WAEC SSCE - General Mathematics - 1998

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The logarithm of a number to a given base is the exponent to which the base must be raised to obtain the number. So, if log\(_{10}\) q = 2.7078, then 10\(^{2.7078}\) = q. Evaluating this expression, we get q ≈ 510.2. Therefore, the correct option is 510.2.

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The expression log\(_{10}\) a = 4 can be read as "logarithm of a to base 10 is 4". This means that 10 raised to the power of 4 is equal to a, or simply a = 10\(^4\). Therefore, the value of a is 10,000. The correct option is (e) 10,000.

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To find the percentage of the class that scored above 6, we need to first count the number of students who scored above 6, and then divide that number by the total number of students and multiply by 100 to get the percentage. We can start by counting the number of students who scored above 6. From the given set of scores, we can see that the following students scored above 6: 9, 9, 8, 7, 8, 7, 7, 11. Counting these students, we get a total of 8 students who scored above 6. To find the percentage of the class that scored above 6, we divide the number of students who scored above 6 (8) by the total number of students (32), and then multiply by 100: 8/32 * 100 = 25% So, the percentage of the class that scored above 6 is 25%, which is closest to 34% when rounded to the nearest whole number. Therefore, the answer is (a) 34%.

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The area of sector OPQ is a fraction of the area of the whole circle. The whole circle has an area of πr², where r is the radius. In this case, r = 3.5 cm, so the area of the whole circle is: A = π(3.5 cm)² = 38.465 cm² The sector OPQ makes up a fraction of the whole circle. The fraction is equal to the central angle of the sector divided by 360 degrees. In this case, the central angle is 50 degrees, so the fraction is: 50/360 = 5/36 So the area of sector OPQ is: A = (5/36)×38.465 cm² ≈ 5.350 cm² Therefore, the correct option is (b) 5.350cm².

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The sum of the exterior angles of any polygon is always equal to 360 degrees. Therefore, we can write an equation as: y + (y + 5) + (y + 10) + (y + 25) = 360 Simplifying the equation, we get: 4y + 40 = 360 Subtracting 40 from both sides, we get: 4y = 320 Dividing both sides by 4, we get: y = 80° Therefore, the value of y is 80^o.

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Let's assume the cost price of the article is $100. Since the trader incurred a loss of 15%, the selling price will be 85% of the cost price. Therefore, Selling price = 85/100 * cost price = 85/100 * 100 = $85. The ratio of the selling price to the cost price is: Selling price : Cost price = $85 : $100 To simplify the ratio, we can divide both sides by the highest common factor, which is 5. This gives: Selling price : Cost price = 17 : 20 Therefore, the ratio of selling price to cost price is 17 : 20. Option C, 17:20, is the correct answer.

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The sum of the interior angles of a hexagon is given by the formula (n-2)×180^o, where n is the number of sides/angles in the polygon. Since a hexagon has six sides/angles, the sum of its interior angles is (6-2)×180^o = 4×180^o = 720^o. Let's assume that three angles of the hexagon are each X^o. Therefore, the sum of these three angles is 3X^o. The other three angles are each 3X^o, so their sum is 9X^o. Thus, the total sum of the six angles is: 3X^o + 9X^o = 12X^o But we also know that the sum of the interior angles of a hexagon is 720^o. Therefore, we can equate the two expressions: 12X^o = 720^o Dividing both sides by 12, we get: X^o = 60^o Therefore, the answer is X = 60^o.

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If both cosine and sine are negative, it means the angle x is in the third quadrant of the unit circle where both coordinates x and y are negative. Therefore, the possible range for x is from 180 degrees to 270 degrees. Hence, the correct option is: - 180^o < x < 270^o

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To convert the number 35 to a number in base two, we need to divide 35 by 2 repeatedly until we get a quotient of 0, taking note of the remainder at each step. The remainders, read from bottom to top, will give us the binary equivalent of 35. 35 divided by 2 gives us a quotient of 17 and a remainder of 1. 17 divided by 2 gives us a quotient of 8 and a remainder of 1. 8 divided by 2 gives us a quotient of 4 and a remainder of 0. 4 divided by 2 gives us a quotient of 2 and a remainder of 0. 2 divided by 2 gives us a quotient of 1 and a remainder of 0. 1 divided by 2 gives us a quotient of 0 and a remainder of 1. Reading the remainders from bottom to top, we get 100011 as the binary equivalent of 35. Therefore, we can conclude that 35 in base two is 100011_two.

