WAEC SSCE - General Mathematics - 2011…

Tambaya 1 Rahoto

Find the size of the angle marked x in the diagram.

108^o

112^o

128^o Amsar da ta dace

142^o

Bayanin Amsa

Karanta bayanin darasi akan Coordinate Geometry (WAEC)

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

Karanta bayanin darasi akan Angles (WAEC)

Coordinate Geometry

Angles And Intercepts On Parallel Lines

Angles

Duba Amsa

Tambaya 2 Rahoto

Simplify \(\frac{m}{n} + \frac{(m - 1)}{5n} = \frac{(m - 2)}{10n}\) where n \(\neq\) 0

\(\frac{m - 3}{10n}\)

\(\frac{11m}{10n}\) Amsar da ta dace

\(\frac{m + 1}{10n}\)

\(\frac{11m + 4}{10n}\)

Bayanin Amsa

Karanta bayanin darasi akan Algebraic Fractions (WAEC)

Algebraic Fractions

Duba Amsa

Tambaya 3 Rahoto

From the equation whose roots are x = \(\frac{1}{2}\) and -\(\frac{2}{3}\)

6x² - x + 2 = 0

6x² - x - 2 = 0

6x² + x + 2 = 0

6x² + x - 2 = 0 Amsar da ta dace

Bayanin Amsa

When a quadratic equation has roots at x = a and x = b, it can be written in factored form as (x-a)(x-b) = 0. Therefore, from the given roots, the factors are (x - \(\frac{1}{2}\)) and (x + \(\frac{2}{3}\)). To get the quadratic equation, we can expand the factors by multiplying them together, which gives us: (x - \(\frac{1}{2}\))(x + \(\frac{2}{3}\)) = x² - \(\frac{1}{2}\)x + \(\frac{2}{3}\)x - \(\frac{1}{2}\)\(\frac{2}{3}\) = x² + \(\frac{1}{6}\)x - \(\frac{1}{3}\) Therefore, the correct option is 6x² + x - 2 = 0.

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Quadratic Equations

Duba Amsa

Tambaya 4 Rahoto

A cylindrical container has a base radius of 14cm and height 18cm. How many litres of liquid can it hold? correct to the nearest litre [Take \(\pi = \frac{22}{7}\)]

11 Amsar da ta dace

Bayanin Amsa

The volume of a cylinder can be calculated using the formula: V = πr²h, where r is the radius of the base and h is the height of the cylinder. Substituting the given values, we have: V = π(14)²(18) V = 11088 cm³ To convert cm³ to litres, we divide by 1000: V = 11088/1000 V = 11.088 litres Rounding to the nearest litre, we get: V ≈ 11 litres Therefore, the answer is 11.

Karanta bayanin darasi akan Volumes (WAEC)

Karanta bayanin darasi akan Inequalities (JAMB)

Volumes

Inequalities

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Tambaya 5 Rahoto

What is the value of 3 in the number 42.7531?

\(\frac{3}{10000}\)

\(\frac{3}{1000}\) Amsar da ta dace

\(\frac{3}{100}\)

\(\frac{1}{10}\)

Bayanin Amsa

The number 42.7531 can be written in expanded form as: 42 + 0.7 + 0.05 + 0.003 + 0.0001 The digit 3 is located in the thousandths place, which represents the decimal value of 0.001. So, the value of 3 in the number 42.7531 is equal to: 3 x 0.001 = 0.003 Therefore, the correct answer is option (A), which is 3 divided by 10000.

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

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Positive And Negative Integers, Rational Numbers

Modular Arithmetic

Duba Amsa

Tambaya 6 Rahoto

Given that cos x^o = \(\frac{1}{r}\), express tan x in terms of r

\(\frac{1}{\sqrt{r}}\)

\(\sqrt{r}\)

\(\sqrt{r^2 + 1}\)

\(\sqrt{r^2 - 1}\) Amsar da ta dace

Bayanin Amsa

We know that: cos x = adjacent side/hypotenuse So, if cos x = 1/r, then adjacent side = 1 and hypotenuse = r. Using the Pythagorean theorem, we can find the opposite side: opposite side = √(hypotenuse^2 - adjacent side^2) = √(r^2 - 1) Finally, we can find the value of tan x: tan x = opposite side/adjacent side = √(r^2 - 1)/1 = √(r^2 - 1) Therefore, the answer is (d) \(\sqrt{r^2 - 1}\).

