WAEC SSCE - General Mathematics - 2008
Tambaya 2 Rahoto
In the figure shown, PQs is a straight line. What is the value of < PRQ?
Tambaya 3 Rahoto
If p-2g + 1 = g + 3p and p - 2 = 0, find g
Bayanin Amsa
Substitute p = 2 in the equation p - 2g + 1 = g + 3p to get: 2 - 2g + 1 = g + 3(2) -2g = 3 - 2 -2g = 1 g = -1/2 Therefore, the answer is -1.
Tambaya 5 Rahoto
P naira invested for 4 years invested for 4 years at r% simple interest per annum yields 0.36 p naira interest. Find the value of r
Bayanin Amsa
We can use the formula for simple interest, which is: I = P * r * t Where I is the interest earned, P is the principal (initial amount invested), r is the annual interest rate, and t is the time in years. In this problem, we know that the interest earned is 0.36P, the principal is P, and the time is 4 years. Substituting these values into the formula, we get: 0.36P = P * r * 4 Simplifying the equation, we can divide both sides by P * 4: 0.36 = r * 4 r = 0.36 / 4 r = 0.09 or 9% Therefore, the answer is option (C) 9.
Tambaya 6 Rahoto
If xo is obtuse, which of the following is true?
Tambaya 7 Rahoto
Solve the inequality 1 - 2x < - \(\frac{1}{3}\)
Bayanin Amsa
The given inequality is: 1 - 2x < -\(\frac{1}{3}\) To solve this inequality, we need to isolate the variable, which in this case is x. First, we can simplify the inequality by adding \(\frac{1}{3}\) to both sides: 1 - 2x + \(\frac{1}{3}\) < 0 Next, we can combine like terms: \(\frac{4}{3}\) - 2x < 0 Now, we can isolate x by subtracting \(\frac{4}{3}\) from both sides: -2x < -\(\frac{4}{3}\) Finally, we can solve for x by dividing both sides by -2, remembering to flip the inequality because we are dividing by a negative number: x > \(\frac{2}{3}\) Therefore, the solution to the inequality is x > \(\frac{2}{3}\).
Tambaya 8 Rahoto
If 4y is 9 greater than the sum of y and 3x, by how much is greater than x?
Bayanin Amsa
Let's start by translating the sentence into an equation: 4y = 9 + y + 3x We can simplify this equation by subtracting y from both sides: 3y = 9 + 3x Now, let's isolate x by subtracting 9 from both sides and then dividing by 3: x = (3y - 9)/3 = y - 3 So we see that x is 3 less than y. Therefore, the answer is (A) 3.
Tambaya 10 Rahoto
In the diagram, < QPR = 60o
< PQR = 50o
< QRS = 2xo
< SRP = 3xo
< UQP = yo and RS//TU
calculate y
Bayanin Amsa
Tambaya 11 Rahoto
Given the sets A = [2, 4, 6, 8] and B = [2, 3, 5, 9]. If a number is picked at random from each of the two sets, what is the probability that their difference is 6 or 7?
Bayanin Amsa
Tambaya 12 Rahoto
In the diagram, /PQ/ = /QR/ and /PR/ = /RS/ = /SP/, calculate the side of < QRS
Bayanin Amsa
Tambaya 13 Rahoto
The angles of a quadrilateral are (x + 10)o, 2yo, 90o and (100 - y)o, Find y in terms of x
Bayanin Amsa
The sum of the angles of a quadrilateral is 360o. Therefore, (x + 10)o + 2yo + 90o + (100 - y)o = 360o. Simplifying the equation, we have: x + 10 + 2y + 90 + 100 - y = 360 x + 200 + y = 360 y = 160 - x Therefore, y is equal to 160 subtracted by x. Hence, the correct option is y = 160 - x.
Tambaya 14 Rahoto
PQR is a sector of a circle centre O, radius 4cm. If PQR = 30o, find, correct to 3 significant figures, the area of sector PQR. [Take \(\pi = \frac{22}{7}\)]
Bayanin Amsa
To find the area of the sector PQR, we need to know the angle at the center of the circle that the sector subtends. Since the radius of the circle is 4 cm and PQR = 30o, the circumference of the circle is 2πr = 8π cm. The fraction of the circle that the sector PQR subtends is 30/360, or 1/12. Therefore, the area of the sector PQR is 1/12 of the total area of the circle: Area of sector PQR = (1/12) × πr2 = (1/12) × π × (4 cm)2 ≈ 3.347 cm2 Rounded to 3 significant figures, the area of sector PQR is approximately 3.35 cm2, which corresponds to the first option: 4.19 cm2.
Tambaya 15 Rahoto
If x \(\alpha\) (45 + \(\frac{1}{2}y\)), which of the following is true>?
Bayanin Amsa
Tambaya 16 Rahoto
If the perimeter of \(\bigtriangleup\)PQR in thr diagram is 24cm, what is the area of \(\bigtriangleup\)PRS?
Tambaya 17 Rahoto
A train travels 60km in M minutes. If its average speed is 400km per hour, find the value of M
Bayanin Amsa
The average speed of the train is given as 400km/hr. This means that in one hour, the train travels 400km. We can convert this to km/minute by dividing by 60 (the number of minutes in an hour). Therefore, the speed of the train in km/minute is: 400km/hr ÷ 60 = 6.67km/min (rounded to two decimal places) We are given that the train travels 60km in M minutes. We can use the formula: speed = distance/time to find the value of M. Rearranging the formula gives: time = distance/speed Substituting the given values, we get: M = 60km ÷ 6.67km/min M = 9 minutes (rounded to the nearest minute) Therefore, the value of M is 9.
Tambaya 18 Rahoto
If the volume of a cube is 343cm3, find the length of its side
Bayanin Amsa
To find the length of a side of a cube, we need to take the cube root of its volume. In this case, the volume is given as 343cm3. Taking the cube root of 343, we get: ∛343 = 7 Therefore, the length of the side of the cube is 7cm. So the correct answer is 7cm.
Tambaya 19 Rahoto
Given the sets A = [2, 4, 6, 8] and B = [2, 3, 5, 9]. If a number is picked at random from each of the two sets, what is the probability that their difference is odd?
Bayanin Amsa
Tambaya 20 Rahoto
Simplify \(\frac{\frac{1}{x} + \frac{1}{y}}{x + y}\)
Bayanin Amsa
To simplify the expression, we can start by getting a common denominator for the two fractions in the numerator. The common denominator is xy. So, we can write: \begin{align*} \frac{\frac{1}{x} + \frac{1}{y}}{x + y} &= \frac{\frac{y}{xy} + \frac{x}{xy}}{x + y} \\ &= \frac{\frac{x + y}{xy}}{x + y} \\ &= \frac{1}{xy} \end{align*} Therefore, the simplified expression is \(\frac{1}{xy}\).
