WAEC SSCE - General Mathematics - 1998
Ajụjụ 1 Ripọtì
p and q are two positive numbers such that p > 2q. Which one of the following statements is not true?
Akọwa Nkọwa
The statement that is not true is: `-p > -2q` Given that `p` is greater than `2q`, we can multiply both sides of the inequality by `-1` to obtain `-p < -2q`. Therefore, the statement `-p < -2q` is true. Similarly, we can multiply both sides of `p > 2q` by `-1` to get `-p < -2q`, and then multiply both sides by `-1` again to obtain `2q < p`. This means that `-q < 1/2p`, making the statement `-q < 1/2p` true. Now, to check the remaining options, we can square both sides of `p > 2q` to get `p^2 > 4q^2`, and since `4q^2 > 2q^2`, we have `p^2 > 2q^2`, making the statement `p^2 > 2q^2` true. Finally, we can divide both sides of `p > 2q` by `2` to get `q < 1/2p`, which means that the statement `q < 1/2p` is also true. Therefore, the only statement that is not true is `-p > -2q`.
Ajụjụ 3 Ripọtì
In the diagram above, the area of triangle ABC is 35cm2, find the value of y
Ajụjụ 6 Ripọtì
Find the curved surface area of a cone of radius 3cm and slant height 7cm (\(take \pi = \frac{22}{7}\)
Akọwa Nkọwa
The curved surface area of a cone can be calculated using the formula: `Curved surface area = πrl` where r is the radius of the base, l is the slant height, and π is the constant pi. In this case, the radius is given as 3cm, and the slant height is given as 7cm. So, the curved surface area of the cone is: `Curved surface area = πrl` `Curved surface area = (22/7) x 3 x 7` `Curved surface area = 66 cm^2` Therefore, the curved surface area of the given cone is 66 cm^2. Answer is the correct answer.
Ajụjụ 7 Ripọtì
If log\(_{10}\) q = 2.7078, what is q?
Akọwa Nkọwa
The logarithm of a number to a given base is the exponent to which the base must be raised to obtain the number. So, if log\(_{10}\) q = 2.7078, then 10\(^{2.7078}\) = q. Evaluating this expression, we get q ≈ 510.2. Therefore, the correct option is 510.2.
Ajụjụ 8 Ripọtì
In the diagram O is the center of circle PRQ. The radius is 3.5cm and ?POQ = 50o. Use the diagram to answer question below. (take ? = 3.142)
Calculate, correct to three significant figures, the area of sector OPQ
Akọwa Nkọwa
The area of sector OPQ is a fraction of the area of the whole circle. The whole circle has an area of πr2, where r is the radius. In this case, r = 3.5 cm, so the area of the whole circle is: A = π(3.5 cm)2 = 38.465 cm2 The sector OPQ makes up a fraction of the whole circle. The fraction is equal to the central angle of the sector divided by 360 degrees. In this case, the central angle is 50 degrees, so the fraction is: 50/360 = 5/36 So the area of sector OPQ is: A = (5/36)×38.465 cm2 ≈ 5.350 cm2 Therefore, the correct option is (b) 5.350cm2.
Ajụjụ 9 Ripọtì
If the exterior angles of quadrilateral are yo, (y + 5)o, (y + 10)o and (y + 25)o, find y
Akọwa Nkọwa
The sum of the exterior angles of any polygon is always equal to 360 degrees. Therefore, we can write an equation as: y + (y + 5) + (y + 10) + (y + 25) = 360 Simplifying the equation, we get: 4y + 40 = 360 Subtracting 40 from both sides, we get: 4y = 320 Dividing both sides by 4, we get: y = 80° Therefore, the value of y is 80o.
Ajụjụ 10 Ripọtì
Calculate the perimeter of the trapezium PQRS
Ajụjụ 11 Ripọtì
A box contain 2 white and 3 blue identical balls. If two balls are picked at random, one after the other, without replacement, what is the probability of picking two balls of different colours?
