WAEC SSCE - General Mathematics - 2009…

Ibeere 1 Ìròyìn

Simplify (x - 3y)² - (x + 3y)²

2(x + 3y)

(2x - 3y)

-12xy Ìdáhùn tó tọ́

6xy

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Ka ìwé ẹ̀kọ́ nípa Quadratic Equations (WAEC)

Algebraic Expressions

Quadratic Equations

Wo Idahun

Ibeere 2 Ìròyìn

What is the median of the following scores: 22 35 41 63 74 82

52 Ìdáhùn tó tọ́

Awọn alaye Idahun

To find the median of a set of numbers, we first need to arrange them in ascending or descending order. In this case, arranging the numbers in ascending order gives us: 22, 35, 41, 63, 74, 82 The median is the middle number in the set when the numbers are arranged in order. Since we have an even number of numbers in this set, there are two middle numbers: 41 and 63. To find the median, we take the average of these two numbers: Median = (41 + 63) / 2 = 104 / 2 = 52 Therefore, the median of the scores is 52.

Ka ìwé ẹ̀kọ́ nípa Euclidean Geometry (JAMB)

Euclidean Geometry

Wo Idahun

Ibeere 3 Ìròyìn

The diameter of a bicycle wheel is 42cm. If the wheel makes 16 complete revolution, what will be the total distance covered by the wheel? [Take $\pi \frac{22}{7}$

672cm

1056cm

2112cm Ìdáhùn tó tọ́

4224cm

Awọn alaye Idahun

The distance covered by the wheel is equal to the circumference of the wheel multiplied by the number of revolutions made. The circumference of the wheel can be calculated using the formula: circumference = diameter x pi circumference = 42cm x 22/7 circumference = 132cm Therefore, the distance covered by the wheel in 16 complete revolutions is: distance = circumference x number of revolutions distance = 132cm x 16 distance = 2112cm So, the answer is (c) 2112cm.

Ka ìwé ẹ̀kọ́ nípa Volumes (WAEC)

Ka ìwé ẹ̀kọ́ nípa Areas (WAEC)

Volumes

Areas

Wo Idahun

Ibeere 4 Ìròyìn

Factorize x² - 2x - 3xy completely

(x - 2)(3x - y)

(x - 3y)(x - 2) Ìdáhùn tó tọ́

(x - 3y)(x - 3y)

(3x + y)(x - 2)

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Algebraic Expressions

Wo Idahun

Ibeere 5 Ìròyìn

PQRS is a trapezuim. QR//PS, /PQ/ = 5cm, /OR/ = 6cm, /PS/ = 10cm and angle QPS = 42^o. Calculate the perpendicular distance between the parallel sides

3.35cm Ìdáhùn tó tọ́

3.73cm

4.50cm

4.62cm

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Algebraic Expressions

Wo Idahun

Ibeere 6 Ìròyìn

In a particular year, the exchange rate of naira (N) varies directly with the Dollars{$}. If N1122 is equivalent to $8, find the Naira equivalent of $36

N8976

N5049 Ìdáhùn tó tọ́

N140.25

N31.17

Awọn alaye Idahun

The exchange rate of naira (N) varies directly with the Dollars($). This means that the exchange rate is constant, and we can set up a proportion to solve the problem. If N1122 is equivalent to $8, we can write: N1122/$8 = x/$36 where x is the Naira equivalent of $36. To solve for x, we can cross-multiply and simplify: N1122 x 36 = $8 x x 40392 = 8x x = 40392/8 = 5049 Therefore, the Naira equivalent of $36 is N5049. So, the correct option is (B) N5049.

Ka ìwé ẹ̀kọ́ nípa Variation (WAEC)

Ka ìwé ẹ̀kọ́ nípa Variation (JAMB)

Variation

Wo Idahun

Ibeere 7 Ìròyìn

If 85% of x is N3230, what is the value of x?

