WAEC SSCE - General Mathematics - 1995

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To solve this inequality, we need to isolate x on one side of the inequality sign. We can start by multiplying both sides by 3, which gives: 2x - 1 < 15 Next, we can add 1 to both sides: 2x < 16 Finally, we can divide both sides by 2 to get: x < 8 Therefore, the solution to the inequality is x < 8. Option (d) is the correct answer.

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In the given diagram, we can see that the two lines AB and DC are parallel. Hence, we can apply the intercept theorem or Thales' theorem which states that if a line is drawn parallel to one side of a triangle to intersect the other two sides, then it divides those sides proportionally. So, we can use this theorem to find the length of |EC| as follows: First, we can notice that the triangles ABE and DCE are similar since they are both right triangles and share the same angle at E. Therefore, we can set up a proportion between their corresponding sides: \begin{align*} \frac{|EC|}{|DC|} &= \frac{|AE|}{|AB|} \\ \frac{|EC|}{9\text{ cm}} &= \frac{8\text{ cm}}{12\text{ cm}} \\ |EC| &= \frac{8\text{ cm}}{12\text{ cm}} \times 9\text{ cm} \\ |EC| &= 6\text{ cm} \end{align*} Therefore, the length of |EC| is 6cm. Thus, the correct option is (E) 6cm.

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To find (101₂)², we need to square the binary number 101₂. We can use the standard multiplication method taught in primary school, but we only need to remember two things: 1) 0 x anything = 0, and 2) 1 x anything = the same anything. Here's the calculation:

  101   (this is 1012)
x 101   (this is also 1012)
-----
 101
101
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11001   (this is the result of (1012)2 in binary)

Therefore, the answer is 11001, which is (25₁₀)₂.

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The total surface area of a closed cylinder consists of the area of its curved surface and the area of its two circular ends. The area of the curved surface of a cylinder is given by 2πrh, where r is the radius of the base and h is the height. The area of each circular end is given by πr^2. In this case, the radius of the cylinder is 7cm and the height is 5cm. So, Area of curved surface = 2πrh = 2 × (22/7) × 7 × 5 = 220cm^2 Area of each circular end = πr^2 = (22/7) × 7^2 = 154cm^2 Total surface area = 2 × area of circular end + area of curved surface = 2 × 154 + 220 = 308 + 220 = 528cm^2 Therefore, the total surface area of the cylinder is 528cm^2. So, the correct option is (D) 528cm^2.

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If the dealer makes a 50% profit, it means that the selling price is 150% (100% cost price + 50% profit) of the cost price. Let the cost price of the crates of soft drinks be x. Then, 150% of x is equal to N600.00. Mathematically, we can write it as: 150/100 * x = N600.00 Simplifying the equation, we have: x = N600.00 * 100/150 x = N400.00 Therefore, the dealer paid N400.00 for the crates of soft drinks.

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To solve the simultaneous equations x + y = 3/2 and x - y = 5/2, we can use the method of elimination. First, we need to eliminate y. We can do this by adding the two equations: (x + y) + (x - y) = 3/2 + 5/2 Simplifying, we get: 2x = 4 Dividing both sides by 2, we get: x = 2 Now that we have solved for x, we can substitute this value into one of the original equations to solve for y. Let's use the first equation: 2 + y = 3/2 Subtracting 2 from both sides, we get: y = -1/2 So the solution to the simultaneous equations is x = 2 and y = -1/2. To find the value of 2y + x, we can simply substitute the values we found: 2y + x = 2(-1/2) + 2 = 1 Therefore, the answer is 1.

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An exterior angle of a polygon is an angle formed by extending one of the sides of the polygon. For a regular polygon, all the exterior angles have the same measure, which can be found by dividing the total sum of the exterior angles (which is always 360 degrees) by the number of sides. In this case, we have a list of angles and we need to determine which one is an exterior angle of a regular polygon. We can start by dividing 360 degrees by each of the answer choices and see if any of the results match the definition of an exterior angle. - 360/95 = 3.789, which is not a whole number and therefore cannot be an exterior angle of a regular polygon. - 360/85 = 4.235, which is not a whole number and therefore cannot be an exterior angle of a regular polygon. - 360/78 = 4.615, which is not a whole number and therefore cannot be an exterior angle of a regular polygon. - 360/75 = 4.8, which is a whole number and therefore could be an exterior angle of a regular polygon. We can confirm this by checking if the sum of this angle and any of the interior angles of a regular polygon add up to 180 degrees. For example, a regular pentagon has interior angles of 108 degrees, and 75 + 108 = 183, which is not 180 degrees. Therefore, 75 degrees is not an exterior angle of a regular pentagon. - 360/72 = 5, which is a whole number and therefore could be an exterior angle of a regular polygon. Again, we can confirm this by checking if the sum of this angle and any of the interior angles of a regular polygon add up to 180 degrees. For example, a regular pentagon has interior angles of 108 degrees, and 72 + 108 = 180 degrees. Therefore, 72 degrees is an exterior angle of a regular pentagon. Therefore, the answer is: 72^o.