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

Karanta bayanin darasi akan Lengths And Perimeters (WAEC)

Sine, Cosine And Tangent Of An Angle

Lengths And Perimeters

Duba Amsa

Tambaya 7 Rahoto

The perimeter of a sector of a circle of radius 4cm is (\(\pi + 8\))cm. Calculate the anle of the sector

45^o Amsar da ta dace

60^o

75^o

90^o

Bayanin Amsa

Karanta bayanin darasi akan Circles (WAEC)

Karanta bayanin darasi akan Triangles And Polygons (WAEC)

Circles

Triangles And Polygons

Duba Amsa

Tambaya 8 Rahoto

In the diagram, /MN/, /OP/, /QOP/ = 125^o. What is the size of < MQR?

11^o

120^o Amsar da ta dace

130^o

160^o

Bayanin Amsa

Karanta bayanin darasi akan Angles (WAEC)

Karanta bayanin darasi akan Triangles And Polygons (WAEC)

Angles

Triangles And Polygons

Duba Amsa

Tambaya 9 Rahoto

The graph represents the relation y = x^o2 - 3x - 3. Find the value of x for which x² - 3x = 7

-1.55, 4.44 Amsar da ta dace

1.55, -4.55

-1.55, -4.55

1.55, 4.55

Bayanin Amsa

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Solution Of Linear Equations

Quadratic Equations

Duba Amsa

Tambaya 10 Rahoto

Simplify \(\frac{\log \sqrt{27}}{\log \sqrt{81}}\)

\(\frac{3}{2}\)

\(\frac{3}{4}\) Amsar da ta dace

Bayanin Amsa

Using the property that \(\log_{a}b = \frac{\log{b}}{\log{a}}\), we can simplify the given expression as follows: \[\frac{\log \sqrt{27}}{\log \sqrt{81}} = \frac{\log 27^{\frac{1}{2}}}{\log 81^{\frac{1}{2}}} = \frac{\frac{1}{2}\log 27}{\frac{1}{2}\log 81} = \frac{\log 3^3}{\log 3^4} = \frac{3\log 3}{4\log 3} = \frac{3}{4}\] Therefore, the simplified form of \(\frac{\log \sqrt{27}}{\log \sqrt{81}}\) is \(\frac{3}{4}\), and the correct option is (D).

Karanta bayanin darasi akan Logarithms (WAEC)

Logarithms

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Tambaya 11 Rahoto

The venn diagram shows the number of students in a class who like reading(R), dancing(D) and swimming(S). How many students like dancing and swimming?

7 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Circles (WAEC)

Karanta bayanin darasi akan Sets (JAMB)

Circles

Sets

Duba Amsa

Tambaya 12 Rahoto

from the diagram, Which of the following statements are true? i. m = q ii. n = q iii. n + p = 180^o iv. p + m = 180^o

i and iii Amsar da ta dace

i and iv

ii and iii

ii and ic

Bayanin Amsa

In the given diagram, we can see that lines n and q are parallel and m is a transversal cutting them. Therefore, angles n and q are alternate interior angles and are equal, i.e., statement i is true. Also, we can see that lines n and p are parallel and q is a transversal cutting them. Therefore, angles n and p are corresponding angles and are equal. As the sum of the corresponding angles is equal to 180 degrees, we have n + p = 180 degrees, i.e., statement iii is also true. However, we cannot determine whether statement ii and iv are true or not based on the given information and the diagram. Therefore, the correct answer is (a) i and iii.