N2745.50

N3714.50

N3800.00 Ìdáhùn tó tọ́

N4845.00

Awọn alaye Idahun

We can solve this problem using a proportion. If 85% of x is N3230, that means we can write: 0.85x = N3230 To solve for x, we need to isolate x on one side of the equation. We can do this by dividing both sides by 0.85: x = N3230 ÷ 0.85 Using a calculator, we get: x ≈ N3800.00 Therefore, the value of x is N3800.00.

Ka ìwé ẹ̀kọ́ nípa Percentages (WAEC)

Ka ìwé ẹ̀kọ́ nípa Positive And Negative Integers, Rational Numbers (WAEC)

Percentages

Positive And Negative Integers, Rational Numbers

Wo Idahun

Ibeere 8 Ìròyìn

From the diagram which of the following statements are correct? i. XQ is a radius of a circle centre Q. ii. /XQ/ = /QY/. iii. /QX/ = /XY/

i only

inand ii only Ìdáhùn tó tọ́

ii and iii only

i, ii and iii

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Ka ìwé ẹ̀kọ́ nípa Statistics (WAEC)

Circles

Statistics

Wo Idahun

Ibeere 9 Ìròyìn

If $\frac{3}{2x} - \frac{2}{3x} = 4$, solve for x

$\frac{4}{5}$

$\frac{4}{13}$

$\frac{5}{24}$ Ìdáhùn tó tọ́

$\frac{13}{24}$

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Algebraic Fractions (WAEC)

Algebraic Fractions

Wo Idahun

Ibeere 10 Ìròyìn

An arc of a circle subtends an angle of 60^o at the centre. If the radius of the circle is 3cm, find , in terms of $\pi$, the length of the arc

$\pi$cm Ìdáhùn tó tọ́

2$\pi$cm

3$\pi$cm

6$\pi$cm

Awọn alaye Idahun

When an angle in degrees is subtended at the center of a circle, the length of the arc it cuts out is given by: $$\text{Length of arc} = \frac{\text{angle}}{360^\circ} \times 2\pi r$$ where r is the radius of the circle. In this case, the angle is 60° and the radius is 3 cm. Substituting these values into the formula above, we get: $$\text{Length of arc} = \frac{60}{360^\circ} \times 2\pi (3\text{ cm}) = \frac{1}{6} \times 6\pi = \pi \text{ cm}$$ Therefore, the length of the arc, in terms of π, is π cm. Hence, the correct option is: $\pi$cm.

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Circles

Wo Idahun

Ibeere 11 Ìròyìn

Using the diagram, find the bearing of X from Y

300^o

240^o Ìdáhùn tó tọ́

120^o

60^o

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Bearings (WAEC)

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Bearings

Circles

Wo Idahun

Ibeere 12 Ìròyìn

The angles of triangle are (x + 10)^o, (2x - 40)^o and (3x - 90)^o. Which of the following accurately describes the triangle?

it is a scalene triangle

it is right angled isosceles triangle

t is an equilateral triangle Ìdáhùn tó tọ́

It is an isosceles triangle but not right angled

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Triangles And Polygons (WAEC)

Triangles And Polygons

Wo Idahun

Ibeere 13 Ìròyìn

If a positive integer, list the values of x which satisfy the equation 3x - 4 < 6 and x - 1 > 0

{1, 2, 3}

{2, 3} Ìdáhùn tó tọ́

{2, 3, 4}

{2, 3, 4, 5}

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Linear Inequalities (WAEC)

Ka ìwé ẹ̀kọ́ nípa Surds (radicals) (WAEC)

Linear Inequalities

Surds (radicals)

Wo Idahun

Ibeere 14 Ìròyìn

Simplify: (3$\frac{1}{2} + 4\frac{1}{3}) \div (\frac{5}{6} - \frac{2}{3}$)

