WAEC SSCE - General Mathematics - 2006…

Ajụjụ 1 Ripọtì

In the diagram |XY| = 12cm, |XZ| = 9cm, |ZN| = 3cm and ZY||NM, calculate |MY|

3cm

4cm Azịza ziri ezi

5cm

6cm

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Triangles And Polygons

Lee Azịza

Ajụjụ 2 Ripọtì

The wheel of a tractor has a diameter 1.4m. What distance does it cover in 100 complete revolutions? [Take $\pi = \frac{22}{7}$]

140m

220m

280m

440m Azịza ziri ezi

Akọwa Nkọwa

The circumference of the wheel is the distance it covers in one revolution. Circumference of the wheel = π × diameter Substituting the given values, Circumference of the wheel = π × 1.4m = 4.4m (rounded to one decimal place) Therefore, the distance covered in 100 revolutions is: Distance covered = Circumference of the wheel × Number of revolutions Substituting the given values, Distance covered = 4.4m × 100 = 440m Therefore, the wheel covers a distance of 440m in 100 complete revolutions. Hence, the correct option is (D) 440m.

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Gụọ akwụkwọ nkuzi na Coordinate Geometry (WAEC)

Gụọ akwụkwọ nkuzi na Number Bases (JAMB)

Representation Of Data

Coordinate Geometry

Number Bases

Lee Azịza

Ajụjụ 3 Ripọtì

In the diagram, P, Q and R are three points in a plane such that the bearing of R from Q is 110^o and the bearing of Q from P is 050^o. Find angle PQR.

60^o

70^o

120^o Azịza ziri ezi

160^o

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Introductory Calculus (WAEC)

Gụọ akwụkwọ nkuzi na Bearings (WAEC)

Gụọ akwụkwọ nkuzi na Matrices And Determinants (JAMB)

Introductory Calculus

Bearings

Matrices And Determinants

Lee Azịza

Ajụjụ 4 Ripọtì

Simplify: 11011$_2$ - 1101$_2$

101000$_2$

1100$_2$

1110$_2$ Azịza ziri ezi

1011$_2$

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Number Bases

Lee Azịza

Ajụjụ 5 Ripọtì

If the simple interest on a sum of money invested at 3% per annum for 2$\frac{1}{2}$ years is N123, find the principal.

N676.50

N820

N1,640 Azịza ziri ezi

N4,920

Akọwa Nkọwa

The formula for simple interest is: I = PRT Where: I = Interest P = Principal R = Rate of Interest T = Time From the question: R = 3% per annum T = 2.5 years I = N123 We can substitute the values given into the formula and solve for P: 123 = P x 0.03 x 2.5 123 = 0.075P P = 123 ÷ 0.075 P = N1,640 Therefore, the principal is N1,640. Answer option (C).

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

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

Modular Arithmetic

Ratio, Proportions And Rates

Lee Azịza

Ajụjụ 6 Ripọtì

In the diagram, PR is a diameter of the circle centre O. RS is a tangent at R and QPR = 58^o. Find < QRS

112^o

116^o

122^o Azịza ziri ezi

148^o

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Circles

Lee Azịza

Ajụjụ 7 Ripọtì

Which of these equations best describes the points of intersection of the curve and the line?

$x^2 + 5x - 6 = 0$

x² - 5x - 6 = 0

x² + x - 6 = 0 Azịza ziri ezi

x² - x - 6 = 0

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Quadratic Equations

Lee Azịza

Ajụjụ 8 Ripọtì

A seller allows 20% discount for cash payment on the marked price of his goods. What is the ratio of the cash payment to the marked price

1:4

1:5

3:4

4:5 Azịza ziri ezi

Akọwa Nkọwa

If the seller allows a 20% discount for cash payment, then the customer will only have to pay 80% of the marked price. Let MP be the marked price and CP be the cash payment. CP = 80% of MP CP/MP = 80/100 CP/MP = 4/5 Therefore, the ratio of the cash payment to the marked price is 4:5. So the correct answer is option (D) 4:5.

