WAEC SSCE - General Mathematics - 2000
Tambaya 1 Rahoto
Given that p varies directly as q while q varies inversely as r, which of the following statements is true?
Bayanin Amsa
The statement "p varies inversely as r" is true. This means that as r increases, p decreases, and as r decreases, p increases. We know that p varies directly as q, so as q increases, p increases, and as q decreases, p decreases. Combining these two statements, we can say that as q increases, r decreases, and as q decreases, r increases. Therefore, p and r have an inverse relationship, meaning that p varies inversely as r. The other options are not true based on the information given.
Tambaya 2 Rahoto
Given that the logarithm of a number is \(\bar{1}.8732\), find, correct to 2 significant figures the square root of the number.
Bayanin Amsa
Tambaya 3 Rahoto
The arc of a circle 50 cm long, subtends angle of 75° at the center of the circle. Find correct to 3 significant figures, the radius of the circle. Take \(\pi = \frac{22}{7}\)
Bayanin Amsa
To find the radius of the circle, we need to use the formula: $$ \text{length of arc} = \theta \frac{\pi r}{180} $$ where $\theta$ is the angle subtended by the arc at the center of the circle, $r$ is the radius of the circle, and $\pi$ is the mathematical constant pi. Substituting the given values, we get: $$ 50 = 75 \times \frac{\pi r}{180} $$ Simplifying the equation, we get: $$ r = \frac{50 \times 180}{75 \times \pi} \approx 38.2 \text{ cm (to 3 significant figures)} $$ Therefore, the correct answer is 38.2cm.
Tambaya 4 Rahoto
Given that \(log_4 x = -3\), find x.
Bayanin Amsa
The given equation is \(\log_4 x = -3\). We know that \(\log_a b = c\) is equivalent to \(a^c = b\). Using this property, we have: $$4^{-3} = x$$ Simplifying the right-hand side, we get: $$\frac{1}{4^3} = x$$ Since \(4^3 = 64\), we have: $$\frac{1}{64} = x$$ Therefore, the answer is \(\frac{1}{64}\).
Tambaya 5 Rahoto
In the diagram, PQS is a circle with center O. RST is a tangent at S and ?SOP = 96o. Find ?PST
Bayanin Amsa
Tambaya 6 Rahoto
The area of a circle is 38.5cm2. Find its diameter [take \(\pi = \frac{22}{7}\)]
Bayanin Amsa
The formula for the area of a circle is given by A = πr2, where A is the area and r is the radius of the circle. To find the diameter, we need to first find the radius. Given A = 38.5 cm2 and π = 22/7, we can write: A = πr2 38.5 = (22/7) * r2 Multiplying both sides by 7/22, we get: r2 = (38.5 * 7) / 22 r2 = 12.25 Taking the square root of both sides, we get: r = 3.5 The diameter is twice the radius, so: diameter = 2 * radius = 2 * 3.5 = 7 cm Therefore, the answer is option (C) 7cm.
Tambaya 7 Rahoto
Each side of a regular convex polygon subtends an angle of 30° at its center. Calculate each interior angle
Bayanin Amsa
A regular convex polygon has equal angles and equal sides. Therefore, to calculate the interior angle of a regular convex polygon, we can use the formula: Interior angle = (n - 2) x 180 / n where n is the number of sides of the polygon. In this case, we know that each side subtends an angle of 30° at the center, which means that there are 12 sides in the polygon (since 360° / 30° = 12). Substituting this value into the formula, we get: Interior angle = (12 - 2) x 180 / 12 = 10 x 180 / 12 = 150° Therefore, each interior angle of the regular convex polygon is 150°. The answer is.
Tambaya 8 Rahoto
Calculate the standard deviation of the following marks; 2, 3, 6, 2, 5, 0, 4, 2
Bayanin Amsa
To calculate the standard deviation of a set of data, we need to follow these steps: 1. Find the mean (average) of the data. 2. For each data point, subtract the mean and square the result. 3. Find the average of the squared differences (this is called the variance). 4. Take the square root of the variance to find the standard deviation. So, for the given data set {2, 3, 6, 2, 5, 0, 4, 2}, we can first find the mean: (mean) = (2+3+6+2+5+0+4+2)/8 = 24/8 = 3 Next, we can find the squared differences from the mean for each data point: (2-3)^2 = 1 (3-3)^2 = 0 (6-3)^2 = 9 (2-3)^2 = 1 (5-3)^2 = 4 (0-3)^2 = 9 (4-3)^2 = 1 (2-3)^2 = 1 Then we can find the average of these squared differences: (variance) = (1+0+9+1+4+9+1+1)/8 = 26/8 = 3.25 Finally, we take the square root of the variance to find the standard deviation: (standard deviation) = sqrt(3.25) ≈ 1.8 Therefore, the standard deviation of the given marks is approximately 1.8. So the correct option is (c) 1.8.
