WAEC SSCE - General Mathematics - 2003
Tambaya 1 Rahoto
Evaluate \((111_{two})^2 - (101_{two})^2\)
Bayanin Amsa
To solve this problem, we need to convert the given binary numbers to decimal, then subtract the square of the second number from the square of the first number, and finally convert the result back to binary. (111two)2 in decimal is equal to (7)2 = 49. (101two)2 in decimal is equal to (5)2 = 25. Therefore, (111two)2 - (101two)2 = 49 - 25 = 24. Finally, we convert 24 to binary, which is 11000two. Therefore, the answer is option (D) 11000two.
Tambaya 2 Rahoto
Calculate the value of y in the diagram
Bayanin Amsa
Sum of interior angle of the diagram equals 360o
180o - 5yo + 136o + 180o + 180o - 3yo = 360o
-8yo + 136o = 0
-8yo = -136;
y = 17
Tambaya 3 Rahoto
A student bought 3 notebooks and 1 pen for N35. After misplacing these items, she again bought 2 notebooks and 2 pens, all of the same type for
N30. What is the cost of a pen?
Bayanin Amsa
Let the cost of a notebook be N and the cost of a pen be P. From the first transaction, we have: 3N + 1P = 35 ...(1) From the second transaction, we have: 2N + 2P = 30 ...(2) We want to find the value of P. We can rearrange equation (2) as follows: 2P = 30 - 2N P = (30 - 2N)/2 Substituting this value of P into equation (1), we get: 3N + 1[(30 - 2N)/2] = 35 Multiplying both sides by 2, we get: 6N + (30 - 2N) = 70 Simplifying the expression, we get: 4N = 40 N = 10 Substituting this value of N into equation (2), we get: 2(10) + 2P = 30 Simplifying the expression, we get: 2P = 10 P = 5 Therefore, the cost of a pen is N5.00. Answer: N5.00
Tambaya 5 Rahoto
In the diagram /PS/ = 9 cm, /OR/ = 5cm, \(P\hat{S}R = 63^o\) and \(S\hat{P}Q = P\hat{Q}R = 90^o\). Find, correct to the nearest whole number, the area of the trapezium,
Bayanin Amsa
To find the area of the trapezium, we need to first find the length of side /PQ/ and /SR/. Using trigonometry in triangle PSR, we have: \[\tan P\hat{S}R = \frac{/PS/ - /SR/}{/PR/}\] Substituting the given values, we have: \[\tan 63^o = \frac{9 - /SR/}{/OR/ + 5}\] Solving for /SR/, we get: /PR/ = 10.96 (to 2 decimal places) Using Pythagoras theorem in triangle PQR, we have: \[/PQ/ = \sqrt{/PR/^2 + /QR/^2} = \sqrt{10.96^2 + 9^2} \approx 13.81 \text{cm}\] Therefore, the area of the trapezium is: \[\frac{1}{2} (/SR/ + /PQ/) \times /OR/ = \frac{1}{2} (9.19 + 13.81) \times 5 \approx 62.00 \approx 62 \text{cm}^2\] Therefore, the answer is 55cm2 rounded to the nearest whole number.
Tambaya 6 Rahoto
Find the value of x in the equation 3x\(^2\) - 8x - 3 = 0
Tambaya 7 Rahoto
Simplify \(\left(\frac{3}{4} - \frac{1}{3}\right)\times 4\frac{1}{3}\div 3\frac{1}{4}\)
Bayanin Amsa
Tambaya 8 Rahoto
The table below gives the distribution of marks obtained by a number of pupils in a class test.
How many pupils scored at least 2 marks?
Bayanin Amsa
Tambaya 11 Rahoto
A ladder, 6m long, leans against a vertical wall at an angle 53o to the horizontal. How high up the wall does the ladder reach?
Bayanin Amsa
Tambaya 12 Rahoto
Which of the following is not a measure of dispersion?
