WAEC SSCE - General Mathematics - 2015…

Tambaya 1 Rahoto

A trader bought an engine for $15,000.00 outside Nigeria. If the exchange rate is $0.075 to N1.00, how much did the engine cost in Naira?

N250,000.00

N200,000.00 Amsar da ta dace

N150,000.00

N100,000.00

Bayanin Amsa

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 2 Rahoto

Factorize completely: 6ax - 12by - 9ay + 8bx

(2a - 3b)(4x + 3y)

(3a + 4b)(2x - 3y) Amsar da ta dace

(3a - 4b)(2x + 3y)

(2a + 3b)(4x -3y)

Bayanin Amsa

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

Tambaya 3 Rahoto

Ada draws the graph of y = x² - x - 2 and y = 2x - 1 on the same axes. Which of these equations is she solving?

x² - x - 3 = 0

x² - 3x - 1 = 0 Amsar da ta dace

x² - 3x - 3 = 0

x² + 3x - 1 = 0

Bayanin Amsa

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Karanta bayanin darasi akan Graphs Of Linear And Quadratic Functions (WAEC)

Quadratic Equations

Graphs Of Linear And Quadratic Functions

Duba Amsa

Tambaya 4 Rahoto

The area of a rhombus is 110 cm$^2$. If the diagonals are 20 cm and (2x + 1) cm long, find the value of x.

5.0 Amsar da ta dace

4.0

3.0

2.5

Bayanin Amsa

Karanta bayanin darasi akan Areas (WAEC)

Karanta bayanin darasi akan Angles (WAEC)

Areas

Angles

Duba Amsa

Tambaya 5 Rahoto

A letter is selected from the letters of the English alphabet. What is the probability that the letter selected is from the word MATHEMATICS?

$\frac{9}{13}$

$\frac{11}{26}$

$\frac{4}{13}$ Amsar da ta dace

$\frac{1}{26}$

Bayanin Amsa

The word MATHEMATICS contains the letters A, C, E, H, I, M, S, and T. Since there are 26 letters in the English alphabet, the probability of selecting any letter from the alphabet is 1 out of 26. Out of the letters in the word MATHEMATICS, there are 8 letters, so the probability of selecting a letter from the word MATHEMATICS is 8 out of 26. Therefore, the probability of selecting a letter from the word MATHEMATICS is: $$\frac{8}{26}=\frac{4}{13}$$ Therefore, the correct answer is option (C).

Karanta bayanin darasi akan Probability (WAEC)

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Probability

Surds (radicals)

Duba Amsa

Tambaya 6 Rahoto

In the diagram, PTR is a tangent to the centre O. If angles TON = 108°, Calculate the size of angle PTN

132^o

126^o Amsar da ta dace

108^o

102^o

Bayanin Amsa

Karanta bayanin darasi akan Circles (WAEC)

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Karanta bayanin darasi akan Euclidean Geometry (JAMB)

Karanta bayanin darasi akan Triangles And Polygons (WAEC)

Circles

Algebraic Expressions

Euclidean Geometry

Triangles And Polygons

Duba Amsa

Tambaya 7 Rahoto

In the diagram, $\bar{OX}$ bisects < YXZ and $\bar{OZ}$ bisects < YZX. If < XYZ = 68^o, calculate the value of < XOZ

68^o

72^o

112^o

124^o Amsar da ta dace

Bayanin Amsa

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

Angles And Intercepts On Parallel Lines

Duba Amsa

Tambaya 8 Rahoto

In the diagram, O is the centre, $\bar{RT}$ is a diameter, < PQT = 33$^o$ and < TOS = 76$^o$. Using the diagram, find the size of angle PRS.

