WAEC SSCE - General Mathematics - 2001
Tambaya 1 Rahoto
Express 25° 45' in decimal (Hint: 1° = 60')
Bayanin Amsa
To convert 25° 45' to decimal, we first note that there are 60 minutes in a degree. So, to convert minutes to degrees, we divide by 60. Thus, 25° 45' = 25 + 45/60 degrees Simplifying, 25° 45' = 25 + 0.75 degrees Combining the whole and fractional parts, we get: 25° 45' = 25.75 degrees Therefore, the answer is: 25.75°.
Tambaya 2 Rahoto
A solid cylinder of radius 7cm is 10 cm long. Find its total surface area.
Bayanin Amsa
A solid cylinder has two circular bases and a curved lateral surface. To find its total surface area, we need to find the area of both circular bases and the area of the lateral surface, and then add them together. The area of one circular base is \(\pi r^2\), where r is the radius of the cylinder. Since the radius is given as 7 cm, the area of one circular base is: \begin{align*} \text{Area of one circular base} &= \pi \times (7 \text{ cm})^2 \\ &= 49 \pi \text{ cm}^2 \\ \end{align*} Since there are two circular bases, the total area of both circular bases is: \begin{align*} \text{Total area of both circular bases} &= 2 \times \text{Area of one circular base} \\ &= 2 \times 49 \pi \text{ cm}^2 \\ &= 98 \pi \text{ cm}^2 \\ \end{align*} The area of the lateral surface is the curved surface area of the cylinder, which is given by \(2\pi rh\), where r is the radius of the cylinder and h is the height (or length) of the cylinder. Since the radius is given as 7 cm and the length is given as 10 cm, the area of the lateral surface is: \begin{align*} \text{Area of lateral surface} &= 2\pi rh \\ &= 2\pi \times (7 \text{ cm}) \times (10 \text{ cm}) \\ &= 140 \pi \text{ cm}^2 \\ \end{align*} Therefore, the total surface area of the cylinder is the sum of the area of both circular bases and the area of the lateral surface: \begin{align*} \text{Total surface area} &= \text{Area of both circular bases} + \text{Area of lateral surface} \\ &= 98 \pi \text{ cm}^2 + 140 \pi \text{ cm}^2 \\ &= 238 \pi \text{ cm}^2 \\ \end{align*} Therefore, the total surface area of the cylinder is 238\(\pi\) cm\(^2\).
Tambaya 4 Rahoto
The diagram shows a triangular prism of length 7cm. The right - angled triangle PQR is a cross section of the prism |PR| = 5cm and |RQ| = 3cm. What is the area of the cross-section?
Bayanin Amsa
The area of a triangle is given by the formula: $$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $$ In triangle PQR, the base is |RQ| = 3cm and the height is |PS|, where S is the foot of the perpendicular from point P to line QR. Since the cross section is a right-angled triangle, we can use Pythagoras theorem to find |PS| as follows: $$ \begin{align*} |PS|^2 &= |PR|^2 - |RS|^2 \\ &= 5^2 - 3^2 \\ &= 16 \\ \end{align*} $$ Therefore, |PS| = 4cm. Substituting the values for base and height into the area formula, we get: $$ \text{Area} = \frac{1}{2} \times 3\text{cm} \times 4\text{cm} = 6\text{cm}^2 $$ Hence, the area of the cross-section is 6 cm2.
Tambaya 6 Rahoto
In the diagram, PQR is a triangle.|PQ| = |PR|, |PS| = |SQ| and PRS = 50°. What is the size of ∠PSQ?
Tambaya 7 Rahoto
Given that sin \(P = \frac{5}{13}\), where p is acute, find the value of cos p - tan p
Bayanin Amsa
We know that \(\sin p = \dfrac{5}{13}\) and $p$ is acute. Using Pythagorean identity, we can find $\cos p$: \begin{align*} \cos^2 p &= 1 - \sin^2 p \\ \cos^2 p &= 1 - \left(\dfrac{5}{13}\right)^2 \\ \cos^2 p &= \dfrac{144}{169} \\ \cos p &= \dfrac{12}{13} \end{align*} Using the definition of tangent, we can find $\tan p$: \begin{align*} \tan p &= \dfrac{\sin p}{\cos p} \\ \tan p &= \dfrac{\frac{5}{13}}{\frac{12}{13}} \\ \tan p &= \dfrac{5}{12} \end{align*} Therefore, \begin{align*} \cos p - \tan p &= \dfrac{12}{13} - \dfrac{5}{12} \\ &= \dfrac{144}{156} - \dfrac{65}{156} \\ &= \dfrac{79}{156} \end{align*} Hence, the answer is \(\frac{79}{156}\).
Tambaya 8 Rahoto
The diagram shows a triangular prism of length 7cm. The right - angled triangle PQR is a cross section of the prism |PR| = 5cm and |RQ| = 3cm. What is the volume of the prism?
Bayanin Amsa
Tambaya 9 Rahoto
What is the mode?
Bayanin Amsa
The mode is the most frequently occurring value in a set of data. In this case, the data set is {22, 23, 24, 25, 27, 30}. The value that appears most frequently in this set is 23 years, which appears twice, making it the mode. Therefore, the answer is 23.0 years.
Tambaya 10 Rahoto
If 2x : (x +1) = 3:2, what is the value of x?
Bayanin Amsa
To solve the equation 2x : (x + 1) = 3 : 2, we need to cross-multiply the ratios. That means we multiply the numerator of the left-hand side ratio by the denominator of the right-hand side ratio, and we multiply the denominator of the left-hand side ratio by the numerator of the right-hand side ratio. So we have: 2x * 2 = 3 * (x + 1) Simplifying the right-hand side: 4x = 3x + 3 Subtracting 3x from both sides: x = 3 Therefore, the value of x is 3. To check, we can substitute x = 3 back into the original equation and see if both sides are equal: 2(3) : (3 + 1) = 6 : 4 = 3 : 2 So the equation is true when x = 3, and that is our final answer.
Tambaya 11 Rahoto
Given that P = {b, d, e, f} and Q = {a, c, f, g} are subsets of the universal set U = {a,b, c, d, e, f, g}. Find P' ∩ Q
Tambaya 12 Rahoto
In a class of 80 students, every students studies Economics or Geography or both. If 65 students study Economics and 50 study Geography, how many study both subjects?
Bayanin Amsa
We can solve this problem using a Venn diagram. Let's draw two circles, one for Economics and one for Geography, with some overlap between them. We know that 65 students study Economics, so we'll put 65 in the circle for Economics. Similarly, we know that 50 students study Geography, so we'll put 50 in the circle for Geography. Now we need to figure out how many students study both subjects. We can call this number "x" for now. We'll put "x" in the overlap between the two circles. We also know that every student in the class studies either Economics or Geography or both. So we need to account for all 80 students in the class. To do this, we can add up the numbers we've put in the circles and subtract the overlap once (since the students in the overlap were counted twice). In other words, Number of students studying Economics + Number of students studying Geography - Number of students studying both = Total number of students in the class Or, 65 + 50 - x = 80 Simplifying, 115 - x = 80 x = 35 So 35 students study both Economics and Geography. Therefore, the answer is option (C) 35.