Tambaya 21 Rahoto
In the diagram, find the size of the angle marked a°
Tambaya 22 Rahoto
XY is a chord of circle centre O and radius 7cm. The chord XY which is 8cm long subtends an angle of 120o at the centre of the circle. Calculate the perimeter of the minor segment. [Take \(\pi = \frac{22}{7}\)]
Bayanin Amsa
Tambaya 23 Rahoto
Every staff in an office owns either a Mercedes and/or a Toyota car. 20 own Mercedes, 15 own Toyota and 5 own both. How many staff are there in the office?
Bayanin Amsa
To find the total number of staff in the office, we need to add the number of staff who own a Mercedes and the number of staff who own a Toyota, but we must subtract the number of staff who own both cars to avoid counting them twice. Using the given information, we know that there are 20 staff who own a Mercedes, 15 staff who own a Toyota, and 5 staff who own both. Therefore, the number of staff who own either a Mercedes or a Toyota is: 20 + 15 - 5 = 30 Therefore, there are 30 staff in the office who own either a Mercedes or a Toyota.
Tambaya 24 Rahoto
If cos (x + 25)o = sin 45o, find the value of x
Bayanin Amsa
We know that cos (90o - A) = sin A. So, cos (x + 25o) = cos (90o - 45o) = cos 45o. Since cos A = cos B implies A = ±B + 360on, we have: x + 25o = ±45o + 360on x = -25o ± 45o + 360on x = 20o + 360on or x = 70o + 360on Since we are given the range of x, we take x = 20o. Therefore, the value of x is 20o.
Tambaya 26 Rahoto
In the diagram, O is the centre of the circle and PQRS is a cyclic quadrilateral. Find the value of x.
Bayanin Amsa
Tambaya 27 Rahoto
Given that x = 2 and y = -\(\frac{1}{4}\), evaluate \(\frac{x^2y - 2xy}{5}\)
Bayanin Amsa
Tambaya 28 Rahoto
If tan x = 1, evaluate sin x + cos x, leaving your answer in the surd form
Bayanin Amsa
If \(\tan x = 1\), then we can find the values of sin x and cos x using the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\). Since \(\tan x = \frac{\sin x}{\cos x} = 1\), we can say that \(\sin x = \cos x\). So we have, \[\sin^2 x + \cos^2 x = 2\cos^2 x = 1 \implies \cos x = \pm\frac{1}{\sqrt{2}}\] Using \(\sin x = \cos x\), we can say that \(\sin x = \pm\frac{1}{\sqrt{2}}\) Now, \[\sin x + \cos x = \pm\frac{1}{\sqrt{2}} \pm \frac{1}{\sqrt{2}} = \pm\sqrt{2}\] Since \(\tan x = 1\), x must be in the first quadrant, where both \(\sin x\) and \(\cos x\) are positive. Therefore, we have \[\sin x + \cos x = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}\] Hence, the answer is \(\sqrt{2}\). Therefore, the option that represents this answer is: $\sqrt{2}$.
Tambaya 29 Rahoto
Factorize 5y2 + 2ay - 3a2
Bayanin Amsa
To factorize 5y^2 + 2ay - 3a^2, we need to find two numbers that multiply to give -15a^2 and add up to 2a. These numbers are 5a and -3a. We can then rewrite the expression as: 5y^2 + 5ay - 3ay - 3a^2 Factorising by grouping: (5y^2 + 5ay) - (3ay + 3a^2) Factorising out the common factor in each group: 5y(y + a) - 3a(y + a) Factorising out the common factor (y + a): (5y - 3a)(y + a) Therefore, the answer is (5y - 3a)(y + a).
Tambaya 31 Rahoto
Simplify: \(\frac{2x^2 - 5x - 12}{4x^2 - 9}\)
Bayanin Amsa
To simplify the given expression, we need to factor the numerator and the denominator and then simplify the fraction by canceling out any common factors. The numerator can be factored as follows: \begin{align*} 2x^2 - 5x - 12 &= 2x^2 - 8x + 3x - 12 \\ &= 2x(x - 4) + 3(x - 4) \\ &= (2x + 3)(x - 4) \end{align*} The denominator can be factored using the difference of squares formula as: \begin{align*} 4x^2 - 9 &= (2x)^2 - 3^2 \\ &= (2x - 3)(2x + 3) \end{align*} Now we can simplify the fraction as follows: \begin{align*} \frac{2x^2 - 5x - 12}{4x^2 - 9} &= \frac{(2x + 3)(x - 4)}{(2x - 3)(2x + 3)} \\ &= \frac{x - 4}{2x - 3} \end{align*} Therefore, the simplified form of the expression is \(\frac{x - 4}{2x - 3}\). So, option (B) is the correct answer.
Tambaya 32 Rahoto
In the diagram, PX is a tangent to the circle and RST is an equilateral triangle. Calculate < PTS
Bayanin Amsa
Tambaya 33 Rahoto
What is the length of an edge of a cube whose total surface area is X cm2 and whose total surface area is \(\frac{X}{2}\)cm3?
Bayanin Amsa
Tambaya 34 Rahoto
If p = \(\frac{3}{5} \sqrt{\frac{q}{r}}\), express q in terms of p and r
Bayanin Amsa
Given that p = \(\frac{3}{5} \sqrt{\frac{q}{r}}\), we need to express q in terms of p and r. Let's first isolate the term containing q on one side of the equation: \begin{align*} p &= \frac{3}{5} \sqrt{\frac{q}{r}} \\ \frac{5}{3}p &= \sqrt{\frac{q}{r}} \\ \left(\frac{5}{3}p\right)^2 &= \frac{q}{r} \\ \frac{25}{9} p^2r &= q \\ \end{align*} Therefore, q is equal to \(\frac{25}{9} p^2r\). So the correct option is: \(\frac{25}{9} p^2r\)
Tambaya 35 Rahoto
An arc of a circle, radius 14cm, is 18.33cm long. Calculate to the nearest degree, the angle which the arc subtends at the centre of the circle. [T = \(\frac{22}{7}\)]
Bayanin Amsa
The formula for the length of an arc is L = rθ where L is the length of the arc, r is the radius of the circle, and θ is the angle subtended by the arc at the centre of the circle in radians. To find the angle in degrees, we need to convert radians to degrees by multiplying by 180/π. In this problem, the radius of the circle is 14cm and the length of the arc is 18.33cm. Plugging these values into the formula, we get: 18.33 = 14θ Solving for θ, we get: θ = 18.33/14 θ = 1.3093 radians To convert to degrees, we multiply by 180/π: θ = 1.3093 × 180/π θ ≈ 75 degrees Therefore, the angle which the arc subtends at the centre of the circle is approximately 75 degrees. Answer: 75o
Tambaya 36 Rahoto
A trader bought 100 tubers at 5 for N350.00. She sold them in sets of 4 for N290.00. Find her gain percent.