Gụọ akwụkwọ nkuzi na Percentages (WAEC)

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

Percentages

Ratio, Proportions And Rates

Lee Azịza

Ajụjụ 9 Ripọtì

A baker used 40% of a 50kg bag of flour. If $\frac{1}{8}$ of the amount used was for the cake, how many kilogram of flour was used for the cake?

2$\frac{1}{2}$ Azịza ziri ezi

6$\frac{1}{2}$

15$\frac{3}{8}$

17$\frac{1}{2}$

Akọwa Nkọwa

The baker used 40% of a 50kg bag of flour, which is equivalent to: $$ 40\% \times 50kg = \frac{40}{100} \times 50kg = 20kg $$ If $\frac{1}{8}$ of the amount used was for the cake, then the amount of flour used for the cake is: $$ \frac{1}{8} \times 20kg = \frac{20}{8} kg = 2\frac{1}{2} kg $$ Therefore, the answer is \boxed{2\frac{1}{2}}.

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

Positive And Negative Integers, Rational Numbers

Lee Azịza

Ajụjụ 10 Ripọtì

What is the place value of 9 in the number 3.0492?

$\frac{9}{10000}$

$\frac{9}{1000}$ Azịza ziri ezi

$\frac{9}{100}$

$\frac{9}{10}$

Akọwa Nkọwa

In the number 3.0492, 9 is in the thousandth place. The places in a decimal number are counted from left to right starting from the tenths place, which is immediately to the right of the decimal point. Each place value to the right of the tenths place is ten times smaller than the one to its left. So the first digit to the right of the decimal point is in the tenths place, the second digit is in the hundredths place, and the third digit is in the thousandths place, and so on. Therefore, the place value of 9 in the number 3.0492 is $\frac{9}{1000}$.

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Number Bases

Lee Azịza

Ajụjụ 11 Ripọtì

In the diagram, $\frac{PQ}{RS}$, find x^o + y^o

360^o

300^o

270^o

180^o Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Triangles And Polygons

Lee Azịza

Ajụjụ 12 Ripọtì

If 2x + y = 10, and y $\neq$ 0, which of the following is not a possible value of x?

5 Azịza ziri ezi

Akọwa Nkọwa

The equation 2x + y = 10 can be rearranged as 2x = 10 - y. Dividing both sides by 2 gives x = 5 - (y/2). Therefore, the value of x will depend on the value of y. Now, if we substitute each of the given values of x into the equation 2x + y = 10, we can determine which ones are not possible. For x = 4, we get 2(4) + y = 10, which simplifies to y = 2. This is a possible value of y, so x = 4 is a possible value of x. For x = 5, we get 2(5) + y = 10, which simplifies to y = 0. But the question states that y ≠ 0, so x = 5 is not a possible value of x. For x = 8, we get 2(8) + y = 10, which simplifies to y = -6. This is a possible value of y, so x = 8 is a possible value of x. For x = 10, we get 2(10) + y = 10, which simplifies to y = -10. This is also a possible value of y, so x = 10 is a possible value of x. Therefore, the answer is x = 5, which is not a possible value of x.

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 13 Ripọtì

If the pass mark was 4. What percentage of the pupils failed the test?

20.0% Azịza ziri ezi

37%

62.5%

80.0%

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Percentages (WAEC)

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

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Percentages

Positive And Negative Integers, Rational Numbers

Modular Arithmetic

Lee Azịza

Ajụjụ 14 Ripọtì

The temperature in a chemical plant was -5^oC at 2.00 am. The temperature fell by 6^oC and then rose again by 7^oC. What was the final temperature?

-6^oC

-4^oC Azịza ziri ezi

4^oC

8^oC

Akọwa Nkọwa

The temperature in the chemical plant was -5^oC at 2.00 am. The temperature fell by 6^oC, so the temperature became (-5 - 6) = -11^oC. The temperature then rose by 7^oC, which means the temperature increased from -11^oC to (-11 + 7) = -4^oC. Therefore, the final temperature was -4^oC. Hence, the correct option is (-4^oC).