Bayanin Amsa
The measure of dispersion tells how the data is spread out from the central tendency. Mean and Standard deviation are both measures of dispersion. Range and Mean deviation are also measures of dispersion. Therefore, the answer is Mean.
Tambaya 13 Rahoto
Simplify \(\frac{4}{a-3}-\frac{1}{a+3}\)
Tambaya 14 Rahoto
Find the values of x for which \( \frac{1}{2x^2 - 13x +15} \) is not defined,
Bayanin Amsa
The expression \( \frac{1}{2x^2 - 13x +15} \) is undefined when the denominator is equal to zero. Thus, we need to solve the equation: $$2x^2 - 13x + 15 = 0$$ We can factorize the quadratic as follows: $$2x^2 - 10x - 3x + 15 = 0$$ $$2x(x-5) - 3(x-5) = 0$$ $$(2x-3)(x-5) = 0$$ Thus, the values of x for which the denominator is zero (and the expression is undefined) are 3/2 and 5. Therefore, the correct answer is option A: 5 or 3/2.
Tambaya 15 Rahoto
Which of the following is/are not the interior angle(s) of a regular polygon? I.108° II. 116° III. 120°
Bayanin Amsa
A regular polygon is a polygon with all sides and angles equal. Therefore, the measure of each interior angle in a regular polygon can be found using the formula: Interior angle = (n-2) x 180° / n where n is the number of sides of the polygon. Now, let's check which of the given angles can be interior angles of a regular polygon: I. 108° = (n-2) x 180° / n --> n = 5 II. 116° = (n-2) x 180° / n --> n ≈ 7.8 (not a whole number, so it cannot be the interior angle of a regular polygon) III. 120° = (n-2) x 180° / n --> n = 5 Therefore, the answer is (II only).
Tambaya 16 Rahoto
If \(P = \sqrt{QR\left(1+\frac{3t}{R}\right)}\), make R the subject of the formula.
Bayanin Amsa
To make R the subject of the formula, we need to isolate R on one side of the equation. Starting with the given equation: $$P = \sqrt{QR\left(1+\frac{3t}{R}\right)}$$ Squaring both sides: $$P^2 = QR\left(1+\frac{3t}{R}\right)$$ Expanding the right side: $$P^2 = QR + 3tQ/R$$ Subtracting 3tQ/R from both sides: $$P^2 - 3tQ/R = QR$$ Dividing both sides by Q: $$\frac{P^2 - 3tQ}{Q} = R$$ Therefore, the answer is: $$R = \frac{P^2 - 3tQ}{Q}$$ So the correct option is: - \(R = \frac{P^2-3Qt}{Q}\)
Tambaya 17 Rahoto
Find the equation whose roots are \(-\frac{2}{3}\) and 3
Bayanin Amsa
The equation whose roots are -2/3 and 3 is of the form: a(x + 2/3)(x - 3) = 0 where 'a' is a constant. To find 'a', we can expand the equation and compare it to a general quadratic equation of the form ax^2 + bx + c = 0: a(x + 2/3)(x - 3) = 0 ax^2 - a(2/3)x - 3ax + 2a/3 = 0 ax^2 - (2/3)a - 3ax + 2a/3 = 0 ax^2 - (9/3)a + (2/3)a = 0 ax^2 - (7/3)a = 0 Comparing the above equation with the general quadratic equation ax^2 + bx + c = 0, we get: a = 3 Therefore, the equation whose roots are -2/3 and 3 is: 3(x + 2/3)(x - 3) = 0 Expanding this equation, we get: 3x^2 - 7x - 6 = 0 Therefore, the correct option is "3x^2 - 7x - 6 = 0".
