WAEC SSCE - General Mathematics - 1995
Tambaya 1 Rahoto
In the diagram above, AB//CD, the bisector of ?BAC and ?ACD meet at E. Find the value of ?AEC
Bayanin Amsa
Since AB is parallel to CD, then we have: \begin{align*} \angle AEC &= \angle AED + \angle DEC \\ &= \frac{1}{2} \angle BAC + \frac{1}{2} \angle ACD \\ &= \frac{1}{2}(\angle BAC + \angle ACD) \\ &= \frac{1}{2}(180^\circ) \qquad \text{(because } \angle BAC + \angle ACD = 180^\circ)\\ &= 90^\circ \end{align*} Therefore, the value of $\angle AEC$ is $\boxed{90^\circ}$.
Tambaya 2 Rahoto
The bearing of two points Q and R from a point P are 030° and 120° respectively, lf /PQ/ = 12 m and /PR/ = 5 m, find the distance QR.
Bayanin Amsa
Tambaya 3 Rahoto
In the diagram above, the value of angles b + c is
Bayanin Amsa
Since angles b and c are alternate angles, they are equal. Therefore, b + c = 2b or 2c. But since a straight line forms a straight angle of 180 degrees, we have that angle a + angle b + angle c = 180 degrees. Solving for b + c, we get: b + c = 180 - a Therefore, the answer is (A) 180o.
Tambaya 4 Rahoto
Two fair dice are tossed together once. Find the probability that the sum of the outcome is at least 10.
Bayanin Amsa
When two dice are tossed together, there are 6 x 6 = 36 possible outcomes. We need to find the probability that the sum of the outcome is at least 10. The only possible pairs of outcomes that sum to at least 10 are: (4, 6), (5, 5), (6, 4). Each of these pairs can occur in two different ways: (4, 6) and (6, 4), (5, 5), and (6, 4). So, there are a total of 6 possible outcomes that sum to at least 10. Therefore, the probability of getting a sum of at least 10 is 6/36, which simplifies to 1/6. Thus, the correct option is (c) 1/6.
Tambaya 5 Rahoto
lf 16/9 , x, 1, y are in Geometric Progression (GP), find the product of x and y.
Bayanin Amsa
Tambaya 6 Rahoto
Solve the following simultaneous equations: x+ y = 3/2; x - y = 5/2 and use your result to find the value of 2y + x
Bayanin Amsa
To solve the simultaneous equations x + y = 3/2 and x - y = 5/2, we can use the method of elimination. First, we need to eliminate y. We can do this by adding the two equations: (x + y) + (x - y) = 3/2 + 5/2 Simplifying, we get: 2x = 4 Dividing both sides by 2, we get: x = 2 Now that we have solved for x, we can substitute this value into one of the original equations to solve for y. Let's use the first equation: 2 + y = 3/2 Subtracting 2 from both sides, we get: y = -1/2 So the solution to the simultaneous equations is x = 2 and y = -1/2. To find the value of 2y + x, we can simply substitute the values we found: 2y + x = 2(-1/2) + 2 = 1 Therefore, the answer is 1.
Tambaya 7 Rahoto
Factorize the expression x(a - c) + y(c - a)
Bayanin Amsa
Given expression is x(a - c) + y(c - a). Rearranging this expression, we get: x(a - c) - y(a - c) We can factorize (a - c) from both terms to get: (a - c)(x - y) Therefore, the factorization of the expression x(a - c) + y(c - a) is (a - c)(x - y). The correct option is (b) (a - c)(x - y).
Tambaya 8 Rahoto
Find the nth term Un of the A.P., 11, 4, -3,....... .
Bayanin Amsa
The given sequence is an arithmetic progression (AP) since each term is obtained by subtracting 7 from the previous term. Let the first term be a = 11, and the common difference be d = -7. The nth term of an arithmetic progression can be found using the formula: Un = a + (n - 1)d Substituting the values of a and d, we get: Un = 11 + (n - 1)(-7) Simplifying, we get: Un = 11 - 7n + 7 Un = 18 - 7n Therefore, the nth term Un of the given AP is 18 - 7n. The answer is (d) Un= 18-7n.