1$\frac{1}{4}$

8$\frac{1}{2}$

47 Ìdáhùn tó tọ́

Awọn alaye Idahun

We will start by simplifying the expression inside the parenthesis first. 3$\frac{1}{2}$ + 4$\frac{1}{3}$ = (7/2) + (13/3) To add these two fractions, we need a common denominator. Multiplying the denominators together gives us 6, so: (7/2) + (13/3) = (21/6) + (26/6) = 47/6 Now, let's simplify the expression in the denominator: $\frac{5}{6} - \frac{2}{3}$ = $\frac{5}{6} - \frac{4}{6}$ = $\frac{1}{6}$ Finally, we can substitute these values into the original expression: (3$\frac{1}{2}$ + 4$\frac{1}{3}$) ÷ ($\frac{5}{6}$ - $\frac{2}{3}$) = (47/6) ÷ (1/6) When dividing fractions, we can multiply the first fraction by the reciprocal of the second: (47/6) ÷ (1/6) = (47/6) x (6/1) = 47 Therefore, the answer is 47.

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Ka ìwé ẹ̀kọ́ nípa Modular Arithmetic (WAEC)

Number Bases

Modular Arithmetic

Wo Idahun

Ibeere 15 Ìròyìn

Two sets are disjoint if

they are both empty

their union is an empty set

their intersection is an empty set Ìdáhùn tó tọ́

one of them is a subset of the other

Awọn alaye Idahun

Two sets are said to be disjoint if their intersection is an empty set. In other words, if there are no common elements between the sets, they are said to be disjoint. For example, the sets {1,2,3} and {4,5,6} are disjoint since their intersection is an empty set {} or ∅. On the other hand, the sets {1,2,3} and {2,3,4} are not disjoint since they have elements in common, namely 2 and 3.

Ka ìwé ẹ̀kọ́ nípa Sets (WAEC)

Ka ìwé ẹ̀kọ́ nípa Sets (JAMB)

Sets

Wo Idahun

Ibeere 16 Ìròyìn

The pie chart shows the distribution of 4320 students who graduated from four departments in a university. If a student is picked at random from the four departments, what id the probability that he is not from the education department?

$\frac{3}{10}$

$\frac{2}{9}$

$\frac{5}{18}$

$\frac{7}{10}$ Ìdáhùn tó tọ́

Awọn alaye Idahun

To find the probability that a student picked at random is not from the education department, we need to find the total number of students who are not from the education department and divide it by the total number of students in all departments. From the pie chart, we can see that the education department has 30% of the total students. Therefore, the remaining three departments have a total of 70% of the total students. To find the probability of picking a student who is not from the education department, we divide the percentage of students who are not in the education department by 100%: Probability = $\frac{70}{100}$ = $\frac{7}{10}$ Therefore, the probability that a student picked at random is not from the education department is $\frac{7}{10}$. The correct option is: $\frac{7}{10}$.

Ka ìwé ẹ̀kọ́ nípa Variation (JAMB)

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Ka ìwé ẹ̀kọ́ nípa Mensuration (JAMB)

Ka ìwé ẹ̀kọ́ nípa Surds (radicals) (WAEC)

Variation

Probability

Mensuration

Surds (radicals)

Wo Idahun

Ibeere 17 Ìròyìn

Find the value of x in the diagram

50^o

60^o

70^o Ìdáhùn tó tọ́

80^o

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Ka ìwé ẹ̀kọ́ nípa Angles (WAEC)

Ka ìwé ẹ̀kọ́ nípa Triangles And Polygons (WAEC)