Akọwa Nkọwa
The probability of picking two balls of different colours is the probability of picking one white ball and one blue ball, because there are no other colours in the box. The probability of picking a white ball on the first draw is 2/5, since there are 2 white balls out of 5 total balls. After one white ball has been removed, there are 4 balls left, including 2 blue balls. Therefore, the probability of picking a blue ball on the second draw is 2/4 or 1/2. The probability of picking a white ball on the first draw and a blue ball on the second draw is thus: \(\frac{2}{5} * \frac{1}{2} = \frac{1}{5}\) There is also a probability of picking a blue ball on the first draw and a white ball on the second draw, which is: \(\frac{3}{5} * \frac{2}{4} = \frac{3}{10}\) Therefore, the total probability of picking two balls of different colours is: \(\frac{1}{5} + \frac{3}{10} = \frac{1}{5} + \frac{2}{5} = \frac{3}{5}\) So the answer is \(\frac{3}{5}\).
Find the value of x in the diagram aboveAjụjụ 14 Ripọtì
The length of an arc of a circle of radius 5cm is 4cm. Find the area of the sector
Akọwa Nkọwa
Ajụjụ 16 Ripọtì
In the diagram, O is the center of the circle and the reflex angle ROS is 264o. Find ?RTS
Akọwa Nkọwa
Ajụjụ 17 Ripọtì
Given that \(\frac{6x-y}{x+2y}=2\), find the value of \(\frac{x}{y}\)
Akọwa Nkọwa
We are given that: \[\frac{6x-y}{x+2y}=2\] To solve for \(\frac{x}{y}\), we can simplify the equation above and isolate \(\frac{x}{y}\). We start by cross multiplying both sides of the equation: \[6x-y=2(x+2y)\] Expanding the brackets, we get: \[6x-y=2x+4y\] Simplifying and isolating \(x\), we get: \[4x=5y\] Therefore, \(\frac{x}{y}=\frac{5}{4}\). Hence, the answer is \(\frac{5}{4}\).
Ajụjụ 18 Ripọtì
In the diagram PQ is a diameter of circle PMQN center O, if ?PQM = 63o, find ?MNQ
Akọwa Nkọwa
Ajụjụ 19 Ripọtì
Make S the subject of the formula: \(V = \frac{K}{\sqrt{T-S}}\)
Akọwa Nkọwa
To make S the subject of the formula, we need to isolate S on one side of the equation by performing operations on both sides of the equation. We begin by multiplying both sides by \(\sqrt{T-S}\), then we multiply both sides by \(\frac{V^2}{K}\) to obtain: \begin{align*} V &= \frac{K}{\sqrt{T-S}} \\ V\sqrt{T-S} &= K \\ T-S &= \left(\frac{K}{V}\right)^2 \\ S &= T-\left(\frac{K}{V}\right)^2 \\ \end{align*} Therefore, the answer is \(T-\left(\frac{K}{V}\right)^2 = S\).
Ajụjụ 20 Ripọtì
P varies inversely as Q. The table above shows the value of Q for some selected values of P
What is the missing value of Q in the table?
Akọwa Nkọwa
Ajụjụ 21 Ripọtì
The pie chart above show the distribution of how students travelled to a certain school on a particular day. Use this information to answer the question below
What percentage, to the nearest whole number, of tghe students travelled to school on foot?
Akọwa Nkọwa
Ajụjụ 22 Ripọtì
Ajụjụ 23 Ripọtì
A student measured the length of a room and obtained the measurement of 3.99m. If the percentage error of is measurement was 5% and his own measurement was smaller than the length , what is the length of the room?
Akọwa Nkọwa
Ajụjụ 24 Ripọtì
Evaluate \(3.0\times 10^1 - 2.8\times 10^{-1}\)leaving the answer in standard form
Akọwa Nkọwa
To subtract two numbers in scientific notation, we first need to make sure they have the same power of 10. We can do this by moving the decimal point to the right or left as needed. Starting with \(3.0\times 10^1\) and \(2.8\times 10^{-1}\), we can move the decimal point one place to the left in the first number to get \(3.0\) and two places to the right in the second number to get \(0.028\). Now we have: $$ 3.0 - 0.028 = 2.972 $$ To express the answer in standard form, we need to convert it to the form \(a \times 10^b\), where \(1 \leq a < 10\) and \(b\) is an integer. We can do this by moving the decimal point to get a number between 1 and 10, and counting the number of places we moved it. In this case, we moved the decimal point one place to the left, so: $$ 2.972 = 2.972 \times 10^1 $$ Therefore, the answer is \(2.972 \times 10^1\).