Tambaya 9 Rahoto
The ages of three men are in the ratio 3:4:5. If the difference between the ages of the oldest and youngest is 18 years, find the sum of the ages of the three men
Bayanin Amsa
Let the ages of the three men be 3x, 4x, and 5x (since their ages are in the ratio 3:4:5). We know that the difference between the ages of the oldest and youngest is 18 years, so: 5x - 3x = 18 2x = 18 x = 9 Therefore, the ages of the three men are: - 3x = 27 years - 4x = 36 years - 5x = 45 years The sum of their ages is: 27 + 36 + 45 = 108 years. Therefore, the correct answer is 108 years.
Tambaya 10 Rahoto
The bar chart shows the distribution of marks scored by a group of students in a test. Use the chart to answer the question below
How many students took the test?
Bayanin Amsa
Tambaya 11 Rahoto
The length of the parallel sides of a trapezium are 5cm and 7cm. If its area is 120cm\(^2\), find the perpendicular distance between the parallel sides
Bayanin Amsa
The area of a trapezium is given by the formula: $$\text{Area} = \frac{1}{2}(a+b)h$$ where $a$ and $b$ are the parallel sides of the trapezium and $h$ is the perpendicular distance between them. In this question, we are given that $a = 5\text{ cm}$, $b = 7\text{ cm}$, and $\text{Area} = 120\text{ cm}^2$. We need to find $h$. Using the formula above, we can rearrange it to get $h$ as the subject: $$h = \frac{2\text{Area}}{a+b}$$ Substituting the given values, we get: $$h = \frac{2(120)}{5+7} = \frac{240}{12} = 20$$ Therefore, the perpendicular distance between the parallel sides is $20\text{ cm}$. The answer is not among the options given, so it is likely there was an error in one of the values or options presented.
Tambaya 13 Rahoto
Which of the following bearings is equivalent to S50°W?
Bayanin Amsa
S50°W is the same as 220°, so the correct answer is. The bearing S50°W represents an angle of 50 degrees west of south, which is equivalent to a bearing of 180° + 50° = 230°. Therefore, is the correct answer.
Tambaya 14 Rahoto
Given that \(27^{(1+x)}=9,)\ find x
Bayanin Amsa
Taking the natural logarithm of both sides, we have: \begin{align*} \ln(27^{1+x}) &= \ln 9 \\ (1+x)\ln 27 &= \ln 9 \\ (1+x)\ln(3^3) &= \ln(3^2) \\ (1+x)(3\ln 3) &= 2\ln 3 \\ 1+x &= \frac{2\ln 3}{3\ln 3} \\ 1+x &= \frac{2}{3} \\ x &= \frac{2}{3} - 1 \\ x &= -\frac{1}{3} \end{align*} Therefore, the answer is (b) \(\frac{-1}{3}\).
Tambaya 15 Rahoto
In the diagram, POS and ROT are straight lines, OPQR is a parallelogram. |OS| = |OT| and ∠OST = 50°. Calculate ∠OPQ.
Bayanin Amsa
Tambaya 16 Rahoto
Given that \(x = -\frac{1}{2}and \hspace{1mm} y = 4 \hspace{1mm} evaluate \hspace{1mm} 3x^2y+xy^2\)
Bayanin Amsa
Substituting the given values of x and y into the expression, we have: \begin{align*} 3x^2y+xy^2 &= 3\left(-\frac{1}{2}\right)^2(4) + \left(-\frac{1}{2}\right)(4)^2\\ &= 3\left(\frac{1}{4}\right)(4) + \left(-\frac{1}{2}\right)(16)\\ &= 3 + (-8)\\ &= -5 \end{align*} Therefore, the answer is -5.