Bayanin Amsa
Tambaya 38 Rahoto
Simplify \(\frac{\log \sqrt{8}}{\log 4 - \log 2}\)
Bayanin Amsa
We can start by simplifying the numerator first: \begin{align*} \log \sqrt{8} &= \log \left(\sqrt{4}\cdot\sqrt{2}\right) \\ &= \log 4 + \log \sqrt{2} \\ &= 2 + \frac{1}{2} \log 2 \end{align*} Similarly, we can simplify the denominator: \begin{align*} \log 4 - \log 2 &= \log\frac{4}{2} \\ &= \log 2 \end{align*} Now we can substitute these simplified expressions back into the original expression: \begin{align*} \frac{\log \sqrt{8}}{\log 4 - \log 2} &= \frac{2 + \frac{1}{2} \log 2}{\log 2} \\ &= \frac{2}{\log 2} + \frac{1}{2} \\ &= \boxed{\frac{3}{2}} \end{align*} Therefore, the correct option is (c) \(\frac{3}{2}\).
Tambaya 39 Rahoto
Given the sets A = [2, 4, 6, 8] and B = [2, 3, 5, 9]. If a number is selected at random from set B, what is the probability that the number is prime?
Tambaya 40 Rahoto
If p = \(\frac{1}{2}\) and \(\frac{1}{p - 1} = \frac{2}{p + x}\), find the value of x
Bayanin Amsa
Given that p = \(\frac{1}{2}\) and \(\frac{1}{p - 1} = \frac{2}{p + x}\), we want to find the value of x. We can start by substituting the value of p in the second equation: \begin{aligned} \frac{1}{p - 1} &= \frac{2}{p + x} \\ \\ \frac{1}{\frac{1}{2} - 1} &= \frac{2}{\frac{1}{2} + x} && \text{(Substitute } p = \frac{1}{2} \text{)}\\ \\ \frac{1}{-\frac{1}{2}} &= \frac{2}{\frac{1}{2} + x} && \text{(Simplify)}\\ \\ -2 &= \frac{2}{\frac{1}{2} + x} && \text{(Simplify)}\\ \\ -2\left(\frac{1}{2} + x\right) &= 2 && \text{(Cross-multiply)}\\ \\ -1 - 2x &= 1 && \text{(Simplify)}\\ \\ -2x &= 2 \\ \\ x &= -1 \end{aligned} Therefore, the value of x is -1.5, which corresponds to option B: -1\(\frac{1}{2}\).
Tambaya 42 Rahoto
Find the mean deviation of these numbers 10, 12, 14, 15, 17, 19.
Bayanin Amsa
To find the mean deviation of a set of numbers, we first find the mean (average) of the numbers. The mean of the numbers 10, 12, 14, 15, 17, and 19 is: (mean) = (10 + 12 + 14 + 15 + 17 + 19) / 6 = 87 / 6 = 14.5 Next, we find the deviation of each number from the mean by subtracting the mean from each number: 10 - 14.5 = -4.5 12 - 14.5 = -2.5 14 - 14.5 = -0.5 15 - 14.5 = 0.5 17 - 14.5 = 2.5 19 - 14.5 = 4.5 We take the absolute value of each deviation to ensure that they are all positive: |-4.5| = 4.5 |-2.5| = 2.5 |-0.5| = 0.5 |0.5| = 0.5 |2.5| = 2.5 |4.5| = 4.5 Then we find the mean of the absolute deviations: (mean deviation) = (4.5 + 2.5 + 0.5 + 0.5 + 2.5 + 4.5) / 6 = 2.5 Therefore, the mean deviation of the numbers is 2.5. Therefore, the correct option is (a) 2.5.
Tambaya 43 Rahoto
If x% of 240 equals 12, find x
Bayanin Amsa
We can start solving the problem by setting up an equation based on the information given. The problem states that "x% of 240 equals 12". In other words, we can translate this into an equation: x/100 * 240 = 12 To solve for x, we can isolate it by dividing both sides of the equation by 240/100, which simplifies to 2.4: x/100 * 240 = 12 x = 12 / 2.4 x = 5 Therefore, the value of x that satisfies the given equation is 5.
Tambaya 44 Rahoto
Find the quadratic equation whose roots are c and -c
Bayanin Amsa
If the roots of the quadratic equation are c and -c, then the equation can be expressed in the factored form as: (x - c)(x + c) = 0 Expanding the above expression: x^2 - c^2 = 0 Therefore, the quadratic equation whose roots are c and -c is x^2 - c^2 = 0. Hence, the correct option is: x2 - c2 = 0.
Tambaya 46 Rahoto
The table below shows how a man spends his income in a month.
| Items | Amount Spent |
| Food | N4500 |
| House Rent | N3000 |
| Provisions | N2500 |
| Electricity | N2000 |
| Transportation | N5000 |
| Others | N3000 |
(a) Represent the information on a pie chart.
(b) What percentage of his income is spent on transportation?
Total monthly income
\[4500+3000+2500+2000+5000+3000=N20\,000\]
(a) Pie chart
The angle for each sector is:
\[\text{Sector angle}=\frac{\text{amount spent}}{20\,000}\times360^\circ\]
| Item | Amount (₦) | Sector angle |
|---|---|---|
| Food | 4,500 | \(\frac{4500}{20000}\times360^\circ=81^\circ\) |
| House Rent | 3,000 | \(54^\circ\) |
| Provisions | 2,500 | \(45^\circ\) |
| Electricity | 2,000 | \(36^\circ\) |
| Transportation | 5,000 | \(90^\circ\) |
| Others | 3,000 | \(54^\circ\) |
| Total | 20,000 | \(360^\circ\) |
The required pie chart is:
(b)
\[\text{Percentage spent on transportation}=\frac{5000}{20000}\times100\%=25\%\]
Therefore, 25% of his income is spent on transportation.
Bayanin Amsa
Total monthly income
\[4500+3000+2500+2000+5000+3000=N20\,000\]
(a) Pie chart
The angle for each sector is:
\[\text{Sector angle}=\frac{\text{amount spent}}{20\,000}\times360^\circ\]
| Item | Amount (₦) | Sector angle |
|---|---|---|
| Food | 4,500 | \(\frac{4500}{20000}\times360^\circ=81^\circ\) |
| House Rent | 3,000 | \(54^\circ\) |
| Provisions | 2,500 | \(45^\circ\) |
| Electricity | 2,000 | \(36^\circ\) |
| Transportation | 5,000 | \(90^\circ\) |
| Others | 3,000 | \(54^\circ\) |
| Total | 20,000 | \(360^\circ\) |
The required pie chart is:
(b)
\[\text{Percentage spent on transportation}=\frac{5000}{20000}\times100\%=25\%\]
Therefore, 25% of his income is spent on transportation.