Tambaya 13 Rahoto
In the diagram, PQ is a diameter of the circle and ∠PRS = 58°. Find ∠STQ.
Bayanin Amsa
Tambaya 14 Rahoto
The graph of the curve \(y = 2x^2 - 5x - 1\) and a straight line PQ were drawn to solve the equation \(2x^2 - 5x + 2 = 0\)
What is the equation of the straight line PQ?
Bayanin Amsa
Tambaya 15 Rahoto
A boy estimated his transport fare for a journey as N190 instead of N200. Find the percentage error in his estimate
Bayanin Amsa
The error in the boy's estimate is the difference between what he estimated and the actual cost of the journey, which is N200 - N190 = N10. To find the percentage error, we divide the error by the actual cost and then multiply by 100%. Percentage error = (error / actual cost) x 100% = (10 / 200) x 100% = 5% Therefore, the percentage error in the boy's estimate is 5%. Option (d) is the correct answer.
Tambaya 17 Rahoto
If N varies directly as M and N = 8 when M = 20 find M when N = 7
Bayanin Amsa
When two variables are said to vary directly with each other, it means that as one increases or decreases, the other also increases or decreases by the same factor or proportion. This relationship can be expressed as: N = kM where k is the constant of proportionality. To find the value of k, we can use the initial values given: N = 8 when M = 20. 8 = k(20) Solving for k, we get: k = 8/20 = 0.4 Now that we know the value of k, we can use it to find M when N = 7. 7 = 0.4M Solving for M, we get: M = 7/0.4 = 17.5 Therefore, M is equal to 17.5 when N is equal to 7. The answer is \(17\frac{1}{2}\).
Tambaya 19 Rahoto
If \(2x^2 + kx - 14 = (x+2)(2x-7)\), find the value of K
Bayanin Amsa
To find the value of k, we need to expand the right-hand side of the equation, which is equal to the left-hand side: \begin{align*} (x+2)(2x-7) &= 2x^2 - 3x - 14 \\ \end{align*} Now we can compare the coefficients of the terms on both sides of the equation: \begin{align*} \text{Coefficient of } x^2: \quad &2 = 2 \\ \text{Coefficient of } x: \quad &-3 = k \\ \text{Coefficient of the constant term:} \quad &-14 = -14 \\ \end{align*} Therefore, we have found that the value of k is -3.
Tambaya 20 Rahoto
Simplify \(\frac{2x-1}{3}-\frac{x+3}{2}\)
Bayanin Amsa
To subtract two fractions, they need to have a common denominator. The common denominator of 3 and 2 is 6, so we can write: \begin{align*} \frac{2x-1}{3}-\frac{x+3}{2} &= \frac{2(2x-1)}{2 \cdot 3}-\frac{3(x+3)}{3 \cdot 2} \\ &= \frac{4x-2}{6}-\frac{3x+9}{6} \\ &= \frac{4x-3x-2-9}{6} \\ &= \frac{x-11}{6} \end{align*} Therefore, the answer is \(\frac{x-11}{6}\).
Tambaya 21 Rahoto
If the probability that an event will occur is p and the probability that it will not occur is q, which of the following is true?
Bayanin Amsa
The sum of the probabilities of an event occurring and not occurring is always equal to 1. So, if p is the probability that an event will occur, then the probability that it will not occur is 1-p (since the total probability is 1). Therefore, we have: p + (1-p) = 1 which simplifies to: p - p + 1 = 1-p + p 1 = 1 So, the correct option is: p + q = 1.
Tambaya 23 Rahoto
If \(\frac{y-3}{2}<\frac{2y-1}{3}\), which of the following is true?
Bayanin Amsa
To solve this inequality, we can start by simplifying both sides. First, we can multiply both sides by 6 to eliminate the denominators: 3(y - 3) < 4(2y - 1) Expanding the right side gives: 3(y - 3) < 8y - 4 Simplifying and collecting like terms, 3y - 9 < 8y - 4 Subtracting 3y from both sides, -9 < 5y - 4 Adding 4 to both sides, -5 < 5y Dividing both sides by 5, -1 < y So the inequality is true for all values of y greater than -1. Therefore, the correct answer is: y > -7.
Tambaya 24 Rahoto
A sector of a circle radius 14 cm subtends an angle 135° at the center of the circle. What is the perimeter of the sector? Take \(\pi = \frac{22}{7}\)
Bayanin Amsa
The perimeter of a sector is given by the sum of the length of the two radii and the arc length between them. In this question, we are given the radius of the sector to be 14cm and the central angle to be 135°. The arc length can be found using the formula for the circumference of a circle, C = 2πr, where r is the radius of the circle. The central angle of 135° is equivalent to \(\frac{135}{360}\) of the full circle, so the arc length of the sector is: \(\frac{135}{360} \times 2\pi \times 14 \approx 32.91cm\) The two radii have the same length and are equal to 14cm each. Therefore, the perimeter of the sector is: 14cm + 14cm + 32.91cm ≈ 60.91cm Rounding this to the nearest whole number gives us 61cm. Hence, the correct answer is 61cm.
Tambaya 25 Rahoto
The nth term of a sequence is \(2^{2n-1}\). Which term of the sequence is \(2^9?\)
Bayanin Amsa
To find which term of the sequence is \(2^9\), we need to solve for n in the equation \(2^{2n-1}=2^9\). First, we can simplify the equation by dividing both sides by \(2^9\), giving us \(2^{2n-1-9}=2^{2n-10}=1\). Next, we can solve for n by taking the logarithm of both sides of the equation. Since any logarithm base can be used, we can use the natural logarithm, denoted as ln: \begin{align*} 2n-10 &= \ln 1 \\ 2n-10 &= 0 \\ 2n &= 10 \\ n &= 5 \end{align*} Therefore, the fifth term of the sequence is \(2^9\).
Tambaya 26 Rahoto
If \(y \propto \frac{1}{\sqrt{x}}\) and x = 16 when y = 2, find x when y = 24
Bayanin Amsa
Tambaya 27 Rahoto
A box contains 5 red, 3 green and 4 blue balls. A boy is allowed to take away two balls from the box. What is the probability that the two balls are red?
Bayanin Amsa
There are 12 balls in total in the box, out of which 5 are red. When the boy takes out the first ball, there are 11 balls left, out of which 4 are red. Therefore, the probability of the first ball being red is 5/12, and the probability of the second ball being red is 4/11. To find the probability of both events happening together, we multiply the probabilities: \(\frac{5}{12}\times\frac{4}{11}=\frac{5}{33}\) Therefore, the probability that the two balls taken out are both red is 5/33. Answer: \(\frac{5}{33}\)
Tambaya 28 Rahoto
Factorize \(6x^2 + 7x - 20\)
Bayanin Amsa
To factorize \(6x^2 + 7x - 20\), we need to find two numbers that multiply to -120 (the product of the leading coefficient, 6, and the constant, -20) and add up to the coefficient of the middle term, 7. These numbers are 15 and -8. Therefore, we can write: \begin{align*} 6x^2 + 7x - 20 &= 6x^2 + 15x - 8x - 20 \\ &= 3x(2x+5) - 4(2x+5) \\ &= (3x-4)(2x+5) \end{align*} So the factorization of \(6x^2 + 7x - 20\) is \((3x-4)(2x+5)\).