Gụọ akwụkwọ nkuzi na Indices (WAEC)

Indices

Lee Azịza

Ajụjụ 15 Ripọtì

In the diagram, the two circles have a common centre O. If the area of the larger circle is 100$\pi$ and that of the smaller circle is 49$\pi$, find x

3 Azịza ziri ezi

Akọwa Nkọwa

Let the radius of the smaller circle be r, and the radius of the larger circle be R. The area of a circle is given by the formula A = $\pi r^2$. Therefore, the radius of the smaller circle is $\sqrt{\frac{49\pi}{\pi}} = 7$. Similarly, the radius of the larger circle is $\sqrt{\frac{100\pi}{\pi}} = 10$. Since O is the centre of both circles, the distance between O and the point where the two circles intersect is R - r = 10 - 7 = 3. From the diagram, we can see that x is also equal to this distance, so x = 3. Therefore, the answer is 3.

Gụọ akwụkwọ nkuzi na Volumes (WAEC)

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Volumes

Circles

Lee Azịza

Ajụjụ 16 Ripọtì

If log_a270 - log_a10 + log_a $\frac{1}{3}$ = 2, what is the value of a?

3 Azịza ziri ezi

Akọwa Nkọwa

Using the logarithmic property that log_a(b) - log_a(c) = log_a(b/c), we can simplify the given expression as: log_a(270/10) + log_a(1/3) = 2 Simplifying further, we get: log_a(27) + log_a(1/3) = 2 Using the logarithmic property that log_a(b) + log_a(c) = log_a(bc), we can write: log_a(27 x 1/3) = 2 log_a(9) = 2 Now, using the definition of logarithm, we can write: a^2 = 9 Taking the square root on both sides, we get: a = 3 or -3 However, a cannot be negative, as the base of a logarithm must be positive. Therefore, the value of a is 3. Hence, the value of a is 3.

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Logarithms

Lee Azịza

Ajụjụ 17 Ripọtì

$\begin{array}{c|c} x & 1 & 4 & p \\ \hline y & 0.5 & 1 & 2.5\end{array}$. The table below satisfies the relation y - k$\sqrt{x}$, where k is a positive constant. Find the value of P,

25 Azịza ziri ezi

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Positive And Negative Integers, Rational Numbers

Surds (radicals)

Lee Azịza

Ajụjụ 18 Ripọtì

Evaluate (0.13)$^3$correct to three significant figures

0.00219

0.00220 Azịza ziri ezi

0.00300

0.00390

Akọwa Nkọwa

To evaluate $(0.13)^3$ correct to three significant figures, we need to first cube the number and then round the answer to three significant figures. $(0.13)^3 = 0.002197$ Rounding to three significant figures, we get 0.00220. Therefore, the correct answer is 0.00220.

Gụọ akwụkwọ nkuzi na Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

Lee Azịza

Ajụjụ 19 Ripọtì

If p = [$\frac{Q(R - T)}{15}$]$^ \frac{1}{3}$, make T the subject of the relation

T = R + $\frac{P^3}{15Q}$

T = R - $\frac{15P^3}{Q}$ Azịza ziri ezi

T = R + $\frac{15P^3}{Q}$

T = 15R - $\frac{Q}{P^3}$

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Lee Azịza

Ajụjụ 20 Ripọtì

In the diagram P, Q, R, S are points on the circle RQS = 30^o. PRS = 50^o and PSQ = 20^o. What is the value of x^o + y^o?

260^o

130^o Azịza ziri ezi

100^o

80^o

Akọwa Nkọwa

Lee Azịza

Ajụjụ 21 Ripọtì

Simplify $\frac{20}{5\sqrt{28} - 2 \sqrt{63}}$

$\frac{5\sqrt{7}}{7}$ Azịza ziri ezi

$\frac{5\sqrt{5}}{7}$

$\frac{7\sqrt{5}}{7}$

$\frac{7\sqrt{7}}{5}$

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Fractions, Decimals, Approximations And Percentage (JAMB)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Positive And Negative Integers, Rational Numbers

Fractions, Decimals, Approximations And Percentage

Surds (radicals)