Circles

Angles

Triangles And Polygons

Wo Idahun

Ibeere 18 Ìròyìn

Find the quadratic equation whose roots are -$\frac{1}{2}$ and 3

2x² - 2x + 3 = 0

2x² - 2x - 3 = 0

2x² - 5x - 3 = 0 Ìdáhùn tó tọ́

3x² - 5x - 3 = 0

Awọn alaye Idahun

To find the quadratic equation given its roots, we use the fact that for a quadratic equation of the form ax² + bx + c = 0, the roots are given by the formula: x = (-b ± √(b² - 4ac)) / 2a If the roots are given as α and β, then the quadratic equation can be written as: (x - α)(x - β) = 0 Expanding the above equation gives: x² - (α + β)x + αβ = 0 Therefore, to find the quadratic equation whose roots are -$\frac{1}{2}$ and 3, we substitute α = -$\frac{1}{2}$ and β = 3 into the equation: x² - (α + β)x + αβ = 0 x² - (-$\frac{1}{2}$ + 3)x + (-$\frac{1}{2}$ × 3) = 0 Simplifying the above equation, we get: 2x² - 5x - 3 = 0 Therefore, the quadratic equation whose roots are -$\frac{1}{2}$ and 3 is 2x² - 5x - 3 = 0. The correct option is (C) 2x² - 5x - 3 = 0.

Ka ìwé ẹ̀kọ́ nípa Quadratic Equations (WAEC)

Quadratic Equations

Wo Idahun

Ibeere 19 Ìròyìn

Find the value to which N3000.00 will amount in 5 years at 6% per annum simple interest

N3,900.00 Ìdáhùn tó tọ́

N3,750.00

N3,600.00

N3,300

Awọn alaye Idahun

Simple interest is calculated as the product of the principal, the rate of interest, and the time duration. From the question, we have a principal of N3000.00, an interest rate of 6%, and a duration of 5 years. Using the formula for simple interest, we can find the interest accrued over the 5 years as: Interest = (P * R * T) / 100 = (3000 * 6 * 5) / 100 = N900.00 The total value to which N3000.00 will amount to after 5 years is the sum of the principal and the interest, which is: Total = Principal + Interest = 3000 + 900 = N3900.00 Therefore, N3000.00 will amount to N3900.00 after 5 years at 6% per annum simple interest. So the correct answer is option A, N3,900.00.

Ka ìwé ẹ̀kọ́ nípa Modular Arithmetic (WAEC)

Ka ìwé ẹ̀kọ́ nípa Ratio, Proportions And Rates (WAEC)

Modular Arithmetic

Ratio, Proportions And Rates

Wo Idahun

Ibeere 20 Ìròyìn

The interior angles of a pentagon are (2^x + 5)^o, (x + 20)^o, x^o, (3x - 20)^o and (x + 15)^o. Find the value of x

80^o

70^o

65^o Ìdáhùn tó tọ́

40^o

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Solution Of Linear Equations (WAEC)

Ka ìwé ẹ̀kọ́ nípa Indices (WAEC)

Solution Of Linear Equations

Indices

Wo Idahun

Ibeere 21 Ìròyìn

Given that sin 60^o = $\frac{\sqrt{3}}{2}$ and cos 60^o = $\frac{1}{2}$, evaluate $\frac{1 - sin 60^o}{1 + cos 60^o}$

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

$\frac{1 - \sqrt{3}}{3}$

$\frac{1 + \sqrt{3}}{3}$

$\frac{2 - \sqrt{3}}{3}$ Ìdáhùn tó tọ́

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Sine, Cosine And Tangent Of An Angle (WAEC)

Sine, Cosine And Tangent Of An Angle

Wo Idahun

Ibeere 22 Ìròyìn

solve $\frac{2x + 1}{6} - \frac{3x - 1}{4}$ = 0

1 Ìdáhùn tó tọ́

$\frac{1}{5}$

-$\frac{1}{5}$

-1

Awọn alaye Idahun

To solve the equation $\frac{2x + 1}{6} - \frac{3x - 1}{4} = 0$, we need to simplify the left-hand side and solve for x. First, we need to find a common denominator for the two fractions. The smallest common multiple of 6 and 4 is 12, so we can rewrite the equation as: \[\frac{2x+1}{6}\cdot \frac{2}{2} - \frac{3x-1}{4}\cdot \frac{3}{3} = 0\] Simplifying, we get: \[\frac{4x+2}{12} - \frac{9x-3}{12} = 0\] Combining the fractions, we get: \[\frac{4x+2-(9x-3)}{12} = 0\] Simplifying further, we get: \[\frac{-5x+5}{12} = 0\] Multiplying both sides by 12, we get: \[-5x+5=0\] Adding 5 to both sides, we get: \[-5x=-5\] Dividing both sides by -5, we get: \[x=1\] Therefore, the solution to the equation is x = 1.