76^o

71^o Amsar da ta dace

38^o

33^o

Bayanin Amsa

Karanta bayanin darasi akan Circles (WAEC)

Circles

Duba Amsa

Tambaya 9 Rahoto

A farmer uses $\frac{2}{5}$ of his land to grow cassava, $\frac{1}{3}$ of the remaining for yam and the rest for maize. Find the part of the land used for maize

$\frac{2}{15}$

$\frac{2}{5}$ Amsar da ta dace

$\frac{2}{3}$

$\frac{4}{5}$

Bayanin Amsa

Let's assume the farmer has 15 parts of land. The fraction of land used for cassava is $\frac{2}{5}$ of 15, which is 6 parts. The remaining land is 15 - 6 = 9 parts. The fraction of the remaining land used for yam is $\frac{1}{3}$ of 9, which is 3 parts. Therefore, the part of the land used for maize is the remaining land after cassava and yam are planted, which is 9 - 3 = 6 parts. So the fraction of the land used for maize is $\frac{6}{15}$, which simplifies to $\frac{2}{5}$. Therefore, the answer is option (B) $\frac{2}{5}$.

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

Karanta bayanin darasi akan Number Bases (WAEC)

Fractions, Decimals And Approximations

Number Bases

Duba Amsa

Tambaya 10 Rahoto

$\begin{array}{c|c}
Scores & 0 - 4 & 5 - 9 & 10 - 14\\
\hline Frequency & 2 & 1 & 2\end{array}$

The table shows the distribution of the scores of some students in a test. Calculate the mean scores.

5.6

6.2

6.6

7.0 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Integration (JAMB)

Karanta bayanin darasi akan Statistics (WAEC)

Integration

Statistics

Duba Amsa

Tambaya 11 Rahoto

The rate of consumption of petrol by a vehicle varies directly as the square of the distance covered. If 4 litres of petrol is consumed on a distance of 15km. how far would the vehicle go on 9 litres of petrol?

22$\frac{1}{2}$km Amsar da ta dace

30km

33$\frac{1}{2}$km

45km

Bayanin Amsa

Since the rate of consumption of petrol varies directly as the square of the distance covered, let's assume that the rate is k$\frac{\text{litres}}{\text{km}^2}$. Then we can write the relationship as: rate of consumption = k(distance covered)$^2$ We are given that when the distance covered is 15 km, the rate of consumption is 4 litres. Substituting these values in the above equation, we get: 4 = k(15)$^2$ k = $\frac{4}{15^2}$ Now, we need to find how far the vehicle can go on 9 litres of petrol. Let's assume that the vehicle can travel a distance of x km on 9 litres of petrol. Using the same equation as before, we get: 9 = $\frac{4}{15^2}$(x)$^2$ Simplifying this equation, we get: x$^2$ = $\frac{9}{\frac{4}{15^2}}$ x$^2$ = 506.25 x = 22.5 Therefore, the vehicle can travel 22.5 km on 9 litres of petrol. So the answer is 22$\frac{1}{2}$km.

Karanta bayanin darasi akan Sequence And Series (WAEC)

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

Sequence And Series

Ratio, Proportions And Rates

Duba Amsa

Tambaya 12 Rahoto

Make K the subject of the relation T = $\sqrt{\frac{TK - H}{K - H}}$

K = $\frac{H(T^2 - 1)}{T^2 - T}$ Amsar da ta dace

K = $\frac{HT}{(T - 1)^2}$

K = $\frac{H(T^2 + 1)}{T}$

K = $\frac{H(T - 1)}{T}$

Bayanin Amsa

Karanta bayanin darasi akan Algebraic Fractions (WAEC)

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Algebraic Fractions

Quadratic Equations

Duba Amsa

Tambaya 13 Rahoto

The scores of twenty students in a test are as follows: 44, 47, 48, 49, 50, 51, 52, 53, 53, 54, 58, 59, 60, 61, 63, 65, 67, 70, 73, 75. Find the third quartile.

64 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Measures Of Dispersion (JAMB)

Measures Of Dispersion

Duba Amsa

Tambaya 14 Rahoto

If $\frac{27^x \times 3^{1 - x}}{9^{2x}} = 1$, find the value of x.