Tambaya 17 Rahoto
Given that (2x + 7) is a factor of \(2x^2 + 3x - 14\), find the other factor
Bayanin Amsa
To find the other factor, we need to use polynomial long division or synthetic division. But since the problem already tells us that (2x + 7) is a factor, we can use this information to simplify the problem. If (2x + 7) is a factor of \(2x^2 + 3x - 14\), then we know that when we divide \(2x^2 + 3x - 14\) by (2x + 7), the remainder is zero. This gives us the equation: \(\frac{2x^2 + 3x - 14}{2x + 7} = x - 2\) Therefore, the other factor is (x - 2).
Tambaya 18 Rahoto
A man bought a television set on hire purchase for N25,000, out of which he paid N10,000, if he is allowed to pay the balance in eight equal installments, find the value of each installment.
Bayanin Amsa
The man bought a television set on hire purchase for N25,000 and paid N10,000 initially. Therefore, the balance he needs to pay is N25,000 - N10,000 = N15,000. Since he is allowed to pay the balance in eight equal installments, the value of each installment will be: N15,000 ÷ 8 = N1875. Therefore, the value of each installment is N1875. The correct option is (c) N1875.
Tambaya 19 Rahoto
Make t the subject of formula \(k = m\sqrt{\frac{t-p}{r}}\)
Bayanin Amsa
To make t the subject of the given formula, we need to isolate t on one side of the equation. Let's begin by squaring both sides of the equation: \begin{align*} k &= m\sqrt{\frac{t-p}{r}}\\ k^2 &= m^2\frac{t-p}{r}\\ k^2r &= m^2(t-p)\\ \end{align*} Next, we'll isolate the t term by dividing both sides of the equation by m^2 and adding p: \begin{align*} k^2r &= m^2(t-p)\\ \frac{k^2r}{m^2}+p &= t\\ \end{align*} Therefore, we have: $$t = \frac{k^2r}{m^2}+p$$ Hence, the answer is option $\mathbf{(B)}$ $\frac{rk^2+pm^2}{m^2}$.
Tambaya 20 Rahoto
Find the missing number in the addition of the following numbers, in base seven
\(\begin{matrix}
4 & 3 & 2 & 1\\
1 & 2 & 3 & 4\\
* & * & * & *\\
1&2&3&4&1
\end{matrix}\)
Bayanin Amsa
Tambaya 21 Rahoto
If the interior angles of hexagon are 107°, 2x°, 150°, 95°, (2x-15)° and 123°, find x.
Bayanin Amsa
The sum of the interior angles of a hexagon is given by the formula (n-2) x 180, where n is the number of sides of the polygon. Therefore, for a hexagon (6 sides), the sum of the interior angles is (6-2) x 180 = 720 degrees. We are given five of the six interior angles of the hexagon: 107°, 2x°, 150°, 95°, and (2x-15)°. We can use these angles to set up an equation and solve for x: 107° + 2x° + 150° + 95° + (2x-15)° + sixth angle = 720° Simplifying, we get: 339° + 4x = 720° Subtracting 339 from both sides, we get: 4x = 381° Dividing both sides by 4, we get: x = 95.25° None of the answer choices match this exact value, but we can round it to the nearest degree to get 95°, which corresponds to choice (b). Therefore, the answer is: - \(65^{\circ}\)
Tambaya 23 Rahoto
The height of a pyramid on square base is 15cm. if the volume is 80cm^3, find the area of the square base.
Bayanin Amsa
The volume of a pyramid is given by the formula V = (1/3)Bh, where B is the area of the base and h is the height of the pyramid. In this question, we are given the volume of the pyramid and its height, and we are asked to find the area of the square base. Let's substitute the given values into the formula for the volume of a pyramid: 80 = (1/3)B(15) Multiplying both sides by 3 gives: 240 = 15B Dividing both sides by 15 gives: 16 = B So the area of the square base is 16 cm2. Therefore, the correct answer is: 16cm2.
Tambaya 24 Rahoto
Which of the following statement is not true about a rectangle? I.Each diagonal cuts the rectangle into two congruent triangles. II. A rectangle has four lines of symmetry III. The diagonals intersect at right angles
Bayanin Amsa
Tambaya 25 Rahoto
In the diagram, |PQ| = |PS| Which of the following statements is true?
Bayanin Amsa
Since |PQ| = |PS|, we know that triangle PQS is an isosceles triangle. Therefore, the angles opposite the equal sides are equal, which means that ?QPS = ?PSQ.