Ka ìwé ẹ̀kọ́ nípa Solution Of Linear Equations (WAEC)

Ka ìwé ẹ̀kọ́ nípa Algebraic Fractions (WAEC)

Ka ìwé ẹ̀kọ́ nípa Simple Operations On Algebraic Expressions (WAEC)

Solution Of Linear Equations

Algebraic Fractions

Simple Operations On Algebraic Expressions

Wo Idahun

Ibeere 23 Ìròyìn

in the diagram, angle 20^o is subtended at the centre of the circle, find the value of x

30^o

50^o Ìdáhùn tó tọ́

80^o

100^o

Awọn alaye Idahun

In the given diagram, we have a circle with center O and angle 20^o subtended at the center. We need to find the value of x. Firstly, we know that the angle subtended at the center of a circle is twice the angle subtended at the circumference by the same arc. Therefore, angle AOB = 2 × 20^o = 40^o Also, we know that angle in a semicircle is a right angle. So, angle AOC = 90^o. Using the fact that the angles in a triangle add up to 180^o, we can find angle BOC as follows: angle BOC = 180^o - angle AOB - angle AOC = 180^o - 40^o - 90^o = 50^o Since angle BOC is an angle at the circumference that subtends the arc BC, which is equal to x degrees, we have: x = angle BOC = 50^o Therefore, the value of x is 50^o. Answer is correct.

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Circles

Wo Idahun

Ibeere 24 Ìròyìn

Simplify; $\frac{1}{2}\sqrt{32} - \sqrt{18} \sqrt{2}$

zero Ìdáhùn tó tọ́

$\sqrt{2}$

2$\sqrt{2}$

4\(\sqrt{2}\0

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Positive And Negative Integers, Rational Numbers (WAEC)

Ka ìwé ẹ̀kọ́ nípa Surds (radicals) (WAEC)

Positive And Negative Integers, Rational Numbers

Surds (radicals)

Wo Idahun

Ibeere 25 Ìròyìn

The capacity of a water tank is 1,800 litres. If the tank is in form of a cuboid with base 600cm by 150 cm. Find the height of the tank

2cm

20cm Ìdáhùn tó tọ́

200cm

2000cm

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Euclidean Geometry (JAMB)

Ka ìwé ẹ̀kọ́ nípa Areas (WAEC)

Euclidean Geometry

Areas

Wo Idahun

Ibeere 26 Ìròyìn

Express 2.7864 x 10^-3 to 2 significant figures

0.28

0.27

0.0028 Ìdáhùn tó tọ́

0.0027

Awọn alaye Idahun

To express 2.7864 x 10^-3 to 2 significant figures, we need to round off the number to the second significant digit from the left. Since the first significant digit is 2, and the second significant digit is 7, we will look at the third digit, which is 8. Since 8 is greater than or equal to 5, we round up the second significant digit (7) by 1. Therefore, the number becomes 0.0028. So, the answer is 0.0028.