Tambaya 9 Rahoto
Solve the equation 3x\(^2\) + 25x -18 = 0
Tambaya 10 Rahoto
The positions of two countries P and Q are (15°N, 12°E) and (65°N, 12°E) respectively. What is the difference in latitude?
Bayanin Amsa
Latitude is a measure of the north-south position of a point on the Earth's surface, measured in degrees from the Equator. In this case, country P is located at 15°N and country Q is located at 65°N. To find the difference in latitude between the two countries, we simply subtract the latitude of P from the latitude of Q: Difference in latitude = 65°N - 15°N = 50°N Therefore, the answer is 50°.
Tambaya 12 Rahoto
Find (101\(_2\))\(^2\), expressing the answer in base 2.
Bayanin Amsa
To find (1012)2, we need to square the binary number 1012. We can use the standard multiplication method taught in primary school, but we only need to remember two things: 1) 0 x anything = 0, and 2) 1 x anything = the same anything. Here's the calculation:
101 (this is 1012) x 101 (this is also 1012) ----- 101 101 --- 11001 (this is the result of (1012)2 in binary)
Therefore, the answer is 11001, which is (2510)2.
Tambaya 14 Rahoto
Express 0.000834 in standard form
Bayanin Amsa
To express a number in standard form, we write it in the form \(a \times 10^b\), where \(1 \leq a < 10\) and \(b\) is an integer. To express 0.000834 in standard form, we need to move the decimal point to the right until we obtain a number between 1 and 10. We have: $$0.000834 = 8.34 \times 10^{-4}$$ Therefore, the answer is (a) 8.34 x 10-4.
Tambaya 15 Rahoto
Find the median of the following numbers 2.64, 2.50, 2.72, 2.91 and 2.35.
Bayanin Amsa
To find the median, we first need to arrange the numbers in order from smallest to largest: 2.35, 2.50, 2.64, 2.72, 2.91 Now, we can see that the middle number is 2.64, which is the median. Therefore, the answer is: 2.64.
Tambaya 16 Rahoto
By selling some crates of soft drinks for N600.00, a dealer makes a profit of 50%. How much did the dealer pay for the drinks?
Bayanin Amsa
If the dealer makes a 50% profit, it means that the selling price is 150% (100% cost price + 50% profit) of the cost price. Let the cost price of the crates of soft drinks be x. Then, 150% of x is equal to N600.00. Mathematically, we can write it as: 150/100 * x = N600.00 Simplifying the equation, we have: x = N600.00 * 100/150 x = N400.00 Therefore, the dealer paid N400.00 for the crates of soft drinks.
Tambaya 17 Rahoto
If R = [2, 4, 6, 7] and S = [1, 2, 4, 8], then R∪S equal
Bayanin Amsa
In set theory, the union of two sets is a set containing all distinct elements from both sets. Therefore, to find R∪S, we need to combine all the elements in R and S without duplication. R = [2, 4, 6, 7] and S = [1, 2, 4, 8], so combining them without duplication gives us: R∪S = [1, 2, 4, 6, 7, 8] Therefore, the correct answer is (a) [1, 2, 4, 6, 7, 8].
Tambaya 18 Rahoto
The diagram above shows a cone with the dimensions of its frustrum indicated. Calculate the height of the cone.
Bayanin Amsa
Tambaya 19 Rahoto
In the diagram above, ?PRQ = 90°, ?QPR = 30° and /PQ/ = 10 cm. Find y.
Bayanin Amsa
We can use trigonometry to solve for y. Since we know that ?QPR = 30°, then we can determine that: sin 30° = y/PQ Solving for y: y = PQ * sin 30° y = 10 * (1/2) y = 5 Therefore, y is 5 cm.
Tambaya 20 Rahoto
In the diagram above; O is the centre of the circle and |BD| = |DC|. If ?DCB = 35°, find ?BAO.