Tambaya 47 Rahoto
(a) Solve, correct to two decimal places, the equation \(4x^{2} = 11x + 21\).
(b) A man invests £1500 for two years at compound interest. After one year, his money amounts to £1560. Find the :
(i) rate of interest ; (ii) interest for the second year.
(c) A car costs N300,000.00. It depreciates by 25% in the first year and 20% in the second year. Find its value after 2 years.
(a) \(4x^{2}=11x+21\Rightarrow 4x^{2}-11x-21=0\).
Using the formula with \(a=4,\ b=-11,\ c=-21\):
\[x=\frac{11\pm\sqrt{(-11)^{2}-4(4)(-21)}}{2(4)}=\frac{11\pm\sqrt{121+336}}{8}=\frac{11\pm\sqrt{457}}{8}\]\(\sqrt{457}=21.378\). So \(x=\dfrac{11+21.378}{8}=4.047\) or \(x=\dfrac{11-21.378}{8}=-1.297\).
\(x\approx 4.05\) or \(x\approx -1.30\)
(b) Principal \(=\pounds1500\); after one year the amount is \(\pounds1560\).
(i) First-year interest \(=1560-1500=\pounds60\).
\[\text{Rate}=\frac{60}{1500}\times100\%=4\%\](ii) Interest for the second year is charged on \(\pounds1560\):
\[\frac{4}{100}\times1560=\pounds62.40\](c) Car costs \(N300{,}000\), depreciating \(25\%\) then \(20\%\).
After year 1: \(300000\times(1-0.25)=300000\times0.75=N225{,}000\).
After year 2: \(225000\times(1-0.20)=225000\times0.80=N180{,}000\).
Value after 2 years \(=N180{,}000.00\)
Bayanin Amsa
(a) \(4x^{2}=11x+21\Rightarrow 4x^{2}-11x-21=0\).
Using the formula with \(a=4,\ b=-11,\ c=-21\):
\[x=\frac{11\pm\sqrt{(-11)^{2}-4(4)(-21)}}{2(4)}=\frac{11\pm\sqrt{121+336}}{8}=\frac{11\pm\sqrt{457}}{8}\]\(\sqrt{457}=21.378\). So \(x=\dfrac{11+21.378}{8}=4.047\) or \(x=\dfrac{11-21.378}{8}=-1.297\).
\(x\approx 4.05\) or \(x\approx -1.30\)
(b) Principal \(=\pounds1500\); after one year the amount is \(\pounds1560\).
(i) First-year interest \(=1560-1500=\pounds60\).
\[\text{Rate}=\frac{60}{1500}\times100\%=4\%\](ii) Interest for the second year is charged on \(\pounds1560\):
\[\frac{4}{100}\times1560=\pounds62.40\](c) Car costs \(N300{,}000\), depreciating \(25\%\) then \(20\%\).
After year 1: \(300000\times(1-0.25)=300000\times0.75=N225{,}000\).
After year 2: \(225000\times(1-0.20)=225000\times0.80=N180{,}000\).
Value after 2 years \(=N180{,}000.00\)
Tambaya 48 Rahoto
(a) A rectangular field is l metres long and b metres wide. Its perimeter is 280 metres. If the length is two and a half times the breadth, find the values of l and b.
(b) The base of a pyramid is a 4.5 metres rectangle. The height of the pyramid is 4 metres. Calculate its volume.
Bayanin Amsa
None
Tambaya 49 Rahoto
(a) If \(2^{x + y} = 16\) and \(4^{x - y} = \frac{1}{32}\), find the value of x and y.
(b) P, Q and R are related in such a way that \(P \propto \frac{Q^{2}}{R}\). When P = 36, Q = 3 and R = 4. Calculate Q when P = 200 and R = 2.
(a) Express both equations to base 2.
\(2^{x+y}=16=2^{4}\Rightarrow x+y=4\)
Since \(4=2^{2}\) and \(\frac{1}{32}=2^{-5}\): \(4^{x-y}=2^{2(x-y)}=2^{-5}\Rightarrow 2(x-y)=-5\Rightarrow x-y=-2.5\)
Solve the pair:
\[x+y=4,\qquad x-y=-2.5\]Adding: \(2x=1.5\Rightarrow x=0.75\). Then \(y=4-0.75=3.25\).
\(x=0.75,\ y=3.25\)
(b) \(P\propto\dfrac{Q^{2}}{R}\Rightarrow P=\dfrac{kQ^{2}}{R}\).
Using \(P=36,\ Q=3,\ R=4\): \(36=\dfrac{k(3)^{2}}{4}=\dfrac{9k}{4}\Rightarrow k=16\).
When \(P=200,\ R=2\): \(200=\dfrac{16Q^{2}}{2}=8Q^{2}\Rightarrow Q^{2}=25\Rightarrow Q=5\).
\(Q=5\)
Bayanin Amsa
(a) Express both equations to base 2.
\(2^{x+y}=16=2^{4}\Rightarrow x+y=4\)
Since \(4=2^{2}\) and \(\frac{1}{32}=2^{-5}\): \(4^{x-y}=2^{2(x-y)}=2^{-5}\Rightarrow 2(x-y)=-5\Rightarrow x-y=-2.5\)
Solve the pair:
\[x+y=4,\qquad x-y=-2.5\]Adding: \(2x=1.5\Rightarrow x=0.75\). Then \(y=4-0.75=3.25\).
\(x=0.75,\ y=3.25\)
(b) \(P\propto\dfrac{Q^{2}}{R}\Rightarrow P=\dfrac{kQ^{2}}{R}\).
Using \(P=36,\ Q=3,\ R=4\): \(36=\dfrac{k(3)^{2}}{4}=\dfrac{9k}{4}\Rightarrow k=16\).
When \(P=200,\ R=2\): \(200=\dfrac{16Q^{2}}{2}=8Q^{2}\Rightarrow Q^{2}=25\Rightarrow Q=5\).
\(Q=5\)
Tambaya 50 Rahoto
(a) A pentagon is such that one of its exterior sides is 60°. Two others are (90 - m)° each while the remaining angles are (30 + 2m)° each. Find the value of m.
(b) 
In the diagram, PQR is a straight line, \(\overline{QR} = \sqrt{3} cm\) and \(\overline{SQ} = 2 cm\). Calculate, correct to one decimal place, < PQS.
