = 27cm Therefore, the base of the parallelogram is 27cm. The answer is, 27cm.

Karanta bayanin darasi akan Areas (WAEC)

Areas

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Tambaya 48 Rahoto

If cos θ = 5/13, what is the value of tan $\theta$ for 0 < θ < 90° ?

13/5

12/5 Amsar da ta dace

5/12

Bayanin Amsa

We can use the following trigonometric identity: \[\tan \theta = \frac{\sin \theta}{\cos \theta}\] We are given that $\cos \theta = 5/13$. To find $\sin \theta$, we can use the Pythagorean identity: \[\sin^2 \theta + \cos^2 \theta = 1\] Substituting $\cos \theta = 5/13$, we get: \[\sin^2 \theta + \left(\frac{5}{13}\right)^2 = 1\] Simplifying, we get: \[\sin^2 \theta = 1 - \frac{25}{169} = \frac{144}{169}\] Taking the square root of both sides, we get: \[\sin \theta = \frac{12}{13}\] Substituting into the formula for $\tan \theta$, we get: \[\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{12}{13}}{\frac{5}{13}} = \boxed{\frac{12}{5}}\]

Karanta bayanin darasi akan Trigonometry (JAMB)

Karanta bayanin darasi akan Positive And Negative Integers, Rational Numbers (WAEC)

Trigonometry

Positive And Negative Integers, Rational Numbers

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Latsa & Riƙe don Ja Shi Gabaɗaya
Danna nan don rufewa

WAEC SSCE - General Mathematics - 1992 (Objective)