Karanta bayanin darasi akan Polynomials (JAMB)

Karanta bayanin darasi akan Measures Of Dispersion (JAMB)

Polynomials

Measures Of Dispersion

Duba Amsa

Tambaya 13 Rahoto

In the diagram /Pq//TS//TU, reflex angle QPS = 245^o angle PST = 115^o, , STU = 65^o and < RPS = x. Find the value of x

80^o

70^o

60^o

50^o Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Indices (WAEC)

Indices

Duba Amsa

Tambaya 14 Rahoto

The height of a cylinder is equal to its radius. If the volume is 0.216 \(\pi m^3\) Calculate the radius.

0.46m Amsar da ta dace

0.60m

0.87m

1.80m

Bayanin Amsa

Karanta bayanin darasi akan Representation Of Data (JAMB)

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Representation Of Data

Modular Arithmetic

Duba Amsa

Tambaya 15 Rahoto

The diagram shows a cyclic quadrilateral PQRS with its diagonals intersecting at K. Which of the following triangles is similar to triangle QKR?

\(\bigtriangleup\) PQK

\(\bigtriangleup\) PKS

\(\bigtriangleup\) SKR Amsar da ta dace

\(\bigtriangleup\) PSR

Bayanin Amsa

Karanta bayanin darasi akan Statistics (WAEC)

Karanta bayanin darasi akan Angles (WAEC)

Karanta bayanin darasi akan Triangles And Polygons (WAEC)

Statistics

Angles

Triangles And Polygons

Duba Amsa

Tambaya 16 Rahoto

a boy looks through a window of a building and sees a mango fruit on the ground 50m away from the foot of the building. If the window is 9m from the ground, calculate, correct to the nearest degree, the angle of depression of the mango from the window

9^o

10^o Amsar da ta dace

11^o

12^o

Bayanin Amsa

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

Angles Of Elevation And Depression

Duba Amsa

Tambaya 17 Rahoto

John pours 96 litres of red oil into a rectangular container with length 220cm and breadth 40cm. Calculate, correct to the nearest cm, the height of the oil in the container

11cm Amsar da ta dace

18cm

21cm

34cm

Bayanin Amsa

To solve this problem, we need to use the formula for the volume of a rectangular container: Volume = Length x Breadth x Height We are given the length and breadth of the container, as well as the volume of the oil. We need to find the height of the oil. First, we need to convert the volume of the oil from litres to cubic centimeters, since the dimensions of the container are given in centimeters. 1 litre = 1000 cubic centimeters Therefore, 96 litres = 96 x 1000 = 96,000 cubic centimeters Now, we can plug in the values we have into the formula for the volume of the container: Volume = Length x Breadth x Height 96,000 = 220 x 40 x Height Simplifying, we get: Height = 96,000 / (220 x 40) Height = 2.18 To round off to the nearest centimeter, we need to look at the first decimal place. If the value in the first decimal place is 5 or more, we round up. If it is less than 5, we round down. In this case, the value in the first decimal place is 1, which is less than 5. Therefore, we round down to 2. So, the height of the oil in the container is approximately 2cm. Therefore, the correct answer is (a) 11cm.

Karanta bayanin darasi akan Lengths And Perimeters (WAEC)

Lengths And Perimeters

Duba Amsa

Tambaya 18 Rahoto

The histogram shows the age distribution of members of a club. What is their modal age?

44.5 Amsar da ta dace

42.5

41.5

40.5

Bayanin Amsa

Karanta bayanin darasi akan Representation Of Data (JAMB)

Representation Of Data

Duba Amsa

Tambaya 19 Rahoto

The length of a piece of stick is 1.75m. A girl measured it as 1.80m. Find the percentage error

\(\frac{28}{7}\)%

\(\frac{29}{7}\)%

\(\frac{20}{7}\)% Amsar da ta dace

Bayanin Amsa

The actual length of the stick is 1.75m and the measured length is 1.80m. The error is the difference between the actual and measured length: 1.80m - 1.75m = 0.05m To find the percentage error, we divide the error by the actual length and multiply by 100%: \frac{0.05}{1.75} \times 100\% \approx 2.857\% \approx \frac{20}{7}\% Therefore, the percentage error is approximately \frac{20}{7}\%.