Ka ìwé ẹ̀kọ́ nípa Positive And Negative Integers, Rational Numbers (WAEC)

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Positive And Negative Integers, Rational Numbers

Number Bases

Wo Idahun

Ibeere 27 Ìròyìn

Each of the interior angles of a regular polygon is 140^o. Calculate the sum of all the interior angles of the polygon

1080^o

1260^o Ìdáhùn tó tọ́

1800^o

2160^o

Awọn alaye Idahun

In a regular polygon with n sides, each interior angle measures: \begin{align*} 180^\circ - \frac{360^\circ}{n} \end{align*} Since each interior angle of this polygon is 140^o, we can equate the above formula to 140^o: \begin{align*} 180^\circ - \frac{360^\circ}{n} &= 140^\circ\\ \frac{360^\circ}{n} &= 40^\circ\\ n &= \frac{360^\circ}{40^\circ}\\ n &= 9 \end{align*} Therefore, the given polygon is a nonagon (a polygon with nine sides). The sum of the interior angles of a polygon with n sides can be calculated using the formula: \begin{align*} S &= (n-2)180^\circ \end{align*} Substituting n=9 into this formula, we get: \begin{align*} S &= (9-2)180^\circ\\ &= 7\cdot180^\circ\\ &= 1260^\circ \end{align*} Therefore, the sum of all the interior angles of the given polygon is 1260^o. Hence, the correct option is: - 1260^o

Ka ìwé ẹ̀kọ́ nípa Indices (WAEC)

Indices

Wo Idahun

Ibeere 28 Ìròyìn

The following is the graph of a quadratic friction, find the value of x when y = 0

1, 2

1, 4 Ìdáhùn tó tọ́

2, 3

1, 6

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Quadratic Equations (WAEC)

Quadratic Equations

Wo Idahun

Ibeere 29 Ìròyìn

In the diagram, \SQ\ = 4cm, \PT\ = 7cm. /TR/ = 5cm and ST//OR. If /SP/ = xcm, find the value of x

5.6 Ìdáhùn tó tọ́

6.5

6.6

6.8

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Polynomials (JAMB)

Ka ìwé ẹ̀kọ́ nípa Sets (JAMB)

Polynomials

Sets

Wo Idahun

Ibeere 30 Ìròyìn

In an examination, Kofi scored x% in Physics, 50% in Chemistry and 70% in Biology. If his mean score for the three subjects was 55%, find x

45 Ìdáhùn tó tọ́

Awọn alaye Idahun

Kofi's mean score for the three subjects was 55%, so the total percentage score for the three subjects is 3 x 55 = 165%. Let's assume that Kofi scored x% in Physics. Then, the total percentage score for Physics, Chemistry and Biology would be: x + 50 + 70 = 165 Simplifying the equation, we get: x = 165 - 50 - 70 x = 45 Therefore, Kofi scored 45% in Physics. So the correct option is (B) 45.

Ka ìwé ẹ̀kọ́ nípa Change Of Subject Of A Formula/Relation (WAEC)

Ka ìwé ẹ̀kọ́ nípa Modular Arithmetic (WAEC)

Change Of Subject Of A Formula/Relation

Modular Arithmetic

Wo Idahun

Ibeere 31 Ìròyìn

Express the square root of 0.000144 in the standard form

1.2 x 10^-4

1.2 x 10^-3

1.2 x 10^-2 Ìdáhùn tó tọ́

1.2 x 10^-1

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Ka ìwé ẹ̀kọ́ nípa Number Bases (JAMB)

Number Bases

Wo Idahun

Ibeere 32 Ìròyìn

If M and N are two fixed points in a plane. Find the locus L = [P : PM = PN]

a line equal to MN

line parallel to MN

Perpendicular bisector of MN Ìdáhùn tó tọ́

A circle centre P, radius MN

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Loci (JAMB)

Ka ìwé ẹ̀kọ́ nípa Loci (WAEC)

Ka ìwé ẹ̀kọ́ nípa Coordinate Geometry (JAMB)

Loci

Coordinate Geometry

Wo Idahun

Ibeere 33 Ìròyìn

A box contains black, white and red identical balls. The probability of picking a black ball at random from the box is $\frac{3}{10}$ and the probability of picking a white ball at random is $\frac{2}{5}$. If there are 30 balls in the box, how many of them are red?