Tambaya 31 Rahoto
A man is four times as old as his son. The difference between their ages is 36 years Find the sum of their ages
Bayanin Amsa
Let the age of the son be x. Then, the age of the man will be 4x (as he is four times as old as his son). The difference in their ages is 36 years, so we have: 4x - x = 36 Simplifying this, we get: 3x = 36 x = 12 So, the age of the son is 12 and the age of the man is 4 times that, which is 48. Therefore, the sum of their ages is: 12 + 48 = 60 years. Hence, the correct option is (c) 60 years.
Tambaya 32 Rahoto
Which of the following is not the the size of an exterior angle of a regular polygon?
Bayanin Amsa
Tambaya 33 Rahoto
Three observation posts P,Q and R are such that Q is due east of P and R is due north of Q. If |PQ| = 5km and |PR| = 10km, find |QR|
Tambaya 34 Rahoto
Which of the following correctly expresses 48 as a product of prime factors?
Bayanin Amsa
To express 48 as a product of prime factors, we need to keep dividing it by prime numbers until we get to 1. Let's start with 2, which is the smallest prime number that can divide 48: 48 ÷ 2 = 24 We can continue dividing by 2 until we can't divide by 2 anymore: 24 ÷ 2 = 12 12 ÷ 2 = 6 6 ÷ 2 = 3 Now, we can see that we have reached a prime number, 3, so we stop dividing. Therefore: 48 = 2 x 2 x 2 x 2 x 3 So the correct option is 2 x 2 x 2 x 2 x 3 or.
Tambaya 36 Rahoto
Express 0.0462 in standard form
Tambaya 37 Rahoto
Evaluate \(\frac{x^2 + x - 2}{2x^2 + x -3}\) when x = -1
Bayanin Amsa
Substituting x = -1, we get: \[\frac{(-1)^2 + (-1) - 2}{2(-1)^2 + (-1) - 3}\] \[=\frac{1 - 1 - 2}{2 - 1 - 3}\] \[=\frac{-2}{-2}\] \[=1\] Therefore, the value of the expression is 1 when x = -1. Hence the correct option is (d) 1.
Tambaya 38 Rahoto
PQRS is a trapezium in which |PS| = 9cm, |QR| = 15cm, |PQ| = \(2\sqrt{3}, \angle PQR = 90^o and \angle QRS = 30^o\). Calculate the area of the trapezium
Bayanin Amsa
Tambaya 40 Rahoto
A chord of length 30cm is 8cm away from the center of the circle. What is the radius of the circle. What is the perimeter of the sector? Take \(\pi =\frac{22}{7}\)
Bayanin Amsa
Tambaya 41 Rahoto
A box contains 5 red, 3 green and 4 blue balls. A boy is allowed to take away two balls from the box. What is the probability that one is green and the other is blue?
Bayanin Amsa
Tambaya 42 Rahoto
Evaluate \(\frac{1}{2}+\frac{3}{4}of\frac{2}{5}\div 1\frac{3}{5}\)
Bayanin Amsa
To evaluate the given expression, we follow the order of operations or BODMAS (Brackets, Order, Division, Multiplication, Addition, and Subtraction) and simplify the expression step by step. First, we simplify the fraction in the middle of the expression, \(\frac{3}{4}of\frac{2}{5} = \frac{3}{4} \times \frac{2}{5} = \frac{3\times 2}{4\times 5} = \frac{3}{10}\) Now, we can rewrite the expression as \(\frac{1}{2}+\frac{3}{10}\div 1\frac{3}{5}\) Next, we simplify the mixed number in the denominator by converting it to an improper fraction, \(1\frac{3}{5} = \frac{8}{5}\) Now, we can rewrite the expression as \(\frac{1}{2}+\frac{3}{10}\div \frac{8}{5}\) To divide by a fraction, we multiply by its reciprocal, \(\frac{1}{2}+\frac{3}{10}\times \frac{5}{8} = \frac{1}{2}+\frac{3\times 5}{10\times 8} = \frac{1}{2}+\frac{3}{16} = \frac{8}{16}+\frac{3}{16} = \frac{11}{16}\) Therefore, the value of the given expression is \(\frac{11}{16}\). Thus, option (B) is the correct answer.
Tambaya 43 Rahoto
The variance of a given distribution is 25. What is the standard deviation?
Bayanin Amsa
The standard deviation is the square root of the variance. Therefore, to find the standard deviation when given the variance, we simply take the square root of the variance. In this case, the variance is given as 25. Taking the square root of 25 gives us: sqrt(25) = 5 So the standard deviation of the distribution is 5. To summarize, the formula for finding the standard deviation from the variance is to take the square root of the variance. In this case, the variance is 25, so the standard deviation is 5.
Tambaya 44 Rahoto
Subtract (-y + 3x + 5z) from (4y - x - 2z)
Bayanin Amsa
To subtract (-y + 3x + 5z) from (4y - x - 2z), we can just subtract the corresponding coefficients of each variable. So, - the coefficient of y is 4 - (-1) = 5 - the coefficient of x is -1 - 3 = -4 - the coefficient of z is -2 - 5 = -7 Therefore, the result is: 5y - 4x - 7z. So the correct option is (a) 5y - 4x - 7z.
Tambaya 45 Rahoto
The bearing S40°E is the same as
Bayanin Amsa
The bearing S40°E is equivalent to an angle that is 40 degrees east of the southern direction. The direction opposite to south is north, so if we draw a line going north and another line going 40 degrees to the east of south, the angle formed between these two lines is the direction that corresponds to the bearing S40°E. This angle is equal to 180° - 40° = 140°. Therefore, the correct answer is 140o.
Tambaya 46 Rahoto
Evaluate \(log_{10}5 + log_{10}20\)
Bayanin Amsa
To evaluate \(log_{10}5 + log_{10}20\), we can use the logarithmic rule that states: $log_{a}x + log_{a}y = log_{a}(xy)$ Applying this rule, we get: $log_{10}5 + log_{10}20 = log_{10}(5 \times 20)$ Simplifying the right-hand side: $log_{10}(5 \times 20) = log_{10}100$ Finally, we know that \(log_{10}100 = 2\), so: \(log_{10}5 + log_{10}20 = log_{10}100 = 2\) Therefore, the answer is 2.
Tambaya 47 Rahoto
which of the following is not quadratic expression?
Bayanin Amsa
A quadratic expression is a polynomial of degree two. This means that the highest exponent in the expression is 2. Therefore, the expression that is not quadratic is y = 5(x-1) which is a linear expression of degree 1.
Tambaya 48 Rahoto
In a bag of oranges, the ratio of the good ones to the bad ones is 5:4. If the number of bad oranges in the bag is 36, how many oranges are there in the altogether?
Bayanin Amsa
The ratio of good to bad oranges is 5:4, which means that for every 5 good oranges, there are 4 bad ones. If we know there are 36 bad oranges, we can use this ratio to find the number of good oranges as follows: 5/4 = x/36 where x is the number of good oranges. We can solve for x by cross-multiplying: 4x = 5*36 x = 45 So there are 45 good oranges in the bag. To find the total number of oranges, we add the number of good and bad ones: 45 + 36 = 81 Therefore, there are 81 oranges in the bag altogether.