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Given expression is x(a - c) + y(c - a). Rearranging this expression, we get: x(a - c) - y(a - c) We can factorize (a - c) from both terms to get: (a - c)(x - y) Therefore, the factorization of the expression x(a - c) + y(c - a) is (a - c)(x - y). The correct option is (b) (a - c)(x - y).

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To find the mean of the remaining observations, we need to subtract the largest value from the sum of all the observations and then divide it by the number of observations minus one. Number of observations = 20 Mean of 20 observations = 4 Largest value = 23 Sum of all observations = Mean x Number of observations = 4 x 20 = 80 Sum of remaining observations = Sum of all observations - Largest value = 80 - 23 = 57 Number of remaining observations = Number of observations - 1 = 20 - 1 = 19 Mean of remaining observations = Sum of remaining observations / Number of remaining observations = 57 / 19 = 3 Therefore, the mean of the remaining observations is 3. Hence, the correct option is (b) 3.

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When two dice are tossed together, there are 6 x 6 = 36 possible outcomes. We need to find the probability that the sum of the outcome is at least 10. The only possible pairs of outcomes that sum to at least 10 are: (4, 6), (5, 5), (6, 4). Each of these pairs can occur in two different ways: (4, 6) and (6, 4), (5, 5), and (6, 4). So, there are a total of 6 possible outcomes that sum to at least 10. Therefore, the probability of getting a sum of at least 10 is 6/36, which simplifies to 1/6. Thus, the correct option is (c) 1/6.

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Latitude is a measure of the north-south position of a point on the Earth's surface, measured in degrees from the Equator. In this case, country P is located at 15°N and country Q is located at 65°N. To find the difference in latitude between the two countries, we simply subtract the latitude of P from the latitude of Q: Difference in latitude = 65°N - 15°N = 50°N Therefore, the answer is 50°.

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To find the median, we first need to arrange the numbers in order from smallest to largest: 2.35, 2.50, 2.64, 2.72, 2.91 Now, we can see that the middle number is 2.64, which is the median. Therefore, the answer is: 2.64.

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The given equation is a quadratic equation in standard form, which is \(ax^2 + bx + c = 0\), where \(a = 1\), \(b = -3\), and \(c = 2\). We can solve for the roots of the quadratic equation by using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Substituting the values of \(a\), \(b\), and \(c\) into the formula, we get: \[x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(2)}}{2(1)}\] \[x = \frac{3 \pm \sqrt{1}}{2}\] Thus, the solutions are: \[x_1 = \frac{3 - 1}{2} = 1 \text{ and } x_2 = \frac{3 + 1}{2} = 2\] Therefore, the smaller value of \(x\) is \(1\). Hence, the answer is 1.

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In a rhombus, the diagonals are perpendicular bisectors of each other and they divide the rhombus into four congruent right-angled triangles. Therefore, each triangle has a base of 6cm and a height of 8cm (half the length of the diagonals). The area of each triangle is given by: Area = 1/2 x base x height = 1/2 x 6cm x 8cm = 24cm^2 Since there are four such triangles in the rhombus, the total area of the rhombus is given by: Area = 4 x 24cm^2 = 96cm^2 Therefore, the area of the rhombus is 96cm^2, which corresponds to the last option.

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There are a total of 2 + 6 + 5 = 13 balls in the box. Out of these, 5 are black. The probability of selecting a black ball is the number of black balls divided by the total number of balls, which is 5/13. Therefore, the answer is 5/13.

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We can use trigonometry to solve for y. Since we know that ?QPR = 30°, then we can determine that: sin 30° = y/PQ Solving for y: y = PQ * sin 30° y = 10 * (1/2) y = 5 Therefore, y is 5 cm.

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In set theory, the union of two sets is a set containing all distinct elements from both sets. Therefore, to find R∪S, we need to combine all the elements in R and S without duplication. R = [2, 4, 6, 7] and S = [1, 2, 4, 8], so combining them without duplication gives us: R∪S = [1, 2, 4, 6, 7, 8] Therefore, the correct answer is (a) [1, 2, 4, 6, 7, 8].

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To express a number in standard form, we write it in the form \(a \times 10^b\), where \(1 \leq a < 10\) and \(b\) is an integer. To express 0.000834 in standard form, we need to move the decimal point to the right until we obtain a number between 1 and 10. We have: $$0.000834 = 8.34 \times 10^{-4}$$ Therefore, the answer is (a) 8.34 x 10^-4.

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Since AB is parallel to CD, then we have: \begin{align*} \angle AEC &= \angle AED + \angle DEC \\ &= \frac{1}{2} \angle BAC + \frac{1}{2} \angle ACD \\ &= \frac{1}{2}(\angle BAC + \angle ACD) \\ &= \frac{1}{2}(180^\circ) \qquad \text{(because } \angle BAC + \angle ACD = 180^\circ)\\ &= 90^\circ \end{align*} Therefore, the value of $\angle AEC$ is $\boxed{90^\circ}$.