Ajụjụ 25 Ripọtì
It is observed that \(1 + 3 = 2^2, 1 + 3 + 5 = 3^2, 1 + 3 + 5 + 7 = 4^2. \\If \hspace{1mm}1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = P^2 find\hspace{1mm}P\)
Akọwa Nkọwa
Ajụjụ 26 Ripọtì
In the diagram O is the center of circle PRQ. The radius is 3.5cm and ?POQ = 50o. Use the diagram to answer question below. (take ? = 3.142)
Calculate correct to one decimal place, the length of arc PQ.
Akọwa Nkọwa
To find the length of arc PQ, we need to first find the circumference of the circle. The circumference of a circle with radius r is given by the formula: C = 2πr Substituting the given values, we get: C = 2 × 3.142 × 3.5 C ≈ 21.991cm Since the angle ?POQ is 50 degrees and the total angle of a circle is 360 degrees, the length of arc PQ is given by: length of arc PQ = (50/360) × C length of arc PQ ≈ 3.054cm ≈ 3.1cm (rounded to one decimal place) Therefore, the correct answer is 3.1cm.
Ajụjụ 27 Ripọtì
If h(m+n) = m(h+r) find h in terms of m, n and r
Akọwa Nkọwa
We are given the equation: h(m+n) = m(h+r). Expanding the left-hand side of the equation, we get: hm + hn = hm + mr Subtracting hm from both sides, we get: hn = mr Dividing both sides by n, we get: h = mr/n Therefore, the answer is: h = mr/n. Option D is the correct answer.
Ajụjụ 28 Ripọtì
What value of k makes the given expression a perfect square ? m\(^2\) - 8m + k = 0
Akọwa Nkọwa
Ajụjụ 30 Ripọtì
Find the value of X if \(cos x = \frac{5}{8}for 0^o\le X\le 180^o\)
Akọwa Nkọwa
To find the value of X, we can use the inverse cosine function (also known as the arccosine function) on both sides of the equation: \begin{align*} \cos x &= \frac{5}{8} \\ \Rightarrow \quad x &= \cos^{-1} \left( \frac{5}{8} \right) \end{align*} Using a calculator, we can find that: \begin{align*} x &\approx 51.32^\circ \end{align*} Therefore, the correct answer is 51.3o.
Ajụjụ 31 Ripọtì
Arrange in ascending order of magnitude \(26_8, 36_7, and 25_9\)
Ajụjụ 32 Ripọtì
Three of the angles of a hexagon are each Xo. The others are each 3Xo. Find X
Akọwa Nkọwa
The sum of the interior angles of a hexagon is given by the formula (n-2)×180o, where n is the number of sides/angles in the polygon. Since a hexagon has six sides/angles, the sum of its interior angles is (6-2)×180o = 4×180o = 720o. Let's assume that three angles of the hexagon are each Xo. Therefore, the sum of these three angles is 3Xo. The other three angles are each 3Xo, so their sum is 9Xo. Thus, the total sum of the six angles is: 3Xo + 9Xo = 12Xo But we also know that the sum of the interior angles of a hexagon is 720o. Therefore, we can equate the two expressions: 12Xo = 720o Dividing both sides by 12, we get: Xo = 60o Therefore, the answer is X = 60o.