9 Ìdáhùn tó tọ́

Awọn alaye Idahun

The probability of picking a black ball at random from the box is $\frac{3}{10}$ and the probability of picking a white ball at random is $\frac{2}{5}$. Let's assume that there are x red balls in the box. Since there are a total of 30 balls, we can write: number of black balls + number of white balls + number of red balls = total number of balls $\frac{3}{10}(30) + \frac{2}{5}(30) + x = 30$ Simplifying the above equation gives: 9 + 12 + x = 30 x = 30 - 9 - 12 x = 9 Therefore, there are 9 red balls in the box.

Wo Idahun

Ibeere 34 Ìròyìn

If /XY/ = 50m, how far cast of X is Y?

25.5m

40.6m

40.8m

43.3m Ìdáhùn tó tọ́

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Ka ìwé ẹ̀kọ́ nípa Triangles And Polygons (WAEC)

Circles

Triangles And Polygons

Wo Idahun

Ibeere 35 Ìròyìn

Find the value of x in the diagram

50^o

30^o

22^o

17^o Ìdáhùn tó tọ́

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Ka ìwé ẹ̀kọ́ nípa Number Bases (JAMB)

Circles

Number Bases

Wo Idahun

Ibeere 36 Ìròyìn

If 9^2x = $\frac{1}{3}$(27^x), find x

-1 Ìdáhùn tó tọ́

-2

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Solution Of Linear Equations (WAEC)

Ka ìwé ẹ̀kọ́ nípa Indices (WAEC)

Solution Of Linear Equations

Indices

Wo Idahun

Ibeere 37 Ìròyìn

Arrange the following numbers in descending orders of magnitude: 22_three, 34_five, 21_six

21_six, 22_three, 34_five,

21_six, 34_five, 22_three

22_three, 34_five, 21_six

34_five, 21_six, 22_three Ìdáhùn tó tọ́

Awọn alaye Idahun

To compare these numbers, we need to convert all of them to the same base. Let's convert them all to base 10 for simplicity: - 22_three = 2\*3¹ + 2\*3⁰ = 6 + 2 = 8 - 34_five = 3\*5¹ + 4\*5⁰ = 15 + 4 = 19 - 21_six = 2\*6¹ + 1\*6⁰ = 12 + 1 = 13 So, in base 10, the numbers are 8, 19, and 13. To arrange them in descending order of magnitude, we simply sort them from largest to smallest: - 19, 13, 8 Therefore, the correct answer is: 34_five, 21_six, 22_three

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Number Bases

Wo Idahun

Ibeere 38 Ìròyìn

The following is the graph of a quadratic friction, find the co-ordinates of point P

(0, 4) Ìdáhùn tó tọ́

(4, 0)

(0, -4)

(-4, 0)

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Quadratic Equations (WAEC)

Ka ìwé ẹ̀kọ́ nípa Graphs Of Linear And Quadratic Functions (WAEC)

Quadratic Equations

Graphs Of Linear And Quadratic Functions

Wo Idahun

Ibeere 39 Ìròyìn

A bucket holds 10 litres of water. How many buckets of water will fill a reservoir of size 8m x 7m x 5m.(1 litre = 1000cm³)`

280

2800

28000 Ìdáhùn tó tọ́

Awọn alaye Idahun

The volume of the reservoir can be found by multiplying its dimensions: 8m x 7m x 5m = 280 m^3 Since 1 liter is equal to 1000 cubic centimeters (cm^3), 1 cubic meter is equal to 1,000,000 cubic centimeters. Therefore, the reservoir has a volume of: 280 m^3 x 1,000,000 cm^3/m^3 = 280,000,000 cm^3 Each bucket can hold 10 liters of water or 10,000 cubic centimeters (cm^3) of water since 1 liter is equal to 1000 cm^3. Therefore, the number of buckets needed to fill the reservoir is: 280,000,000 cm^3 ÷ 10,000 cm^3/bucket = 28,000 buckets Therefore, the answer is 28,000.