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In the given diagram, there is a straight line that intersects two parallel lines, forming several angles. By the properties of parallel lines, we know that the alternate interior angles are equal. Thus, angle x is equal to the alternate interior angle formed by the transversal and the parallel line. We can see that angle x is adjacent to the angle of 130^o, which is one of the interior angles of the triangle formed by the intersecting lines. Therefore, the sum of angle x and 130^o should be equal to 180^o (the sum of angles in a triangle). Thus, we have: angle x + 130^o = 180^o angle x = 180^o - 130^o angle x = 50^o Therefore, the answer is (E) 100^o.

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The logarithm of a number to base 10 is the power to which 10 is raised to give the number. Therefore, if log₁₀x = y, then x = 10^y. In this case, the logarithm to base 10 is given as 2.6025. Therefore, the number is x = 10^2.6025. Using a calculator, we get x ≈ 400.4. Therefore, the number whose logarithm to base 10 is 2.6025 is approximately 400.4.

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The probability of picking two balls of different colours is the probability of picking one white ball and one blue ball, because there are no other colours in the box. The probability of picking a white ball on the first draw is 2/5, since there are 2 white balls out of 5 total balls. After one white ball has been removed, there are 4 balls left, including 2 blue balls. Therefore, the probability of picking a blue ball on the second draw is 2/4 or 1/2. The probability of picking a white ball on the first draw and a blue ball on the second draw is thus: \(\frac{2}{5} * \frac{1}{2} = \frac{1}{5}\) There is also a probability of picking a blue ball on the first draw and a white ball on the second draw, which is: \(\frac{3}{5} * \frac{2}{4} = \frac{3}{10}\) Therefore, the total probability of picking two balls of different colours is: \(\frac{1}{5} + \frac{3}{10} = \frac{1}{5} + \frac{2}{5} = \frac{3}{5}\) So the answer is \(\frac{3}{5}\).

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To subtract two numbers in scientific notation, we first need to make sure they have the same power of 10. We can do this by moving the decimal point to the right or left as needed. Starting with \(3.0\times 10^1\) and \(2.8\times 10^{-1}\), we can move the decimal point one place to the left in the first number to get \(3.0\) and two places to the right in the second number to get \(0.028\). Now we have: $$ 3.0 - 0.028 = 2.972 $$ To express the answer in standard form, we need to convert it to the form \(a \times 10^b\), where \(1 \leq a < 10\) and \(b\) is an integer. We can do this by moving the decimal point to get a number between 1 and 10, and counting the number of places we moved it. In this case, we moved the decimal point one place to the left, so: $$ 2.972 = 2.972 \times 10^1 $$ Therefore, the answer is \(2.972 \times 10^1\).

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We are given that: \[\frac{6x-y}{x+2y}=2\] To solve for \(\frac{x}{y}\), we can simplify the equation above and isolate \(\frac{x}{y}\). We start by cross multiplying both sides of the equation: \[6x-y=2(x+2y)\] Expanding the brackets, we get: \[6x-y=2x+4y\] Simplifying and isolating \(x\), we get: \[4x=5y\] Therefore, \(\frac{x}{y}=\frac{5}{4}\). Hence, the answer is \(\frac{5}{4}\).

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To make S the subject of the formula, we need to isolate S on one side of the equation by performing operations on both sides of the equation. We begin by multiplying both sides by \(\sqrt{T-S}\), then we multiply both sides by \(\frac{V^2}{K}\) to obtain: \begin{align*} V &= \frac{K}{\sqrt{T-S}} \\ V\sqrt{T-S} &= K \\ T-S &= \left(\frac{K}{V}\right)^2 \\ S &= T-\left(\frac{K}{V}\right)^2 \\ \end{align*} Therefore, the answer is \(T-\left(\frac{K}{V}\right)^2 = S\).