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

-$\frac{1}{2}$

-1

Bayanin Amsa

Karanta bayanin darasi akan Number Bases (WAEC)

Karanta bayanin darasi akan Indices (WAEC)

Number Bases

Indices

Duba Amsa

Tambaya 15 Rahoto

Describe the shaded portion in the diagram

P' $\cap$ Q $\cap$ R' Amsar da ta dace

(P' $\cap$ R)' $\cap$ R

P' $\cap$ Q $\cap$ R

(P $\cap$ Q)' $\cap$ R

Bayanin Amsa

The shaded portion in the diagram represents the set of elements that belong to the complement of P and the intersection of Q and R. In other words, it represents the set of elements that are not in P, but are in both Q and R. Therefore, the correct answer is P' $\cap$ Q $\cap$ R'.

Karanta bayanin darasi akan Logical Reasoning (WAEC)

Karanta bayanin darasi akan Sets (JAMB)

Logical Reasoning

Sets

Duba Amsa

Tambaya 16 Rahoto

In the given diagram, $\bar{QT}$ and $\bar{PR}$ are straight lines, < ROS = (3n - 20), < SOT = n, < POL = m and < QOL is a right angle. Find the value of n.

35^o Amsar da ta dace

40^o

55^o

60^o

Bayanin Amsa

Since < QOL is a right angle, we have: < QOT + < SOT = 90^o Since < QOT = 180^o - < POL, we have: < QOT = 180^o - m Substituting into the first equation, we get: 180^o - m + n = 90^o n - m = 90^o - 180^o = -90^o Also, from the diagram, we have: < QOT + < ROS + < SOT = 180^o Substituting < QOT = 180^o - < POL and simplifying, we get: 180^o - m + (3n - 20) + n = 180^o 4n - m = 20 We now have a system of two equations: n - m = -90^o 4n - m = 20 Solving simultaneously, we get: n = 35^o

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

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

Algebraic Expressions

Angles And Intercepts On Parallel Lines

Duba Amsa

Tambaya 17 Rahoto

In the diagram, VW//YZ, |WX| = 6cm, |XY| = 16cm, |YZ| = 20cm and |ZX| = 12cm. Calculate |VX|

3cm

4cm

6cm

8cm Amsar da ta dace

Bayanin Amsa

Duba Amsa

Tambaya 18 Rahoto

Simplify: $\frac{3x - y}{xy} - \frac{2x + 3y}{2xy} + \frac{1}{2}$

$\frac{4x + 5y - xy}{2xy}$

$\frac{5y - 4x + xy}{2xy}$

$\frac{5x + 4y - xy}{2xy}$

$\frac{4x - 5y + xy}{2xy}$ Amsar da ta dace

Bayanin Amsa

To simplify the expression, we first need to find a common denominator for the fractions. The lowest common multiple of the denominators xy and 2xy is 2xy. $\frac{3x - y}{xy} - \frac{2x + 3y}{2xy} + \frac{1}{2} = \frac{3x\cdot2 - y\cdot2}{xy\cdot2} - \frac{2x\cdot1 + 3y\cdot1}{2xy\cdot1} + \frac{1\cdot xy}{2\cdot xy}$ Simplifying further, we get: $\frac{6x - 2y}{2xy} - \frac{2x + 3y}{2xy} + \frac{xy}{2xy}$ Combining like terms, we get: $\frac{4x - 5y + xy}{2xy}$ Therefore, the answer is option D: $\frac{4x - 5y + xy}{2xy}$.

Duba Amsa

Tambaya 19 Rahoto

Which of the following is used to determine the mode of a grouped data?

Bar chart

Frequency polygon

Ogive

Histogram Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Representation Of Data (JAMB)

Karanta bayanin darasi akan Integration (JAMB)

Representation Of Data

Integration

Duba Amsa

Tambaya 20 Rahoto

The dimension of a rectangular tank are 2m by 7m by 11m. If its volume is equal to that of a cylindrical tank of height 4cm, calculate the base radius of the cylindrical tank. [Take $\pi = \frac{22}{7}$]

14cm

3$\frac{1}{2}$m Amsar da ta dace

1$\frac{3}{4}$m

Bayanin Amsa

Karanta bayanin darasi akan Volumes (WAEC)

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

Volumes

Fractions, Decimals And Approximations

Duba Amsa

Tambaya 21 Rahoto

In the diagram, O is the centre. If PQ//RS and ∠ONS = 140°, find the size of ∠POM.