Lee Azịza

Ajụjụ 22 Ripọtì

P is a point on the same plane with a fixed point A. If P moves such that it is always equidistant from A, the locus of P is

a straight line joining A and P

the perpendicular bisector of AP

a circle with centre A Azịza ziri ezi

the triangle with centre P

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Loci (JAMB)

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Sine, Cosine And Tangent Of An Angle

Loci

Circles

Lee Azịza

Ajụjụ 23 Ripọtì

The ratio of boys to girls in a class is 5:3. Find the probability of selecting at random, a girl from the class

$\frac{1}{8}$

$\frac{1}{3}$

$\frac{3}{8}$ Azịza ziri ezi

$\frac{3}{5}$

Akọwa Nkọwa

The ratio of boys to girls in the class is 5:3. This means that for every 5 boys, there are 3 girls in the class. Let B be the number of boys in the class and G be the number of girls in the class. Then, we can write: B:G = 5:3 We can simplify this ratio by dividing both sides by the greatest common factor of 5 and 3, which is 1. B:G = 5/1 : 3/1 B:G = 5 : 3 This means that there are 5 parts for boys and 3 parts for girls in the class. The total number of parts is 5+3 = 8. To find the probability of selecting a girl at random, we need to find the number of parts that represent the girls and divide it by the total number of parts. The number of parts that represent the girls is 3, since there are 3 parts for girls out of a total of 8 parts. Therefore, the probability of selecting a girl at random is: P(selecting a girl) = number of parts that represent the girls / total number of parts P(selecting a girl) = 3/8 Therefore, the probability of selecting a girl at random from the class is 3/8.

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Surds (radicals)

Lee Azịza

Ajụjụ 24 Ripọtì

Simplify $\frac{x - 4}{4} - \frac{x - 3}{6}$

$\frac{x - 18}{12}$

$\frac{x - 6}{12}$ Azịza ziri ezi

$\frac{x - 18}{24}$

$\frac{x - 6}{24}$

Akọwa Nkọwa

To simplify the given expression, we first need to find a common denominator for the two fractions. In this case, the least common multiple of 4 and 6 is 12. So, we can write: \[\frac{x - 4}{4} - \frac{x - 3}{6} = \frac{3(x - 4)}{12} - \frac{2(x - 3)}{12}\] Now, we can combine the two fractions by subtracting the numerators: \[\frac{3(x - 4)}{12} - \frac{2(x - 3)}{12} = \frac{3x - 12 - 2x + 6}{12}\] Simplifying the numerator, we get: \[\frac{3x - 2x - 12 + 6}{12} = \frac{x - 6}{12}\] Therefore, the simplified expression is $\frac{x - 6}{12}$. So, the answer is option (B).

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 25 Ripọtì

In the diagram, |QR| = 5cm, PQR = 60^o and PSR = 45^o. Find |PS|, leaving your answe in surd form.

4$\sqrt{5}$cm

3$\sqrt{7}$cm

4$\sqrt{6}$cm

5$\sqrt{6}$cm Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Circles

Surds (radicals)

Lee Azịza

Ajụjụ 26 Ripọtì

Given that y = 1 - $\frac{2x}{4x - 3}$, find the value of x for which y is undefined

$\frac{3}{4}$ Azịza ziri ezi

$\frac{-3}{4}$

-3

Akọwa Nkọwa

To find the value of x for which y is undefined, we need to look for values of x that make the denominator of the expression for y equal to zero, since division by zero is undefined. The denominator of the expression for y is 4x - 3. Therefore, we need to find the value of x that makes 4x - 3 equal to zero. Solving the equation 4x - 3 = 0, we get: 4x = 3 x = 3/4 Therefore, the value of x for which y is undefined is x = 3/4. Explanation: To understand why we need to look for values of x that make the denominator of the expression for y equal to zero, we need to remember that division by zero is undefined. When we divide a number by zero, there is no answer because it is impossible to divide any number into equal parts of zero size. In the given expression for y, the denominator is 4x - 3. If we plug in x = 3/4, the denominator becomes: 4x - 3 = 4(3/4) - 3 = 3 - 3 = 0 This means that when x = 3/4, the denominator becomes zero, and therefore division by zero occurs, which makes the value of y undefined. Hence, the value of x for which y is undefined is x = 3/4.

