WAEC SSCE - General Mathematics - 2016

Akọwa Nkọwa

To simplify the expression \(\frac{2}{1-x} - \frac{1}{x}\), we need to first find a common denominator. The denominator of the first fraction is \((1-x)\) and the denominator of the second fraction is \(x\). The common denominator of these two fractions is \(x(1-x)\). Now, we need to rewrite each fraction with this common denominator. For the first fraction, we can multiply the numerator and denominator by \(x\), giving us \(\frac{2x}{x(1-x)}\). For the second fraction, we can multiply the numerator and denominator by \((1-x)\), giving us \(\frac{-(1-x)}{x(1-x)}\). Putting these two fractions together, we get: \[\frac{2x}{x(1-x)} - \frac{1-x}{x(1-x)} = \frac{2x - (1-x)}{x(1-x)} = \frac{3x-1}{x(1-x)}\] Therefore, the answer is \(\frac{3x-1}{x(1-x)}\).

Akọwa Nkọwa

Akọwa Nkọwa

The given sequence is the Fibonacci sequence, where the first two terms are 0 and 1, and each subsequent term is the sum of the two preceding it. Therefore, the next three terms are: - 13 (8 + 5) - 21 (13 + 8) - 34 (21 + 13) Hence, the answer is 13, 21, 34.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

If 1cm represents 5km on the map, then x cm will represent 100km² on the map. To find x, we can use the formula for area of a square, which is A = s². In this case, we want to solve for s, where A = 100 and s represents the side length on the map in centimeters. So, s² = A s² = 100 s = √100 s = 10cm Therefore, 10cm on the map represents 100km² on the ground. To find the area on the map that represents 100km², we need to find the area of a square with a side length of 10cm. Area = s² Area = 10cm x 10cm Area = 100cm² So, the area on the map that represents 100km² is 100cm², which is equal to 1cm x 1cm, 4cm², or.

Akọwa Nkọwa

Akọwa Nkọwa

To simplify the expression (\(\frac{3}{4} - \frac{2}{3}\)) x 1\(\frac{1}{5}\), we first need to evaluate the subtraction inside the parentheses: \begin{align*} \frac{3}{4} - \frac{2}{3} &= \frac{9}{12} - \frac{8}{12} \\ &= \frac{1}{12} \end{align*} So now we have: \begin{align*} (\frac{3}{4} - \frac{2}{3})\times 1\frac{1}{5} &= \frac{1}{12} \times \frac{6}{5} \\ &= \frac{1 \times 6}{12 \times 5} \\ &= \frac{1}{10} \end{align*} Therefore, the answer is \(\frac{1}{10}\).

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

The volume of a sphere of radius r is given by the formula: \(\frac{4}{3}\pi r^3\). The volume of a cylinder of radius 3cm and height 4cm is given by the formula: \(\pi (3cm)^2(4cm) = 36\pi cm^3\). According to the question, the volume of the sphere is equal to the volume of the cylinder. Therefore: \[\frac{4}{3}\pi r^3 = 36\pi\] Dividing both sides by \(\frac{4}{3}\pi\), we get: \[r^3 = 27\] Taking the cube root of both sides, we get: \[r = 3\] Therefore, the value of r is 3cm. Answer is correct.

Akọwa Nkọwa

Akọwa Nkọwa

The problem can be solved using the formula for direct variation: d = kt^2 where d is the distance, t is the time, and k is the constant of variation. We can solve for k using the given information that the stone falls 45cm in 3 seconds: 45 = k(3)^2 45 = 9k k = 5 Now that we know k, we can use the formula to find how far the stone will fall in 6 seconds: d = 5(6)^2 d = 5(36) d = 180cm Therefore, the answer is 180cm.