Bayanin Amsa
We can start by using the fact that the angle at the center of the circle is twice the angle at the circumference that subtends the same arc. Therefore, since |BD| = |DC|, then angle CBD = angle BDC. Also, angle COB is equal to 2*angle BDC because it subtends the same arc as angle BDC. Therefore, angle COB = 2*angle CBD. Since triangle AOB is isosceles (|OA| = |OB|), then angle BAO = angle BOA. Using the fact that the angles in a triangle sum to 180 degrees, we can write: angle BAO + angle BOA + angle COB = 180 Substituting in the expressions we derived earlier, we get: angle BAO + angle BAO + 2*angle CBD = 180 Simplifying, we get: 2*angle BAO + 2*angle CBD = 180 angle BAO + angle CBD = 90 Now we can use the fact that angle DCB = 35 degrees and |BD| = |DC| to find angle CBD: angle DCB + angle BDC + angle CBD = 180 35 + 35 + angle CBD = 180 angle CBD = 110 Substituting this value back into the earlier equation, we get: angle BAO + 110 = 90 angle BAO = 90 - 110 = -20 Since angle BAO cannot be negative, we made an error along the way. Double-checking our work, we realize that we made a mistake when substituting in the value for angle COB. It should be equal to 2*angle DCB, not 2*angle BDC. Therefore: angle COB = 2*angle DCB = 2*35 = 70 Substituting this value back into the earlier equation, we get: angle BAO + 70 = 90 angle BAO = 20 Therefore, the answer is 20 degrees.
Tambaya 23 Rahoto
Use the graph of y = 3x\(^2\) + x - 7 above to answer the question
What is the minimum value of y?
Tambaya 24 Rahoto
A cylindrical container, closed at both ends, has a radius of 7cm and height 5cm [Take π = 22/7]
Find the total surface area of the container
Bayanin Amsa
The total surface area of a closed cylinder consists of the area of its curved surface and the area of its two circular ends. The area of the curved surface of a cylinder is given by 2πrh, where r is the radius of the base and h is the height. The area of each circular end is given by πr^2. In this case, the radius of the cylinder is 7cm and the height is 5cm. So, Area of curved surface = 2πrh = 2 × (22/7) × 7 × 5 = 220cm^2 Area of each circular end = πr^2 = (22/7) × 7^2 = 154cm^2 Total surface area = 2 × area of circular end + area of curved surface = 2 × 154 + 220 = 308 + 220 = 528cm^2 Therefore, the total surface area of the cylinder is 528cm^2. So, the correct option is (D) 528cm^2.
Tambaya 25 Rahoto
Solve the inequality: \(\frac{1}{3}(2x - 1) < 5\)
Bayanin Amsa
To solve this inequality, we need to isolate x on one side of the inequality sign. We can start by multiplying both sides by 3, which gives: 2x - 1 < 15 Next, we can add 1 to both sides: 2x < 16 Finally, we can divide both sides by 2 to get: x < 8 Therefore, the solution to the inequality is x < 8. Option (d) is the correct answer.
Tambaya 26 Rahoto
The mean of 20 observations in an experiment is 4, lf the observed largest value is 23, find the mean of the remaining observations.
Bayanin Amsa
To find the mean of the remaining observations, we need to subtract the largest value from the sum of all the observations and then divide it by the number of observations minus one. Number of observations = 20 Mean of 20 observations = 4 Largest value = 23 Sum of all observations = Mean x Number of observations = 4 x 20 = 80 Sum of remaining observations = Sum of all observations - Largest value = 80 - 23 = 57 Number of remaining observations = Number of observations - 1 = 20 - 1 = 19 Mean of remaining observations = Sum of remaining observations / Number of remaining observations = 57 / 19 = 3 Therefore, the mean of the remaining observations is 3. Hence, the correct option is (b) 3.
Tambaya 27 Rahoto
From a box containing 2 red, 6 white and 5 black balls, a ball is randomly selected. What is the probability that the selected ball is black?
Bayanin Amsa
There are a total of 2 + 6 + 5 = 13 balls in the box. Out of these, 5 are black. The probability of selecting a black ball is the number of black balls divided by the total number of balls, which is 5/13. Therefore, the answer is 5/13.