40^o Amsar da ta dace

50^o

60^o

80^o

Bayanin Amsa

Karanta bayanin darasi akan Euclidean Geometry (JAMB)

Euclidean Geometry

Duba Amsa

Tambaya 22 Rahoto

The population of students in a school is 810. If this is represented on a pie chart, calculate the sectoral angle for a class of 7 students

32^o Amsar da ta dace

45^o

60^o

75^o

Bayanin Amsa

To find the sectoral angle for a class of 7 students, we first need to calculate what percentage of the total population they represent: Percentage of class of 7 students = (Number of students in the class / Total population) x 100% Percentage of class of 7 students = (7 / 810) x 100% Percentage of class of 7 students = 0.864% (rounded to three decimal places) Next, we need to convert this percentage to an angle of the pie chart. We know that a full circle is 360 degrees, and the percentage of the class of 7 students represents a certain portion of this circle. We can use the following formula to calculate the sectoral angle: Sectoral angle = Percentage of class of 7 students x 360 degrees Plugging in the percentage we calculated earlier, we get: Sectoral angle = 0.864% x 360 degrees Sectoral angle ≈ 3.11 degrees (rounded to two decimal places) Therefore, the sectoral angle for a class of 7 students in the pie chart is approximately 3.11 degrees. So, the correct option is (a) 32^o.

Karanta bayanin darasi akan Representation Of Data (JAMB)

Karanta bayanin darasi akan Sets (JAMB)

Karanta bayanin darasi akan Mensuration (JAMB)

Representation Of Data

Sets

Mensuration

Duba Amsa

Tambaya 23 Rahoto

Calculate the mean deviation of 5, 3, 0, 7, 2, 1

0.0

2.0 Amsar da ta dace

2.5

3.0

Bayanin Amsa

To find the mean deviation, we need to first calculate the mean of the given set of numbers: Mean = (5 + 3 + 0 + 7 + 2 + 1) / 6 = 18 / 6 = 3 Next, we need to find the absolute deviation of each number from the mean: |5 - 3| = 2 |3 - 3| = 0 |0 - 3| = 3 |7 - 3| = 4 |2 - 3| = 1 |1 - 3| = 2 To calculate the mean deviation, we add up these absolute deviations and divide by the number of numbers: Mean deviation = (2 + 0 + 3 + 4 + 1 + 2) / 6 = 12 / 6 = 2 Therefore, the answer is option (B) 2.0.

Karanta bayanin darasi akan Measures Of Location (JAMB)

Measures Of Location

Duba Amsa

Tambaya 24 Rahoto

The probability that kebba, Ebou and Omar will hit a target are $\frac{2}{3}$, $\frac{3}{4}$ and $\frac{4}{5}$ respectively. Find the probability that only Kebba will hit the target.

$\frac{2}{5}$

$\frac{7}{60}$

$\frac{1}{30}$ Amsar da ta dace

$\frac{1}{60}$

Bayanin Amsa

The probability that only Kebba will hit the target means that Kebba hits the target and the other two miss the target. We can find this probability by multiplying the probability of Kebba hitting the target with the probability of Ebou missing the target and the probability of Omar missing the target. Probability that only Kebba hits the target = Probability of Kebba hitting the target × Probability of Ebou missing the target × Probability of Omar missing the target = $\frac{2}{3}$ × $\frac{1}{4}$ × $\frac{1}{5}$ = $\frac{2}{3\times4\times5}$ = $\frac{1}{30}$ Therefore, the probability that only Kebba will hit the target is $\frac{1}{30}$.