Akọwa Nkọwa

Akọwa Nkọwa

Using Pythagoras theorem, we can find that |UW| = 2cm. Next, we notice that \(\Delta\)UTW is an isosceles triangle since |TX| = |WX|. Therefore, o. Using the fact that the angles in a triangle add up to 180^o, we can find that o - o - (60^o + o. Therefore, we can substitute this into the previous equation to get: 90^o + o - (60^o + o = 2o, and since \(\Delta\)UTW is an isosceles triangle, we have that o. Hence, the answer is (C) 75^o.

Akọwa Nkọwa

Akọwa Nkọwa

The formula for the volume of a pyramid is given as: V = (1/3) * base_area * height Let the length of the rectangle be x, then the base area of the pyramid is given as: base_area = x * 6 From the question, we are told that the volume of the pyramid is 90cm³ and its height is 15cm. Substituting into the formula for the volume of a pyramid, we have: 90 = (1/3) * (x * 6) * 15 Multiplying both sides by 3 gives: 270 = 90 * 6x Dividing both sides by 90 gives: 3 = 2x Therefore, x = 3/2 = 1.5 Hence, the value of x is 3. Answer option A, 3, is the correct answer.

Akọwa Nkọwa

Akọwa Nkọwa

The curved surface area of a cylinder is given by the formula: 2\(\pi\)rh. Given that the cylinder is 5cm high and its curved surface area is 110cm², we can write: 2\(\pi\)rh = 110, where h = 5 Substituting the value of \(\pi = \frac{22}{7}\) and h = 5, we get: 2 x \(\frac{22}{7}\) x r x 5 = 110 Simplifying this expression, we get: r = \(\frac{7}{2}\) r = 3.5cm (to one decimal place) Therefore, the radius of the cylinder is approximately 3.5cm. Hence, the answer is 3.5cm.

Akọwa Nkọwa

Akọwa Nkọwa

We are given that log₂(3x - 1) = 5. Using the definition of logarithms, we know that 2⁵ = 32 is equal to the expression inside the logarithm. That is, 3x - 1 = 32 Adding 1 to both sides, we get 3x = 33 Dividing by 3, we get x = 11 Therefore, the value of x is 11. Answer: 11.

Akọwa Nkọwa

To make s the subject of the relation P = S + \(\frac{sm^2}{nr}\), we need to isolate s on one side of the equation. First, we can start by moving the \(\frac{sm^2}{nr}\) term to the other side of the equation by subtracting it from both sides: P - \(\frac{sm^2}{nr}\) = S Next, we can solve for s by multiplying both sides of the equation by \(\frac{nr}{m^2}\): s = \(\frac{nr}{m^2}\)(P - \(\frac{sm^2}{nr}\)) Simplifying the right-hand side, we get: s = \(\frac{nrp}{m^2}\) - \(\frac{s}{m}\) To isolate s, we can add \(\frac{s}{m}\) to both sides of the equation: s + \(\frac{sm}{m^2}\) = \(\frac{nrp}{m^2}\) Simplifying the left-hand side, we get: s(\(\frac{m + 1}{m^2}\)) = \(\frac{nrp}{m^2}\) Finally, we can solve for s by dividing both sides of the equation by \(\frac{m+1}{m^2}\): s = \(\frac{nrp}{nr + m^2}\) Therefore, the answer is s = \(\frac{nrp}{nr + m^2}\).

Akọwa Nkọwa

We know that the relation y = x² + 2x + k passes through the point (2,0), which means that when x=2, y=0. So, substituting these values in the relation, we get: 0 = 2² + 2(2) + k 0 = 4 + 4 + k 0 = 8 + k Therefore, k = -8. Hence, the value of k is -8.

Akọwa Nkọwa

We can use the distance formula to find the length PQ, which is the distance between points P and Q on the line. The distance formula is: distance = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) where (x1, y1) = P and (x2, y2) = Q. Plugging in the values: distance = \(\sqrt{(5-1)^2 + (8-2)^2}\) = \(\sqrt{16 + 36}\) = \(\sqrt{52}\) = \(2\sqrt{13}\) Therefore, the length PQ is \(2\sqrt{13}\). Answer: \(2\sqrt{13}\).