Tambaya 49 Rahoto
Given that y = px + q and y = 5 when x = 3, while y = 4 when x = 2, find the value of p and q.
Bayanin Amsa
To find the values of p and q, we need to use the information given in the problem. We are given that: y = px + q When x = 3, y = 5 5 = 3p + q --- (Equation 1) When x = 2, y = 4 4 = 2p + q --- (Equation 2) We can solve these two equations simultaneously to get the values of p and q. We can do this by subtracting equation 2 from equation 1: 5 = 3p + q - (4 = 2p + q) 1 = p Substituting p = 1 into equation 2, we get: 4 = 2(1) + q 4 = 2 + q q = 2 Therefore, the values of p and q are p = 1 and q = 2. Therefore, the answer is (B) p = 1, q = 2.
Tambaya 50 Rahoto
The number of child births recorded in 50 maternity centres of a local government in August 1993 are as follows :
50 99 81 86 69 85 93 63 92 65 77 74 76 71 90 74 81 94 67 75 95 81 68 105 99 68 75 75 76 73 79 74 80 69 74 62 74 80 79 68 79 75 75 71 83 75 80 85 81 82
(a) Construct a frequency distribution table, using class intervals 45 - 54, 55 - 64, etc.
(b) Draw the histogram for the distribution
(c) Use your histogram to estimate the mode.
(d) Calculate the mean number of births.
(a) Frequency distribution
| Number of births | Class midpoint, \(x\) | Frequency, \(f\) | \(fx\) |
|---|---|---|---|
| 45–54 | 49.5 | 1 | 49.5 |
| 55–64 | 59.5 | 2 | 119.0 |
| 65–74 | 69.5 | 15 | 1042.5 |
| 75–84 | 79.5 | 21 | 1669.5 |
| 85–94 | 89.5 | 7 | 626.5 |
| 95–104 | 99.5 | 3 | 298.5 |
| 105–114 | 109.5 | 1 | 109.5 |
| Total | 50 | 3915 |
(b) Histogram
The continuous class boundaries are \(44.5\text{–}54.5,\ 54.5\text{–}64.5,\ldots,\ 104.5\text{–}114.5\). Since all class widths are 10, the heights of the adjacent bars are their frequencies.
(c) Estimated mode
The modal class is \(75\text{–}84\). From the histogram, or by interpolation within this class,
\[\text{Mode}=L+\frac{f_1-f_0}{(f_1-f_0)+(f_1-f_2)}\times c\]
\[=74.5+\frac{21-15}{(21-15)+(21-7)}\times10\]
\[=74.5+\frac{6}{20}\times10=\boxed{77.5\text{ births (approximately)}}.\]
(d) Mean number of births
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{3915}{50}=\boxed{78.3\text{ births}}.\]
Bayanin Amsa
(a) Frequency distribution
| Number of births | Class midpoint, \(x\) | Frequency, \(f\) | \(fx\) |
|---|---|---|---|
| 45–54 | 49.5 | 1 | 49.5 |
| 55–64 | 59.5 | 2 | 119.0 |
| 65–74 | 69.5 | 15 | 1042.5 |
| 75–84 | 79.5 | 21 | 1669.5 |
| 85–94 | 89.5 | 7 | 626.5 |
| 95–104 | 99.5 | 3 | 298.5 |
| 105–114 | 109.5 | 1 | 109.5 |
| Total | 50 | 3915 |
(b) Histogram
The continuous class boundaries are \(44.5\text{–}54.5,\ 54.5\text{–}64.5,\ldots,\ 104.5\text{–}114.5\). Since all class widths are 10, the heights of the adjacent bars are their frequencies.
(c) Estimated mode
The modal class is \(75\text{–}84\). From the histogram, or by interpolation within this class,
\[\text{Mode}=L+\frac{f_1-f_0}{(f_1-f_0)+(f_1-f_2)}\times c\]
\[=74.5+\frac{21-15}{(21-15)+(21-7)}\times10\]
\[=74.5+\frac{6}{20}\times10=\boxed{77.5\text{ births (approximately)}}.\]
(d) Mean number of births
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{3915}{50}=\boxed{78.3\text{ births}}.\]
Tambaya 51 Rahoto
(a) In a market survey, 100 traders sell fruits, 40 sell apples, 46 oranges, 50 mangoes, 14 apples and oranges, 15 apples and mangoes and 10 sell the three types of fruits. Each of the 100 traders sells at least one of the three fruits.
(i) Represent the information in a Venn diagram ; (ii) Find the number that sell oranges and mangoes only.
(b) Find the value of x for which \(312_{four} + 52_{x} = 96_{ten}\)
(a) Apples \(A=40\), Oranges \(O=46\), Mangoes \(M=50\). \(A\cap O=14\), \(A\cap M=15\), all three \(=10\); every trader sells at least one, total \(=100\).
(i) Venn diagram. Three intersecting circles; centre (all three) \(=10\).
(ii) Let \(O\cap M\) (total) \(=y\). By the inclusion-exclusion principle:
\(|A|+|O|+|M|-|A\cap O|-|A\cap M|-|O\cap M|+|A\cap O\cap M|=100\).
\(40+46+50-14-15-y+10=100 \Rightarrow 117-y=100 \Rightarrow y=17\).
Oranges and mangoes only \(=y-10=17-10=\mathbf{7}\).
(b) \(312_{four}+52_x=96_{ten}\).
\(312_{four}=3(16)+1(4)+2=54\). So \(52_x=96-54=42\).
\(52_x=5x+2=42 \Rightarrow 5x=40 \Rightarrow x=\mathbf{8}\).
Bayanin Amsa
(a) Apples \(A=40\), Oranges \(O=46\), Mangoes \(M=50\). \(A\cap O=14\), \(A\cap M=15\), all three \(=10\); every trader sells at least one, total \(=100\).
(i) Venn diagram. Three intersecting circles; centre (all three) \(=10\).
(ii) Let \(O\cap M\) (total) \(=y\). By the inclusion-exclusion principle:
\(|A|+|O|+|M|-|A\cap O|-|A\cap M|-|O\cap M|+|A\cap O\cap M|=100\).
\(40+46+50-14-15-y+10=100 \Rightarrow 117-y=100 \Rightarrow y=17\).
Oranges and mangoes only \(=y-10=17-10=\mathbf{7}\).
(b) \(312_{four}+52_x=96_{ten}\).
\(312_{four}=3(16)+1(4)+2=54\). So \(52_x=96-54=42\).
\(52_x=5x+2=42 \Rightarrow 5x=40 \Rightarrow x=\mathbf{8}\).
Tambaya 52 Rahoto
(a) Simplify : \(\frac{1}{3^{5n}} \times 9^{n - 1} \times 27^{n + 1}\)
(b) The sum of the ages of a woman and her daughter is 46 years. In 4 years' time, the ratio of their ages will be 7 : 2. Find their present ages.