(a) The exterior angles of any polygon add up to \(360^\circ\). The pentagon has five exterior angles: one is \(60^\circ\), two are \((90-m)^\circ\) each, and the remaining two are \((30+2m)^\circ\) each. Therefore:
\[60+2(90-m)+2(30+2m)=360\]\[60+180-2m+60+4m=360\]\[300+2m=360\]\[2m=60\Rightarrow m=30\](b) From the diagram PQR is a straight line, \(\overline{QR}=\sqrt{3}\text{ cm}\), \(\overline{SQ}=2\text{ cm}\), and the right-angle mark at R shows \(SR\perp QR\). So triangle SQR is right-angled at R, with SQ the hypotenuse.
In right-angled triangle SQR, using \(\angle SQR\):
\[\cos(\angle SQR)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{\overline{QR}}{\overline{SQ}}=\frac{\sqrt{3}}{2}\]\[\angle SQR=\cos^{-1}\!\left(\tfrac{\sqrt{3}}{2}\right)=30^\circ\]Since PQR is a straight line, \(\angle PQS\) and \(\angle SQR\) are angles on a straight line at Q:
\[\angle PQS=180^\circ-\angle SQR=180^\circ-30^\circ=150.0^\circ\]Bayanin Amsa
(a) The exterior angles of any polygon add up to \(360^\circ\). The pentagon has five exterior angles: one is \(60^\circ\), two are \((90-m)^\circ\) each, and the remaining two are \((30+2m)^\circ\) each. Therefore:
\[60+2(90-m)+2(30+2m)=360\]\[60+180-2m+60+4m=360\]\[300+2m=360\]\[2m=60\Rightarrow m=30\](b) From the diagram PQR is a straight line, \(\overline{QR}=\sqrt{3}\text{ cm}\), \(\overline{SQ}=2\text{ cm}\), and the right-angle mark at R shows \(SR\perp QR\). So triangle SQR is right-angled at R, with SQ the hypotenuse.
In right-angled triangle SQR, using \(\angle SQR\):
\[\cos(\angle SQR)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{\overline{QR}}{\overline{SQ}}=\frac{\sqrt{3}}{2}\]\[\angle SQR=\cos^{-1}\!\left(\tfrac{\sqrt{3}}{2}\right)=30^\circ\]Since PQR is a straight line, \(\angle PQS\) and \(\angle SQR\) are angles on a straight line at Q:
\[\angle PQS=180^\circ-\angle SQR=180^\circ-30^\circ=150.0^\circ\]Tambaya 51 Rahoto
(a) The triangle ABC has sides AB = 17m, BC = 12m and AC = 10m. Calculate the :
(i) largest angle of the triangle ; (ii) area of the triangle.
(b) From a point T on a horizontal ground, the angle of elevation of the top R of a tower RS, 38m high is 63°. Calculate, correct to the nearest metre, the distance between T and S.
(a)(i) Largest angle. The largest angle faces the longest side \(AB=17\) m, so it is angle \(C\), between sides \(CA=10\) and \(CB=12\). By the cosine rule:
\[\cos C=\frac{CA^{2}+CB^{2}-AB^{2}}{2\,(CA)(CB)}=\frac{10^{2}+12^{2}-17^{2}}{2(10)(12)}=\frac{100+144-289}{240}=\frac{-45}{240}=-0.1875\]\(C=\cos^{-1}(-0.1875)\approx 100.8^{\circ}\).
Largest angle \(\approx 100.8^{\circ}\).
(ii) Area.
\[\text{Area}=\frac{1}{2}(CA)(CB)\sin C=\frac{1}{2}(10)(12)\sin 100.8^{\circ}=60\times0.9823\approx 58.9\text{ m}^{2}\](b) The tower \(RS\) is vertical with \(RS=38\) m; from \(T\) on the ground the elevation of the top \(R\) is \(63^{\circ}\). \(TS\) is the horizontal distance sought.
\[\tan 63^{\circ}=\frac{RS}{TS}=\frac{38}{TS}\Rightarrow TS=\frac{38}{\tan 63^{\circ}}=\frac{38}{1.9626}\approx 19.4\text{ m}\]Distance \(TS\approx 19\) m (nearest metre).
Bayanin Amsa
(a)(i) Largest angle. The largest angle faces the longest side \(AB=17\) m, so it is angle \(C\), between sides \(CA=10\) and \(CB=12\). By the cosine rule:
\[\cos C=\frac{CA^{2}+CB^{2}-AB^{2}}{2\,(CA)(CB)}=\frac{10^{2}+12^{2}-17^{2}}{2(10)(12)}=\frac{100+144-289}{240}=\frac{-45}{240}=-0.1875\]\(C=\cos^{-1}(-0.1875)\approx 100.8^{\circ}\).
Largest angle \(\approx 100.8^{\circ}\).
(ii) Area.
\[\text{Area}=\frac{1}{2}(CA)(CB)\sin C=\frac{1}{2}(10)(12)\sin 100.8^{\circ}=60\times0.9823\approx 58.9\text{ m}^{2}\](b) The tower \(RS\) is vertical with \(RS=38\) m; from \(T\) on the ground the elevation of the top \(R\) is \(63^{\circ}\). \(TS\) is the horizontal distance sought.
\[\tan 63^{\circ}=\frac{RS}{TS}=\frac{38}{TS}\Rightarrow TS=\frac{38}{\tan 63^{\circ}}=\frac{38}{1.9626}\approx 19.4\text{ m}\]Distance \(TS\approx 19\) m (nearest metre).
Tambaya 52 Rahoto
(a) Without using calculator or tables, find the value of \(\log 3.6\) given that \(\log 2 = 0.3010, \log 3 = 0.4771\) and \(\log 5 = 0.6990\).
(b) If all numbers in the equation \(\frac{y}{y + 101} = \frac{11}{10010}\) are in base two, solve for y.
(a) Write \(3.6 = \dfrac{36}{10} = \dfrac{2^2 \times 3^2}{10}\):
\[\log 3.6 = 2\log 2 + 2\log 3 - \log 10\]
\[= 2(0.3010) + 2(0.4771) - 1 = 0.6020 + 0.9542 - 1 = 0.5562\]
(b) All numbers are in base two. Convert to base ten: \(101_2 = 5\), \(11_2 = 3\), \(10010_2 = 18\). The equation \(\dfrac{y}{y + 101} = \dfrac{11}{10010}\) becomes
\[\frac{y}{y + 5} = \frac{3}{18} = \frac16\]
\[6y = y + 5 \;\Rightarrow\; 5y = 5 \;\Rightarrow\; y = 1_{10}\]
Converting back to base two, \(y = 1_2\).
Bayanin Amsa
(a) Write \(3.6 = \dfrac{36}{10} = \dfrac{2^2 \times 3^2}{10}\):
\[\log 3.6 = 2\log 2 + 2\log 3 - \log 10\]
\[= 2(0.3010) + 2(0.4771) - 1 = 0.6020 + 0.9542 - 1 = 0.5562\]
(b) All numbers are in base two. Convert to base ten: \(101_2 = 5\), \(11_2 = 3\), \(10010_2 = 18\). The equation \(\dfrac{y}{y + 101} = \dfrac{11}{10010}\) becomes
\[\frac{y}{y + 5} = \frac{3}{18} = \frac16\]
\[6y = y + 5 \;\Rightarrow\; 5y = 5 \;\Rightarrow\; y = 1_{10}\]
Converting back to base two, \(y = 1_2\).