Tambaya 29 Rahoto
What is the smaller value of x for which x\(^2\) - 3x + 2= 0?
Bayanin Amsa
The given equation is a quadratic equation in standard form, which is \(ax^2 + bx + c = 0\), where \(a = 1\), \(b = -3\), and \(c = 2\). We can solve for the roots of the quadratic equation by using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Substituting the values of \(a\), \(b\), and \(c\) into the formula, we get: \[x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(2)}}{2(1)}\] \[x = \frac{3 \pm \sqrt{1}}{2}\] Thus, the solutions are: \[x_1 = \frac{3 - 1}{2} = 1 \text{ and } x_2 = \frac{3 + 1}{2} = 2\] Therefore, the smaller value of \(x\) is \(1\). Hence, the answer is 1.
Tambaya 30 Rahoto
Which of the following angles is an exterior angle of a regular polygon?
Bayanin Amsa
An exterior angle of a polygon is an angle formed by extending one of the sides of the polygon. For a regular polygon, all the exterior angles have the same measure, which can be found by dividing the total sum of the exterior angles (which is always 360 degrees) by the number of sides. In this case, we have a list of angles and we need to determine which one is an exterior angle of a regular polygon. We can start by dividing 360 degrees by each of the answer choices and see if any of the results match the definition of an exterior angle. - 360/95 = 3.789, which is not a whole number and therefore cannot be an exterior angle of a regular polygon. - 360/85 = 4.235, which is not a whole number and therefore cannot be an exterior angle of a regular polygon. - 360/78 = 4.615, which is not a whole number and therefore cannot be an exterior angle of a regular polygon. - 360/75 = 4.8, which is a whole number and therefore could be an exterior angle of a regular polygon. We can confirm this by checking if the sum of this angle and any of the interior angles of a regular polygon add up to 180 degrees. For example, a regular pentagon has interior angles of 108 degrees, and 75 + 108 = 183, which is not 180 degrees. Therefore, 75 degrees is not an exterior angle of a regular pentagon. - 360/72 = 5, which is a whole number and therefore could be an exterior angle of a regular polygon. Again, we can confirm this by checking if the sum of this angle and any of the interior angles of a regular polygon add up to 180 degrees. For example, a regular pentagon has interior angles of 108 degrees, and 72 + 108 = 180 degrees. Therefore, 72 degrees is an exterior angle of a regular pentagon. Therefore, the answer is: 72o.
Tambaya 31 Rahoto
The locus of a point which is equidistant from two given fixed points is the
Bayanin Amsa
The locus of a point which is equidistant from two given fixed points is the perpendicular bisector of the straight line joining them. This means that the locus of the point is the line that is perpendicular to the straight line joining the two given points and passes through the midpoint of that line.
Tambaya 32 Rahoto
A water tank of height \(\frac{1}{2}\) m has a square base of side \(1\frac{1}{2}\) m. lf it is filled with water from a water tanker holding 1500 litres, how many litres of water are left in the water tanker? [1000 litres = 1m\(^3\)]
Bayanin Amsa
The volume of the water tank can be calculated as follows: Volume of tank = Base area x Height = (1.5m x 1.5m) x 0.5m = 1.125 m\(^3\) Since 1000 litres = 1 m\(^3\), the volume of the tank is equivalent to 1,125 litres. Therefore, the amount of water left in the water tanker after filling the tank is: 1500 - 1125 = 375 litres. Hence, the answer is 375 litres.
Tambaya 33 Rahoto
Solve the equation (x +2)(x - 7) = 0
Bayanin Amsa
The given equation is a quadratic equation which can be solved by factoring the left-hand side of the equation: (x + 2)(x - 7) = 0 The product of two factors is equal to zero if and only if at least one of the factors is zero. So we can set each factor to zero and solve for x: x + 2 = 0 or x - 7 = 0 Solving for x in each case gives: x = -2 or x = 7 Therefore, the solutions to the equation (x + 2)(x - 7) = 0 are x = -2 and x = 7.