Karanta bayanin darasi akan Percentages (WAEC)

Percentages

Duba Amsa

Tambaya 20 Rahoto

in a quiz competition, a student answers n questions correctly and was given D(n + 50) for each question correctly answered. If he gets D600.00 altogether, how many questions did he answer correctly?

10 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Number Bases (WAEC)

Karanta bayanin darasi akan Financial Arithmetic (WAEC)

Number Bases

Financial Arithmetic

Duba Amsa

Tambaya 21 Rahoto

John pours 96 litres of red oil into a rectangular container with length 220cm and breadth 40cm. Calculate, correct to the nearest cm, the height of the oil in the container

11cm Amsar da ta dace

18cm

21cm

34cm

Bayanin Amsa

To calculate the height of the oil in the container, we need to use the formula for the volume of a rectangular prism: Volume = length x breadth x height First, we need to convert the given volume from liters to cubic centimeters, since the dimensions of the container are in centimeters. 96 liters = 96,000 cubic centimeters Next, we can plug in the given values into the formula: 96,000 = 220 x 40 x height Solving for height, we get: height = 96,000 / (220 x 40) height ≈ 11.0 cm (rounded to the nearest cm) Therefore, the height of the oil in the container is approximately 11 cm. Note: When working with volume, it's important to make sure the units are consistent throughout the problem. In this case, we converted liters to cubic centimeters to match the dimensions of the container.

Duba Amsa

Tambaya 22 Rahoto

If 27^x = 9^y. Find the value of \(\frac{x}{y}\)

\(\frac{1}{3}\)

\(\frac{2}{3}\) Amsar da ta dace

1\(\frac{1}{2}\)

Bayanin Amsa

If we can find the value of x and y, then we can calculate x/y by dividing x by y. Given: 27^x = 9^y We can rewrite 27 as 3³ and 9 as 3² to get: (3³)^x = (3²)^y Applying the power of a power rule, we get: 3^3x = 3^2y For two exponential expressions to be equal, their bases must be equal. Therefore: 3^3x = 3^2y implies 3x = 2y Dividing both sides by y, we get: \(\frac{x}{y} = \frac{2}{3}\) Therefore, the answer is \(\frac{2}{3}\).

Karanta bayanin darasi akan Indices (WAEC)

Indices

Duba Amsa

Tambaya 23 Rahoto

If N112.00 exchanges for D14.95, calculate the value of D1.00 in naira

0.13

7.49 Amsar da ta dace

8.00

13.00

Bayanin Amsa

To calculate the value of D1.00 in naira, we can use the given exchange rate of N112.00 to D14.95. We can find the value of 1 D in Naira by dividing N112.00 by the equivalent value of D14.95. So, 1 D = N112.00/D14.95 To simplify this, we can first convert D14.95 to its decimal equivalent by dividing by 100: D14.95 = 14.95/100 = 0.1495 Now we can substitute this value into the equation: 1 D = N112.00/0.1495 Simplifying this expression, we get: 1 D = N748.16 Therefore, the value of D1.00 in Naira is N748.16. Answer: 7.49.

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

Karanta bayanin darasi akan Areas (WAEC)

Positive And Negative Integers, Rational Numbers

Areas

Duba Amsa

Tambaya 24 Rahoto

In the diagram, STUV is a straight line. < TSY = < UXY = 40^o and < VUW = 110^o. Calculate < TYW

150^o Amsar da ta dace

140^o

130^o

120^o

Bayanin Amsa

Karanta bayanin darasi akan Angles (WAEC)

Karanta bayanin darasi akan Triangles And Polygons (WAEC)

Angles

Triangles And Polygons

Duba Amsa

Tambaya 25 Rahoto

What must be added to (2x - 3y) to get (x - 2y)?

5y - x

y - x Amsar da ta dace

x - 5x

x - y

Bayanin Amsa

To get from (2x - 3y) to (x - 2y), we need to subtract x from 2x and add 2y to -3y. Therefore, we need to add (x - 2y) - (2x - 3y) to (2x - 3y) to get (x - 2y). Simplifying (x - 2y) - (2x - 3y), we have: (x - 2y) - (2x - 3y) = x - 2y - 2x + 3y = -x + y Therefore, we need to add (-x + y) to (2x - 3y) to get (x - 2y). Simplifying (2x - 3y) + (-x + y), we have: (2x - 3y) + (-x + y) = 2x - 3y - x + y = x - 2y So, we need to add (-x + y) to (2x - 3y) to get (x - 2y). Therefore, the answer is (B) y - x.