Tambaya 18 Rahoto
Find the value of x which satisfies the equation
5(x-7)=7-2x
Bayanin Amsa
We can start by expanding the brackets and simplifying the equation. 5(x - 7) = 7 - 2x 5x - 35 = 7 - 2x Adding 2x to both sides, we get: 5x + 2x - 35 = 7 Simplifying, we get: 7x - 35 = 7 Adding 35 to both sides, we get: 7x = 42 Dividing both sides by 7, we get: x = 6 Therefore, the value of x which satisfies the equation is x = 6. Answer: x = 6
Tambaya 19 Rahoto
A pyramid of volume 120cm3 has a rectangular base which measures 5cm by 6cm. Calculate the height of the pyramid
Bayanin Amsa
The volume of a pyramid is given by the formula: V = (1/3)Ah where V is the volume, A is the area of the base, and h is the height of the pyramid. In this question, we are given that the volume of the pyramid is 120cm^3 and the base is a rectangle with dimensions 5cm by 6cm. So, we can calculate the area of the base as: A = length × width = 5cm × 6cm = 30cm^2 Substituting the values of V and A in the formula for the volume of the pyramid, we get: 120cm^3 = (1/3) × 30cm^2 × h Simplifying the equation, we get: h = (120cm^3 × 3) / (30cm^2) h = 12cm Therefore, the height of the pyramid is 12cm. So, the correct option is "(3) 12cm".
Tambaya 20 Rahoto
What is the size of angle x in the diagram
Tambaya 21 Rahoto
The table below gives the distribution of marks obtained by a number of pupils in a class test.
Find the median of the distribution
Bayanin Amsa
Tambaya 22 Rahoto
The locus of points equidistant from two intersecting straight lines PQ and PR is
Bayanin Amsa
The locus of points equidistant from two intersecting straight lines PQ and PR is the point of intersection of the perpendicular bisectors of PQ and PR. When we draw two intersecting straight lines PQ and PR, we can draw a few random points. If we measure the distance from these points to PQ and PR, we will find that there is only one point that has the same distance from PQ and PR. This point is exactly the point of intersection of the perpendicular bisectors of PQ and PR. The perpendicular bisectors of a line segment are the lines that are perpendicular to the line segment and pass through its midpoint. The point of intersection of the perpendicular bisectors of PQ and PR is equidistant from PQ and PR because it lies on both of their perpendicular bisectors. Therefore, the locus of points equidistant from two intersecting straight lines PQ and PR is the point of intersection of the perpendicular bisectors of PQ and PR. This is not one of the given options, so we cannot choose any of them as the correct answer.
Tambaya 23 Rahoto
From the Venn diagram below, how many elements are in P∩Q?
Bayanin Amsa
The intersection of two sets P and Q is the set of elements that are in both P and Q. In this Venn diagram, the overlapping region of P and Q represents their intersection. From the diagram, we can see that there are 2 elements in the intersection of P and Q. Therefore, the answer is 2.
Tambaya 24 Rahoto
The square root of a number is 2k. What is half of the number
Bayanin Amsa
Let's start by using some algebra to solve for the number in terms of its square root: If the square root of a number is 2k, then we can write: √x = 2k Squaring both sides of the equation, we get: x = (2k)^2 Simplifying the right-hand side, we have: x = 4k^2 Now, we're asked to find half of the number, which is simply half of x: 1/2 * x = 1/2 * 4k^2 Simplifying, we get: 1/2 * x = 2k^2 So the correct answer is option D: 2k^2. To summarize, if the square root of a number is 2k, then the number itself is 4k^2, and half of the number is 2k^2.
Tambaya 25 Rahoto
Simplify \(\frac{4\sqrt{18}}{\sqrt{8}}\)
Bayanin Amsa
We can simplify the given expression as follows: \begin{align*} \frac{4\sqrt{18}}{\sqrt{8}} &= \frac{4\sqrt{9\cdot 2}}{\sqrt{4\cdot 2}}\\ &= \frac{4\cdot 3\sqrt{2}}{2}\\ &= 2\cdot 3\sqrt{2}\\ &= 6\sqrt{2} \end{align*} Therefore, the simplified form of \(\frac{4\sqrt{18}}{\sqrt{8}}\) is \(6\sqrt{2}\). So the correct option is (c) 6.