Ka ìwé ẹ̀kọ́ nípa Logical Reasoning (WAEC)

Ka ìwé ẹ̀kọ́ nípa Representation Of Data (JAMB)

Logical Reasoning

Representation Of Data

Wo Idahun

Ibeere 40 Ìròyìn

If (2x + 3)³ = 125, find the value of x

1 Ìdáhùn tó tọ́

Awọn alaye Idahun

We are given that (2x + 3)³ = 125. We can rewrite 125 as 5³. Therefore, we have: (2x + 3)³ = 5³ Taking the cube root of both sides, we get: 2x + 3 = 5 Solving for x, we get: 2x = 2 x = 1 Therefore, the value of x is 1. Answer: (a) 1.

Ka ìwé ẹ̀kọ́ nípa Indices (WAEC)

Indices

Wo Idahun

Ibeere 41 Ìròyìn

PQRS is a trapezium. QR//PS, /PQ/ = 5cm, /OR/ = 6cm, /PS/ = 10cm and angle QPS = 42^o. Calculate, correct to the nearest cm², the area of the trapezium (h = 3.35cm² )

27cm² Ìdáhùn tó tọ́

30cm²

36cm²

37cm²

Awọn alaye Idahun

Wo Idahun

Ibeere 42 Ìròyìn

In the diagram, /TP/ = 12cm and it is 6cm from O, the centre of the circle, Calculate < TOP

120^o

90^o Ìdáhùn tó tọ́

60^o

45^o

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Circles

Wo Idahun

Ibeere 43 Ìròyìn

In the diagram, IG is parallel to JE, JEF = 120^o and FHG = 130^o, fins the angle marked t

40^o

70^o Ìdáhùn tó tọ́

80^o

100^o

Awọn alaye Idahun

Wo Idahun

Ibeere 44 Ìròyìn

The pie chart shows the distribution of 4320 students who graduated from four departments in a university. How many students graduated from the science department?

840

864 Ìdáhùn tó tọ́

920

960

Awọn alaye Idahun

To find the number of students who graduated from the science department, we need to look at the pie chart and determine the percentage of students that belong to that department. From the pie chart, we can see that the science department occupies 20% of the entire circle. To find the number of students in the science department, we need to calculate 20% of 4320, which can be done by multiplying 4320 by 0.20. 20% of 4320 = 0.20 x 4320 = 864 Therefore, the number of students who graduated from the science department is 864. So, the correct answer is option B: 864.

Ka ìwé ẹ̀kọ́ nípa Mensuration (JAMB)

Mensuration

Wo Idahun

Ibeere 45 Ìròyìn

If log 2 = x, log 3 = y and log 7 = z, find, in terms of x, y and z, the value of log ($\frac{28}{3}$)

2x + y - z

2x + z - y Ìdáhùn tó tọ́

x + y - 2x

x + x - y

Awọn alaye Idahun

We can use logarithmic rules to solve this question. Let's first express log($\frac{28}{3}$) in terms of x, y, and z. log($\frac{28}{3}$) = log(28) - log(3) We know that: log(28) = log(4 × 7) = log(4) + log(7) = 2x + z log(3) = y Therefore: log($\frac{28}{3}$) = 2x + z - y So the correct answer is 2x + z - y.

Ka ìwé ẹ̀kọ́ nípa Logarithms (WAEC)

Logarithms

Wo Idahun

Ibeere 46 Ìròyìn

If c and k are the roots of 6 - x - x² = 0, find c + k

-1 Ìdáhùn tó tọ́

-3

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Modular Arithmetic (WAEC)

Ka ìwé ẹ̀kọ́ nípa Quadratic Equations (WAEC)

Modular Arithmetic

Quadratic Equations

Wo Idahun






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WAEC SSCE - General Mathematics - 2009 (Objective)