Now, let's look at the answer choices:
- ?QPS = QRS: We cannot determine this from the information given. We do not know anything about the angle QRS.
- |PO| = |RO|: We cannot determine this from the information given. We do not have any information about points P, O, and R.
- QR||PS: We cannot determine this from the information given. We do not have any information about whether QR and PS are parallel or not.
- ?PQR = ?PSR: Since |PQ| = |PS|, we know that triangle PQS is an isosceles triangle. Therefore, ?PQS = ?PSQ. Also, since |PQ| = |PS| and PQRS is a parallelogram, we know that |QR| = |PS|. Therefore, triangle PSR is also an isosceles triangle, and ?SPR = ?PSR. Finally, since PQRS is a parallelogram, we know that ?PQR = ?SPR. Therefore, ?PQR = ?PSR.
So the correct answer is ?PQR = ?PSR.
Tambaya 26 Rahoto
Simplify \(\left(\frac{16}{81}\right)^{-\frac{3}{4}}\times \sqrt{\frac{100}{81}}\)
Bayanin Amsa
We can simplify the given expression using the rules of exponents and radicals as follows: \begin{align*} \left(\frac{16}{81}\right)^{-\frac{3}{4}}\times \sqrt{\frac{100}{81}} &= \left(\frac{81}{16}\right)^{\frac{3}{4}}\times \frac{10}{9} \\ &= \left[\left(\frac{3^4}{2^4}\right)^{\frac{1}{4}}\right]^3 \times \frac{10}{9} \\ &= \left(\frac{3}{2}\right)^3 \times \frac{10}{9} \\ &= \frac{27 \times 10}{8 \times 9} \\ &= \frac{15}{4} \end{align*} Therefore, the simplified expression is \(\frac{15}{4}\). So, the answer is option (D).
Tambaya 27 Rahoto
A tap leaks at the rate of 2cm\(^3\) per seconds. How long will it take the tap to fill a container of 45 liters capacity? (1 liters = 1000cm\(^3\))
Bayanin Amsa
The container has a capacity of 45 liters = 45,000 cm\(^3\). The tap leaks at the rate of 2cm\(^3\) per second. Therefore, the time taken to fill the container can be found by dividing the volume of the container by the rate of the tap: Time = Volume / Rate Time = 45,000 cm\(^3\) / 2 cm\(^3\)/s Time = 22,500 seconds We can convert the seconds to hours and minutes as follows: 22,500 seconds = 6 hours 15 minutes Therefore, the answer is 6hr 15min.
Tambaya 28 Rahoto
Form an inequality for a distance d meters which is more than 18m, but not more than 23m
Bayanin Amsa
The inequality for a distance d meters which is more than 18m, but not more than 23m can be written as: 18 < d ≤ 23 Here, the lower limit is exclusive because the distance should be more than 18m and the upper limit is inclusive because the distance should not be more than 23m. This means that any value of d that is greater than 18m and less than or equal to 23m satisfies this inequality. Therefore, the correct option is: 18 < d ≤ 23
Tambaya 29 Rahoto
In the diagram, AB is a vertical pole and BC is horizontal. If |AC| = 10m and |BC| = 5m, calculate the angle of depression of C from A
Bayanin Amsa
The angle of depression of C from A is the angle formed between the line of sight from A to C and the horizontal line. This angle can be found using trigonometry. First, we can find the length of AB by using the Pythagorean theorem: |AB| = √(|AC|² - |BC|²) = √(10² - 5²) = √75 = 5√3 Next, we can find the tangent of the angle of depression: tan(θ) = |BC| / |AB| = 5 / (5√3) = √3 / 3 Finally, we can find the angle itself by taking the inverse tangent (or arctangent) of this value: θ = tan⁻¹(√3 / 3) ≈ 30° Therefore, the angle of depression of C from A is approximately 30°. The answer closest to this value is, 60°.
Tambaya 30 Rahoto
A tree is 8km due south of a building. Kofi is standing 8km west of the tree. Find the bearing of Kofi from the building
Bayanin Amsa
To find the bearing of Kofi from the building, we need to draw a diagram and use trigonometry.
K
|
|
|
T
/
/
/
B
We know that the tree is 8km due south of the building, so we can draw a line segment from B to T, and label it 8km.
K
|
|
|
T
/ \
/ \
/ \
B-------T
8km
We also know that Kofi is standing 8km west of the tree, so we can draw a line segment from K to T, and label it 8km.