Tambaya 34 Rahoto
The angle of a sector of a circle is 108°. If the radius of the circle is 31/2cm, find the perimeter of the sector
Bayanin Amsa
A sector of a circle is a region bounded by two radii and an arc. The perimeter of a sector is the sum of the arc length and the lengths of the two radii. In this case, we are given that the angle of the sector is 108° and the radius of the circle is 31/2 cm. To find the arc length, we need to know the circumference of the entire circle, which is given by 2πr, where r is the radius of the circle. So, the circumference of the circle is 2π(31/2) cm = 31π cm. The angle of the sector is 108°, which is 108/360 or 3/10 of the entire circle. So, the arc length of the sector is (3/10) × 31π cm = 9.3π cm. The two radii of the sector have length 31/2 cm each. Therefore, the perimeter of the sector is the sum of the arc length and the two radii, which is 9.3π cm + 31/2 cm + 31/2 cm = (9.3π + 31) / 2 cm. This is approximately equal to 13.6 cm. Therefore, the answer is (E) 13 3/5 cm.
Tambaya 35 Rahoto
Find the value(s) of x for which the expression is undefined: \(\frac{6x - 1}{x^2 + 4x - 5}\)
Bayanin Amsa
The given expression is undefined when the denominator is equal to zero because division by zero is undefined. Therefore, we need to find the values of x that make the denominator zero. The denominator of the expression is \(x^2 + 4x - 5\). We can factor this quadratic expression as \((x + 5)(x - 1)\). So the expression is undefined when either \(x + 5\) or \(x - 1\) equals zero, since division by zero is undefined. Therefore, the values of x that make the expression undefined are \(x = -5\) or \(x = 1\). So the answer is: - -5 or 1
Tambaya 36 Rahoto
If three children shares N10.50 among themselves in ratio 6:7:8, how much is the largest share?
Bayanin Amsa
To solve this problem, we need to first add up the parts of the ratio, which gives us 6+7+8 = 21. Next, we divide the total amount shared (N10.50) by the sum of the ratio parts (21) to find the value of one part: N10.50 ÷ 21 = N0.50 Now that we know the value of one part, we can find the value of each part of the ratio by multiplying the value of one part by the corresponding number in the ratio: - 6 parts: 6 × N0.50 = N3.00 - 7 parts: 7 × N0.50 = N3.50 - 8 parts: 8 × N0.50 = N4.00 So the largest share is N4.00, which corresponds to the 8 parts of the ratio. Therefore, the correct answer is option (C) N4.00.
Tambaya 37 Rahoto
Which of the following could be the inequality illustrated in the sketch graph above?
Bayanin Amsa
Tambaya 38 Rahoto
What is the mode of the numbers 8, 10, 9, 9, 10, 8, 11, 8, 10, 9, 8 and 14?
Bayanin Amsa
The mode is the number that appears most frequently in the given set of numbers. In this case, we can see that the number 8 appears 4 times, while the other numbers appear either 2 or 1 times. Therefore, the mode of the set is 8. So, the answer is 8.
Tambaya 39 Rahoto
In the diagram above, AB//CD. What is the size of the angle marked x?
Bayanin Amsa
Tambaya 40 Rahoto
The diagonals AC and BD of a rhombus ABCD are 16cm and 12cm long respectively. Calculate the area of the rhombus.
Bayanin Amsa
In a rhombus, the diagonals are perpendicular bisectors of each other and they divide the rhombus into four congruent right-angled triangles. Therefore, each triangle has a base of 6cm and a height of 8cm (half the length of the diagonals). The area of each triangle is given by: Area = 1/2 x base x height = 1/2 x 6cm x 8cm = 24cm^2 Since there are four such triangles in the rhombus, the total area of the rhombus is given by: Area = 4 x 24cm^2 = 96cm^2 Therefore, the area of the rhombus is 96cm^2, which corresponds to the last option.