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 25 Rahoto

Adding 42 to a given positive number gives the same result as squaring the number. Find the number

7 Amsar da ta dace

Bayanin Amsa

Let's call the unknown number "x". According to the problem, adding 42 to x gives the same result as squaring x. We can write this as an equation: x + 42 = x^2 To solve for x, we can rearrange this equation to get it in standard quadratic form: x^2 - x - 42 = 0 Now we can factor this quadratic equation: (x - 7)(x + 6) = 0 This gives us two possible solutions: x = 7 and x = -6. However, we were given that x is a positive number, so we can discard the negative solution. Therefore, the answer is x = 7. To check our answer, we can substitute x = 7 into the original equation: 7 + 42 = 49 7^2 = 49 So adding 42 to 7 does indeed give the same result as squaring 7.

Duba Amsa

Tambaya 26 Rahoto

Tom will be 25 years old in n years' time. If he is 5 years younger than Bade's present age.

(30 - n)years Amsar da ta dace

(20 - n)years

(25 - n)years

(30 + n)years

Bayanin Amsa

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Karanta bayanin darasi akan Circles (WAEC)

Solution Of Linear Equations

Circles

Duba Amsa

Tambaya 27 Rahoto

In the diagram, PQ//RT, QR//Su,

134^o

132^o Amsar da ta dace

96^o

48^o

Bayanin Amsa

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

Karanta bayanin darasi akan Triangles And Polygons (WAEC)

Angles And Intercepts On Parallel Lines

Triangles And Polygons

Duba Amsa

Tambaya 28 Rahoto

Given that logx 64 = 3, evaluate x log$_2$8

12 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Karanta bayanin darasi akan Logarithms (WAEC)

Modular Arithmetic

Logarithms

Duba Amsa

Tambaya 29 Rahoto

Find the equation whose roots are $\frac{3}{4}$ and -4

4x² - 13x + 12 = 0

4x² - 13x - 12 = 0

4x² + 13x - 12 = 0 Amsar da ta dace

4x² + 13x + 12 = 0

Bayanin Amsa

If a quadratic equation has roots p and q, then it can be expressed in the factored form as:

(x - p)(x - q) = 0

Expanding this equation, we get:

x² - (p + q)x + pq = 0

In this case, the roots are given as $\frac{3}{4}$ and -4. Thus, the equation can be expressed as:

(x - \(\frac{3}{4}\))(x + 4) = 0

Expanding this equation, we get:

x² + (4 - \(\frac{3}{4}\))x - 3 = 0

Multiplying throughout by 4 to eliminate fractions, we get:

4x² + 13x - 12 = 0

Therefore, the equation whose roots are $\frac{3}{4}$ and -4 is 4x² + 13x - 12 = 0.

Hence, the answer is 4x² + 13x - 12 = 0.

Duba Amsa

Tambaya 30 Rahoto

In the diagram, the shaded part is carpet laid in a room with dimensions 3.5m by 2.2m leaving a margin of 0.5m round it. Find area of the margin

4.7m² Amsar da ta dace

4.9m²

5.7m²

5.9m²

Bayanin Amsa

Karanta bayanin darasi akan Lengths And Perimeters (WAEC)

Lengths And Perimeters

Duba Amsa

Tambaya 31 Rahoto

Find the 7^th term of the sequence: 2, 5, 10, 17, 6,...

50 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Sequence And Series (WAEC)

Sequence And Series

Duba Amsa

Tambaya 32 Rahoto

In a circle radius rcm, a chord 16$\sqrt{3}cm$ long is 10cmfrom the centre of the circle. Find, correct to the nearest cm, the value of r

22cm

17cm Amsar da ta dace

16cm

15cm

Bayanin Amsa

In a circle, a radius is a line segment that connects the center of the circle to any point on the circle. A chord is a line segment that connects two points on a circle. Given that a chord of length 16$\sqrt{3}cm$ is 10cm from the center of the circle, we can use the Pythagorean theorem to find the length of the radius. First, we draw a line from the center of the circle perpendicular to the chord, creating a right triangle with the radius, the perpendicular line, and half the chord length (which is $8\sqrt{3}$ cm). Using the Pythagorean theorem: \begin{align*} r^2 &= (8\sqrt{3})^2 + 10^2 \\ r^2 &= 192 + 100 \\ r^2 &= 292 \\ r &\approx \sqrt{292} \\ r &\approx 17 \text{ cm} \end{align*} Rounding to the nearest whole number, we get the answer of 17cm. Therefore, the correct option is (b) 17cm.