(a) \(\dfrac{1}{3^{5n}}\times 9^{n-1}\times 27^{n+1}\). Convert everything to base 3:
\(9^{n-1}=3^{2(n-1)}=3^{2n-2}\); \(27^{n+1}=3^{3(n+1)}=3^{3n+3}\); \(\dfrac{1}{3^{5n}}=3^{-5n}\).
Add exponents: \(-5n+(2n-2)+(3n+3)= (-5n+2n+3n)+(-2+3)=0+1=1\).
Result \(=3^{1}=\mathbf{3}\).
(b) Let woman \(=w\), daughter \(=d\). \(w+d=46\).
In 4 years: \(\dfrac{w+4}{d+4}=\dfrac{7}{2}\Rightarrow 2(w+4)=7(d+4)\Rightarrow 2w+8=7d+28\Rightarrow 2w-7d=20\).
Substitute \(w=46-d\): \(2(46-d)-7d=20\Rightarrow 92-9d=20\Rightarrow 9d=72\Rightarrow d=8\), \(w=38\).
Woman is 38 years, daughter is 8 years. Check: in 4 years \(42:12=7:2\). \(\checkmark\)
Bayanin Amsa
(a) \(\dfrac{1}{3^{5n}}\times 9^{n-1}\times 27^{n+1}\). Convert everything to base 3:
\(9^{n-1}=3^{2(n-1)}=3^{2n-2}\); \(27^{n+1}=3^{3(n+1)}=3^{3n+3}\); \(\dfrac{1}{3^{5n}}=3^{-5n}\).
Add exponents: \(-5n+(2n-2)+(3n+3)= (-5n+2n+3n)+(-2+3)=0+1=1\).
Result \(=3^{1}=\mathbf{3}\).
(b) Let woman \(=w\), daughter \(=d\). \(w+d=46\).
In 4 years: \(\dfrac{w+4}{d+4}=\dfrac{7}{2}\Rightarrow 2(w+4)=7(d+4)\Rightarrow 2w+8=7d+28\Rightarrow 2w-7d=20\).
Substitute \(w=46-d\): \(2(46-d)-7d=20\Rightarrow 92-9d=20\Rightarrow 9d=72\Rightarrow d=8\), \(w=38\).
Woman is 38 years, daughter is 8 years. Check: in 4 years \(42:12=7:2\). \(\checkmark\)
Tambaya 53 Rahoto
(a) Draw the table of values for the relation \(y = x^{2}\) for the interval \(-3 \leq x \leq 4\).
(b) Using a scale of 2 cm to 1 unit on the x- axis and 2 cm to 2 units on the y- axis, draw the graphs of : (i) \(y = x^{2}\) ; (ii) \(y = 2x + 3\) for \(-3 \leq x \leq 4\).
(c) Use your graph to find : (i) the roots of the equation \(x^{2} = 2x + 3\) ; (ii) the gradient of \(y = x^{2}\) at x = -2.
(a) Table of values for \(y=x^2\)
| \(x\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|---|---|---|
| \(y=x^2\) | 9 | 4 | 1 | 0 | 1 | 4 | 9 | 16 |
(b) Values for the straight line \(y=2x+3\)
| \(x\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|---|---|---|
| \(y=2x+3\) | \(-3\) | \(-1\) | 1 | 3 | 5 | 7 | 9 | 11 |
Using the stated scales, plot the points and draw smooth curve \(y=x^2\) and the straight line \(y=2x+3\). A tangent to the curve at \((-2,4)\) is also shown for part (c)(ii).
(c)(i) The graphs intersect at \((-1,1)\) and \((3,9)\). Hence, the roots of \(x^2=2x+3\) are
\[\boxed{x=-1\text{ and }x=3}.\]
(c)(ii) From the tangent at \(x=-2\), take the points \((-3,8)\) and \((-1,0)\):
\[\text{gradient}=\frac{0-8}{-1-(-3)}=\frac{-8}{2}=\boxed{-4}.\]
Bayanin Amsa
(a) Table of values for \(y=x^2\)
| \(x\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|---|---|---|
| \(y=x^2\) | 9 | 4 | 1 | 0 | 1 | 4 | 9 | 16 |
(b) Values for the straight line \(y=2x+3\)
| \(x\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|---|---|---|
| \(y=2x+3\) | \(-3\) | \(-1\) | 1 | 3 | 5 | 7 | 9 | 11 |
Using the stated scales, plot the points and draw smooth curve \(y=x^2\) and the straight line \(y=2x+3\). A tangent to the curve at \((-2,4)\) is also shown for part (c)(ii).
(c)(i) The graphs intersect at \((-1,1)\) and \((3,9)\). Hence, the roots of \(x^2=2x+3\) are
\[\boxed{x=-1\text{ and }x=3}.\]
(c)(ii) From the tangent at \(x=-2\), take the points \((-3,8)\) and \((-1,0)\):
\[\text{gradient}=\frac{0-8}{-1-(-3)}=\frac{-8}{2}=\boxed{-4}.\]
Tambaya 54 Rahoto
(a) A boy blew his rubber balloon to a spherical shape. The balloon burst when its diameter was 15 cm. Calculate, correct to the nearest whole number, the volume of air in the balloon at the point of bursting. [Take \(\pi = \frac{22}{7}\)]
(b) A point X is on latitude 28°N and longitude 105°W. Y is another point on the same latitude as X but on longitude 35°E. (i) Calculate, correct to three significant figures, the distance between X and Y along latitude 28°N ; (ii) How far is X from the equator? [Take \(\pi = \frac{22}{7}\) and radius of the earth = 6,400km].
(a) Volume of the balloon. Diameter \(=15\text{ cm}\Rightarrow r=7.5\text{ cm}\).
\(V=\dfrac{4}{3}\pi r^{3}=\dfrac{4}{3}\times\dfrac{22}{7}\times 7.5^{3}=\dfrac{88}{21}\times 421.875=\dfrac{37125}{21}=1767.86\).
Volume \(\approx \mathbf{1768\ cm^{3}}\).
(b) X: \(28^{\circ}\)N, \(105^{\circ}\)W; Y: \(28^{\circ}\)N, \(35^{\circ}\)E. Difference in longitude \(=105+35=140^{\circ}\).
(i) Distance X to Y along latitude \(28^{\circ}\)N:
\(d=\dfrac{140}{360}\times 2\pi R\cos 28^{\circ}=\dfrac{140}{360}\times 2\times\dfrac{22}{7}\times 6400\times\cos 28^{\circ}\).
\(=\dfrac{140}{360}\times 40228.57\times 0.8829\approx 13{,}800\text{ km}\ (\mathbf{1.38\times 10^{4}\ km})\).
(ii) Distance of X from the equator (along its meridian):
\(d=\dfrac{28}{360}\times 2\pi R=\dfrac{28}{360}\times 40228.57\approx \mathbf{3130\ km}\).
Bayanin Amsa
(a) Volume of the balloon. Diameter \(=15\text{ cm}\Rightarrow r=7.5\text{ cm}\).
\(V=\dfrac{4}{3}\pi r^{3}=\dfrac{4}{3}\times\dfrac{22}{7}\times 7.5^{3}=\dfrac{88}{21}\times 421.875=\dfrac{37125}{21}=1767.86\).
Volume \(\approx \mathbf{1768\ cm^{3}}\).