Tambaya 41 Rahoto
Without using tables, find the value of \(\frac{\sin 20°}{\cos 70°} + \frac{\cos 25°}{\sin 65°}\)
Bayanin Amsa
Tambaya 42 Rahoto
If log\(_{10}\) x = \(\bar{2}.3675\) and log\(_{10}\) y = \(\bar{2}.9738\), what is the value of x + y, correct lo three significant figures?
Bayanin Amsa
Tambaya 43 Rahoto
Use the graph of y = 3x\(^2\) + x - 7 above to answer the question
Find the roots of the equation y=3x\(^2\) + x - 7
Bayanin Amsa
Tambaya 44 Rahoto
In the diagram above. |AB| = 12cm, |AE| = 8cm, |DCl = 9cm and AB||DC. Calculate |EC|
Bayanin Amsa
In the given diagram, we can see that the two lines AB and DC are parallel. Hence, we can apply the intercept theorem or Thales' theorem which states that if a line is drawn parallel to one side of a triangle to intersect the other two sides, then it divides those sides proportionally. So, we can use this theorem to find the length of |EC| as follows: First, we can notice that the triangles ABE and DCE are similar since they are both right triangles and share the same angle at E. Therefore, we can set up a proportion between their corresponding sides: \begin{align*} \frac{|EC|}{|DC|} &= \frac{|AE|}{|AB|} \\ \frac{|EC|}{9\text{ cm}} &= \frac{8\text{ cm}}{12\text{ cm}} \\ |EC| &= \frac{8\text{ cm}}{12\text{ cm}} \times 9\text{ cm} \\ |EC| &= 6\text{ cm} \end{align*} Therefore, the length of |EC| is 6cm. Thus, the correct option is (E) 6cm.
Tambaya 45 Rahoto
A cylindrical container, closed at both ends, has a radius of 7cm and height 5cm [Take π = 22/7]
What is the volume of the container?
Bayanin Amsa
The volume of a cylinder can be calculated using the formula: V = πr2h, where V is the volume, r is the radius, and h is the height. Substituting the given values, we have: V = (22/7) x 72 x 5 V = 770 cm3 Therefore, the volume of the container is 770cm3. The answer is option (E).
Tambaya 46 Rahoto
A 120° sector of a circle of radius 21cm is bent to form a cone. What is the base radius of the cone?
Bayanin Amsa
Tambaya 48 Rahoto
If \(y \propto \frac{1}{x^2}\) and x = 3 when y = 4, find y when x = 2.
Bayanin Amsa
If we have that y is proportional to \(\frac{1}{x^2}\), then we can write: y = k\(\frac{1}{x^2}\) where k is a constant of proportionality. To find k, we can use the fact that when x = 3, y = 4: 4 = k\(\frac{1}{3^2}\) Simplifying: k = 36 Now we can use k to find y when x = 2: y = k\(\frac{1}{2^2}\) y = 9 Therefore, the answer is y = 9.
Tambaya 49 Rahoto
The frequency table shows the marks scored by 32 students in a test.
| Marks scored | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| No of students | 2 | 3 | 4 | 4 | 4 | 4 | 5 | 3 | 2 | 1 |
Find the :
(a)(i) mean ; (ii) median ; (iii) mode of the marks;
(b) percentage of the students who scored at least 8 marks.
None
Bayanin Amsa
None
Tambaya 50 Rahoto
The table shows the scores of 2000 candidates in an entrance examination into a private secondary school.
| % Mark | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 | 81-90 |
| No of pupils |
68 | 184 | 294 | 402 | 480 | 310 | 164 | 98 |
(a) Prepare a cumulative frequency table and draw the cumulative frequency curve for the distribution.
(b) Use your curve to estimate the : (i) cut off mark, if 300 candidates are to be offered admission ; (ii) probability that a candidate picked at random, scored at least 45%.
Bayanin Amsa
None
Tambaya 51 Rahoto
(a) In the diagram, PQSR and SRYZ are parallelograms and PQYZ is a straight line. If /QY/ = 2cm and /RS/ = 3cm, find /PZ/.