Tambaya 26 Rahoto
The probabilities of a boy passing English and Mathematics test are x and y respectively. Find the probability of the boy failing both tests
Bayanin Amsa
Tambaya 27 Rahoto
The four interior angles of a quadrilateral are (x + 20) o, (x+ 10) o (2x - 45) o and (x - 25) o. Find the value of x
Bayanin Amsa
Tambaya 28 Rahoto
Given, that \(4P4_5 = 119_{10}\), find the value of P
Bayanin Amsa
To solve this problem, we need to first understand what \(4P4_5\) means. Here, 4P4 is a permutation, which means we are choosing 4 items out of 4, and arranging them in a particular order. The subscript 5 means we are working in base 5. We can calculate the value of \(4P4_5\) as follows: \(4P4_5 = 4! = 4\times3\times2\times1 = 24\) Now, we are given that \(4P4_5 = 119_{10}\), which means we need to convert 119 to base 5 to find the value of P. 119 in base 5 is equal to 1444. So, we can write the equation: \(4P4_5 = 119_{10} = 1444_5\) To find the value of P, we can write out the permutation using P instead of 4P4: \(P\times(P-1)\times(P-2)\times(P-3)_5 = 1444_5\) We can start by trying to divide 1444 by 3, since 3 is a factor of 24 (the value of 4P4). \(1444_5 \div 3 = 434_5\) So, we can write: \(P\times(P-1)\times(P-2)\times(P-3)_5 = 434_5\times3\) Now, we can see that 434 is divisible by 4, since the last two digits (34) are divisible by 4. \(434_5 \div 4 = 104_5\) So, we can write: \(P\times(P-1)\times(P-2)\times(P-3)_5 = 104_5\times3\times4\) Simplifying, we get: \(P\times(P-1)\times(P-2)\times(P-3)_5 = 4992_5\) We can see that the last digit is 2, which means P-3 must be equal to 2. So, P = 5 - 2 = 3. Therefore, the value of P is 3.
Tambaya 29 Rahoto
The table below gives the distribution of marks obtained by a number of pupils in a class test.
The mode of the distribution is
Tambaya 30 Rahoto
PQRS is a cyclic quadrilateral. If ∠QPS = 75°, what is the size of ∠QRS?
Bayanin Amsa
A cyclic quadrilateral is a quadrilateral whose vertices all lie on a single circle. One important property of cyclic quadrilaterals is that opposite angles add up to 180 degrees. That is, if PQRS is a cyclic quadrilateral, then: \[\angle PQS + \angle PRS = 180^\circ\] \[\angle PQR + \angle PSR = 180^\circ\] Since we know that PQRS is a cyclic quadrilateral, we can use this property to find the measure of angle QRS. Let x be the measure of angle QRS. Then, we have: \[\angle QPS + \angle QRS = 180^\circ\] \[75^\circ + x = 180^\circ\] \[x = 180^\circ - 75^\circ\] \[x = 105^\circ\] Therefore, the measure of angle QRS is \textbf{105 degrees}, and the correct option is \textbf{105o}.}
Tambaya 31 Rahoto
Given that x ≅ 0.0102 correct to 3 significant figures, which of the following cannot be the actual value of x?
Bayanin Amsa
Tambaya 33 Rahoto
The bearing of P from Q is x, where 270o < x < 360o. Find the bearing of Q from P
Bayanin Amsa
The bearing of P from Q is the angle measured clockwise from the north direction to the line PQ. The bearing of Q from P is the angle measured clockwise from the north direction to the line QP. Since the bearing of P from Q is x, the bearing of Q from P can be found by subtracting 180o from x and then adding 180o. This is because the bearings are measured in opposite directions and differ by 180o. Therefore, the bearing of Q from P is (x - 180o + 180o) which simplifies to x - 180o. Thus, the answer is (x - 180)o.
Tambaya 34 Rahoto
In\( ∆ PQR, P\hat{Q}P = 84^°, |Q\hat{P}R |= 43^°\) and |PQ| = 5cm. Find /QR/ in cm, correct to 1 decimal place.