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Gụọ akwụkwọ nkuzi na Algebraic Fractions (WAEC)

Algebraic Expressions

Algebraic Fractions

Lee Azịza

Ajụjụ 27 Ripọtì

$\begin{array}{c|c} x & 1 & 4 & p \\ \hline y & 0.5 & 1 & 2.5\end{array}$. The table above satisfies the relation y = k$\sqrt{x}$, where k is a positive constant. Find the value of K.

0.5 Azịza ziri ezi

1.5

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Surds (radicals)

Lee Azịza

Ajụjụ 28 Ripọtì

Given that (2x - 1)(x + 5) = 2x² - mx - 5, what is the value of m

-9 Azịza ziri ezi

-10

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 29 Ripọtì

If tan y = 0.404, where y is acute, find cos 2y

0.035

0.719 Azịza ziri ezi

0.808

0.927

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Trigonometry (JAMB)

Sine, Cosine And Tangent Of An Angle

Trigonometry

Lee Azịza

Ajụjụ 30 Ripọtì

The diagram shows an arc MN of a circle, centre O, with 10cm. If < MON = 72^o. calculate the length of the arc, correct to three significant figures. [Take $\pi = \frac{22}{7}$]

6.28cm

12.6cm Azịza ziri ezi

22.6cm

62.8cm

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Circles

Lee Azịza

Ajụjụ 31 Ripọtì

a conical water-jug is 7cm in diameter and 6cm deep. find the volume of water it can hold. [Take $\pi = \frac{22}{7}$]

22cm²

44cm³

77cm³ Azịza ziri ezi

308²

Akọwa Nkọwa

The volume of a cone can be calculated using the formula: V = (1/3)πr²h where V is the volume of the cone, r is the radius of the circular base, h is the height of the cone, and π is a mathematical constant approximately equal to 22/7. Given that the diameter of the circular base is 7cm, the radius r can be calculated as half of the diameter: r = 7/2 = 3.5cm. The height of the cone is given as 6cm. Substituting these values into the formula for the volume of a cone, we get: V = (1/3) x (22/7) x (3.5cm)² x 6cm Simplifying this expression, we get: V = (1/3) x (22/7) x 12.25cm² x 6cm V = (22/21) x 74.25cm³ V = 77cm³ Therefore, the volume of water the conical water-jug can hold is 77cm³. Answer option (C) is correct.

Gụọ akwụkwọ nkuzi na Volumes (WAEC)

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Gụọ akwụkwọ nkuzi na Angles (WAEC)

Volumes

Areas

Angles

Lee Azịza

Ajụjụ 32 Ripọtì

If p = {1, 3, 5, 7, 9} and Q = {2, 4, 6, 8, 10} are subsets of a universal set. U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. What are the elements of P¹ $\cap$ Q¹?

{1, 3, 5, 7, 9}

[2, 4, 6, 8, 10}

$\varnothing$ Azịza ziri ezi

{1, 2, 6, 8, 10}

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Sets (WAEC)

Gụọ akwụkwọ nkuzi na Algebraic Fractions (WAEC)

Gụọ akwụkwọ nkuzi na Sets (JAMB)

Sets

Algebraic Fractions

Sets

Lee Azịza

Ajụjụ 33 Ripọtì

A ladder 16m long leans against an electric pole. If the ladder makes an angle of 65^o with the ground, how far up the electric pole does its top reach

6.8m

14.5m Azịza ziri ezi

17.7m

34.3m

Akọwa Nkọwa

We can use trigonometry to solve this problem. Let x be the distance from the bottom of the ladder to the electric pole. Then, we can create a right triangle with the ladder as the hypotenuse, x as one leg, and the distance we are trying to find as the other leg. Using trigonometric functions, we know that: sin(65^o) = opposite/hypotenuse sin(65^o) = (distance up the pole)/16 (distance up the pole) = 16*sin(65^o) (distance up the pole) ≈ 14.5m Therefore, the answer is 14.5m.