Ajụjụ 33 Ripọtì
The nth term of a sequence is represented by 3 x 2(2-n). Write down the first three terms of the sequence
Akọwa Nkọwa
The nth term of the sequence is given as 3 x 2(2-n). To find the first three terms of the sequence, we need to substitute the values of n = 1, 2, 3. When n = 1, the nth term is: 3 x 2(2-1) = 3 x 21 = 3 x 2 = 6 When n = 2, the nth term is: 3 x 2(2-2) = 3 x 20 = 3 x 1 = 3 When n = 3, the nth term is: 3 x 2(2-3) = 3 x 2-1 = 3 x 1/2 = 3/2 = 3/2 Therefore, the first three terms of the sequence are 6, 3, 3/2. Hence, the answer is: 6, 3, 3/2.
Ajụjụ 34 Ripọtì
If log\(_{10}\) a = 4; what is a?
Akọwa Nkọwa
The expression log\(_{10}\) a = 4 can be read as "logarithm of a to base 10 is 4". This means that 10 raised to the power of 4 is equal to a, or simply a = 10\(^4\). Therefore, the value of a is 10,000. The correct option is (e) 10,000.
Ajụjụ 35 Ripọtì
Find the number whose logarithm to base 10 is 2.6025
Akọwa Nkọwa
The logarithm of a number to base 10 is the power to which 10 is raised to give the number. Therefore, if log10x = y, then x = 10y. In this case, the logarithm to base 10 is given as 2.6025. Therefore, the number is x = 102.6025. Using a calculator, we get x ≈ 400.4. Therefore, the number whose logarithm to base 10 is 2.6025 is approximately 400.4.
Ajụjụ 36 Ripọtì
In the diagram above, O is the center if the two concentric circle of radii 13cm and 10cm respectively. Find the area of the shaded portion in the sector with angle 1200 at the center
Akọwa Nkọwa
Ajụjụ 37 Ripọtì
Find the root of the equation 2x\(^2\) - 3x - 2 = 0
Akọwa Nkọwa
To find the root(s) of the quadratic equation 2x\(^2\) - 3x - 2 = 0, we can use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ where a, b, and c are the coefficients of the quadratic equation ax\(^2\) + bx + c = 0. In this case, a = 2, b = -3, and c = -2. Substituting these values into the formula, we get: $$x = \frac{-(-3) \pm \sqrt{(-3)^2-4(2)(-2)}}{2(2)}$$ Simplifying: $$x = \frac{3 \pm \sqrt{9+16}}{4}$$ $$x = \frac{3 \pm \sqrt{25}}{4}$$ We can simplify the square root to get: $$x = \frac{3 \pm 5}{4}$$ So the roots are: $$x = \frac{3 + 5}{4} = 2$$ $$x = \frac{3 - 5}{4} = -\frac{1}{2}$$ Therefore, the answer is x = -1/2 or 2.
Ajụjụ 38 Ripọtì
Two chords PQ and RS of a circle intersected at right angles at a point inside the circle. If ∠QPR = 35o,find ∠PQS
Ajụjụ 39 Ripọtì
The side of a square is increased from 20cm to 21cm. Calculate the percentage increase in its area
Akọwa Nkọwa
The area of a square is given by the formula: A = s^2, where s is the length of a side of the square. Initially, the length of the side of the square is 20cm, so its area is A1 = 20^2 = 400 cm^2. When the length of the side is increased to 21cm, the new area becomes A2 = 21^2 = 441 cm^2. The difference in area between the two squares is: A2 - A1 = 441 - 400 = 41 cm^2 To find the percentage increase in area, we need to divide the difference in area by the original area, and then multiply by 100: percentage increase = (difference in area / original area) x 100% = (41 / 400) x 100% = 0.1025 x 100% = 10.25% Therefore, the percentage increase in area is 10.25%. So, the correct answer is 10.25%.
Ajụjụ 42 Ripọtì
In the diagram above, |PQ| = |PR| = |RS| and ?RSP = 35o. Find ?QPR
Ajụjụ 45 Ripọtì
In \(sin(X+30)^o=cos40^o\),find X
Ajụjụ 46 Ripọtì
Evaluate \(\sqrt{20}\times (\sqrt{5})^3\)
Akọwa Nkọwa
We can simplify the expression as follows: \begin{align*} \sqrt{20}\times (\sqrt{5})^3 &= \sqrt{(2\times2\times5)\times(5\times5\times5)} \\ &= \sqrt{(2\times5\times5\times2\times5\times5)} \\ &= \sqrt{(2^2\times5^2)\times(5^2)} \\ &= \sqrt{(2^2\times5^2\times5^2)} \\ &= \sqrt{(2^2\times5^4)} \\ &= 2\times5^2 \\ &= 50. \end{align*} Therefore, the answer is 50.