Bayanin Amsa
Tambaya 35 Rahoto
In constructing an angle, Olu draws line OX. With centre O and a convenient radius, he draws an arc intersecting OX at P. With centre P and the same radius, he draws an arc intersecting the first arc at Q and finally joins OQ. What is the size of angle POQ so constructed?
Bayanin Amsa
Tambaya 36 Rahoto
In a ∆ XYZ, /YZ/ = 6cm YXZ = 60o and XYZ is a right angle. Calculate /XZ/in cm, leaving your answer in surd form
Bayanin Amsa
Tambaya 39 Rahoto
A right circular cone is such that its radius r is twice its height h. Find its volume in terms of h
Bayanin Amsa
The formula for the volume of a right circular cone is: V = (1/3) * π * r^2 * h In this problem, we are given that the radius r is twice the height h. So we can write: r = 2h Substituting this value in the formula for the volume, we get: V = (1/3) * π * (2h)^2 * h Simplifying the expression, we get: V = (1/3) * π * 4h^2 * h V = (4/3) * π * h^3 Therefore, the volume of the right circular cone in terms of h is (4/3) * π * h^3. Hence, the correct option is: - \(\frac{4}{3}\pi h^3\)
Tambaya 40 Rahoto
Evaluate Cos 45o Cos 30o - Sin 45o Sin 30o leaving the answer in surd form
Tambaya 41 Rahoto
A man made a loss of 15% by selling an article for N595. Find the cost price of the article
Bayanin Amsa
To find the cost price of the article, we need to use the formula: Selling price = Cost price - Loss where the loss is given as a percentage of the cost price. In this case, the loss is 15%. Let's first convert the loss percentage to a decimal: 15% = 0.15 Now we can substitute the given values into the formula: N595 = Cost price - 0.15(Cost price) Simplifying this equation, we get: N595 = Cost price - 0.15Cost price N595 = 0.85Cost price To solve for the cost price, we can divide both sides of the equation by 0.85: Cost price = N595 ÷ 0.85 Cost price = N700 Therefore, the cost price of the article is N700.00.
Tambaya 42 Rahoto
Out of 60 members of an Association, 15 are Doctors and 9 are Lawyers. If a member is selected at random from the Association, what is the probability that the member is neither a Doctor Nor a Lawyer
Bayanin Amsa
We know that there are 15 doctors and 9 lawyers out of a total of 60 members in the Association. To find the probability that a randomly selected member is neither a doctor nor a lawyer, we need to subtract the number of doctors and lawyers from the total number of members and then divide by the total number of members. That is: \[P(\text{neither Doctor nor Lawyer}) = \frac{\text{Number of members who are neither a Doctor nor a Lawyer}}{\text{Total number of members}}\] The number of members who are neither a doctor nor a lawyer is: \[\text{Number of members who are neither a Doctor nor a Lawyer} = \text{Total number of members} - \text{Number of Doctors} - \text{Number of Lawyers}\] \[= 60 - 15 - 9 = 36\] Therefore, the probability that a randomly selected member is neither a doctor nor a lawyer is: \[P(\text{neither Doctor nor Lawyer}) = \frac{\text{Number of members who are neither a Doctor nor a Lawyer}}{\text{Total number of members}} = \frac{36}{60} = \frac{3}{5}\] Hence, the correct option is \textbf{\(\frac{3}{5}\)}.