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Karanta bayanin darasi akan Algebraic Fractions (WAEC)

Algebraic Expressions

Algebraic Fractions

Duba Amsa

Tambaya 26 Rahoto

If \(\sqrt{72} + \sqrt{32} - 3 \sqrt{18} = x \sqrt{8}\), Find the value of x

\(\frac{3}{4}\)

\(\frac{1}{2}\) Amsar da ta dace

\(\frac{1}{4}\)

Bayanin Amsa

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Surds (radicals)

Duba Amsa

Tambaya 27 Rahoto

Using the cumulative frequency curve. What is the 80th percentile?

45.5

46.5

47.5 Amsar da ta dace

48.5

Bayanin Amsa

Karanta bayanin darasi akan Integration (JAMB)

Integration

Duba Amsa

Tambaya 28 Rahoto

A regular polygon of n sides has each exterior angle to 45^o. Find the value of n

8 Amsar da ta dace

Bayanin Amsa

In a regular polygon with n sides, each exterior angle measures 360/n degrees. We are given that in this polygon, each exterior angle is 45 degrees. Therefore, we can set up an equation: 360/n = 45 To solve for n, we can cross-multiply and simplify: 360 = 45n n = 360/45 n = 8 Therefore, the regular polygon in question has 8 sides. Answer: 8.

Duba Amsa

Tambaya 29 Rahoto

A rectangular garden measures 18.6m by 12.5m. Calculate, correct to three significant figures, the area of the garden

230m²

231m²

232m²

233m² Amsar da ta dace

Bayanin Amsa

The area of a rectangle is given by multiplying the length by the width. Therefore, the area of the garden is: Area = length × width Area = 18.6m × 12.5m Area = 232.5m² Rounding to three significant figures gives 233m². Therefore, the answer is (d) 233m².

Karanta bayanin darasi akan Lengths And Perimeters (WAEC)

Lengths And Perimeters

Duba Amsa

Tambaya 30 Rahoto

One of the factors of (mn - nq - n2 + mq) is (m - n). The other factor is?

(n - q)

(q - n)

(n + q) Amsar da ta dace

(q - m)

Bayanin Amsa

Karanta bayanin darasi akan Number Bases (WAEC)

Number Bases

Duba Amsa

Tambaya 31 Rahoto

Solve the equation; 3x - 2y = 7, x + 2y = -3

x = 1, y = -2 Amsar da ta dace

x = 1, y = 3

x = -2, y = -1

x = 4, y = -3

Bayanin Amsa

To solve this system of equations, we can use the method of elimination. We will add the two equations together, which will eliminate the y variable: (3x - 2y) + (x + 2y) = 7 + (-3) Simplifying, we get: 4x = 4 Dividing both sides by 4, we get: x = 1 Substituting x = 1 into one of the equations, we get: 1 + 2y = -3 Solving for y, we get: y = -2 Therefore, the solution to the system of equations is x = 1, y = -2. So, the correct option is (a) x = 1, y = -2.

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Duba Amsa

Tambaya 32 Rahoto

G varies directly as the square of H, If G is 4 when H is 3, find H when G = 100

15 Amsar da ta dace

225

Bayanin Amsa

In this problem, we are given that G varies directly as the square of H. This means that if H is multiplied by some factor, then G will be multiplied by the square of that factor. Mathematically, we can write this as: G ∝ H^2 where the symbol "∝" means "varies directly as". We are also given that G is 4 when H is 3. Using this information, we can write: 4 ∝ 3^2 To find H when G = 100, we can use the same relationship: G ∝ H^2 If we let the constant of proportionality be k, we can write: G = kH^2 To solve for k, we can use the initial condition where G is 4 when H is 3: 4 = k(3^2) Simplifying, we get: k = 4/9 Now we can use this value of k to find H when G is 100: 100 = (4/9)H^2 Multiplying both sides by 9/4, we get: 225 = H^2 Taking the square root of both sides, we get: H = 15 Therefore, the correct answer is (a) 15. In summary, we used the direct variation relationship between G and H^2 to find the constant of proportionality, and then used that constant and the given value of G to solve for H.