K
|
|
| 8km
T
/ \
/ \
/ \
B-------T
8km
Now, we can use trigonometry to find the bearing of Kofi from the building. We know that Kofi is west of the tree, so the bearing we want is the angle between the line segment BT and the line segment BK. We can use the tangent function to find this angle:
tan(theta) = opposite / adjacent
tan(theta) = BT / BK
tan(theta) = 8km / 8km
theta = tan-1(1)
theta ≈ 45°
So the angle between BT and BK is approximately 45 degrees. Since Kofi is west of the tree, we know that the bearing of Kofi from the building is 180 degrees plus 45 degrees, or 225 degrees.
Therefore, the answer is (c) 225o.
Tambaya 31 Rahoto
The bar chart shows the distribution of marks scored by a group of students in a test. Use the chart to answer the question below
How many students scored 4 marks and above?
Bayanin Amsa
Looking at the bar chart, we can see that the horizontal axis represents the marks scored by the students, and the vertical axis represents the number of students who scored those marks. To find out how many students scored 4 marks and above, we need to add up the number of students who scored in the bars corresponding to 4, 5, 6, 7, and 8 marks. Adding up the heights of these bars, we get: 2 + 4 + 5 + 4 + 2 = 17 Therefore, 17 students scored 4 marks and above in the test. Hence, the answer is option D: 17.
Tambaya 32 Rahoto
In the diagram, PQRS is a circle center O. PQR is a diameter and ∠PRQ = 40°. Calculate ∠QSR.
Bayanin Amsa
Since PQR is a diameter, it follows that ∠PQR = 90°. Since the sum of the angles in a triangle is 180°, then ∠QRP = 180° - 90° - 40° = 50°. Angles in the same segment of a circle are equal. Therefore, ∠QSR = ∠QRP = 50°. Hence, the answer is option (D) 50°.
Tambaya 33 Rahoto
If \(104_x = 68\), find the value of x
Bayanin Amsa
To find the value of x, we need to convert the given number from base-x to base-10, and then solve for x. First, let's write out what the given number means in expanded form: \begin{align*} 104_x &= 1 \times x^2 + 0 \times x^1 + 4 \times x^0 \\ &= x^2 + 4 \end{align*} We're also told that this equals 68, so we can set up an equation: $$x^2 + 4 = 68$$ Solving for x, we get: \begin{align*} x^2 &= 64 \\ x &= \pm 8 \end{align*} Since x is the base of a number system, it must be a positive integer. Therefore, the only possible solution is x = 8. Therefore, the value of x is 8.
Tambaya 34 Rahoto
What fraction must be subtracted from the sum of \(2\frac{1}{6}\) and \(2\frac{7}{12}\) to give \(3\frac{1}{4}\)?
Tambaya 35 Rahoto
Find the equation whose roots are -8 and 5
Tambaya 36 Rahoto
Which of the following numbers is perfect cube?
Bayanin Amsa
To determine which of the given numbers is a perfect cube, we need to find the cube root of each number and check if it is a whole number or not. The cube root of a number is a value that, when cubed, gives the original number. For example, the cube root of 27 is 3, because 3 x 3 x 3 = 27. Using a calculator or by hand, we can find that: - The cube root of 350 is approximately 7.327, which is not a whole number. - The cube root of 504 is approximately 8.005, which is not a whole number. - The cube root of 950 is approximately 9.685, which is not a whole number. - The cube root of 1728 is exactly 12, which is a whole number. Therefore, the only number that is a perfect cube from the options given is 1728.
Tambaya 37 Rahoto
Express \(\frac{7}{19}\) as a percentage, correct to one decimal place
Bayanin Amsa
To express a fraction as a percentage, we can multiply the fraction by 100. So, \begin{align*} \frac{7}{19} \times 100 &= 36.842 \\ &\approx 36.8 \quad \text{(to one decimal place)}. \end{align*} Therefore, \(\frac{7}{19}\) as a percentage, correct to one decimal place, is 36.8%. Therefore, the correct option is (c) 36.8%.