(b) X: \(28^{\circ}\)N, \(105^{\circ}\)W; Y: \(28^{\circ}\)N, \(35^{\circ}\)E. Difference in longitude \(=105+35=140^{\circ}\).
(i) Distance X to Y along latitude \(28^{\circ}\)N:
\(d=\dfrac{140}{360}\times 2\pi R\cos 28^{\circ}=\dfrac{140}{360}\times 2\times\dfrac{22}{7}\times 6400\times\cos 28^{\circ}\).
\(=\dfrac{140}{360}\times 40228.57\times 0.8829\approx 13{,}800\text{ km}\ (\mathbf{1.38\times 10^{4}\ km})\).
(ii) Distance of X from the equator (along its meridian):
\(d=\dfrac{28}{360}\times 2\pi R=\dfrac{28}{360}\times 40228.57\approx \mathbf{3130\ km}\).
Tambaya 55 Rahoto
Given that \(\log_{10} 2 = 0.3010\) and \(\log_{10} 3 = 0.4771\), calculate without using mathematical tables or calculator, the value of :
(a) \(\log_{10} 54\) ;
(b) \(\log_{10} 0.24\).
Given \(\log_{10}2=0.3010,\ \log_{10}3=0.4771\).
(a) \(54=2\times 27=2\times 3^{3}\).
\(\log_{10}54=\log_{10}2+3\log_{10}3=0.3010+3(0.4771)=0.3010+1.4313=\mathbf{1.7323}\).
(b) \(0.24=\dfrac{24}{100}=\dfrac{2^{3}\times 3}{100}\).
\(\log_{10}24=3\log_{10}2+\log_{10}3=0.9030+0.4771=1.3801\).
\(\log_{10}0.24=\log_{10}24-\log_{10}100=1.3801-2=\mathbf{-0.6199}\) (i.e. \(\bar{1}.3801\)).
Bayanin Amsa
Given \(\log_{10}2=0.3010,\ \log_{10}3=0.4771\).
(a) \(54=2\times 27=2\times 3^{3}\).
\(\log_{10}54=\log_{10}2+3\log_{10}3=0.3010+3(0.4771)=0.3010+1.4313=\mathbf{1.7323}\).
(b) \(0.24=\dfrac{24}{100}=\dfrac{2^{3}\times 3}{100}\).
\(\log_{10}24=3\log_{10}2+\log_{10}3=0.9030+0.4771=1.3801\).
\(\log_{10}0.24=\log_{10}24-\log_{10}100=1.3801-2=\mathbf{-0.6199}\) (i.e. \(\bar{1}.3801\)).
Tambaya 56 Rahoto
(a)
In the diagram, MN || ST, NP || QT and < STQ = 70°. Find x.
(b)
In the diagram above, AC is a straight line, |BC| = |BD|, \(\stackrel\frown{BCD} = 50°\) and \(\stackrel\frown{BAD} = 55°\). Find \(\stackrel\frown{BDA}\).
(a) Find \(x\).
From the diagram, \(MN\parallel ST\) and \(NP\parallel QT\), with \(\angle STQ=70^{\circ}\) at \(T\), and \(x\) is the angle at \(N\) between \(NM\) and \(NP\).
Extend \(QT\) to cut \(MN\) at a point \(W\). Because \(MN\parallel ST\) and \(QT\) is a transversal, corresponding angles give
\[\angle MWQ=\angle STQ=70^{\circ}\]Now \(MN\) is a transversal of the parallel lines \(NP\) and \(QT\). The ray \(NP\) points to the same side of \(MN\) as \(P\) (above), while \(WQ\) points to the opposite side (below), so \(x\) and \(\angle MWQ\) are co-interior (allied) angles:
\[x+\angle MWQ=180^{\circ}\]\[x=180^{\circ}-70^{\circ}=\mathbf{110^{\circ}}\](b) Find \(\angle BDA\).
From the diagram \(A,B,C\) lie on the straight line \(AC\), with \(D\) above it, \(|BC|=|BD|\), \(\angle BCD=50^{\circ}\) and \(\angle BAD=55^{\circ}\).
Triangle \(BCD\) is isosceles with \(|BC|=|BD|\), so the base angles are equal:
\[\angle BDC=\angle BCD=50^{\circ}\]\[\angle DBC=180^{\circ}-50^{\circ}-50^{\circ}=80^{\circ}\]Since \(A,B,C\) are collinear, \(\angle DBA\) and \(\angle DBC\) are angles on a straight line at \(B\):
\[\angle DBA=180^{\circ}-80^{\circ}=100^{\circ}\]In triangle \(ABD\):
\[\angle BDA=180^{\circ}-\angle BAD-\angle DBA=180^{\circ}-55^{\circ}-100^{\circ}=\mathbf{25^{\circ}}\]Bayanin Amsa
(a) Find \(x\).
From the diagram, \(MN\parallel ST\) and \(NP\parallel QT\), with \(\angle STQ=70^{\circ}\) at \(T\), and \(x\) is the angle at \(N\) between \(NM\) and \(NP\).
Extend \(QT\) to cut \(MN\) at a point \(W\). Because \(MN\parallel ST\) and \(QT\) is a transversal, corresponding angles give
\[\angle MWQ=\angle STQ=70^{\circ}\]Now \(MN\) is a transversal of the parallel lines \(NP\) and \(QT\). The ray \(NP\) points to the same side of \(MN\) as \(P\) (above), while \(WQ\) points to the opposite side (below), so \(x\) and \(\angle MWQ\) are co-interior (allied) angles:
\[x+\angle MWQ=180^{\circ}\]\[x=180^{\circ}-70^{\circ}=\mathbf{110^{\circ}}\](b) Find \(\angle BDA\).
From the diagram \(A,B,C\) lie on the straight line \(AC\), with \(D\) above it, \(|BC|=|BD|\), \(\angle BCD=50^{\circ}\) and \(\angle BAD=55^{\circ}\).
Triangle \(BCD\) is isosceles with \(|BC|=|BD|\), so the base angles are equal:
\[\angle BDC=\angle BCD=50^{\circ}\]\[\angle DBC=180^{\circ}-50^{\circ}-50^{\circ}=80^{\circ}\]Since \(A,B,C\) are collinear, \(\angle DBA\) and \(\angle DBC\) are angles on a straight line at \(B\):
\[\angle DBA=180^{\circ}-80^{\circ}=100^{\circ}\]In triangle \(ABD\):
\[\angle BDA=180^{\circ}-\angle BAD-\angle DBA=180^{\circ}-55^{\circ}-100^{\circ}=\mathbf{25^{\circ}}\]Tambaya 57 Rahoto
(a) The probabilities that three boys pass an examination are \(\frac{2}{3}, \frac{5}{8}\) and \(\frac{3}{4}\) respectively. Find the probability that :
(i) all three boys pass ; (ii) none of the boys pass ; (iii) only two of the boys pass.
(b) A shop-keeper marks a television set for sale at N36,000 so as to make a profit of 20% on the cost price. When he sells it, he allows a discount of 5% of the marked price. Calculate the actual percentage profit.
(a) Pass probabilities \(\tfrac{2}{3},\tfrac{5}{8},\tfrac{3}{4}\); fail probabilities \(\tfrac{1}{3},\tfrac{3}{8},\tfrac{1}{4}\).