Akọwa Nkọwa

Akọwa Nkọwa

The valid conclusion from the premise "Nigeria footballers are good footballers" is "Joseph is a Nigerian footballer therefore he is a good footballer". This is because the premise establishes that Nigeria footballers are good, so anyone who is a Nigeria footballer can be inferred to be a good footballer. Therefore, Joseph, who is a Nigerian footballer, can be concluded to be a good footballer. The other options are not valid because they do not follow logically from the given premise.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

The given expression will be undefined when its denominator is equal to zero since division by zero is undefined. So we need to find the values of y that make the denominator zero. \begin{align*} y^2 + 4y - 21 &= 0\\ (y+7)(y-3) &= 0 \end{align*} The denominator is equal to zero when either y+7=0 or y-3=0. Therefore, the expression is undefined when y=-7 or y=3. So, the answer is (c) 3, -7.

Akọwa Nkọwa

Akọwa Nkọwa

The gradient of a line is the measure of its steepness, or slope. To find the gradient of the line PQ, we need to use the formula: Gradient (m) = change in y / change in x We can calculate the change in y by subtracting the y-coordinate of point P from the y-coordinate of point Q: 8 - 2 = 6 Similarly, we can calculate the change in x by subtracting the x-coordinate of point P from the x-coordinate of point Q: 5 - 1 = 4 Therefore, the gradient of the line PQ is: Gradient (m) = change in y / change in x = 6 / 4 = 3/2 Therefore, the answer is option C: \(\frac{3}{2}\).

Akọwa Nkọwa

When two balls are drawn from the bag with replacement, there are a total of $9\times9=81$ possible outcomes, since there are 9 balls in the bag and we are replacing each ball after drawing. To find the probability that the first ball is red and the second ball is blue, we can use the multiplication rule of probability. The probability that the first ball is red is $\frac{5}{9}$, since there are 5 red balls out of 9 total balls in the bag. After replacing the first ball, there are still 9 balls in the bag, but now 4 of them are blue. So the probability that the second ball is blue, given that the first ball was red, is $\frac{4}{9}$. Therefore, the probability that the first ball is red and the second ball is blue is: $$\frac{5}{9} \times \frac{4}{9} = \frac{20}{81}$$ Hence the answer is $\frac{20}{81}$.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

First, let's simplify the numerator: \begin{align*} (p-r)^2 - r^2 &= (p^2 - 2pr + r^2) - r^2 \\ &= p^2 - 2pr \end{align*} Now, let's factor the denominator: \begin{align*} 2p^2 - 4pr &= 2p(p - 2r) \end{align*} Substituting these results, we get: \begin{align*} \frac{(p-r)^2 - r^2}{2p^2 - 4pr} &= \frac{p^2 - 2pr}{2p(p - 2r)} \\ &= \frac{p(p-2r)}{2p(p-2r)} \\ &= \frac{1}{2} \end{align*} Therefore, the answer is: \boxed{\frac{1}{2}}.

Akọwa Nkọwa

We can simplify this expression by using the properties of exponents and simplifying the numbers. First, we can write 27 as 3³ and 81 as 3⁴. Next, we can simplify the numerator by using the distributive property of exponents: \begin{align*} \frac{3^{n-1} \times 27^{n+1}}{81^n} &= \frac{3^{n-1} \times 3^{3(n+1)}}{3^{4n}} \\ &= \frac{3^{n-1} \times 3^{3n+3}}{3^{4n}} \\ &= \frac{3^{4n-1}}{3^{4n}} \times 3^{3n+3} \\ &= 3^{-1} \times 3^{3n+3} \\ &= 3^{3n+2} \\ &= 3^2 \times 3^{3n} \\ &= 9 \times 3^n \\ \end{align*} Therefore, the simplified expression is 9 * 3ⁿ. So, the answer is 9.