Tambaya 53 Rahoto
(a) Solve the inequality : \(\frac{2}{5}(x - 2) - \frac{1}{6}(x + 5) \leq 0\).
(b) Given that P = \(\frac{x^{2} - y^{2}}{x^{2} + xy}\),
(i) express P in its simplest form ; (ii) find the value of P if x = -4 and y = -6.
(a) \(\dfrac25(x - 2) - \dfrac16(x + 5) \le 0\). Multiply through by the LCM \(30\):
\[12(x - 2) - 5(x + 5) \le 0\]
\[12x - 24 - 5x - 25 \le 0 \;\Rightarrow\; 7x - 49 \le 0 \;\Rightarrow\; 7x \le 49 \;\Rightarrow\; x \le 7\]
(b)(i) Factorise numerator and denominator of \(P = \dfrac{x^2 - y^2}{x^2 + xy}\):
\[P = \frac{(x - y)(x + y)}{x(x + y)} = \frac{x - y}{x}\]
(ii) With \(x = -4,\, y = -6\):
\[P = \frac{-4 - (-6)}{-4} = \frac{2}{-4} = -\frac12\]
Bayanin Amsa
(a) \(\dfrac25(x - 2) - \dfrac16(x + 5) \le 0\). Multiply through by the LCM \(30\):
\[12(x - 2) - 5(x + 5) \le 0\]
\[12x - 24 - 5x - 25 \le 0 \;\Rightarrow\; 7x - 49 \le 0 \;\Rightarrow\; 7x \le 49 \;\Rightarrow\; x \le 7\]
(b)(i) Factorise numerator and denominator of \(P = \dfrac{x^2 - y^2}{x^2 + xy}\):
\[P = \frac{(x - y)(x + y)}{x(x + y)} = \frac{x - y}{x}\]
(ii) With \(x = -4,\, y = -6\):
\[P = \frac{-4 - (-6)}{-4} = \frac{2}{-4} = -\frac12\]
Tambaya 54 Rahoto
(a) If 3, x, y, 18 are the terms of an Arithmetic Progression (A.P), find the values of x and y.
(b)(i) The sum of the second and third terms of a grometric progression is six times the fourth term. Find the two possible values of the common ratio.
(ii) If the second term is 8 and the common ratio is positive, find the first six terms.
(a) \(3,\ x,\ y,\ 18\) are consecutive terms of an A.P. The first term is \(3\) and the fourth term is \(18\):
\[18=3+(4-1)d=3+3d\Rightarrow d=5\]So \(x=3+5=8\) and \(y=8+5=13\).
\(x=8,\ y=13\)
(b)(i) For a G.P. with first term \(a\) and ratio \(r\), terms are \(a,\ ar,\ ar^{2},\ ar^{3},\ldots\)
"Sum of second and third terms \(=6\times\) fourth term":
\[ar+ar^{2}=6ar^{3}\]Divide through by \(ar\ (a,r\neq0)\): \(1+r=6r^{2}\Rightarrow 6r^{2}-r-1=0\).
\[r=\frac{1\pm\sqrt{1+24}}{12}=\frac{1\pm5}{12}\]\(r=\tfrac{1}{2}\) or \(r=-\tfrac{1}{3}\).
(ii) The ratio is positive, so \(r=\tfrac{1}{2}\). The second term is \(ar=8\):
\[a\times\tfrac{1}{2}=8\Rightarrow a=16\]First six terms: \(16,\ 8,\ 4,\ 2,\ 1,\ \tfrac{1}{2}\).
Bayanin Amsa
(a) \(3,\ x,\ y,\ 18\) are consecutive terms of an A.P. The first term is \(3\) and the fourth term is \(18\):
\[18=3+(4-1)d=3+3d\Rightarrow d=5\]So \(x=3+5=8\) and \(y=8+5=13\).
\(x=8,\ y=13\)
(b)(i) For a G.P. with first term \(a\) and ratio \(r\), terms are \(a,\ ar,\ ar^{2},\ ar^{3},\ldots\)
"Sum of second and third terms \(=6\times\) fourth term":
\[ar+ar^{2}=6ar^{3}\]Divide through by \(ar\ (a,r\neq0)\): \(1+r=6r^{2}\Rightarrow 6r^{2}-r-1=0\).
\[r=\frac{1\pm\sqrt{1+24}}{12}=\frac{1\pm5}{12}\]\(r=\tfrac{1}{2}\) or \(r=-\tfrac{1}{3}\).
(ii) The ratio is positive, so \(r=\tfrac{1}{2}\). The second term is \(ar=8\):
\[a\times\tfrac{1}{2}=8\Rightarrow a=16\]First six terms: \(16,\ 8,\ 4,\ 2,\ 1,\ \tfrac{1}{2}\).
Tambaya 55 Rahoto
(a) Using ruler and a pair of compasses only, construct :
(i) a quadrilateral PQRS such that /PQ/ = 7 cm, < QPS = 60°, /PS/ = 6.5 cm, < PQR = 135° and /QS/ = /QR/ ;
(ii) locus, \(l_{1}\) of points equidistant from P and Q ;
(iii) locus, \(l_{2}\) of points equidistant from P and S.
(b)(i) Label the point T where \(l_{1}\) and \(l_{2}\) intersect. (ii) With center T and radius /TP/, construct a circle \(l_{3}\).
(a) Construction
- Draw the line segment \(PQ=7\text{ cm}\).
- At \(P\), construct an angle of \(60^\circ\). On this ray, mark \(S\) such that \(PS=6.5\text{ cm}\).
- Join \(Q\) to \(S\). With centre \(Q\) and radius \(QS\), draw an arc. At \(Q\), construct \(\angle PQR=135^\circ\); let this ray meet the arc at \(R\). Join \(R\) to \(S\). Thus \(QR=QS\).
- Construct the perpendicular bisector of \(PQ\) and label it \(l_1\). This is the locus of points equidistant from \(P\) and \(Q\).
- Construct the perpendicular bisector of \(PS\) and label it \(l_2\). This is the locus of points equidistant from \(P\) and \(S\).
(b)
- The point of intersection of \(l_1\) and \(l_2\) is \(T\).
- With centre \(T\) and radius \(TP\), draw the circle \(l_3\), as shown above. The circle passes through \(P\), \(Q\) and \(S\), since \(TP=TQ=TS\).
Bayanin Amsa
(a) Construction
- Draw the line segment \(PQ=7\text{ cm}\).