Tambaya 43 Rahoto
Simplify \(\left(1\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2\)
Bayanin Amsa
To simplify this expression, we need to first simplify the terms inside the parentheses: \(\left(1\frac{2}{3}\right)^2\) can be rewritten as \(\left(\frac{5}{3}\right)^2\) because 1 whole and 2 thirds is equal to 5 thirds. Similarly, \(\left(\frac{2}{3}\right)^2\) can be simplified to \(\frac{4}{9}\). Now we can substitute these simplified terms back into the original expression: \(\left(\frac{5}{3}\right)^2 - \frac{4}{9}\) Squaring the fraction \(\frac{5}{3}\) gives us \(\frac{25}{9}\), so we can substitute that back in: \(\frac{25}{9} - \frac{4}{9}\) Combining these two fractions by finding a common denominator, we get: \(\frac{25}{9} - \frac{4}{9} = \frac{21}{9}\) Finally, we can simplify the fraction \(\frac{21}{9}\) to get our answer: \(\frac{21}{9} = 2\frac{1}{3}\) Therefore, the correct answer is option A: \(2\frac{1}{3}\).
Tambaya 45 Rahoto
Given that p varies as the square of q and q varies inversely as the square root of r. How does p vary with r?
Bayanin Amsa
We can use the relation p ∝ q^2 and q ∝ 1/√r to find the relation between p and r. Substituting the value of q in terms of r in the relation p ∝ q^2, we get: p ∝ (1/√r)^2 p ∝ 1/r Therefore, p varies inversely as r. So, the answer is: "p varies inversely as r".
Tambaya 46 Rahoto
From the Venn Diagram below, find Q' ? R.
Bayanin Amsa
In a Venn diagram, Q' means everything outside Q, while the intersection of Q' and R means the common elements in the regions outside Q and in R. Thus, Q' ? R is the region(s) marked in R but not in Q. From the diagram, we see that only regions (c), (g), and (h) are in R but not in Q. Therefore, Q' ? R = (c, g, h). So the correct option is (c, g, h).
Tambaya 47 Rahoto
(a) If p varies directly as \(r^{2}\) and p = 3.2 when r = 4, find the value of p when r = 6.5.
(b) Solve the simultaneous equations :
\(\frac{x}{2} + \frac{y}{4} = 1 ; \frac{x}{3} - \frac{y}{4} = \frac{-1}{6}\)
Bayanin Amsa
None
Tambaya 48 Rahoto
An aeroplane flies due west for 3 hours from P (lat. 50°N, long. 60°W) to a point Q at an average speed of 600km/h. The aeroplane then flies due south from Q to a point Y 500km away. Calculate, correct to 3 significant figures,
(a) the longitude of Q ;
(b) the latitude of Y . [Take the radius of the earth = 6400km and \(\pi = \frac{22}{7}\)].
Tambaya 50 Rahoto
(a) Simplify : \(\frac{\frac{1}{3}c^{2} - \frac{2}{3}cd}{\frac{1}{2}d^{2} - \frac{1}{4}cd}\)
(b)
In the diagram, YPF is a straight line. < XPY = 44°, < MPF = 46°, < XYP = < MFP = 90°, /XY/ = 7cm and /MP/ = 9 cm.
(i) Calculate, correct to 3 significant figures, /XM/ and /YF/ ; (ii) Find < XMP.
Bayanin Amsa
None
Tambaya 51 Rahoto
Without using Mathematical tables or a calculator, simplify :
(a) \(\sqrt{50} - 3\sqrt{2}(2\sqrt{2} - 5) - 5\sqrt{32}\)
(b) \(\frac{1}{2} \log_{10} \frac{25}{4} - 2 \log_{10} \frac{4}{5} + \log_{10} \frac{320}{125}\).
Bayanin Amsa
None
Tambaya 52 Rahoto
The table shows the number of suitcases possessed by a group of travellers.
| No. of suitcases | 0 | 1 | 2 | 3 | 4 | 5 |
| Travellers | 2 | 7 | 7 | 2 | 3 | 9 |
(a) Calculate the (i) median (ii) mean, correct to the nearest whole number.
(b) Draw a bar chart to represent the information.
Bayanin Amsa
None
Tambaya 53 Rahoto
The table shows the monthly contributions and expenditure pattern of an employee in 1999.
| Item | Percentage |
| Pension | 5 |
| Income Tax | 25 |
| Food | 40 |
| Transport | 10 |
| Rent | 12.5 |
| Others | 7.5 |
(a) Draw a pie chart to illustrate the data.