Karanta bayanin darasi akan Circles (WAEC)

Karanta bayanin darasi akan Triangles And Polygons (WAEC)

Circles

Triangles And Polygons

Duba Amsa

Tambaya 33 Rahoto

Find the value of p if $\frac{1}{4}$p + 3q = 10 and 2p - q = 7

4 Amsar da ta dace

-3

-4

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 34 Rahoto

The angle of elevation of an aircraft from a point K on the horizontal ground 30$\alpha$. If the aircraft is 800m above the ground, how far is it from K?

400.00m

692.82m

923.76m

1,600.99m Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Introductory Calculus (WAEC)

Karanta bayanin darasi akan Coordinate Geometry (WAEC)

Introductory Calculus

Coordinate Geometry

Duba Amsa

Tambaya 35 Rahoto

If m = 4, n = 9 and r = 16., evaluate $\frac{m}{n}$ - 1$\frac{7}{9}$ + $\frac{n}{r}$

1$\frac{5}{16}$

1$\frac{1}{16}$

$\frac{5}{16}$

- 1$\frac{37}{48}$ Amsar da ta dace

Bayanin Amsa

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

Fractions, Decimals And Approximations

Duba Amsa

Tambaya 36 Rahoto

If {X: 2 d- x d- 19; X integer} and 7 + x = 4 (mod 9), find the highest value of x

15 Amsar da ta dace

Bayanin Amsa

To solve this problem, we need to use the given conditions to find the possible values of x and then determine the highest value among them. From the first condition, we know that x is an integer between 2 and 19 (inclusive) and can be represented as {X: 2 d- x d- 19; X integer}. This means that x can take on any of the following values: 2, 3, 4, ..., 18, 19. The second condition tells us that 7 + x is congruent to 4 modulo 9, which can be written as: 7 + x ≡ 4 (mod 9) To solve for x, we can subtract 7 from both sides of the congruence: x ≡ -3 (mod 9) Since we want x to be between 2 and 19, we can add or subtract multiples of 9 to -3 until we get a value within the range of possible values for x. Doing so, we obtain: x ≡ -3 (mod 9) x ≡ 15 (mod 9) Since 15 is the largest value of x that satisfies both conditions, our answer is 15.

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 37 Rahoto

The sum 11011₂, 1111₂ and 10_m10_n0₂. Find the value of m and n.

m = 0, n = 0

m = 1, n = 0

m = 0, n = 1 Amsar da ta dace

m = 1, n = 1

Bayanin Amsa

Karanta bayanin darasi akan Number Bases (WAEC)

Number Bases

Duba Amsa

Tambaya 38 Rahoto

$\begin{array}{c|c}
x & 0 & 1\frac{1}{4} & 2 & 4\\
\hline
y & 3 & 5\frac{1}{2} & &
\end{array}$
The table given shows some values for a linear graph. Find the gradient of the line

2 Amsar da ta dace

Bayanin Amsa

To find the gradient of a line, we need to calculate the change in y divided by the change in x between two points on the line. Let's choose two points on the line from the given table. We can select the points (0,3) and (2,y), where y is the value of y for x = 2. The change in y is y - 3, and the change in x is 2 - 0 = 2. Therefore, the gradient of the line is: gradient = change in y / change in x = (y - 3) / 2 We don't know the value of y yet, but we can use the other point on the line to find it. Using the points (1.25, 5.5) and (2, y), we get: gradient = (y - 5.5) / (2 - 1.25) = (y - 5.5) / 0.75 Setting the two expressions for the gradient equal to each other, we get: (y - 3) / 2 = (y - 5.5) / 0.75 Solving for y gives: y = 7 Therefore, the value of y for x = 2 is 7, and the gradient of the line is: gradient = (y - 3) / 2 = (7 - 3) / 2 = 2 So the answer is 2, and the gradient of the line is 2. In simple terms, the gradient of a line tells us how steep the line is. If the gradient is positive, the line is sloping upwards from left to right. If the gradient is negative, the line is sloping downwards from left to right. The larger the absolute value of the gradient, the steeper the line.