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We can simplify the expression as follows: \begin{align*} \sqrt{20}\times (\sqrt{5})^3 &= \sqrt{(2\times2\times5)\times(5\times5\times5)} \\ &= \sqrt{(2\times5\times5\times2\times5\times5)} \\ &= \sqrt{(2^2\times5^2)\times(5^2)} \\ &= \sqrt{(2^2\times5^2\times5^2)} \\ &= \sqrt{(2^2\times5^4)} \\ &= 2\times5^2 \\ &= 50. \end{align*} Therefore, the answer is 50.

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To find the root(s) of the quadratic equation 2x\(^2\) - 3x - 2 = 0, we can use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ where a, b, and c are the coefficients of the quadratic equation ax\(^2\) + bx + c = 0. In this case, a = 2, b = -3, and c = -2. Substituting these values into the formula, we get: $$x = \frac{-(-3) \pm \sqrt{(-3)^2-4(2)(-2)}}{2(2)}$$ Simplifying: $$x = \frac{3 \pm \sqrt{9+16}}{4}$$ $$x = \frac{3 \pm \sqrt{25}}{4}$$ We can simplify the square root to get: $$x = \frac{3 \pm 5}{4}$$ So the roots are: $$x = \frac{3 + 5}{4} = 2$$ $$x = \frac{3 - 5}{4} = -\frac{1}{2}$$ Therefore, the answer is x = -1/2 or 2.

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The statement that is not true is: `-p > -2q` Given that `p` is greater than `2q`, we can multiply both sides of the inequality by `-1` to obtain `-p < -2q`. Therefore, the statement `-p < -2q` is true. Similarly, we can multiply both sides of `p > 2q` by `-1` to get `-p < -2q`, and then multiply both sides by `-1` again to obtain `2q < p`. This means that `-q < 1/2p`, making the statement `-q < 1/2p` true. Now, to check the remaining options, we can square both sides of `p > 2q` to get `p^2 > 4q^2`, and since `4q^2 > 2q^2`, we have `p^2 > 2q^2`, making the statement `p^2 > 2q^2` true. Finally, we can divide both sides of `p > 2q` by `2` to get `q < 1/2p`, which means that the statement `q < 1/2p` is also true. Therefore, the only statement that is not true is `-p > -2q`.

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We are given the equation: h(m+n) = m(h+r). Expanding the left-hand side of the equation, we get: hm + hn = hm + mr Subtracting hm from both sides, we get: hn = mr Dividing both sides by n, we get: h = mr/n Therefore, the answer is: h = mr/n. Option D is the correct answer.

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The area of a square is given by the formula: A = s^2, where s is the length of a side of the square. Initially, the length of the side of the square is 20cm, so its area is A1 = 20^2 = 400 cm^2. When the length of the side is increased to 21cm, the new area becomes A2 = 21^2 = 441 cm^2. The difference in area between the two squares is: A2 - A1 = 441 - 400 = 41 cm^2 To find the percentage increase in area, we need to divide the difference in area by the original area, and then multiply by 100: percentage increase = (difference in area / original area) x 100% = (41 / 400) x 100% = 0.1025 x 100% = 10.25% Therefore, the percentage increase in area is 10.25%. So, the correct answer is 10.25%.

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To find the length of arc PQ, we need to first find the circumference of the circle. The circumference of a circle with radius r is given by the formula: C = 2πr Substituting the given values, we get: C = 2 × 3.142 × 3.5 C ≈ 21.991cm Since the angle ?POQ is 50 degrees and the total angle of a circle is 360 degrees, the length of arc PQ is given by: length of arc PQ = (50/360) × C length of arc PQ ≈ 3.054cm ≈ 3.1cm (rounded to one decimal place) Therefore, the correct answer is 3.1cm.

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The curved surface area of a cone can be calculated using the formula: `Curved surface area = πrl` where r is the radius of the base, l is the slant height, and π is the constant pi. In this case, the radius is given as 3cm, and the slant height is given as 7cm. So, the curved surface area of the cone is: `Curved surface area = πrl` `Curved surface area = (22/7) x 3 x 7` `Curved surface area = 66 cm^2` Therefore, the curved surface area of the given cone is 66 cm^2. Answer is the correct answer.