Akọwa Nkọwa

To find out how many students are in the club, we need to add up the frequencies in the table. So, Number of 13-year-olds = 10 Number of 14-year-olds = 24 Number of 15-year-olds = 8 Number of 16-year-olds = 5 Number of 17-year-olds = 3 Total number of students = 10 + 24 + 8 + 5 + 3 = 50 Therefore, there are 50 students in the club.

Akọwa Nkọwa

Akọwa Nkọwa

The range of a set of data is the difference between the maximum and minimum values in the set. In this case, the minimum mark is 3 and the maximum mark is 16. Therefore, the range is 16 - 3 = 13. Hence, the answer is 13.

Akọwa Nkọwa

To find the median, we need to first arrange the ages in order from lowest to highest. Then, we can determine which age lies in the middle of the list. Arranging the ages in order of increasing magnitude, we have: $$13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 17, 17, 17$$ There are a total of 10+24+8+5+3 = 50 ages in the list, which is an even number. To find the median, we need to take the average of the two middle ages. The two middle ages are the 25th and 26th ages in the list, which are both 14. Therefore, the median age is: $$(14 + 14)/2 = 14$$ So the correct answer is 14.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

To find the lower quartile, we need to identify the point on the cumulative frequency curve that corresponds to 25% of the total frequency. From the graph, we see that the total frequency is 40, and 25% of this is 10. The point on the curve that corresponds to a frequency of 10 is at a value of 19.0. Therefore, the lower quartile is 19.0. Note that the cumulative frequency curve shows the cumulative frequency of data values up to and including each data point on the horizontal axis. So, we can read off quartiles and other percentiles directly from the graph.

Akọwa Nkọwa

Akọwa Nkọwa

Construction using ruler and a pair of compasses only

(a)(i) Triangle XYZ. Draw the base \(|XY| = 10\text{ cm}\). At \(X\) construct \(\angle YXZ = 45^{\circ}\): first raise a perpendicular at \(X\) to obtain \(90^{\circ}\), then bisect that right angle to get \(45^{\circ}\). At \(Y\) construct \(\angle XYZ = 30^{\circ}\): first construct a \(60^{\circ}\) angle (an arc of radius equal to \(|XY_1|\) stepped off along the base), then bisect it to get \(30^{\circ}\). The ray from \(X\) and the ray from \(Y\) meet at \(Z\). This fixes \(\triangle XYZ\), and the third angle is \[\angle XZY = 180^{\circ} - 45^{\circ} - 30^{\circ} = 105^{\circ}.\]

(a)(ii) Locus \(l_1\). The set of points equidistant from \(Y\) and \(Z\) is the perpendicular bisector of \(YZ\). With centre \(Y\) and then centre \(Z\), using the same radius (a little more than half of \(|YZ|\)), draw arcs that cross above and below \(YZ\); the straight line through the two crossings is \(l_1\).

(a)(iii) Locus \(l_2\). The line through \(Z\) parallel to \(XY\). Since \(XY\) is the base, drop a perpendicular from \(Z\) to \(XY\) and at \(Z\) construct a right angle to that perpendicular; the resulting line through \(Z\) is parallel to \(XY\) and is \(l_2\).

(b) Mark \(M\), the point where \(l_1\) and \(l_2\) intersect.

The complete construction is shown below (measured values: \(|XZ| \approx 5.2\text{ cm}\), \(|YZ| \approx 7.3\text{ cm}\)).

(c) Measurement of \(\angle ZMY\). Because \(M\) lies on \(l_1\), it is equidistant from \(Y\) and \(Z\), so \(|MY| = |MZ|\) and \(\triangle MYZ\) is isosceles. On measuring the angle at \(M\) with a protractor: \[\boxed{\angle ZMY = 120^{\circ}.}\]

Akọwa Nkọwa

(a) Equation of the line

\(2x+y=6\Rightarrow y=-2x+6\), so its gradient is \(-2\). A parallel line has the same gradient \(m=-2\) and passes through \((2,-3)\):

\[y-(-3)=-2(x-2)\Rightarrow y+3=-2x+4\Rightarrow y=-2x+1\quad\text{i.e. }2x+y=1.\]

(b)(i) With \(x\,\Delta\,y=(x+y+xy)\bmod 8\) on \(T=\{2,3,5,7\}\):

(e.g. \(2\Delta5=(2+5+10)\bmod 8=17\bmod 8=1\).)