Karanta bayanin darasi akan Graphs Of Linear And Quadratic Functions (WAEC)

Graphs Of Linear And Quadratic Functions

Duba Amsa

Tambaya 39 Rahoto

Two angles of a pentagon are in the ratio 2:3. The others are 60^o each. Calculate the smaller of the two angles

72^o

100^o

120^o

144^o Amsar da ta dace

Bayanin Amsa

Let's call the two angles that are in the ratio 2:3 as "2x" and "3x". The sum of the interior angles of a pentagon is given by the formula (5-2) x 180 degrees = 540 degrees. Since the other three angles are all 60 degrees each, we can find the sum of those angles by multiplying 60 by 3, which gives us 180 degrees. Now we can set up an equation to solve for the unknown angle, which is 2x + 3x. 2x + 3x + 180 = 540 Simplifying this equation, we get: 5x + 180 = 540 Subtracting 180 from both sides, we get: 5x = 360 Dividing both sides by 5, we get: x = 72 So the smaller angle, which is 2x, is equal to 2 times 72, which is 144 degrees. Therefore, the correct answer is 144 degrees.

Karanta bayanin darasi akan Angles (WAEC)

Angles

Duba Amsa

Tambaya 40 Rahoto

If $\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3}}$ is simplified as m + n$\sqrt{6}$, find the value of (m + n)

$\frac{1}{3}$

$\frac{2}{3}$

1$\frac{1}{3}$ Amsar da ta dace

1$\frac{2}{3}$

Bayanin Amsa

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Surds (radicals)

Duba Amsa

Tambaya 41 Rahoto

In the diagram, O is the centre, $\bar{RT}$ is a diameter, < PQT = 33$^o$ and <TOS = 76$^o$. Using the diagram, calculate the value of angle PTR.

73^o

67^o

57^o Amsar da ta dace

37^o

Bayanin Amsa

Karanta bayanin darasi akan Number Bases (WAEC)

Karanta bayanin darasi akan Coordinate Geometry (WAEC)

Number Bases

Coordinate Geometry

Duba Amsa

Tambaya 42 Rahoto

The volume of a cone of height 3cm is 38$\frac{1}{2}$cm³. Find the radius of its base. [Take $\pi = \frac{22}{7}$]

3.0cm

3.5cm Amsar da ta dace

4.0cm

4.5cm

Bayanin Amsa

Duba Amsa

Tambaya 43 Rahoto

A trader bought 100 oranges at 5 for N40.00 and 20 for N120.00. Find the profit or loss percent

20% profit

20% loss

25% profit

25% loss Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Number Bases (WAEC)

Number Bases

Duba Amsa

Tambaya 44 Rahoto

PQRT is square. If x is the midpoint of PQ, Calculate correct to the nearest degree, LPXS

53^o

55^o

63^o Amsar da ta dace

65^o

Bayanin Amsa

Duba Amsa

Tambaya 45 Rahoto

If 2ⁿ = y, Find 2$^{(2 + \frac{n}{3})}$

4y$^\frac{1}{3}$ Amsar da ta dace

4y$^-3$

2y$^\frac{1}{3}$

2y$^-3$

Bayanin Amsa

We know that 2$^n$ = y. Therefore, 2$^{(2 + \frac{n}{3})}$ can be written as 2$^2$ × 2$^{\frac{n}{3}}$ = 4 × 2$^{\frac{n}{3}}$. Substituting 2$^n$ = y in the above equation, we get: 2$^{(2 + \frac{n}{3})}$ = 4y$^\frac{1}{3}$ Therefore, the answer is (a) 4y$^\frac{1}{3}$.

Karanta bayanin darasi akan Indices (WAEC)

Karanta bayanin darasi akan Logarithms (WAEC)

Indices

Logarithms

Duba Amsa






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WAEC SSCE - General Mathematics - 2015 (Objective)