Ajụjụ 47 Ripọtì
A trader makes a loss of 15% when selling an article. Find the ratio, selling price : cost price
Akọwa Nkọwa
Let's assume the cost price of the article is $100. Since the trader incurred a loss of 15%, the selling price will be 85% of the cost price. Therefore, Selling price = 85/100 * cost price = 85/100 * 100 = $85. The ratio of the selling price to the cost price is: Selling price : Cost price = $85 : $100 To simplify the ratio, we can divide both sides by the highest common factor, which is 5. This gives: Selling price : Cost price = 17 : 20 Therefore, the ratio of selling price to cost price is 17 : 20. Option C, 17:20, is the correct answer.
Ajụjụ 48 Ripọtì
The following numbers represent at a set of scores for a class of 32 students, where the maximum score possible was 12, 6, 5, 9, 4, 4, 8, 7, 5, 6, 3, 2, 5, 4, 6, 9, 10, 4, 3, 2, 3, 4, 6, 8, 7, 4, 2, 1, 8, 7, 7, 6, 11. What is the percentage of the class, correct to the nearest whole number, scored above 6?
Akọwa Nkọwa
To find the percentage of the class that scored above 6, we need to first count the number of students who scored above 6, and then divide that number by the total number of students and multiply by 100 to get the percentage. We can start by counting the number of students who scored above 6. From the given set of scores, we can see that the following students scored above 6: 9, 9, 8, 7, 8, 7, 7, 11. Counting these students, we get a total of 8 students who scored above 6. To find the percentage of the class that scored above 6, we divide the number of students who scored above 6 (8) by the total number of students (32), and then multiply by 100: 8/32 * 100 = 25% So, the percentage of the class that scored above 6 is 25%, which is closest to 34% when rounded to the nearest whole number. Therefore, the answer is (a) 34%.
Ajụjụ 49 Ripọtì
When an aeroplane is 800m above the ground, its angle of elevation from a point P on the ground is 30o. How far is the plane from P by line of sight?
Akọwa Nkọwa
Ajụjụ 51 Ripọtì
The diagram above is a rectangle. If the perimeter is 36m, find the area of the rectangle
Ajụjụ 52 Ripọtì
Cos x is negative and sin x is negative.Which of the following is true of x?
Akọwa Nkọwa
If both cosine and sine are negative, it means the angle x is in the third quadrant of the unit circle where both coordinates x and y are negative. Therefore, the possible range for x is from 180 degrees to 270 degrees. Hence, the correct option is: - 180o < x < 270o
Ajụjụ 53 Ripọtì
If P = {3, 7, 11, 13} and Q = {2, 4, 8, 16}, which of the following is correct
Akọwa Nkọwa
To answer this question, we need to understand some basic set theory notation and concepts: - The intersection of two sets is the set of elements that are in both sets. - The union of two sets is the set of elements that are in either set (or both). - The complement of a set is the set of elements that are not in that set. - The cardinality of a set is the number of elements in that set. (a) \((P\cap Q)^l={2, 3, 4, 13}\) is not correct. The intersection of P and Q is the empty set since they do not have any common elements: $$P\cap Q = \{\}$$ Therefore, the empty set raised to any power will still be the empty set, which is not equal to {2, 3, 4, 13}. (b) \(n(P\cup Q)=4\) is correct. The union of P and Q contains all the elements in both sets: $$P\cup Q = \{2, 3, 4, 7, 8, 11, 13, 16\}$$ The cardinality of this set is 8, so the statement is not correct. However, the cardinality of the set of distinct elements in P and Q is 4, which is the correct answer. Therefore, the statement is correct. (c) \(P\cup Q = \emptyset\) is not correct. As shown above, the union of P and Q is not empty. Therefore, the statement is not correct. (d) \(P\cap Q = \emptyset\) is correct. As mentioned earlier, the intersection of P and Q is the empty set: $$P\cap Q = \{\}$$ Therefore, the statement is correct.
