WAEC SSCE - General Mathematics - 2011…

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).

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Gụọ akwụkwọ nkuzi na Algebraic Fractions (WAEC)

Algebraic Expressions

Algebraic Fractions

Lee Azịza

Ajụjụ 3 Ripọtì

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 Azịza ziri ezi

0.87m

1.80m

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Coordinate Geometry (WAEC)

Number Bases

Coordinate Geometry

Lee Azịza

Ajụjụ 4 Ripọtì

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

45^o Azịza ziri ezi

60^o

75^o

90^o

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Circles

Triangles And Polygons

Lee Azịza

Ajụjụ 5 Ripọtì

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 Azịza ziri ezi

11^o

12^o

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Angles Of Elevation And Depression (WAEC)

Angles Of Elevation And Depression

Lee Azịza

Ajụjụ 6 Ripọtì

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

26 Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Sets (JAMB)

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

Sets

Ratio, Proportions And Rates

Lee Azịza

Ajụjụ 7 Ripọtì

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

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

\(\frac{1}{2}\) Azịza ziri ezi

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

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Surds (radicals)

Lee Azịza

Ajụjụ 8 Ripọtì

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 Azịza ziri ezi

y = 4.6

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Simple Operations On Algebraic Expressions (WAEC)

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Simple Operations On Algebraic Expressions

Quadratic Equations

Lee Azịza

Ajụjụ 9 Ripọtì

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 Azịza ziri ezi

18cm

21cm

34cm

Akọwa Nkọwa

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.

Lee Azịza

Ajụjụ 10 Ripọtì

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

45.5

46.5

47.5 Azịza ziri ezi

48.5

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Integration (JAMB)

Integration

Lee Azịza

Ajụjụ 11 Ripọtì

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

0.13

7.49 Azịza ziri ezi

8.00

13.00

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Positive And Negative Integers, Rational Numbers (WAEC)

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Positive And Negative Integers, Rational Numbers

Areas

Lee Azịza

Ajụjụ 12 Ripọtì

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

11^o

120^o Azịza ziri ezi

130^o

160^o

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Angles (WAEC)

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Angles

Triangles And Polygons

Lee Azịza

Ajụjụ 13 Ripọtì

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

8 Azịza ziri ezi

Akọwa Nkọwa

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.

Lee Azịza

Ajụjụ 14 Ripọtì

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

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

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

-4 \(\frac{7}{12}\) Azịza ziri ezi

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

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Solution Of Linear Equations

Algebraic Expressions

Lee Azịza

Ajụjụ 15 Ripọtì

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

-5 Azịza ziri ezi

-2

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Quadratic Equations

Lee Azịza

Ajụjụ 16 Ripọtì

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 Azịza ziri ezi

outside the triangle

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Polynomials (JAMB)

Gụọ akwụkwọ nkuzi na Euclidean Geometry (JAMB)

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Polynomials

Euclidean Geometry

Triangles And Polygons

Lee Azịza

Ajụjụ 17 Ripọtì

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 Azịza ziri ezi

18cm

21cm

34cm

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Lengths And Perimeters (WAEC)

Lengths And Perimeters

Lee Azịza

Ajụjụ 18 Ripọtì

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

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

\(\frac{2}{3}\) Azịza ziri ezi

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Indices (WAEC)

Indices

Lee Azịza

Ajụjụ 19 Ripọtì

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

35.5

36.5 Azịza ziri ezi

37.5

38.5

Akọwa Nkọwa

Lee Azịza

Ajụjụ 20 Ripọtì

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

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

\(\sqrt{r}\)

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

\(\sqrt{r^2 - 1}\) Azịza ziri ezi

Akọwa Nkọwa

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) \(\sqrt{r^2 - 1}\).

Gụọ akwụkwọ nkuzi na Sine, Cosine And Tangent Of An Angle (WAEC)

Gụọ akwụkwọ nkuzi na Lengths And Perimeters (WAEC)

Sine, Cosine And Tangent Of An Angle

Lengths And Perimeters

Lee Azịza

Ajụjụ 21 Ripọtì

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 Azịza ziri ezi

\(\bigtriangleup\) PSR

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Gụọ akwụkwọ nkuzi na Angles (WAEC)

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Statistics

Angles

Triangles And Polygons

Lee Azịza

Ajụjụ 22 Ripọtì

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

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

\(\frac{3}{4}\) Azịza ziri ezi

Akọwa Nkọwa

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).

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Logarithms

Lee Azịza

Ajụjụ 23 Ripọtì

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 Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Indices (WAEC)

Indices

Lee Azịza

Ajụjụ 24 Ripọtì

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 Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Gụọ akwụkwọ nkuzi na Sets (JAMB)

Circles

Sets

Lee Azịza

Ajụjụ 25 Ripọtì

What is the value of 3 in the number 42.7531?

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

\(\frac{3}{1000}\) Azịza ziri ezi

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

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

Akọwa Nkọwa

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 \(\frac{3}{1000}\).