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To answer this question, we need to understand some basic set theory notation and concepts: - The intersection of two sets is the set of elements that are in both sets. - The union of two sets is the set of elements that are in either set (or both). - The complement of a set is the set of elements that are not in that set. - The cardinality of a set is the number of elements in that set. (a) \((P\cap Q)^l={2, 3, 4, 13}\) is not correct. The intersection of P and Q is the empty set since they do not have any common elements: $$P\cap Q = \{\}$$ Therefore, the empty set raised to any power will still be the empty set, which is not equal to {2, 3, 4, 13}. (b) \(n(P\cup Q)=4\) is correct. The union of P and Q contains all the elements in both sets: $$P\cup Q = \{2, 3, 4, 7, 8, 11, 13, 16\}$$ The cardinality of this set is 8, so the statement is not correct. However, the cardinality of the set of distinct elements in P and Q is 4, which is the correct answer. Therefore, the statement is correct. (c) \(P\cup Q = \emptyset\) is not correct. As shown above, the union of P and Q is not empty. Therefore, the statement is not correct. (d) \(P\cap Q = \emptyset\) is correct. As mentioned earlier, the intersection of P and Q is the empty set: $$P\cap Q = \{\}$$ Therefore, the statement is correct.

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The nth term of the sequence is given as 3 x 2^(2-n). To find the first three terms of the sequence, we need to substitute the values of n = 1, 2, 3. When n = 1, the nth term is: 3 x 2^(2-1) = 3 x 2¹ = 3 x 2 = 6 When n = 2, the nth term is: 3 x 2^(2-2) = 3 x 2⁰ = 3 x 1 = 3 When n = 3, the nth term is: 3 x 2^(2-3) = 3 x 2^-1 = 3 x 1/2 = 3/2 = ³/₂ Therefore, the first three terms of the sequence are 6, 3, ³/₂. Hence, the answer is: 6, 3, ³/₂.

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To find the value of X, we can use the inverse cosine function (also known as the arccosine function) on both sides of the equation: \begin{align*} \cos x &= \frac{5}{8} \\ \Rightarrow \quad x &= \cos^{-1} \left( \frac{5}{8} \right) \end{align*} Using a calculator, we can find that: \begin{align*} x &\approx 51.32^\circ \end{align*} Therefore, the correct answer is 51.3^o.

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(a) Cumulative frequency table

The total number of chickens is \(N=200\). Plot the upper class boundaries against the cumulative frequencies, beginning with \((44.5,0)\), and join the points with a smooth curve.

(b) Interquartile range

\[Q_1=\frac{N}{4}=\frac{200}{4}=50,\qquad Q_3=\frac{3N}{4}=\frac{3(200)}{4}=150.\]

From the cumulative frequency curve, the corresponding egg numbers are approximately \(Q_1=54\) and \(Q_3=63\).

\[\text{Interquartile range}=Q_3-Q_1=63-54=9\text{ eggs}.\]

(c) Chickens laying at least \(60\) eggs per year:

\[52+28+10=90.\]

\[P(\text{at least }60\text{ eggs})=\frac{90}{200}=\frac{9}{20}=0.45.\]

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(a) \(2Ax - Kx^{2}\)

When \(x = 1\),

\(2A(1) - K(1^{2}) = 7\)

\(2A - K = 7 \quad \ldots (1)\)

When \(x = 2\),

\(2A(2) - K(2^{2}) = 4\)

\(4A - 4K = 4\)

\(A - K = 1 \quad \ldots (2)\)

From (2),

\(A = 1 + K\)

Substitute into (1):

\(2(1 + K) - K = 7\)

\(2 + 2K - K = 7\)

\(K = 5\)

\(A = 1 + 5 = 6\)

\(\therefore A = 6,\quad K = 5\)

(b) \(x^{2} - 3x - 1 = 0\)

\(a = 1,\quad b = -3,\quad c = -1\)