(ii)(a) \(5\Delta7=7\), so \(2\Delta(5\Delta7)=2\Delta7=7\).

(ii)(b) \(5\Delta7=7\), so we need \(2\Delta n=7\). From the \(2\)-row, \(2\Delta7=7\), hence \(n=7\).

Akọwa Nkọwa

(a) Solve \(7(x+4)-\tfrac{2}{3}(x-6)\le 2[x-3(x+5)]\).

Right side: \(x-3(x+5)=x-3x-15=-2x-15\), so \(2(-2x-15)=-4x-30\).

Left side: \(7x+28-\tfrac{2}{3}(x-6)=7x+28-\tfrac{2}{3}x+4=\tfrac{19}{3}x+32\).

Inequality: \(\tfrac{19}{3}x+32\le -4x-30\). Multiply through by 3:

\(x\le -6\)

(b) Let \(t\) = tricycles and \(c\) = taxicabs.

Vehicles: \(t+c=20\).
Passengers: \(2t+4c=66\).

From the first, \(c=20-t\). Substitute:

The company has 7 tricycles (and 13 taxicabs).

Akọwa Nkọwa

(a) Finding \(\angle VUS\).

The points V, R, S, T, U lie on a circle, and line MRQ is a tangent at R. Given \(\angle RTS = 28^\circ\) and \(\angle VRM = 46^\circ\).

Step 1 - use the tangent-chord (alternate segment) property on chord RV. The angle between the tangent RM and the chord RV equals the inscribed angle in the alternate segment standing on RV. Hence the arc VR (the intercepted arc) satisfies

\[\angle VRM = \tfrac{1}{2}(\text{arc } VR) \;\Rightarrow\; \text{arc } VR = 2 \times 46^\circ = 92^\circ.\]

Step 2 - use the inscribed angle on chord RS. \(\angle RTS = 28^\circ\) is the angle at the circumference standing on chord RS, so

\[\text{arc } RS = 2 \times 28^\circ = 56^\circ.\]

Step 3 - find \(\angle VUS\). \(\angle VUS\) is the inscribed angle at U standing on chord VS. It equals half the arc VS that does not contain U; that arc runs from V through R to S:

\[\text{arc } VRS = \text{arc } VR + \text{arc } RS = 92^\circ + 56^\circ = 148^\circ.\]

\[\angle VUS = \tfrac{1}{2}(148^\circ) = 74^\circ.\]

\[\boxed{\angle VUS = 74^\circ.}\]

(b) Radius of the cylindrical tin.

The tin is closed at one end only, so its total surface area is the curved surface plus one circular end:

\[A = 2\pi r h + \pi r^2 = 462,\qquad h = 7,\ \pi = \tfrac{22}{7}.\]

Substitute:

\[2 \times \tfrac{22}{7} \times r \times 7 + \tfrac{22}{7} r^2 = 462.\]

\[44r + \tfrac{22}{7}r^2 = 462.\]

Multiply every term by 7:

\[308r + 22r^2 = 3234.\]

Divide through by 22:

\[14r + r^2 = 147 \;\Rightarrow\; r^2 + 14r - 147 = 0.\]

Solve:

\[r = \frac{-14 \pm \sqrt{14^2 + 4(147)}}{2} = \frac{-14 \pm \sqrt{196 + 588}}{2} = \frac{-14 \pm \sqrt{784}}{2} = \frac{-14 \pm 28}{2}.\]

Taking the positive root: \(r = \dfrac{14}{2} = 7\text{ cm}.\)

Check: \(2(\tfrac{22}{7})(7)(7) + (\tfrac{22}{7})(7^2) = 308 + 154 = 462\text{ cm}^2.\) \(\;\checkmark\)

\[\boxed{r = 7\text{ cm}.}\]

Akọwa Nkọwa

(a) Completing the square.