Duba Amsa

Tambaya 33 Rahoto

The histogram shows the age distribution of members of a club. How many members are in the club?

52 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Representation Of Data (JAMB)

Karanta bayanin darasi akan Mensuration (JAMB)

Representation Of Data

Mensuration

Duba Amsa

Tambaya 34 Rahoto

In the diagram /Pq//TS//TU, reflex angle QPS = 245^o angle PST = 115^o, , STU = 65^o and < RPS = x. Find the value of x

80^o

70^o

60^o

50^o Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Indices (WAEC)

Karanta bayanin darasi akan Triangles And Polygons (WAEC)

Indices

Triangles And Polygons

Duba Amsa

Tambaya 35 Rahoto

The graph represents the relation y = x\(^2\) - 3x - 3. What is the equation of the line of symmetry of the graph?

y = 0.5

x = 1.0

x = 1.5 Amsar da ta dace

y = 4.6

Bayanin Amsa

To find the equation of the line of symmetry of the graph, we need to identify the axis of symmetry. The axis of symmetry is a vertical line that passes through the vertex of the parabola. The vertex of the parabola is the point where the parabola changes direction, and it can be found by using the formula: x = -b / (2a) where a and b are the coefficients of the quadratic equation y = ax\(^2\) + bx + c. In the given equation y = x\(^2\) - 3x - 3, a = 1, b = -3, and c = -3. Substituting these values in the formula, we get: x = -(-3) / (2*1) = 3/2 = 1.5 Therefore, the line of symmetry is a vertical line passing through x = 1.5. So, the correct answer is (C) x = 1.5.

Karanta bayanin darasi akan Simple Operations On Algebraic Expressions (WAEC)

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Simple Operations On Algebraic Expressions

Quadratic Equations

Duba Amsa

Tambaya 36 Rahoto

Using the cumulative frequency curve, estimate the median of the data represented on the graph.

35.5

36.5 Amsar da ta dace

37.5

38.5

Bayanin Amsa

Duba Amsa

Tambaya 37 Rahoto

Given that n(p) = 19, m(P \(\cup\) Q) = 38 and n(P \(\cap\) Q) = 7, Find n(C)

26 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Sets (JAMB)

Karanta bayanin darasi akan Ratio, Proportions And Rates (WAEC)

Sets

Ratio, Proportions And Rates

Duba Amsa

Tambaya 38 Rahoto

The height of a cylinder is equal to its radius. If the volume is 0.216 \(\pi\) m\(^3\). Calculate the radius.

0.46m

0.60m Amsar da ta dace

0.87m

1.80m

Bayanin Amsa

Let's denote the radius of the cylinder as r and its height as h. We are given that the height of the cylinder is equal to its radius, so h = r. We also know the volume of the cylinder, which is given by: V = \(\pi\)r\(^2\)h Substituting h = r, we get: V = \(\pi\)r\(^2\)r = \(\pi\)r\(^3\) We are given that the volume of the cylinder is 0.216 \(\pi\) m\(^3\). So, we can solve for r as follows: 0.216 \(\pi\) = \(\pi\)r\(^3\) r\(^3\) = 0.216 Taking the cube root of both sides, we get: r = 0.6 Therefore, the radius of the cylinder is 0.6 meters. So, the answer is 0.60m.