Using the quadratic formula,

\[x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\]

\[x = \frac{-(-3) \pm \sqrt{(-3)^{2} - 4(1)(-1)}}{2(1)}\]

\[x = \frac{3 \pm \sqrt{9 + 4}}{2}\]

\[x = \frac{3 \pm \sqrt{13}}{2}\]

\[x = \frac{3 \pm 3.606}{2}\]

\[x = \frac{3 + 3.606}{2} = 3.303 \approx 3.3\]

or

\[x = \frac{3 - 3.606}{2} = -0.303 \approx -0.3\]

\(\therefore x = 3.3\) or \(x = -0.3\), correct to 1 decimal place.

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(a) Construction of \(\triangle QRT\)

1. Draw \(QR=8\text{ cm}\).
2. With centre \(Q\) and radius \(4.5\text{ cm}\), draw an arc.
3. With centre \(R\) and radius \(6\text{ cm}\), draw an arc cutting the first arc at \(T\).
4. Join \(QT\) and \(RT\).

(b) Construction of \(QRSP\)

1. Produce \(QT\) through \(T\), and mark \(P\) on it such that \(QP=9\text{ cm}\).
2. Through \(R\), construct a line parallel to \(QP\). Mark \(S\) on this line so that \(RS=4.5\text{ cm}\).
3. Join \(P\) to \(S\). Drop \(RK\perp PQ\), where \(K\) lies on \(PQ\).

(i) From the construction, \[|PS|=6.0\text{ cm}.\]

(ii) The perpendicular distance between \(RS\) and \(PQ\) is \(RK\): \[RK=5.9\text{ cm}.\]

(iii) \(PQ\parallel RS\), while \(PQ=9\text{ cm}\) and \(RS=4.5\text{ cm}\). Thus, \(QRSP\) is a trapezium.

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(a)(i) \(35 = 5\times7\), so by the addition law of logarithms:

\[\log_{10}35 = \log_{10}5 + \log_{10}7 = 0.6990 + 0.8451 = 1.5441.\]

(a)(ii) \(2.8 = \dfrac{28}{10} = \dfrac{4\times7}{10} = \dfrac{2^2\times7}{10}\):

\[\log_{10}2.8 = 2\log_{10}2 + \log_{10}7 - \log_{10}10 = 2(0.3010) + 0.8451 - 1 = 0.4471.\]

(b) Take logarithms of \(N^{0.8942} = 2.8\):

\[0.8942\log_{10}N = \log_{10}2.8 = 0.4471 \Rightarrow \log_{10}N = \frac{0.4471}{0.8942} = 0.5.\]

\[N = 10^{0.5} = \sqrt{10} = 3.16.\]

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(a) Common ratio. For a GP, \(T_3 = ar^2 = 360\) and \(T_6 = ar^5 = 1215\). Dividing:

\[\frac{ar^5}{ar^2} = r^3 = \frac{1215}{360} = \frac{27}{8}\Rightarrow r = \sqrt[3]{\tfrac{27}{8}} = \tfrac{3}{2}.\]

(b) First term.

\[ar^2 = 360 \Rightarrow a\left(\tfrac{3}{2}\right)^2 = 360 \Rightarrow a\times\tfrac{9}{4} = 360 \Rightarrow a = 160.\]

(c) Sum of the first four terms. The terms are \(160, 240, 360, 540\):

\[S_4 = \frac{a(r^4 - 1)}{r - 1} = \frac{160\left(\left(\tfrac{3}{2}\right)^4 - 1\right)}{\tfrac{3}{2} - 1} = \frac{160(5.0625 - 1)}{0.5} = 1300.\]

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(a) For \(y=5-7x-6x^2\), the completed table is:

\(x\)	\(-3\)	\(-2\)	\(-1\)	\(-0.5\)	\(0\)	\(1\)	\(2\)
\(y=5-7x-6x^2\)	\(-28\)	\(-5\)	\(6\)	\(7\)	\(5\)	\(-8\)	\(-33\)

(b) The required graph of \(y=5-7x-6x^2\) and the line \(y=3\) are shown below. The plotted points from the table lie on the smooth curve.