Clear the fractions in \(\dfrac{x-2}{4}=\dfrac{x+2}{2x}\) by cross-multiplying:

\[2x(x-2)=4(x+2)\]

\[2x^{2}-4x=4x+8\]

\[2x^{2}-8x-8=0\]

Divide through by 2 to make the coefficient of \(x^{2}\) equal to 1:

\[x^{2}-4x-4=0 \quad\Rightarrow\quad x^{2}-4x=4\]

Add the square of half the coefficient of \(x\), i.e. \(\left(\tfrac{-4}{2}\right)^{2}=4\), to both sides:

\[x^{2}-4x+4=4+4\]

\[(x-2)^{2}=8\]

\[x-2=\pm\sqrt{8}=\pm 2.8284\]

\[x=2+2.8284=4.83 \quad\text{or}\quad x=2-2.8284=-0.83\]

\[\boxed{x=4.83 \text{ or } x=-0.83}\ \text{(to 2 d.p.)}\]

(b) Circle PQRST, centre O.

From the diagram: PS is a diameter, the tick marks show \(QR=RS\), the arrows show \(RS\parallel QT\), and \(\angle QTS=52^\circ\).

(i) \(\angle SQT\).

\(\angle QTS=52^\circ\) is an angle at the circumference standing on the arc \(QRS\), so

\[\text{arc } QR+\text{arc } RS=2\times 52^\circ=104^\circ.\]

Since \(QR=RS\), the equal chords cut off equal arcs, so \(\text{arc } QR=\text{arc } RS=52^\circ\).

Because \(RS\parallel QT\), the arcs between the parallel chords are equal, giving \(\text{arc } ST=\text{arc } QR=52^\circ\).

\(\angle SQT\) stands at the circumference on arc \(ST\):

\[\angle SQT=\tfrac{1}{2}\,\text{arc } ST=\tfrac{1}{2}\times 52^\circ=\boxed{26^\circ}.\]

(ii) \(\angle PQT\).

PS is a diameter, so \(\angle PQS=90^\circ\) (angle in a semicircle).

The ray \(QT\) lies between \(QP\) and \(QS\), so

\[\angle PQT=\angle PQS-\angle SQT=90^\circ-26^\circ=\boxed{64^\circ}.\]

Akọwa Nkọwa

(a) Finding x.

From the diagram, J-H-G-F is a straight line along the bottom. The line J-K-L is straight (K lies on JL), and the line K-M-G is straight (M lies on KG). The marked angles are \(\angle KLM = x\) at L, \(\angle LMK = y\) at M, \(\angle KJH = r\) at J, and \(\angle KGF = 110^\circ\) at G, with the condition \(2x = r = y\).

Step 1: angle the line KG makes with the base at G. Since J-G-F is straight, \(\angle KGF\) and \(\angle KGJ\) are on a straight line:

Step 2: angle at K in triangle KLM. The angles of \(\triangle KLM\) sum to \(180^\circ\):

Because J-K-L is a straight line, the angle on the other side of K (angle JKG, which is the angle of triangle JKG at K) is supplementary:

Step 3: use triangle JKG. Its vertices are J and G on the base and K above. Its angles are \(\angle KJG = r\), \(\angle JKG = x+y\) and \(\angle KGJ = 70^\circ\), summing to \(180^\circ\):

Step 4: apply \(r = 2x\) and \(y = 2x\):

x = 22\(^\circ\). (Then \(r = 44^\circ\), \(y = 44^\circ\).)

(b) Total amount collected by the two groups.

Let the total collected by the girls be \(G\). Then the boys collected \(G + 600\) (N600 more).