Karanta bayanin darasi akan Number Bases (WAEC)

Karanta bayanin darasi akan Coordinate Geometry (WAEC)

Number Bases

Coordinate Geometry

Duba Amsa

Tambaya 39 Rahoto

Find the smaller value of x that satisfies the equation x₂ + 7x + 10 = 0

-5 Amsar da ta dace

-2

Bayanin Amsa

We are given a quadratic equation x² + 7x + 10 = 0 and we need to find the smaller value of x that satisfies the equation. To solve the equation, we can factorize it by finding two numbers whose product is 10 and whose sum is 7. We can see that the two numbers are 2 and 5, since 2 × 5 = 10 and 2 + 5 = 7. So, we can write the equation as (x + 2)(x + 5) = 0. For this equation to be true, either (x + 2) = 0 or (x + 5) = 0. Therefore, we get x = -2 or x = -5. Since we are asked to find the smaller value of x, we choose x = -5 as the answer. Hence, the smaller value of x that satisfies the equation x² + 7x + 10 = 0 is -5.

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Quadratic Equations

Duba Amsa

Tambaya 40 Rahoto

Solve for x in the equation; \(\frac{3}{5}\)(2x - 1) = \(\frac{1}{4}\)(5x - 3)

zero

3 Amsar da ta dace

Bayanin Amsa

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

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Fractions, Decimals And Approximations

Solution Of Linear Equations

Algebraic Expressions

Duba Amsa

Tambaya 41 Rahoto

If a number is chosen at random from the set {x: 4 \(\leq x \leq 15\)}. Find the probability that it is a multiple of 3 or a multiple of 4

\(\frac{1}{12}\)

\(\frac{5}{12}\)

\(\frac{7}{12}\) Amsar da ta dace

\(\frac{11}{12}\)

Bayanin Amsa

Karanta bayanin darasi akan Probability (WAEC)

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Probability

Surds (radicals)

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Tambaya 42 Rahoto

What is the value of 3 in the number 42.7531?

\(\frac{3}{10000}\)

\(\frac{3}{1000}\) Amsar da ta dace

\(\frac{3}{100}\)

\(\frac{1}{10}\)

Bayanin Amsa

The digit 3 in the number 42.7531 is in the thousandth place (the digit after the decimal point and three places to the right of it). Therefore, its value is \(\frac{3}{1000}\).

Karanta bayanin darasi akan Number Bases (WAEC)

Number Bases

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Tambaya 43 Rahoto

Esther was facing S 20° W. She turned 90° in the clock wise direction. What direction is she facing?

N 70^o W Amsar da ta dace

S 70^o W

N 20^o W

S 20^o E

Bayanin Amsa

If Esther was initially facing S 20° W and then turned 90° clockwise, she would end up facing in a new direction. To determine the new direction, we can add 90° to her initial direction. When we add 90° to S 20° W, we rotate the direction clockwise by 90°, which means the new direction will be to the right of the initial direction. To find the new direction, we need to subtract the initial angle from 90°: 90° - 20° = 70° Therefore, Esther is facing in the direction of N 70° W after turning 90° clockwise from her initial direction of S 20° W.

Karanta bayanin darasi akan Introductory Calculus (WAEC)

Karanta bayanin darasi akan Trigonometry (JAMB)

Introductory Calculus

Trigonometry

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Tambaya 44 Rahoto

Which of these angles can be constructed using ruler and a pair of compasses only?

115^o

125^o

135^o Amsar da ta dace

145^o

Bayanin Amsa

Karanta bayanin darasi akan Euclidean Geometry (JAMB)

Euclidean Geometry

Duba Amsa

Tambaya 45 Rahoto

The cross section section of a uniform prism is a right-angled triangle with sides 3cm. 4cm and 5cm. If its length is 10cm. Calculate the total surface area

142cm²

132cm² Amsar da ta dace

122cm²

112cm²

Bayanin Amsa

Karanta bayanin darasi akan Integration (JAMB)

Karanta bayanin darasi akan Sets (JAMB)

Karanta bayanin darasi akan Mensuration (JAMB)

Integration

Sets

Mensuration

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Tambaya 46 Rahoto

The perpendicular bisectors of the sides of an acute-angled triangle are drawn. Which of these statements is correct? They intersect

on one of the vertices

at a midpoint of a side

inside the triangle Amsar da ta dace

outside the triangle

Bayanin Amsa

The perpendicular bisectors of the sides of an acute-angled triangle intersect at a point inside the triangle. This point is called the circumcenter, which is equidistant from the three vertices of the triangle. To see why this is true, consider two sides of the triangle, and let their perpendicular bisectors intersect at a point O. Since O lies on the perpendicular bisector of each of the two sides, it is equidistant from the endpoints of each of those sides. Therefore, O is equidistant from two vertices of the triangle. Similarly, O is equidistant from the third vertex, so it must be the circumcenter of the triangle. Since the triangle is acute-angled, the circumcenter lies inside the triangle.