(c)(i) The roots of \(2-7x-6x^2=0\) are obtained by writing

\[2-7x-6x^2=0\iff 5-7x-6x^2=3.\]

Thus, they are the \(x\)-coordinates where the curve meets \(y=3\). Reading from the graph gives

\[\boxed{x\approx-1.4\text{ or }x\approx0.2}.\]

(c)(ii) The highest point of the curve is approximately \(7\) units. Hence,

\[\boxed{\text{Maximum value of }y=7\text{ (approximately)}}.\]

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(a) Venn diagram

First, write the sets in roster form:

\[A=\{0,1,2,3,4,5,6,7\},\quad B=\{4,6,8,10,12\},\quad C=\{2,3,5,7\}.\]

(b)(i)

\[B\cup C=\{2,3,4,5,6,7,8,10,12\}.\]

Therefore, relative to \(U=\{0,1,2,\ldots,12\}\),

\[(B\cup C)'=\{0,1,9,11\}.\]

(b)(ii)

\[A'=\{8,9,10,11,12\}.\]

Since \(B\cap C=\varnothing\),

\[A'\cap B\cap C=\varnothing.\]

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(a) One number is chosen from \(\{2,3,4\}\) and one from \(\{1,3,5\}\). There are \(3\times 3 = 9\) equally likely pairs. We need the sum \(S\) to satisfy \(3 < S < 7\), i.e. \(S \in \{4,5,6\}\).

The sums strictly between \(3\) and \(7\) are: \((2,3)=5,\;(3,1)=4,\;(3,3)=6,\;(4,1)=5\) - that is \(4\) favourable outcomes.

(b) \(ABCD\) is a circle. \(DAE\) and \(CBE\) are straight lines meeting at the outer point \(E\) (with \(\angle E = m\)); \(ABF\) and \(DCF\) are straight lines meeting at the outer point \(F\) (with \(\angle F = y\)). Let the four arcs be \(AB = p,\; BC = q,\; CD = r,\; DA = s\), so \(p+q+r+s = 360^\circ\).

Angle at each external point = half the difference of the intercepted arcs.

Given \(y + m = 90^\circ\):

Combine with \(p+q+r+s = 360^\circ\). Adding the two equations gives \(2(r+s) = 540^\circ\), so

At vertex \(B\) the straight lines \(CBE\) and \(ABF\) cross, and \(x\) is the angle \(\angle ABE = \angle FBC\) (vertically opposite), which is the supplement of the interior angle \(\angle ABC\). Now \(\angle ABC\) is inscribed on arc \(ADC\):

Therefore

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(a) Dimensions. Let the shorter side be \(x\) m; the longer side is \(x + 1.5\) m. The area is 13.5 m²:

\[x(x + 1.5) = 13.5 \Rightarrow x^2 + 1.5x - 13.5 = 0.\]

Multiply through by 2: \(2x^2 + 3x - 27 = 0\), which factorises as \((2x + 9)(x - 3) = 0\).

Taking the positive root, \(x = 3\). So the dimensions are

\[3\text{ m by }4.5\text{ m}.\]

(b) Area that can be carpeted. At N250.00 per square metre with N2,000.00 available:

\[\text{Area} = \frac{2000}{250} = 8\text{ m}^2.\]

So 8 m² of the floor can be covered (the floor is 13.5 m², so it cannot all be carpeted).

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Place \(X\) at the origin. \(Y\) is due east, and \(W\) is \(180\,\text{m}\) from \(X\) on bearing \(135^\circ\) (South-East), giving \(W = (180\sin135^\circ,\ 180\cos135^\circ) = (127.28,\ -127.28)\).

(a)(i) Finding \(|XY|\): \(Y = (y, 0)\). From \(W\) the bearing of \(Y\) is \(050^\circ\), so \[\tan50^\circ = \frac{y - 127.28}{0 - (-127.28)} = \frac{y - 127.28}{127.28}.\] \[y - 127.28 = 127.28\tan50^\circ = 151.68 \Rightarrow y = 278.96.\] Therefore \(|XY| \approx \mathbf{279\,\text{m}}\).