• Average of the 12 girls: \(\dfrac{G}{12}\)
• Average of the 10 boys: \(\dfrac{G + 600}{10}\)

The boys' average is N100 more than the girls' average:

Multiply through by 60 (the LCM of 10 and 12):

So the girls collected \(\text{N}2400\) and the boys collected \(2400 + 600 = \text{N}3000\).

Total collected = N5,400.00. (Check: girls' average \(=200\), boys' average \(=300\), difference \(=100\).)

Akọwa Nkọwa

(a) Evaluate \(\dfrac{0.09\times1.21}{3.3\times0.00025}\).

Numerator: \(0.09\times1.21=0.1089\).

Denominator: \(3.3\times0.00025=0.000825\).

In standard form: \(132=1.32\times10^{2}\).

\(=1.32\times10^{2}\)

(b) Principal \(P=\text{GH}\!\cent5600\); interest earned in 3 years \(=1200\), so the amount is \(A=5600+1200=6800\).

Compound interest: \(A=P(1+r)^{3}\).

Rate \(\approx 6.69\%\) per annum (3 s.f.).

Akọwa Nkọwa

(a) Frequency table

Tallying the 50 weights gives:

Total frequency \(=1+5+12+14+10+6+2=50\).

(b)(i) Mean

\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{3253}{50}=65.06\text{ kg}\]

(b)(ii) Standard deviation

\[\text{S.D.}=\sqrt{\frac{\sum fx^{2}}{\sum f}-\left(\frac{\sum fx}{\sum f}\right)^{2}}=\sqrt{\frac{211733}{50}-(65.06)^{2}}\]

\[=\sqrt{4234.66-4232.8036}=\sqrt{1.8564}=1.36\text{ kg (2 d.p.)}\]

Akọwa Nkọwa

(a) Find |QR|.

From the diagram, \(PQST\) is a parallelogram, and \(P,\,Q,\,R\) lie on one straight line. The opposite sides of the parallelogram are equal, so:

\[|PQ|=|TS|=8\text{ cm}\]

Triangle \(PSR\) has its base along the line \(PR\), and \(SM=6\text{ cm}\) is the perpendicular height from \(S\) to that base.

Using the area of a triangle:

\[\text{Area}=\frac{1}{2}\times\text{base}\times\text{height}\]

\[36=\frac{1}{2}\times |PR|\times 6=3\,|PR|\]

\[|PR|=\frac{36}{3}=12\text{ cm}\]

Since \(|PR|=|PQ|+|QR|\):

\[12=8+|QR|\Rightarrow |QR|=4\text{ cm}\]

(b) Height of the flagpole.

Let the horizontal distance from the tree to the flagpole be \(d\). The bird is at the top of the tree, at height \(10.65\text{ m}\).

Bottom of the flagpole (ground level) is seen at an angle of depression of \(60^{\circ}\). The vertical drop is the full \(10.65\text{ m}\):

\[\tan 60^{\circ}=\frac{10.65}{d}\Rightarrow d=\frac{10.65}{\tan 60^{\circ}}=\frac{10.65}{1.7321}=6.149\text{ m}\]

Top of the flagpole is seen at an angle of depression of \(45^{\circ}\). The vertical drop from the bird down to the top of the flagpole is:

\[\text{drop}=d\tan 45^{\circ}=6.149\times 1=6.149\text{ m}\]

The height of the flagpole is the tree height minus this drop:

\[h=10.65-6.149=4.501\text{ m}\]

Correct to 3 significant figures, the height of the flagpole is \(4.50\text{ m}\).

Akọwa Nkọwa

None

Akọwa Nkọwa

Given distribution.

(a) Number of throws. Let \(N\) be the total number of throws. Then

\[N = 25+30+x+28+40+32 = 155 + x.\]

The probability of obtaining a 3 is \(\dfrac{x}{N} = 0.225\), so

\[x = 0.225(155 + x) \Rightarrow x = 34.875 + 0.225x \Rightarrow 0.775x = 34.875 \Rightarrow x = 45.\]

Therefore \(N = 155 + 45 = \mathbf{200}\). The die was thrown 200 times.