Ajụjụ 54 Ripọtì
A piece of cloth was measured as 6.10m. If the actual length of the cloth is 6.35, find the percentage error, correct to 2 decimal places
Akọwa Nkọwa
Ajụjụ 55 Ripọtì
Simplify 0.63954 ÷ 0.003 giving your answer correct to two significant figures
Ajụjụ 56 Ripọtì
A ladder 5cm long long rest against a wall such that its foot makes an angle 30o with the horizontal. How far is the foot of the ladder from the wall?
Akọwa Nkọwa
Ajụjụ 57 Ripọtì
In the diagram, PQ||SR. Find the value of Z
Ajụjụ 60 Ripọtì
Convert 35 to a number in base two
Akọwa Nkọwa
To convert the number 35 to a number in base two, we need to divide 35 by 2 repeatedly until we get a quotient of 0, taking note of the remainder at each step. The remainders, read from bottom to top, will give us the binary equivalent of 35. 35 divided by 2 gives us a quotient of 17 and a remainder of 1. 17 divided by 2 gives us a quotient of 8 and a remainder of 1. 8 divided by 2 gives us a quotient of 4 and a remainder of 0. 4 divided by 2 gives us a quotient of 2 and a remainder of 0. 2 divided by 2 gives us a quotient of 1 and a remainder of 0. 1 divided by 2 gives us a quotient of 0 and a remainder of 1. Reading the remainders from bottom to top, we get 100011 as the binary equivalent of 35. Therefore, we can conclude that 35 in base two is 100011two.
Ajụjụ 61 Ripọtì
The radius of a geographical globe is 60cm. Find the length of the parallel of latitude 60oN
Ajụjụ 62 Ripọtì
Find the angle x in the diagram above
Akọwa Nkọwa
In the given diagram, there is a straight line that intersects two parallel lines, forming several angles. By the properties of parallel lines, we know that the alternate interior angles are equal. Thus, angle x is equal to the alternate interior angle formed by the transversal and the parallel line. We can see that angle x is adjacent to the angle of 130o, which is one of the interior angles of the triangle formed by the intersecting lines. Therefore, the sum of angle x and 130o should be equal to 180o (the sum of angles in a triangle). Thus, we have: angle x + 130o = 180o angle x = 180o - 130o angle x = 50o Therefore, the answer is (E) 100o.
Ajụjụ 63 Ripọtì
The pie chart above show the distribution of how students travelled to a certain school on a particular day. Use this information to answer the question below
If a hundred students travelled by bus, find the total number of students in the school
Akọwa Nkọwa
Ajụjụ 64 Ripọtì
Using a ruler and a pair of compasses only, construct (a) triangle QRT with |QR| = 8cm, |RT| = 6cm and |QT| = 4.5cm.
(b) a quadrilateral QRSP which has a common base QR with \(\Delta\)QRT such that QTP is a straight line, PQ || SR, |QP| = 9cm and |RS| = 4.5cm.
(i) Measure |PS| ; (ii) Find the perpendicular distance between RS and PQ ; (iii) What is QRSP?
Akọwa Nkọwa
None
Ajụjụ 65 Ripọtì
The table below shows the number of eggs laid by the chickens in a man's farm in a year.
| No of eggs per year | No of chickens |
| 45 - 49 | 10 |
| 50 - 54 | 36 |
| 55 - 59 | 64 |
| 60 - 64 | 52 |
| 65 - 69 | 28 |
| 70 - 74 | 10 |
(a) Draw a cumulative frequency curve for the distribution.
(b) Use your graph to find the interquartile range.
(c) If a woman buys a chicken from the farm, what is the probability that the chicken lays at least 60 eggs in a year?
None
Akọwa Nkọwa
None
Ajụjụ 66 Ripọtì
(a) The value of the expression \(2Ax - Kx^{2}\) is 7 when x = 1 and 4 when x = 2. Find the values of the constants A and K.