- At \(P\), construct an angle of \(60^\circ\). On this ray, mark \(S\) such that \(PS=6.5\text{ cm}\).
- Join \(Q\) to \(S\). With centre \(Q\) and radius \(QS\), draw an arc. At \(Q\), construct \(\angle PQR=135^\circ\); let this ray meet the arc at \(R\). Join \(R\) to \(S\). Thus \(QR=QS\).
- Construct the perpendicular bisector of \(PQ\) and label it \(l_1\). This is the locus of points equidistant from \(P\) and \(Q\).
- Construct the perpendicular bisector of \(PS\) and label it \(l_2\). This is the locus of points equidistant from \(P\) and \(S\).
(b)
- The point of intersection of \(l_1\) and \(l_2\) is \(T\).
- With centre \(T\) and radius \(TP\), draw the circle \(l_3\), as shown above. The circle passes through \(P\), \(Q\) and \(S\), since \(TP=TQ=TS\).
Tambaya 56 Rahoto
(a) Copy and complete the table of values for \(y = 3\sin x + 2\cos x\) for \(0° \leq x \leq 360°\).
| x | 0° | 60° | 120° | 180° | 240° | 300° | 360° |
| y | 2.00 | 2.00 |
(b) Using a scale of 2 cm to 60° on x- axis and 2 cm to 1 unit on the y- axis, draw the graph of \(y = 3 \sin x + 2 \cos x\) for \(0° \leq x \leq 360°\).
(c) Use your graph to solve the equation : \(3 \sin x + 2 \cos x = 1.5\).
(d) Find the range of values of x for which \(3\sin x + 2\cos x < -1\).
(a) For \(y=3\sin x+2\cos x\), the completed table is:
| \(x\) | \(0^\circ\) | \(60^\circ\) | \(120^\circ\) | \(180^\circ\) | \(240^\circ\) | \(300^\circ\) | \(360^\circ\) |
|---|---|---|---|---|---|---|---|
| \(y\) | 2.00 | 3.60 | 1.60 | -2.00 | -3.60 | -1.60 | 2.00 |
For example, at \(x=60^\circ\), \(y=3(0.8660)+2(0.5)=3.598\approx3.60\).
(b) The graph, drawn using the stated scales, is shown below.
(c) Draw the horizontal line \(y=1.5\). Its intersections with the curve give
\[x\approx122^\circ\quad\text{or}\quad x\approx351^\circ.\]
(d) Draw the horizontal line \(y=-1\). The curve is below this line between its two points of intersection. Hence,
\[\boxed{162^\circ<x<310^\circ}.\]
Bayanin Amsa
(a) For \(y=3\sin x+2\cos x\), the completed table is:
| \(x\) | \(0^\circ\) | \(60^\circ\) | \(120^\circ\) | \(180^\circ\) | \(240^\circ\) | \(300^\circ\) | \(360^\circ\) |
|---|---|---|---|---|---|---|---|
| \(y\) | 2.00 | 3.60 | 1.60 | -2.00 | -3.60 | -1.60 | 2.00 |
For example, at \(x=60^\circ\), \(y=3(0.8660)+2(0.5)=3.598\approx3.60\).
(b) The graph, drawn using the stated scales, is shown below.
(c) Draw the horizontal line \(y=1.5\). Its intersections with the curve give
\[x\approx122^\circ\quad\text{or}\quad x\approx351^\circ.\]
(d) Draw the horizontal line \(y=-1\). The curve is below this line between its two points of intersection. Hence,
\[\boxed{162^\circ<x<310^\circ}.\]
Tambaya 57 Rahoto
(a) 
In the diagram, AB // CD and BC // FE. \(\stackrel\frown{CDE} = 75°\) and \(\stackrel\frown{DEF} = 26°\). Find the angles marked x and y.
(b) 
The diagram shows a circle ABCD with centre O and radius 7 cm. The reflex angle AOC = 190° and < DAO = 35°. Find :
(i) < ABC ; (ii) < ADC.
(c) Using the diagram in (b) above, calculate, correct to 3 significant figures, the length of : (i) arc ABC ; (ii) the chord AD. [Take \(\pi = 3.142\)].
(a) Finding x and y. From the diagram \(AB\parallel CD\), \(BC\parallel FE\), \(\widehat{CDE}=75^\circ\) and \(\widehat{DEF}=26^\circ\).
Draw a line through \(D\) parallel to \(BC\) and \(FE\). Because \(DE\) is a transversal of the parallel lines \(FE\) and this new line, the angle \(DE\) makes with the parallel family at \(D\) equals the angle it makes at \(E\):
\[\text{angle between } DE \text{ and the } BC\text{-family}=\widehat{DEF}=26^\circ.\]Since \(\widehat{CDE}=75^\circ\) is the angle between \(DC\) and \(DE\), the angle between \(DC\) and the \(BC\)-family is
\[75^\circ-26^\circ=49^\circ.\]Now \(x=\widehat{ABC}\) is the angle at \(B\) between \(BA\) (in the \(CD\)-family, since \(AB\parallel CD\)) and \(BC\). This is exactly that same angle between the two families:
\[\boxed{x=49^\circ.}\]By alternate angles (\(AB\parallel CD\), transversal \(BC\)), \(\widehat{BCD}=\widehat{ABC}=49^\circ\). The marked angle \(y\) is the reflex angle at \(C\) (the loop over the peak):
\[y=360^\circ-49^\circ=\boxed{311^\circ.}\](b) Circle \(ABCD\), centre \(O\), radius \(7\text{ cm}\). Reflex \(\widehat{AOC}=190^\circ\), so the ordinary \(\widehat{AOC}=360^\circ-190^\circ=170^\circ\).
(i) \(\widehat{ABC}\) stands on the arc \(ADC\) (the arc corresponding to the reflex \(190^\circ\)). Angle at circumference \(=\tfrac12\) angle at centre:
\[\widehat{ABC}=\tfrac12(190^\circ)=95^\circ.\](ii) \(\widehat{ADC}\) stands on the arc \(ABC\) (central angle \(170^\circ\)):
\[\widehat{ADC}=\tfrac12(170^\circ)=85^\circ.\](Check: \(ABCD\) is cyclic, \(95^\circ+85^\circ=180^\circ\).)
(c)(i) Arc \(ABC\). Central angle \(=170^\circ\), \(r=7\), \(\pi=3.142\):
\[\text{arc}=\frac{170}{360}\times 2\times 3.142\times 7=\frac{170}{360}\times 43.988=20.77\approx 20.8\text{ cm}.\](c)(ii) Chord \(AD\). In \(\triangle OAD\), \(OA=OD=7\) so it is isosceles with base angles \(\widehat{DAO}=\widehat{ODA}=35^\circ\). Hence \(\widehat{AOD}=180^\circ-2(35^\circ)=110^\circ\).
\[AD=2r\sin\!\left(\tfrac{110^\circ}{2}\right)=2(7)\sin 55^\circ=14\times 0.8192=11.47\approx 11.5\text{ cm}.\]Bayanin Amsa
(a) Finding x and y. From the diagram \(AB\parallel CD\), \(BC\parallel FE\), \(\widehat{CDE}=75^\circ\) and \(\widehat{DEF}=26^\circ\).