The completed frequencies are: 25, 30, 45, 28, 40, 32.

(b) P(even number or prime number). Even scores: \(\{2, 4, 6\}\); prime scores: \(\{2, 3, 5\}\). Their union is \(\{2, 3, 4, 5, 6\}\) (only the score 1 is excluded). Adding their frequencies:

\[30 + 45 + 28 + 40 + 32 = 175.\]

\[P(\text{even or prime}) = \frac{175}{200} = \frac{7}{8} = 0.875.\]

Equivalently, \(1 - P(1) = 1 - \dfrac{25}{200} = 0.875\).

Akọwa Nkọwa

(a) Copy and complete the table of values for \(y = 3\sin x + 2\cos x\), correct to one decimal place.

Each missing value is obtained by direct substitution:

\(x = 0^\circ:\ 3\sin 0 + 2\cos 0 = 3(0) + 2(1) = 2.0\)

\(x = 30^\circ:\ 3(0.500) + 2(0.866) = 1.5 + 1.7 = 3.2\)

\(x = 60^\circ:\ 3(0.866) + 2(0.500) = 2.6 + 1.0 = 3.6\)

\(x = 150^\circ:\ 3(0.500) + 2(-0.866) = 1.5 - 1.7 = -0.2\)

\(x = 210^\circ:\ 3(-0.500) + 2(-0.866) = -1.5 - 1.7 = -3.2\)

\(x = 300^\circ:\ 3(-0.866) + 2(0.500) = -2.6 + 1.0 = -1.6\)

\(x = 330^\circ:\ 3(-0.500) + 2(0.866) = -1.5 + 1.7 = 0.2\)

The completed table is:

(b) Draw the graph of \(y = 3\sin x + 2\cos x\) for \(0^\circ \leq x \leq 360^\circ\).

Using scales of 2 cm to \(30^\circ\) on the x-axis and 2 cm to 1 unit on the y-axis, the thirteen points from the table are plotted and joined with a smooth curve. The curve is a single wave that rises to a maximum of about \(3.6\) near \(x = 60^\circ\) and falls to a minimum of about \(-3.6\) near \(x = 240^\circ\). The dashed horizontal line \(y = -2\) is added for part (c)(ii).

(c) Use the graph to solve:

(i) \(3\sin x + 2\cos x = 0\)

This means \(y = 0\); the solutions are the points where the curve cuts the x-axis. Reading these from the graph gives

\[x \approx 147^\circ \quad \text{and} \quad x \approx 325.5^\circ.\]

(ii) \(2 + 2\cos x + 3\sin x = 0\)

Rearrange so that the graphed expression stands alone:

\[3\sin x + 2\cos x = -2, \quad \text{i.e.} \quad y = -2.\]

Draw the horizontal line \(y = -2\) and read where it meets the curve. The line cuts the curve at

\[x \approx 180^\circ \quad \text{and} \quad x \approx 292.5^\circ.\]

Akọwa Nkọwa

(a) Find \(p:q\).

Cross multiply:

(b) The bridge cross-section.

From the diagram, \(PQRU\) is a rectangle with the two vertical sides \(PU=QR=4\text{ m}\), and a semicircular hollow is cut from the base. Along the base the two flat pieces are \(UT=SR=2\text{ m}\), and \(TS\) is the diameter of the semicircle. Let the radius be \(r\), so \(TS=2r\).

(i) Length \(PQ\).

The outline (perimeter) is the top \(PQ\), the two sides, the two flat base pieces, and the semicircular arc:

Since \(PQRU\) is a rectangle, \(PQ=UR=UT+TS+SR=2+2r+2=2r+4\). With arc \(=\pi r\):

Therefore

(ii) Area of the cross-section.

Area \(=\) rectangle \(-\) semicircle:

(The heights and base pieces read \(4\) and \(2\) from the figure; with the stated perimeter \(34\) this gives a consistent \(r=3.5\).)