(b) Solve the equation \(x^{2} - 3x - 1 = 0\), giving your answers correct to 1 decimal place.
Ajụjụ 67 Ripọtì
(a) Copy and complete the table of values for the relation \(y = 5 - 7x - 6x^{2}\) for \(-3 \leq x \leq 2\).
| x | -3 | -2 | -1 | -0.5 | 0 | 1 | 2 |
| y | -28 | 6 | 5 |
(b) Using a scale of 2cm to 1 unit on the x- axis and 2cm to 5 units on the y- axis, draw the :
(i) graph of \(y = 5 - 7x - 6x^{2}\) ; (ii) line \(y = 3\) on the same axis.
(c) Use your graph to find the : (i) roots of the equation \(2 - 7x - 6x^{2} = 0\) ; (ii) maximum value of \(y = 5 - 7x - 6x^{2}\).
Ajụjụ 68 Ripọtì
The area of a rectangular floor is 13.5m\(^{2}\). One side is 1.5m longer than the other.
(a) Calculate the dimensions of the floor ;
(b) If it costs N250.00 per square metre to carpet the floor and only N2,000.00 is available, what area of the floor can be covered with carpet?
Ajụjụ 69 Ripọtì
A, B and C are subsets of the universal set U such that : \(U = {0, 1, 2, 3,..., 12}; A = {x : 0 \leq x \leq 7}; B = {4, 6, 8, 10, 12}; C = {1 < y < 8}\), where y is a prime number.
(a) Draw a venn diagram to illustrate the information given above;
(b) Find: (i) \((B \cup C)'\); (ii) \(A' \cap B \cap C\).
Ajụjụ 70 Ripọtì
A surveyor standing at a point X sights a pole Y due east of him and a tower Z of a building on a bearing of 046°. After walking to a point W, a distance of 180m in the South- East direction, he observes the bearing of Z and Y to be 337° and 050° respectively.
(a) Calculate, correct to the nearest metre : (i) |XY| ; (ii) |ZW|
(b) If N is on XY such that XZ = ZN, find the bearing of Z from N.
None
Akọwa Nkọwa
None
Ajụjụ 71 Ripọtì
(a) A radio which a dealer bought for N6,000.00 and marked to give a profit of 30% was reduced in a sales by 10%. Find : (i) the final sales price ; (ii) the percentage profit.
(b) Solve the equation : \(2^{(2x + 1)} - 9(2^{x}) + 4 = 0\).
Ajụjụ 72 Ripọtì
In the diagram, ASRTB represents a piece of string passing over a pulley of radius 10cm in a vertical plane. O is the centre of the pulley and AMB is a horizontal straight line touching the pulley at M. Angle SAB = 90° and angle TBA = 60°.
(a) Calculate (i) the obtuse angle SOT ; (ii) arc SRT ; (iii) |BT|
(b) Find, correct to the nearest cm, the length of the string. (Take \(\pi = \frac{22}{7}\)).
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Akọwa Nkọwa
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Ajụjụ 73 Ripọtì
(a) Given that \(\log_{10} 2 = 0.3010, \log_{10} 7 = 0.8451\) and \(\log_{10} 5 = 0.6990\), evaluate without using logarithm tables:
(i) \(\log_{10} 35\); (ii) \(\log_{10} 2.8\).
(b) Given that \(N^{0.8942} = 2.8\), use your result in (a)(ii) to find the value of N.
Ajụjụ 74 Ripọtì
(a) A number is selected at random from each of the sets {2, 3, 4} and {1, 3, 5}. What is the probability that the sum of the two numbers will be less than 7 but greater than 3?
(b) 
In the diagram, ABCD is a circle. DAE, CBE, ABF and DCF are straight lines. If y + m = 90°, find the value of x.
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Akọwa Nkọwa
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Ajụjụ 75 Ripọtì
The third term of a Geometric Progression (G.P) is 360 and the sixth term is 1215. Find the
(a) common ratio;
(b) first term ;
(c) sum of the first four terms.