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

Gụọ akwụkwọ nkuzi na Coordinate Geometry (WAEC)

Angles Of Elevation And Depression

Coordinate Geometry

Lee Azịza

Ajụjụ 34 Ripọtì

Which of the following is not a measure of centre tendency?

mean

mode

range Azịza ziri ezi

median

Akọwa Nkọwa

Range is not a measure of central tendency. Measures of central tendency are statistical measures that represent the typical or central values of a data set. They include the mean, median, and mode. The mean is the arithmetic average of a data set, the median is the middle value when the data is arranged in order, and the mode is the most frequently occurring value in the data set. On the other hand, range is a measure of dispersion, which describes how spread out the data is. It is the difference between the highest and lowest values in the data set.

Gụọ akwụkwọ nkuzi na Measures Of Dispersion (JAMB)

Measures Of Dispersion

Lee Azịza

Ajụjụ 35 Ripọtì

In the diagram, PQUV, PQTU, QRTU and QRST are parallelograms. |UV| = 4.8cm and the perpendicular distance between PR and VS is 5cm. Calculate the area of quadrilateral PRSV

96cm²

72cm²

60cm² Azịza ziri ezi

24cm²

Akọwa Nkọwa

Lee Azịza

Ajụjụ 36 Ripọtì

A machine valued at N20,000 depreciates by 10% every year. What will be the value of the machine at the end of two years?

N16,200 Azịza ziri ezi

N1,4,200

12,000

8000.

Akọwa Nkọwa

At the end of the first year, the value of the machine will be 90% of N20,000 = N18,000. At the end of the second year, the value of the machine will be 90% of N18,000 = N16,200. Therefore, the value of the machine at the end of two years will be N16,200. So the correct option is: - N16,200

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

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

Modular Arithmetic

Ratio, Proportions And Rates

Lee Azịza

Ajụjụ 37 Ripọtì

What must be added to x² - 3x to make it a perfect square?

$\frac{9}{4}$ Azịza ziri ezi

$\frac{9}{2}$

Akọwa Nkọwa

To make a perfect square, we need to add the square of half of the coefficient of the x-term. In this case, the coefficient of x is -3, so we need to add $\left(\frac{-3}{2}\right)^2 = \frac{9}{4}$ to make it a perfect square. Thus, the answer is $\frac{9}{4}$.

Gụọ akwụkwọ nkuzi na Linear Inequalities (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Linear Inequalities

Surds (radicals)

Quadratic Equations

Lee Azịza

Ajụjụ 38 Ripọtì

A fair coin is tossed three times. Find the probability of getting two heads and one tail.

$\frac{1}{2}$

$\frac{3}{8}$ Azịza ziri ezi

$\frac{1}{4}$

$\frac{1}{8}$

Akọwa Nkọwa

When a fair coin is tossed, there are two possible outcomes, heads (H) or tails (T), each with a probability of 1/2. Since the coin is tossed three times, the total number of possible outcomes is 2 × 2 × 2 = 8. We need to find the probability of getting two heads and one tail, which can happen in three different ways: HHT, HTH, or THH. Each of these outcomes has a probability of (1/2) × (1/2) × (1/2) = 1/8. Therefore, the probability of getting two heads and one tail is the sum of the probabilities of the three possible outcomes: P(HHT) + P(HTH) + P(THH) = 1/8 + 1/8 + 1/8 = 3/8. Hence, the answer is 3/8.

Lee Azịza

Ajụjụ 39 Ripọtì

In the diagram, PQR is a straight line, QRST is a parallelogram, < TPQ = 72^o and < RST = 126^o. What type of triangle is $\bigtriangleup$ PQT?

An equilateral triangle

A right-angled triangle

An isosceles triangle Azịza ziri ezi

A scalene triangle

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Triangles And Polygons

Lee Azịza

Ajụjụ 40 Ripọtì

In the diagram what is x + y in terms of z?