(b) P and Q are two towns on the earth's surface on latitude 56°N. Thei longitudes are 25°E and 95°E respectively. Find the distance PQ along their parallel of latitude, correct to the nearest km. [Take radius of the earth as 6400km and \(\pi = \frac{22}{7}\)]
Bayanin Amsa
None
Tambaya 53 Rahoto
(a) A pack of 52 playing cards is shuffled and a card is drawn at random. Calculate the probability that it is either a five or a red nine.
[Hint : There are 4 fives and 2 red nines in a pack of 52 cards]
(b) P, Q and R are points in the same horizontal plane. The bearing of Q from P is 150° and the bearing of R from Q is 060°. If /PQ/ = 5m and /QR/ = 3m, find the bearing of R from P, correct to the nearest degree.
None
Bayanin Amsa
None
Tambaya 54 Rahoto
(a) Using mathematical tables, find ; (i) \(2 \sin 63.35°\) ; (ii) \(\log \cos 44.74°\);
(b) Find the value of K given that \(\log K - \log (K - 2) = \log 5\);
(c) Use logarithm tables to evaluate \(\frac{(3.68)^{2} \times 6.705}{\sqrt{0.3581}}\)
Tambaya 55 Rahoto
(a) Copy and complete the table for the relation \(y = 2 \cos 2x - 1\).
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° |
| \(y = 2\cos 2x - 1\) | 1.0 | 0 | 1.0 |
(b) Using a scale of 2cm = 30° on the x- axis and 2cm = 1 unit on the y- axis, draw the graph of \(y = 2 \cos 2x - 1\) for \(0° \leq x \leq 180°\).
(c) On the same axis, draw the graph of \(y = \frac{1}{180} (x - 360)\)
(d) Use your graphs to find the : (i) values of x for which \(2 \cos 2x + \frac{1}{2} = 0\); (ii) roots of the equation \(2 \cos 2x - \frac{x}{180} + 1 = 0\).
None
Bayanin Amsa
None
Tambaya 56 Rahoto
(a) Given that \(p = x + ym^{3}\), find m in terms of p, x and y.
(b) Using the method of completing the square, find the roots of the equation \(x^{2} - 6x + 7 = 0\), correct to 1 decimal place.
(c) The product of two consecutive positive odd numbers is 195. By constructing a quadratic equation and solving it, find the two numbers.
Bayanin Amsa
None
Tambaya 57 Rahoto
(a) Factorise : \(px - 2px - 4qy + 2py\)
(b) Given that the universal set U = {1, 2, 3, 4,5, 6, 7, 8, 9, 10}, P = {1, 2, 4, 6, 10} and Q = {2, 3, 6, 9}; show that \((P \cup Q)' = P' \cap Q'\)
Bayanin Amsa
None
Tambaya 58 Rahoto
(a) A man travels from a village X on a bearing of 060° to a village Y which is 20km away. From Y, he travels to a village Z, on a bearing of 195°. If Z is directly east of X, calculate, correct to three significant figures, the distance of :
(i) Y from Z ; (ii) Z from X .
(b) An aircraft flies due South from an airfield on latitude 36°N, longitude 138°E to an airfield on latitude 36°S, longitude 138°E.
(i) Calculate the distance travelled, correct to three significant figures ; (ii) if the speed of the aircraft is 800km per hour, calculate the time taken, correct to the nearest hour.
[Take \(\pi = \frac{22}{7}\), R = 6400km].
Tambaya 59 Rahoto
(a) Using a ruler and a pair of compasses only, construct triangle ABC with /AB/ = 7.5cm, /BC/ = 8.1cm and < ABC = 105°.
(b) Locate a point D on BC such that /BD/ : /DC/ is 3 : 2.
(c) Through D, construct a line I perpendicular to BC.
(d) If the line I meets AC at P, measure /BP/.
Bayanin Amsa
None
Tambaya 60 Rahoto
A box contains 5 blue balls, 3 black balls and 2 red balls of the same size. A ball is selected at random from the box and then replaced. A second ball is then selected. Find the probability of obtaining
(a) two red balls ;
(b) two blue balls or two black balls ;
(c) one black and one red ball in any order.
None
Bayanin Amsa
None