(a)(ii) Finding \(|ZW|\): \(Z\) lies on bearing \(046^\circ\) from \(X\): \(Z = t(\sin46^\circ,\cos46^\circ)\). From \(W\), \(Z\) is on bearing \(337^\circ\): \(Z = W + s(\sin337^\circ,\cos337^\circ)\). Solving \[0.7193t + 0.3907s = 127.28,\qquad 0.6947t - 0.9205s = -127.28,\] gives \(t = |XZ| = 72.2\) and \(s = |ZW| \approx \mathbf{193\,\text{m}}\), with \(Z = (51.95,\ 50.17)\).

(b) Bearing of \(Z\) from \(N\): \(N\) is on \(XY\) (the east line) with \(|XZ| = |ZN|\). Since \(Z = (51.95, 50.17)\) and \(|XZ| = 72.2\), \[(n - 51.95)^2 + 50.17^2 = 72.2^2 \Rightarrow n = 103.9\ (\text{taking } n \neq 0).\] So \(N = (103.9, 0)\) and \(\vec{NZ} = (-51.95,\ 50.17)\), which points North-West. \[\text{angle west of north} = \tan^{-1}\!\frac{51.95}{50.17} = 46^\circ \Rightarrow \text{bearing} = 360^\circ - 46^\circ = \mathbf{314^\circ}.\]

Maelezo ya Majibu

(a)(i) Final sales price. Marked price for a 30% profit:

\[\text{MP} = 6000\times1.30 = \text{N}7800.\]

Reduced by 10% in the sales:

\[\text{Sales price} = 7800\times0.90 = \text{N}7020.\]

(a)(ii) Percentage profit.

\[\%\text{ profit} = \frac{7020 - 6000}{6000}\times100 = \frac{1020}{6000}\times100 = 17\%.\]

(b) Let \(u = 2^x\). Then \(2^{(2x+1)} = 2\cdot2^{2x} = 2u^2\), so the equation becomes

\[2u^2 - 9u + 4 = 0 \Rightarrow (2u - 1)(u - 4) = 0.\]

\[u = 4\ \Rightarrow\ 2^x = 4\ \Rightarrow\ x = 2;\qquad u = \tfrac{1}{2}\ \Rightarrow\ 2^x = 2^{-1}\ \Rightarrow\ x = -1.\]

Answer: \(x = 2\) or \(x = -1\).

Maelezo ya Majibu

From the diagram, the pulley has centre O and radius \(r = 10\text{ cm}\). The string runs \(A \to S\) (tangent on the left), over the arc \(S \to R \to T\) at the top, then \(T \to B\) (tangent on the right). \(AMB\) is horizontal, tangent to the pulley at M, with \(\angle SAB = 90^\circ\) and \(\angle TBA = 60^\circ\).

(a)(i) Obtuse angle SOT

From A, the tangents \(AS\) and \(AM\) meet the radii at right angles, so in quadrilateral \(ASOM\):

From B, the tangents \(BT\) and \(BM\) give, in quadrilateral \(BTOM\):

Going from S round the bottom through M to T gives \(90^\circ + 120^\circ = 210^\circ\). The angle on the top (through R), the obtuse \(\angle SOT\), is

(a)(ii) Arc SRT

(a)(iii) |BT|

In right triangle \(OTB\), \(OT = 10\text{ cm}\), \(\angle OTB = 90^\circ\), and \(OB\) bisects \(\angle TBM\) so \(\angle TBO = 30^\circ\):

(b) Length of the string

First find \(AS\). In right triangle \(OSA\), \(OS = 10\text{ cm}\), \(\angle OSA = 90^\circ\), and \(OA\) bisects \(\angle SAM\) so \(\angle SAO = 45^\circ\):

The string length is

Length of the string \(\approx 54\text{ cm}\).

Summary: \(\angle SOT = 150^\circ\), arc \(SRT \approx 26\text{ cm}\), \(|BT| = 10\sqrt{3}\approx 17.32\text{ cm}\), string \(\approx 54\text{ cm}\).