Tambaya 38 Rahoto
simplify \(\frac{10}{\sqrt{32}}\)
Bayanin Amsa
We can simplify \(\frac{10}{\sqrt{32}}\) by rationalizing the denominator, which means we multiply both the numerator and denominator by the same number so that the denominator becomes a rational number (i.e., a number that can be written as a fraction). In this case, we can multiply the numerator and denominator by \(\sqrt{32}\), since \(\sqrt{32} = \sqrt{16 \times 2} = \sqrt{16} \times \sqrt{2} = 4\sqrt{2}\). So we have: $$\frac{10}{\sqrt{32}} \times \frac{\sqrt{32}}{\sqrt{32}} = \frac{10\sqrt{32}}{32} = \frac{5\sqrt{2}}{4}$$ Therefore, the answer is \(\frac{5}{4}\sqrt{2}\).
Tambaya 39 Rahoto
Simplify \(\frac{1}{x-3}-\frac{3(x-1)}{x^2 - 9}\)
Tambaya 40 Rahoto
The probabilities that Kodjo and Adoga pass an examination are \(\frac{3}{4}\) and \(\frac{3}{5}\) respectively. Find the probability of both boys failing the examination
Bayanin Amsa
The probability of Kodjo passing the exam is \(\frac{3}{4}\) and the probability of Adoga passing the exam is \(\frac{3}{5}\). To find the probability of both boys failing the exam, we need to find the probability of Kodjo failing the exam and the probability of Adoga failing the exam, and then multiply these probabilities together. The probability of Kodjo failing the exam is \(1 - \frac{3}{4} = \frac{1}{4}\). Similarly, the probability of Adoga failing the exam is \(1 - \frac{3}{5} = \frac{2}{5}\). Therefore, the probability of both boys failing the exam is: $$ \frac{1}{4} \times \frac{2}{5} = \frac{2}{20} = \frac{1}{10} $$ So, the correct answer is \(\frac{1}{10}\).
Tambaya 41 Rahoto
If x varies inversely as y and \(x = \frac{2}{3}\) when y = 9, find the value of y when \(x=\frac{3}{4}\)
Bayanin Amsa
The problem statement tells us that x and y are inversely proportional, which means that their product is constant. We can write this relationship as: xy = k where k is a constant. We are also given that when y = 9, x = 2/3. Substituting these values into the equation above, we get: (2/3)(9) = k k = 6 Now we can use this value of k to find y when x = 3/4: (3/4)y = 6 y = (6 x 4)/3 y = 8 Therefore, when x = 3/4, y = 8. So the answer is option (D).
Tambaya 42 Rahoto
A car moves at an average speed of 30kmh\(^{-1}\), how long does it take to cover 200 meters?
Bayanin Amsa
To find the time taken by the car to cover 200 meters, we need to use the formula: time = distance ÷ speed Here, the distance is 200 meters and the speed is 30 km/h. However, we need to convert the speed to meters per second (m/s) to get the answer in seconds. We know that 1 km = 1000 meters and 1 hour = 3600 seconds. So, 30 km/h = (30 x 1000) m / (3600 s) = 8.33 m/s (approx.) Now, we can substitute the values into the formula: time = distance ÷ speed = 200 m ÷ 8.33 m/s ≈ 24 seconds Therefore, the car takes 24 seconds to cover 200 meters. Hence, the answer is 24 sec.
Tambaya 43 Rahoto
A tree is 8km due south of a building. Kofi is standing 8km west of the tree. How far is Kofi from the building?
Bayanin Amsa
We can use the Pythagorean theorem to solve this problem. Kofi is 8 km west of the tree, which means he is directly north of the building. We can draw a right-angled triangle with the building at the southeast corner, the tree at the southwest corner, and Kofi at the north corner. The distance between the building and the tree is the hypotenuse of the right-angled triangle, and the distance between Kofi and the tree is one of the legs. Using the Pythagorean theorem, we can find the length of the hypotenuse: hypotenuse² = leg₁² + leg₂² In this case, leg₁ is 8 km (the distance between Kofi and the tree), and leg₂ is 8 km (the distance between the tree and the building): hypotenuse² = 8² + 8² hypotenuse² = 128 hypotenuse = √128 hypotenuse = 8√2 Therefore, Kofi is 8√2 km away from the building. Option C, 8√2km, is the correct answer.