(b) If the employee's gross monthly salary was N10,800.00, calculate (i) the pension contribution of the employee ; (ii) the income tax paid by the employee.
(c) If the pension contribution and income tax were deducted from the gross monthly salary, before payment, calculate the take- home pay of the employee.
Tambaya 54 Rahoto
The table below shows the values of the relation \(y = 11 - 2x - 2x^{2}\) for \(-4 \leq x \leq 3\).
| x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| y | -13 | 11 |
(a) Copy and complete the table.
(b) Using a scale of 2 cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of \(y = 11 - 2x - 2x^{2}\).
(c) Use your graph to find : (i) the roots of the equation \(11 - 2x - 2x^{2} = 0\) ; (ii) the values of x for which \(3 - 2x - 2x^{2} = 0\) ; (iii) the gradient of the curve at x = 1.
Tambaya 55 Rahoto
In the diagram, PQT is a straight line and SQ // RT.
(a) Join QR and show that : (i) < RPS = < QRT ; (ii) < PRS = < QTR.
(b) ABC is a triangle. The sides AB and AC are produced to D and E respectively such that < DBC = 132° and < ECD = 96°. Show that \(\Delta\) ABC is isosceles.
Bayanin Amsa
None
Tambaya 56 Rahoto
(a) Simplify : \(625^{\frac{3}{8}} \times 5^{\frac{1}{2}} \div 25\)
(b) Solve the following equations correct to one decimal place.
(i) \(\tan (\theta + 25)° = 5.145\)
(ii) \(5\cos \theta - 1 = 0\), where \(0° \leq \theta \leq 90°\).
Bayanin Amsa
None
Tambaya 57 Rahoto
(a) A = {1, 2, 5, 7} and B = {1, 3, 6, 7} are subsets of the universal set U = {1, 2, 3,...., 10}. Find (i) \(A'\) ; (ii) \((A \cap B)'\) ; (iii) \((A \cup B)'\) ; (iv) the subsets of B each of which has three elements.
(b) Write down the 15th term of the sequence, \(\frac{2}{1 \times 3}, \frac{2}{2 \times 4}, \frac{4}{3 \times 5}, \frac{5}{4 \times 6},...\).
(c) An Arithmetic Progression (A.P) has 3 as its first term and 4 as the common difference, (i) write an expression in its simplest form for the nth term ; (ii) find the least term of the A.P that is greater than 100.
Tambaya 58 Rahoto
The marks obtained by 40 students in an examination are as follows :
85 77 87 74 77 78 79 89 95 90 78 73 86 83 91 74 84 81 83 75 77 70 81 69 75 63 76 87 61 78 69 96 65 80 84 80 77 74 88 72.
(a) Copy and complete the table for the distribution using the above data.
| Class Boundaries | Tally | Frequency |
| 59.5 - 64.5 | ||
| 64.5 - 69.5 | ||
| 69.5 - 74.5 | ||
| 74.5 - 79.5 | ||
| 79.5 - 84.5 | ||
| 84.5 - 89.5 | ||
| 89.5 - 94.5 | ||
| 94.5 - 99.5 |
(b) Draw a histogram to represent the distribution.
(c) Using your histogram, estimate the modal mark.
(d) If a student is chosen at random, find the probability that the student obtains a mark greater than 79.
Bayanin Amsa
None
Tambaya 59 Rahoto
Using ruler and a pair of compasses only,
(a) construct a quadrilateral PXYQ such that /PX/ = 9.9 cm, /QX/ = 10.2 cm, < QPZ = 75°, /QY/ = 10.4 cm and PQ // XY.
(b) Construct the (i) locus \(l_{1}\) of points equidistant from X and Y ; (ii) locus \(l_{2}\) of points equidistant from QY and YX.
(c) Locate M, the point of intersection of \(l_{1}\) and \(l_{2}\).
(d) Measure /PM/.
Bayanin Amsa
None