Draw a line through \(D\) parallel to \(BC\) and \(FE\). Because \(DE\) is a transversal of the parallel lines \(FE\) and this new line, the angle \(DE\) makes with the parallel family at \(D\) equals the angle it makes at \(E\):
\[\text{angle between } DE \text{ and the } BC\text{-family}=\widehat{DEF}=26^\circ.\]Since \(\widehat{CDE}=75^\circ\) is the angle between \(DC\) and \(DE\), the angle between \(DC\) and the \(BC\)-family is
\[75^\circ-26^\circ=49^\circ.\]Now \(x=\widehat{ABC}\) is the angle at \(B\) between \(BA\) (in the \(CD\)-family, since \(AB\parallel CD\)) and \(BC\). This is exactly that same angle between the two families:
\[\boxed{x=49^\circ.}\]By alternate angles (\(AB\parallel CD\), transversal \(BC\)), \(\widehat{BCD}=\widehat{ABC}=49^\circ\). The marked angle \(y\) is the reflex angle at \(C\) (the loop over the peak):
\[y=360^\circ-49^\circ=\boxed{311^\circ.}\](b) Circle \(ABCD\), centre \(O\), radius \(7\text{ cm}\). Reflex \(\widehat{AOC}=190^\circ\), so the ordinary \(\widehat{AOC}=360^\circ-190^\circ=170^\circ\).
(i) \(\widehat{ABC}\) stands on the arc \(ADC\) (the arc corresponding to the reflex \(190^\circ\)). Angle at circumference \(=\tfrac12\) angle at centre:
\[\widehat{ABC}=\tfrac12(190^\circ)=95^\circ.\](ii) \(\widehat{ADC}\) stands on the arc \(ABC\) (central angle \(170^\circ\)):
\[\widehat{ADC}=\tfrac12(170^\circ)=85^\circ.\](Check: \(ABCD\) is cyclic, \(95^\circ+85^\circ=180^\circ\).)
(c)(i) Arc \(ABC\). Central angle \(=170^\circ\), \(r=7\), \(\pi=3.142\):
\[\text{arc}=\frac{170}{360}\times 2\times 3.142\times 7=\frac{170}{360}\times 43.988=20.77\approx 20.8\text{ cm}.\](c)(ii) Chord \(AD\). In \(\triangle OAD\), \(OA=OD=7\) so it is isosceles with base angles \(\widehat{DAO}=\widehat{ODA}=35^\circ\). Hence \(\widehat{AOD}=180^\circ-2(35^\circ)=110^\circ\).
\[AD=2r\sin\!\left(\tfrac{110^\circ}{2}\right)=2(7)\sin 55^\circ=14\times 0.8192=11.47\approx 11.5\text{ cm}.\]Tambaya 58 Rahoto
The ages, in years, of 50 teachers in a school are given below :
21 37 49 27 49 42 26 33 46 40 50 29 23 24 29 31 36 22 27 38 30 26 42 39 34 23 21 32 41 46 46 31 33 29 28 43 47 40 34 44 26 38 34 49 45 27 25 33 39 40
(a) Form a frequency distribution table of the data using the intervals : 21 - 25, 26 - 30, 31 - 35 etc.
(b) Draw the histogram of the distribution
(c) Use your histogram to estimate the mode
(d) Calculate the mean age.
(a) Frequency distribution
| Age (years) | Tally | Class mark, \(x\) | Frequency, \(f\) | \(fx\) |
|---|---|---|---|---|
| 21 - 25 | ||||| || | 23 | 7 | 161 |
| 26 - 30 | ||||| ||||| | | 28 | 11 | 308 |
| 31 - 35 | ||||| |||| | 33 | 9 | 297 |
| 36 - 40 | ||||| |||| | 38 | 9 | 342 |
| 41 - 45 | ||||| | | 43 | 6 | 258 |
| 46 - 50 | ||||| ||| | 48 | 8 | 384 |
| Total | 50 | 1750 | ||
(b) Histogram
The continuous class boundaries are \(20.5\), \(25.5\), \(30.5\), \(35.5\), \(40.5\), \(45.5\) and \(50.5\). Since all class widths are 5 years, the bar heights are the frequencies.
(c) Modal age
The modal class is \(26\text{ - }30\), for which \(L=25.5\), \(f_1=11\), \(f_0=7\), \(f_2=9\), and class width \(h=5\).
\[\text{Mode}=L+\left(\frac{f_1-f_0}{(f_1-f_0)+(f_1-f_2)}\right)h\]
\[=25.5+\left(\frac{11-7}{(11-7)+(11-9)}\right)\times5=25.5+\frac{4}{6}\times5=28.8\]
Hence, the estimated mode is \(28.8\) years.
(d) Mean age
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{1750}{50}=35\]
Therefore, the mean age is 35 years.
Bayanin Amsa
(a) Frequency distribution
| Age (years) | Tally | Class mark, \(x\) | Frequency, \(f\) | \(fx\) |
|---|---|---|---|---|
| 21 - 25 | ||||| || | 23 | 7 | 161 |
| 26 - 30 | ||||| ||||| | | 28 | 11 | 308 |
| 31 - 35 | ||||| |||| | 33 | 9 | 297 |
| 36 - 40 | ||||| |||| | 38 | 9 | 342 |
| 41 - 45 | ||||| | | 43 | 6 | 258 |
| 46 - 50 | ||||| ||| | 48 | 8 | 384 |
| Total | 50 | 1750 | ||
(b) Histogram
The continuous class boundaries are \(20.5\), \(25.5\), \(30.5\), \(35.5\), \(40.5\), \(45.5\) and \(50.5\). Since all class widths are 5 years, the bar heights are the frequencies.
(c) Modal age
The modal class is \(26\text{ - }30\), for which \(L=25.5\), \(f_1=11\), \(f_0=7\), \(f_2=9\), and class width \(h=5\).
\[\text{Mode}=L+\left(\frac{f_1-f_0}{(f_1-f_0)+(f_1-f_2)}\right)h\]
\[=25.5+\left(\frac{11-7}{(11-7)+(11-9)}\right)\times5=25.5+\frac{4}{6}\times5=28.8\]
Hence, the estimated mode is \(28.8\) years.
(d) Mean age
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{1750}{50}=35\]
Therefore, the mean age is 35 years.