Tambaya 44 Rahoto
Solve the equation \(3y^2 = 27y\)
Bayanin Amsa
To solve the equation \(3y^2 = 27y\), we can begin by factoring out the common factor of 3y from the left side of the equation: $$3y^2 - 27y = 0$$ Next, we can factor out 3y from each term: $$3y(y - 9) = 0$$ This equation can be solved by the Zero Product Property, which states that if the product of two factors is equal to zero, then at least one of the factors must be zero. Therefore, we have: $$3y = 0 \text{ or } y - 9 = 0$$ Solving each equation for y, we get: $$y = 0 \text{ or } y = 9$$ So the solution to the equation \(3y^2 = 27y\) is: $$y = 0 \text{ or } y = 9$$ Therefore, the correct answer is: y = 0 or 9.
Tambaya 45 Rahoto
Find the value of x such that the expression \(\frac{1}{x}+\frac{4}{3x}-\frac{5}{6x}+1\) equals zero
Tambaya 46 Rahoto
A bicycle wheel of radius 42cm is rolled over a distance 66 meters. How many revolutions does it make?[Take \(\pi = \frac{22}{7}\)]
Bayanin Amsa
To calculate the number of revolutions made by the bicycle wheel, we need to find out the distance the wheel travels in one revolution, and then divide the total distance traveled by this distance. The circumference of the wheel is given by the formula C = 2πr, where r is the radius of the wheel. Substituting the given value, we get: C = 2 x (22/7) x 42 cm C = 264 cm Therefore, the distance the wheel travels in one revolution is 264 cm. To find the number of revolutions made by the wheel over a distance of 66 meters, we first need to convert 66 meters to centimeters. 1 meter = 100 centimeters, so 66 meters = 66 x 100 = 6600 centimeters. Now, we can find the number of revolutions made by the wheel by dividing the distance traveled by the distance traveled in one revolution: Number of revolutions = Total distance traveled / Distance traveled in one revolution Number of revolutions = 6600 cm / 264 cm Number of revolutions = 25 Therefore, the bicycle wheel makes 25 revolutions.
Tambaya 47 Rahoto
If the simple interest on N2000 after 9 months is N60, at what rate per annum is the interest charged?
Bayanin Amsa
Given: Simple Interest (SI) = N60, Principal (P) = N2000, Time (T) = 9 months. We can use the formula for simple interest to find the interest rate per annum. Simple Interest = (P * R * T) / 100 Where R is the interest rate per annum. Converting the time in months to years, we get: T = 9/12 = 0.75 years Substituting the values in the formula, we get: 60 = (2000 * R * 0.75) / 100 Simplifying the equation, we get: R = (60 * 100) / (2000 * 0.75) R = 4 Therefore, the interest rate per annum is 4%. Note that this is a simple interest calculation and not a compound interest calculation.
Tambaya 48 Rahoto
Express 398753 correct to three significant figures
Bayanin Amsa
To express 398753 correct to three significant figures, we need to look at the first three digits of the number and round it off accordingly. The first three digits of 398753 are 398. The fourth digit, which is 7, is greater than or equal to 5, so we need to round up the third digit. Therefore, when we round off 398753 to three significant figures, we get: $$ 398753 \approx 399000 $$ So, the correct answer is 399000.
Tambaya 49 Rahoto
(a) Two places X and Y on the equator are on longitudes 67°E and 123°E respectively. (i) What is the distance between them along the equator? (ii) How far from the North pole is X? [Take \(\pi = \frac{22}{7}\) and radius of earth = 6400km].
(b)
I In the diagram, PQR is a circle centre O. N is the mid-point of chord PQ. |PQ| = 8cm, |ON| = 3cm and < ONR = 20°. Calculate the size of < ORN to the nearest degree.
Tambaya 50 Rahoto
(a) A man earns N150,000 per annum. He is allowed a tax free pay on N40,000. If he pays 25 kobo in the naira as tax on his taxable income, how much has he left?
(b) A bookshop has 650 copies of a book for sale. The books were marked at N75 per copy in order to make a profit of 30%. A bookseller bought 300 copies at 5% discount. If the remaining copies are sold at N75 each, calculate the percentage profit the bookshop would make on the whole.
Bayanin Amsa
None
Tambaya 51 Rahoto
(a) A cylindrical well of radius 1 metre is dug out to a depth of 8 metres. (i) calculate, in m\(^{3}\), the volume of soil dug out ; (ii) if the soil is used to raise the level of rectangular floor of a room 4m by 12m, calculate, correct to the nearest cm, the thickness of the new layer of soil. [Take \(\pi = \frac{22}{7}\)].