(i) All three pass \(=\dfrac{2}{3}\times\dfrac{5}{8}\times\dfrac{3}{4}=\dfrac{30}{96}=\dfrac{5}{16}\).
(ii) None passes \(=\dfrac{1}{3}\times\dfrac{3}{8}\times\dfrac{1}{4}=\dfrac{3}{96}=\dfrac{1}{32}\).
(iii) Exactly two pass:
- 1&2 pass, 3 fails: \(\tfrac{2}{3}\cdot\tfrac{5}{8}\cdot\tfrac{1}{4}=\tfrac{10}{96}\)
- 1&3 pass, 2 fails: \(\tfrac{2}{3}\cdot\tfrac{3}{8}\cdot\tfrac{3}{4}=\tfrac{18}{96}\)
- 2&3 pass, 1 fails: \(\tfrac{1}{3}\cdot\tfrac{5}{8}\cdot\tfrac{3}{4}=\tfrac{15}{96}\)
Total \(=\dfrac{10+18+15}{96}=\dfrac{43}{96}\).
(b) Marked price \(=\text{N}36{,}000\) gives \(20\%\) profit, so cost price \(=\dfrac{36000}{1.20}=\text{N}30{,}000\).
Discount of \(5\%\): selling price \(=36000\times 0.95=\text{N}34{,}200\).
Profit \(=34200-30000=\text{N}4{,}200\); actual percentage profit \(=\dfrac{4200}{30000}\times 100=\mathbf{14\%}\).
Bayanin Amsa
(a) Pass probabilities \(\tfrac{2}{3},\tfrac{5}{8},\tfrac{3}{4}\); fail probabilities \(\tfrac{1}{3},\tfrac{3}{8},\tfrac{1}{4}\).
(i) All three pass \(=\dfrac{2}{3}\times\dfrac{5}{8}\times\dfrac{3}{4}=\dfrac{30}{96}=\dfrac{5}{16}\).
(ii) None passes \(=\dfrac{1}{3}\times\dfrac{3}{8}\times\dfrac{1}{4}=\dfrac{3}{96}=\dfrac{1}{32}\).
(iii) Exactly two pass:
- 1&2 pass, 3 fails: \(\tfrac{2}{3}\cdot\tfrac{5}{8}\cdot\tfrac{1}{4}=\tfrac{10}{96}\)
- 1&3 pass, 2 fails: \(\tfrac{2}{3}\cdot\tfrac{3}{8}\cdot\tfrac{3}{4}=\tfrac{18}{96}\)
- 2&3 pass, 1 fails: \(\tfrac{1}{3}\cdot\tfrac{5}{8}\cdot\tfrac{3}{4}=\tfrac{15}{96}\)
Total \(=\dfrac{10+18+15}{96}=\dfrac{43}{96}\).
(b) Marked price \(=\text{N}36{,}000\) gives \(20\%\) profit, so cost price \(=\dfrac{36000}{1.20}=\text{N}30{,}000\).
Discount of \(5\%\): selling price \(=36000\times 0.95=\text{N}34{,}200\).
Profit \(=34200-30000=\text{N}4{,}200\); actual percentage profit \(=\dfrac{4200}{30000}\times 100=\mathbf{14\%}\).
Tambaya 58 Rahoto
The sides of a rectangular floor are xm and (x + 7)m. The diagonal is (x + 8)m. Calculate, in metres :
(a) the value of x ;
(b) the area of the floor.
Sides \(x\) and \((x+7)\), diagonal \((x+8)\). By Pythagoras:
\(x^{2}+(x+7)^{2}=(x+8)^{2}\).
\(x^{2}+x^{2}+14x+49=x^{2}+16x+64\).
\(2x^{2}+14x+49-x^{2}-16x-64=0 \Rightarrow x^{2}-2x-15=0\).
\((x-5)(x+3)=0 \Rightarrow x=5\) (reject \(x=-3\), a length must be positive).
(a) \(x=5\ \text{m}\).
(b) Area \(=x(x+7)=5\times 12=\mathbf{60\ m^{2}}\).
Bayanin Amsa
Sides \(x\) and \((x+7)\), diagonal \((x+8)\). By Pythagoras:
\(x^{2}+(x+7)^{2}=(x+8)^{2}\).
\(x^{2}+x^{2}+14x+49=x^{2}+16x+64\).
\(2x^{2}+14x+49-x^{2}-16x-64=0 \Rightarrow x^{2}-2x-15=0\).
\((x-5)(x+3)=0 \Rightarrow x=5\) (reject \(x=-3\), a length must be positive).
(a) \(x=5\ \text{m}\).
(b) Area \(=x(x+7)=5\times 12=\mathbf{60\ m^{2}}\).
Tambaya 59 Rahoto
(a) The roots of the equation \(2x^{2} + (p + 1)x + 9 = 0\), are 1 and 3, where p and q are constants. Find the values of p and q.
(b) The weight of an object varies inversely as the square of its distance from the centre of the earth. A small satellite weighs 80kg on the earth's surface. Calculate, correct to the nearest whole number, the weight of the satellite when it is 800km above the surface of the earth. [Take the radius of the earth as 6,400km].
(a) The equation is \(2x^{2}+(p+1)x+q=0\) with roots \(1\) and \(3\) (the constant term is the unknown \(q\), not a fixed number).
Sum of roots \(=1+3=4=-\dfrac{p+1}{2}\Rightarrow p+1=-8\Rightarrow p=-9\).
Product of roots \(=1\times 3=3=\dfrac{q}{2}\Rightarrow q=6\).
\(p=-9,\ q=6\).
(b) Weight \(W\propto\dfrac{1}{d^{2}}\), so \(W=\dfrac{k}{d^{2}}\). On the surface, \(d=6400\text{ km}\), \(W=80\text{ kg}\):
\(k=80\times 6400^{2}\).
At \(800\text{ km}\) above the surface, \(d=6400+800=7200\text{ km}\):
\(W=80\times\left(\dfrac{6400}{7200}\right)^{2}=80\times\dfrac{64}{81}=\dfrac{5120}{81}=63.2\).
Weight \(\approx \mathbf{63\ kg}\).
Bayanin Amsa
(a) The equation is \(2x^{2}+(p+1)x+q=0\) with roots \(1\) and \(3\) (the constant term is the unknown \(q\), not a fixed number).
Sum of roots \(=1+3=4=-\dfrac{p+1}{2}\Rightarrow p+1=-8\Rightarrow p=-9\).
Product of roots \(=1\times 3=3=\dfrac{q}{2}\Rightarrow q=6\).
\(p=-9,\ q=6\).
(b) Weight \(W\propto\dfrac{1}{d^{2}}\), so \(W=\dfrac{k}{d^{2}}\). On the surface, \(d=6400\text{ km}\), \(W=80\text{ kg}\):
\(k=80\times 6400^{2}\).
At \(800\text{ km}\) above the surface, \(d=6400+800=7200\text{ km}\):
\(W=80\times\left(\dfrac{6400}{7200}\right)^{2}=80\times\dfrac{64}{81}=\dfrac{5120}{81}=63.2\).