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Number Bases

Lee Azịza

Ajụjụ 26 Ripọtì

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

N 70^o W Azịza ziri ezi

S 70^o W

N 20^o W

S 20^o E

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Introductory Calculus (WAEC)

Gụọ akwụkwọ nkuzi na Trigonometry (JAMB)

Introductory Calculus

Trigonometry

Lee Azịza

Ajụjụ 27 Ripọtì

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

-1.55, 4.44 Azịza ziri ezi

1.55, -4.55

-1.55, -4.55

1.55, 4.55

Akọwa Nkọwa

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 28 Ripọtì

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

60^o Azịza ziri ezi

55^o

50^o

45^o

Akọwa Nkọwa

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°.

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Circles

Triangles And Polygons

Lee Azịza

Ajụjụ 29 Ripọtì

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 Azịza ziri ezi

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Quadratic Equations

Lee Azịza

Ajụjụ 30 Ripọtì

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

(n - q)

(q - n)

(n + q) Azịza ziri ezi

(q - m)

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Number Bases

Lee Azịza

Ajụjụ 31 Ripọtì

What is the value of 3 in the number 42.7531?

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

\(\frac{3}{1000}\) Azịza ziri ezi

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

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

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Positive And Negative Integers, Rational Numbers (WAEC)

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Positive And Negative Integers, Rational Numbers

Modular Arithmetic

Lee Azịza

Ajụjụ 32 Ripọtì

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

230m²

231m²

232m²

233m² Azịza ziri ezi

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Lengths And Perimeters (WAEC)

Lengths And Perimeters

Lee Azịza

Ajụjụ 33 Ripọtì

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

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

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

\(\frac{7}{12}\) Azịza ziri ezi

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

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Probability

Surds (radicals)

Lee Azịza

Ajụjụ 34 Ripọtì

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

5y - x

y - x Azịza ziri ezi

x - 5x

x - y

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Gụọ akwụkwọ nkuzi na Algebraic Fractions (WAEC)

Algebraic Expressions

Algebraic Fractions

Lee Azịza

Ajụjụ 35 Ripọtì

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 Azịza ziri ezi

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Volumes (WAEC)

Gụọ akwụkwọ nkuzi na Inequalities (JAMB)

Volumes

Inequalities

Lee Azịza

Ajụjụ 36 Ripọtì

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 Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Financial Arithmetic (WAEC)

Number Bases

Financial Arithmetic

Lee Azịza

Ajụjụ 37 Ripọtì

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

108^o

112^o

128^o Azịza ziri ezi

142^o

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Coordinate Geometry (WAEC)

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

Gụọ akwụkwọ nkuzi na Angles (WAEC)

Coordinate Geometry

Angles And Intercepts On Parallel Lines

Angles

Lee Azịza

Ajụjụ 38 Ripọtì

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

150^o Azịza ziri ezi

140^o

130^o

120^o

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Angles (WAEC)

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Angles

Triangles And Polygons

Lee Azịza

Ajụjụ 39 Ripọtì

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

115^o

125^o

135^o Azịza ziri ezi

145^o

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Euclidean Geometry (JAMB)

Euclidean Geometry

Lee Azịza

Ajụjụ 40 Ripọtì

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 Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Indices (WAEC)

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Indices

Triangles And Polygons

Lee Azịza

Ajụjụ 41 Ripọtì

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 Azịza ziri ezi

122cm²

112cm²

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Integration (JAMB)

Gụọ akwụkwọ nkuzi na Sets (JAMB)

Gụọ akwụkwọ nkuzi na Mensuration (JAMB)

Integration

Sets

Mensuration

Lee Azịza

Ajụjụ 42 Ripọtì

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

44.5 Azịza ziri ezi

42.5

41.5

40.5

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Representation Of Data

Lee Azịza

Ajụjụ 43 Ripọtì

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

15 Azịza ziri ezi

225

Akọwa Nkọwa

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.

Lee Azịza

Ajụjụ 44 Ripọtì

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

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

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

\(\frac{20}{7}\)% Azịza ziri ezi

Akọwa Nkọwa

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}\%.

Gụọ akwụkwọ nkuzi na Percentages (WAEC)

Percentages

Lee Azịza

Ajụjụ 45 Ripọtì

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

x = 1, y = -2 Azịza ziri ezi

x = 1, y = 3

x = -2, y = -1

x = 4, y = -3

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Lee Azịza

Ajụjụ 46 Ripọtì

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

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

\(\frac{11m}{10n}\) Azịza ziri ezi

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

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

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Algebraic Fractions (WAEC)

Algebraic Fractions

Lee Azịza

Ajụjụ 47 Ripọtì

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

52 Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Gụọ akwụkwọ nkuzi na Mensuration (JAMB)

Representation Of Data

Mensuration

Lee Azịza

Ajụjụ 48 Ripọtì

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

0.46m Azịza ziri ezi

0.60m

0.87m

1.80m

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Representation Of Data

Modular Arithmetic

Lee Azịza

Ajụjụ 49 Ripọtì

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 Azịza ziri ezi

i and iv

ii and iii

ii and ic

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Polynomials (JAMB)

Gụọ akwụkwọ nkuzi na Measures Of Dispersion (JAMB)

Polynomials

Measures Of Dispersion

Lee Azịza






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WAEC SSCE - General Mathematics - 2011 (Objective)