(b) 
The diagram shows a quadrilateral ABCD in which < DAB is a right- angle. |AB| = 3.3 cm, |BC| = 3.9 cm, |CD| = 5.6 cm. (i) find the length of BD. (ii) show that < BCD = 90°.
Bayanin Amsa
None
Tambaya 52 Rahoto
(a) Given the expression \(y = ax^{2} - bx - 12\) , find the values of x when a = 1, b = 2 and y = 3.
(b) If \(\sqrt{x^{2} + 1} = \frac{5}{4}\), find the positive value of x.
Bayanin Amsa
None
Tambaya 53 Rahoto
(a) A surveyor walks 100m up a hill which slopes at an angle of 24° to the horizontal. Calculate, correct to the nearest metre, the height through which he rises.
(b) 
In the diagram, ABC is an isosceles triangle. |AB| = |AC| = 5 cm, and |BC| = 8 cm. Calculate, correct to the nearest degree, < BAC.
(c) Two boats, 70 metres apart and on opposite sides of a light-house, are in a straight line with the light-house. The angles of elevation of the top of the light-house from the two boats are 71.6° and 45°. Find the height of the light-house. [Take \(\tan 71.6° = 3\)].
Bayanin Amsa
None
Tambaya 54 Rahoto
(a) Given that \(\cos x = 0.7431, 0° < x < 90°\), use tables to find the values of : (i) \(2 \sin x\) ; (ii) \(\tan \frac{x}{2}\).
(b) The interior angles of a pentagon are in ratio 2 : 3 : 4 : 4 : 5. Find the value of the largest angle.
Bayanin Amsa
None
Tambaya 55 Rahoto
(a) The mean of 1, 2, x, 11, y, 14, arranged in ascending order, is 8 and the median is 9. Find the values of x and y.
(b) 
In the diagram, MN || PQ, |LM| = 3cm and |LP| = 4cm. If the area of \(\Delta\) LMN is 18\(cm^{2}\), find the area of the quadrilateral MPQN.
None
Bayanin Amsa
None
Tambaya 57 Rahoto
(a) The first term of an Arithmetic Progression (A.P) is 8. The ratio of the 7th term to the 9th term is 5 : 8. Calculate the common difference of the progression.
(b) A sphere of radius 2 cm is of mass 11.2g. Find (i) the volume of the sphere ; (ii) the density of the sphere ; (iii) the mass of a sphere of the same material but with radius 3cm. [Take \(\pi = \frac{22}{7}\)].
Tambaya 58 Rahoto
(a)
| Limes | Apples | |
| Good | 10 | 8 |
| Bad | 6 | 6 |
The table shows the number of limes and apples of the same size in a bag. If two of the fruits are picked at random, one at a time, without replacement, find the probability that : (i) both are good limes ; (ii) both are bad fruits ; (iii) one is a good apple and the other a bad lime.
(b) Solve the equation \(\log_{3} (4x + 1) - \log_{3} (3x - 5) = 2\).
Bayanin Amsa
None
Tambaya 59 Rahoto
(a) Copy and complete the following table of values for the relation \(y = x^{2} - 2x - 5\)
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| y | -2 | -6 | -2 | 3 | 10 |
(b) Draw the graph of the relation \(y = x^{2} - 2x - 5\); using a scale of 2 cm to 1 unit on the x- axis, and 2 cm to 2 units on the y- axis.
(c) Using the same axes, draw the graph of \(y = 2x + 3\).
(d) Obtain in the form \(ax^{2} + bx + c = 0\) where a, b and c are integers, the equation which is satisfied by the x- coordinate of the points of intersection of the two graphs.
(e) From your graphs, determine the roots of the equation obtained in (d) above.
Tambaya 60 Rahoto
(a) Evaluate and express your answer in standard form : \(\frac{4.56 \times 3.6}{0.12}\)
(b) Without using mathematical tables or calculator, evaluate \((73.8)^{2} - (26.2)^{2}\).
(c) Simplify \(\sqrt{1\frac{19}{81}}\), expressing your answer in the form \(\frac{a}{b}\) where a and b are positive integers.
Bayanin Amsa
None
Tambaya 61 Rahoto
The pie chart shows the distribution of marks scored by 200 pupils in a test.
(a) How many pupils scored : (i) between 41 and 50 marks? ; (ii) above 80 marks ?
(b) What fraction of the pupils scored at most 50 marks?
(c) What is the modal class?