Weight \(\approx \mathbf{63\ kg}\).
Tambaya 60 Rahoto
The diagram above shows the bar charts representing the number of vehicles manufactured by a company in January, February and March, 1992.
(a) How many vehicles were produced in February?
(b) What fraction of the vehicles manufactured in February were cars?
(c) How many buses were produced altogether from January to March, 1992?
(d) What is the ratio in the lowest term of the number of lorries produced in February to that in March?
Read the height of every bar against the vertical scale, which is marked in steps of \(5\) vehicles. Match each shading to the key: fine dots are cars, solid black is lorries, diagonal cross-hatching is buses. Reading each of the three bars in each month group gives the table below.
| Month | Cars | Lorries | Buses | Monthly total |
|---|---|---|---|---|
| January | 10 | 15 | 20 | 45 |
| February | 40 | 25 | 30 | 95 |
| March | 5 | 35 | 50 | 90 |
The same data drawn to scale:
(a) Vehicles produced in February. Add the three February bars:
\[ 40 + 25 + 30 = 95 \text{ vehicles.} \]
(b) Fraction of February vehicles that were cars. Cars in February over the February total:
\[ \frac{40}{95} = \frac{8}{19}. \]
(c) Buses produced altogether, January to March. Add the three bus readings:
\[ 20 + 30 + 50 = 100 \text{ buses.} \]
(d) Ratio of lorries in February to lorries in March. The lorry bars read \(25\) and \(35\). Divide both by their highest common factor, \(5\):
\[ 25 : 35 = 5 : 7. \]
The key point when reading a grouped bar chart is to fix the scale first (each square here is \(5\) vehicles) and to match every bar to the correct shading in the key before adding. Miscounting the scale or reading the wrong shading is what leads to a wrong total such as treating February as \(75\) instead of \(95\).
Bayanin Amsa
Read the height of every bar against the vertical scale, which is marked in steps of \(5\) vehicles. Match each shading to the key: fine dots are cars, solid black is lorries, diagonal cross-hatching is buses. Reading each of the three bars in each month group gives the table below.
| Month | Cars | Lorries | Buses | Monthly total |
|---|---|---|---|---|
| January | 10 | 15 | 20 | 45 |
| February | 40 | 25 | 30 | 95 |
| March | 5 | 35 | 50 | 90 |
The same data drawn to scale:
(a) Vehicles produced in February. Add the three February bars:
\[ 40 + 25 + 30 = 95 \text{ vehicles.} \]
(b) Fraction of February vehicles that were cars. Cars in February over the February total:
\[ \frac{40}{95} = \frac{8}{19}. \]
(c) Buses produced altogether, January to March. Add the three bus readings:
\[ 20 + 30 + 50 = 100 \text{ buses.} \]
(d) Ratio of lorries in February to lorries in March. The lorry bars read \(25\) and \(35\). Divide both by their highest common factor, \(5\):
\[ 25 : 35 = 5 : 7. \]
The key point when reading a grouped bar chart is to fix the scale first (each square here is \(5\) vehicles) and to match every bar to the correct shading in the key before adding. Miscounting the scale or reading the wrong shading is what leads to a wrong total such as treating February as \(75\) instead of \(95\).
Tambaya 61 Rahoto
In the diagram, three points A, B and C are on the same horizontal ground. B is 15m from A, on a bearing of 053°, C is 18m from B on a bearing of 161°. A vertical pole with top T is erected at B such that < ATB = 58°. Calculate, correct to three significant figures,
(a) the length of AC.
(b) the bearing of C from A ;
(c) the height of the pole BT.
From the diagram: \(AB=15\text{ m}\) on a bearing of \(053^\circ\), \(BC=18\text{ m}\) on a bearing of \(161^\circ\), the pole BT is vertical at B, and \(\angle ATB=58^\circ\).
(a) Length of AC. First find \(\angle ABC\). The bearing of A from B is the back-bearing of \(053^\circ\):
\[053^\circ+180^\circ=233^\circ\]The bearing of C from B is \(161^\circ\), so
\[\angle ABC=233^\circ-161^\circ=72^\circ\]Apply the cosine rule to triangle ABC:
\[AC^2=AB^2+BC^2-2\,(AB)(BC)\cos(\angle ABC)\]\[AC^2=15^2+18^2-2(15)(18)\cos72^\circ\]\[AC^2=225+324-540(0.30902)=382.13\]\[AC=19.5\text{ m (3 s.f.)}\](b) Bearing of C from A. Use the sine rule to find \(\angle BAC\):
\[\frac{\sin\angle BAC}{BC}=\frac{\sin\angle ABC}{AC}\]\[\sin\angle BAC=\frac{18\sin72^\circ}{19.548}=\frac{17.119}{19.548}=0.87575\]\[\angle BAC=61.1^\circ\]C lies clockwise of B as seen from A, so the bearing of C from A is
\[053^\circ+61.1^\circ=114^\circ\ (\text{3 s.f.})\](c) Height of the pole BT. The pole is vertical and BA is horizontal, so triangle ATB is right-angled at B. With \(\angle ATB=58^\circ\) and \(AB=15\text{ m}\) opposite that angle:
\[\tan(\angle ATB)=\frac{AB}{BT}\Rightarrow BT=\frac{AB}{\tan58^\circ}=\frac{15}{1.6003}\]\[BT=9.37\text{ m (3 s.f.)}\]Bayanin Amsa
From the diagram: \(AB=15\text{ m}\) on a bearing of \(053^\circ\), \(BC=18\text{ m}\) on a bearing of \(161^\circ\), the pole BT is vertical at B, and \(\angle ATB=58^\circ\).
(a) Length of AC. First find \(\angle ABC\). The bearing of A from B is the back-bearing of \(053^\circ\):
\[053^\circ+180^\circ=233^\circ\]The bearing of C from B is \(161^\circ\), so
\[\angle ABC=233^\circ-161^\circ=72^\circ\]Apply the cosine rule to triangle ABC:
\[AC^2=AB^2+BC^2-2\,(AB)(BC)\cos(\angle ABC)\]\[AC^2=15^2+18^2-2(15)(18)\cos72^\circ\]\[AC^2=225+324-540(0.30902)=382.13\]\[AC=19.5\text{ m (3 s.f.)}\](b) Bearing of C from A. Use the sine rule to find \(\angle BAC\):
\[\frac{\sin\angle BAC}{BC}=\frac{\sin\angle ABC}{AC}\]\[\sin\angle BAC=\frac{18\sin72^\circ}{19.548}=\frac{17.119}{19.548}=0.87575\]\[\angle BAC=61.1^\circ\]C lies clockwise of B as seen from A, so the bearing of C from A is
\[053^\circ+61.1^\circ=114^\circ\ (\text{3 s.f.})\](c) Height of the pole BT. The pole is vertical and BA is horizontal, so triangle ATB is right-angled at B. With \(\angle ATB=58^\circ\) and \(AB=15\text{ m}\) opposite that angle:
\[\tan(\angle ATB)=\frac{AB}{BT}\Rightarrow BT=\frac{AB}{\tan58^\circ}=\frac{15}{1.6003}\]\[BT=9.37\text{ m (3 s.f.)}\]