Karanta bayanin darasi akan Polynomials (JAMB)

Karanta bayanin darasi akan Euclidean Geometry (JAMB)

Karanta bayanin darasi akan Triangles And Polygons (WAEC)

Polynomials

Euclidean Geometry

Triangles And Polygons

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Tambaya 47 Rahoto

In the diagram, PO and OR are radii, |PQ| = |QR| and reflex < PQR is 240^o. Calculate the value x

60^o Amsar da ta dace

55^o

50^o

45^o

Bayanin Amsa

In the given diagram, we have a circle with center O and radii OP and OR. The reflex angle PQR is 240° and |PQ| = |QR|. We need to find the value of x. Since |PQ| = |QR|, we know that triangle PQR is an isosceles triangle. Therefore, the angles opposite to PQ and QR are equal. Let's denote the angle PQR by y. Then we have: 2y + 60° = 360° (sum of angles in a triangle) 2y = 300° y = 150° Therefore, each of the angles opposite to PQ and QR is equal to (180° - 150°)/2 = 15°. Now, consider the triangle OQP. We know that the sum of angles in a triangle is 180°. Therefore: ∠OQP + ∠QOP + ∠OPQ = 180° Since OP and OQ are radii, ∠QOP = ∠OPQ. Let's denote this angle by z. Then we have: z + z + 15° = 180° 2z = 165° z = 82.5° Finally, consider the triangle OXR. We know that the sum of angles in a triangle is 180°. Therefore: ∠OXR + ∠ORX + ∠ROX = 180° Since OR and OX are radii, ∠ORX = ∠ROX. Let's denote this angle by x. Then we have: x + x + 60° = 180° 2x = 120° x = 60° Therefore, the value of x is 60°. Answer: 60°.

Karanta bayanin darasi akan Circles (WAEC)

Karanta bayanin darasi akan Triangles And Polygons (WAEC)

Circles

Triangles And Polygons

Duba Amsa

Tambaya 48 Rahoto

Simplify 1\(\frac{3}{4} - (2 \frac{1}{3} + 4)\)

3\(\frac{5}{12}\)

2\(\frac{7}{12}\)

-4 \(\frac{7}{12}\) Amsar da ta dace

-5 \(\frac{7}{12}\)

Bayanin Amsa

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Solution Of Linear Equations

Algebraic Expressions

Duba Amsa

Tambaya 49 Rahoto

Factorize the expression: am + bn - an - bm

(a - b)(m + n)

(a - b)(m - n) Amsar da ta dace

(a + b)(m - n)

(a + b)(m + n)

Bayanin Amsa

We can begin by grouping the first two terms and the last two terms together: am + bn = a(m) + b(n) = (a+b)n - bn an + bm = a(n) + b(m) = (a+b)m - am Now, we can substitute these expressions back into the original equation: am + bn - an - bm = [(a+b)n - bn] - [(a+b)m - am] We can simplify this expression by combining like terms: am + bn - an - bm = (a+b)n - bn - (a+b)m + am am + bn - an - bm = (a+b)n - (a+b)m + am - bn Finally, we can factor out the common factor of (a+b) from the first two terms and the common factor of (-1) from the last two terms: am + bn - an - bm = (a+b)(n-m) - (b-a)(n-m) Therefore, the answer is (a+b)(n-m) - (b-a)(n-m), which can be further simplified to (a-b)(m-n). Thus, the correct option is (a - b)(m - n).

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Karanta bayanin darasi akan Algebraic Fractions (WAEC)

Algebraic Expressions

Algebraic Fractions

Duba Amsa






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WAEC SSCE - General Mathematics - 2011 (Objective)