180^o - z Azịza ziri ezi

180^o + z

z - 180^o

Akọwa Nkọwa

In a triangle, the sum of the angles is always 180 degrees. In the given diagram, we can see that the angles x, y, and z form a triangle. Therefore, we have: x + y + z = 180^o Rearranging this equation, we get: x + y = 180^o - z Hence, the value of x + y in terms of z is 180^o - z. Therefore, the answer is (a) 180^o - z.

Gụọ akwụkwọ nkuzi na Angles (WAEC)

Angles

Lee Azịza

Ajụjụ 41 Ripọtì

How many pupils took the test?

40 Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Surds (radicals)

Lee Azịza

Ajụjụ 42 Ripọtì

If x + y = 12 and 3x - y = 20, find the value of 2x - y

12 Azịza ziri ezi

Akọwa Nkọwa

We are given two equations as follows: x + y = 12 .... (Equation 1) 3x - y = 20 .... (Equation 2) We need to find the value of 2x - y. To solve the problem, we can use the method of substitution or elimination. Here, we will use the method of substitution. From Equation 1, we can write: y = 12 - x Substituting this value of y in Equation 2, we get: 3x - (12 - x) = 20 Simplifying the equation, we get: 4x - 12 = 20 Adding 12 to both sides, we get: 4x = 32 Dividing both sides by 4, we get: x = 8 Now, substituting the value of x in Equation 1, we get: y = 12 - 8 y = 4 Therefore, we have found the values of x and y as 8 and 4, respectively. Finally, substituting these values in the expression 2x - y, we get: 2x - y = 2(8) - 4 = 12 Hence, the value of 2x - y is 12.

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 43 Ripọtì

A rectangular carpet 2.5m long and 2.4m wide covers 5% of a rectangular floor. Calculate the area of the floor

30m²

57m²

120m² Azịza ziri ezi

225m²

Akọwa Nkọwa

Let's assume that the rectangular floor has a length of L meters and a width of W meters. We know that the rectangular carpet covers 5% of the rectangular floor, which means that the area of the carpet is 5% of the area of the floor. We can write this as an equation: 2.5 × 2.4 = 0.05 × L × W Multiplying the left-hand side, we get: 6 = 0.05LW Dividing both sides by 0.05, we get: 120 = LW Therefore, the area of the rectangular floor is LW, which is 120 square meters. Hence, the area of the floor is 120 m², which is the third option.

Gụọ akwụkwọ nkuzi na Inequalities (JAMB)

Gụọ akwụkwọ nkuzi na Lengths And Perimeters (WAEC)

Inequalities

Lengths And Perimeters

Lee Azịza

Ajụjụ 44 Ripọtì

If 30% of y is equal to x, what in terms of x, is 30% of 3y?

$\frac{x}{9}$

$\frac{x}{3}$

3x Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Percentages (WAEC)

Gụọ akwụkwọ nkuzi na Fractions, Decimals, Approximations And Percentage (JAMB)

Percentages

Fractions, Decimals, Approximations And Percentage

Lee Azịza

Ajụjụ 45 Ripọtì

Simplify $\frac{25 \frac{2}{3} \div 25 \frac{1}{6}}{( \frac{1}{5})^{-\frac{7}{6}} \times ( \frac{1}{5})^{\frac{1}{6}}}$

$\frac{1}{5}$

1 Azịza ziri ezi

$\frac{1}{25}$

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

Lee Azịza

Ajụjụ 46 Ripọtì

The values of three angles at a point are 3y - 45^o, y + 25^o and y^o . Find the value of y.

40^o

58^o

68^o

76^o Azịza ziri ezi

Akọwa Nkọwa

When three angles at a point are added together, they sum up to 360 degrees. Therefore, we have: (3y - 45) + (y + 25) + y = 360 Simplifying the equation, we get: 5y - 20 = 360 Adding 20 to both sides, we get: 5y = 380 Dividing both sides by 5, we get: y = 76 Hence, the value of y is 76 degrees, and the correct option is (d) 76^o.

Gụọ akwụkwọ nkuzi na Angles (WAEC)

Angles

Lee Azịza






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WAEC SSCE - General Mathematics - 2006 (Objective)