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The given sequence is an arithmetic progression (AP) since each term is obtained by subtracting 7 from the previous term. Let the first term be a = 11, and the common difference be d = -7. The nth term of an arithmetic progression can be found using the formula: Un = a + (n - 1)d Substituting the values of a and d, we get: Un = 11 + (n - 1)(-7) Simplifying, we get: Un = 11 - 7n + 7 Un = 18 - 7n Therefore, the nth term Un of the given AP is 18 - 7n. The answer is (d) Un= 18-7n.

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The given equation is a quadratic equation which can be solved by factoring the left-hand side of the equation: (x + 2)(x - 7) = 0 The product of two factors is equal to zero if and only if at least one of the factors is zero. So we can set each factor to zero and solve for x: x + 2 = 0 or x - 7 = 0 Solving for x in each case gives: x = -2 or x = 7 Therefore, the solutions to the equation (x + 2)(x - 7) = 0 are x = -2 and x = 7.

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The volume of a cylinder can be calculated using the formula: V = πr²h, where V is the volume, r is the radius, and h is the height. Substituting the given values, we have: V = (22/7) x 7² x 5 V = 770 cm³ Therefore, the volume of the container is 770cm³. The answer is option (E).

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Since angles b and c are alternate angles, they are equal. Therefore, b + c = 2b or 2c. But since a straight line forms a straight angle of 180 degrees, we have that angle a + angle b + angle c = 180 degrees. Solving for b + c, we get: b + c = 180 - a Therefore, the answer is (A) 180^o.

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The volume of the water tank can be calculated as follows: Volume of tank = Base area x Height = (1.5m x 1.5m) x 0.5m = 1.125 m\(^3\) Since 1000 litres = 1 m\(^3\), the volume of the tank is equivalent to 1,125 litres. Therefore, the amount of water left in the water tanker after filling the tank is: 1500 - 1125 = 375 litres. Hence, the answer is 375 litres.

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We can start by using the fact that the angle at the center of the circle is twice the angle at the circumference that subtends the same arc. Therefore, since |BD| = |DC|, then angle CBD = angle BDC. Also, angle COB is equal to 2*angle BDC because it subtends the same arc as angle BDC. Therefore, angle COB = 2*angle CBD. Since triangle AOB is isosceles (|OA| = |OB|), then angle BAO = angle BOA. Using the fact that the angles in a triangle sum to 180 degrees, we can write: angle BAO + angle BOA + angle COB = 180 Substituting in the expressions we derived earlier, we get: angle BAO + angle BAO + 2*angle CBD = 180 Simplifying, we get: 2*angle BAO + 2*angle CBD = 180 angle BAO + angle CBD = 90 Now we can use the fact that angle DCB = 35 degrees and |BD| = |DC| to find angle CBD: angle DCB + angle BDC + angle CBD = 180 35 + 35 + angle CBD = 180 angle CBD = 110 Substituting this value back into the earlier equation, we get: angle BAO + 110 = 90 angle BAO = 90 - 110 = -20 Since angle BAO cannot be negative, we made an error along the way. Double-checking our work, we realize that we made a mistake when substituting in the value for angle COB. It should be equal to 2*angle DCB, not 2*angle BDC. Therefore: angle COB = 2*angle DCB = 2*35 = 70 Substituting this value back into the earlier equation, we get: angle BAO + 70 = 90 angle BAO = 20 Therefore, the answer is 20 degrees.

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The given expression is undefined when the denominator is equal to zero because division by zero is undefined. Therefore, we need to find the values of x that make the denominator zero. The denominator of the expression is \(x^2 + 4x - 5\). We can factor this quadratic expression as \((x + 5)(x - 1)\). So the expression is undefined when either \(x + 5\) or \(x - 1\) equals zero, since division by zero is undefined. Therefore, the values of x that make the expression undefined are \(x = -5\) or \(x = 1\). So the answer is: - -5 or 1

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The locus of a point which is equidistant from two given fixed points is the perpendicular bisector of the straight line joining them. This means that the locus of the point is the line that is perpendicular to the straight line joining the two given points and passes through the midpoint of that line.

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The mode is the number that appears most frequently in the given set of numbers. In this case, we can see that the number 8 appears 4 times, while the other numbers appear either 2 or 1 times. Therefore, the mode of the set is 8. So, the answer is 8.

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A sector of a circle is a region bounded by two radii and an arc. The perimeter of a sector is the sum of the arc length and the lengths of the two radii. In this case, we are given that the angle of the sector is 108° and the radius of the circle is 31/2 cm. To find the arc length, we need to know the circumference of the entire circle, which is given by 2πr, where r is the radius of the circle. So, the circumference of the circle is 2π(31/2) cm = 31π cm. The angle of the sector is 108°, which is 108/360 or 3/10 of the entire circle. So, the arc length of the sector is (3/10) × 31π cm = 9.3π cm. The two radii of the sector have length 31/2 cm each. Therefore, the perimeter of the sector is the sum of the arc length and the two radii, which is 9.3π cm + 31/2 cm + 31/2 cm = (9.3π + 31) / 2 cm. This is approximately equal to 13.6 cm. Therefore, the answer is (E) 13 3/5 cm.

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If we have that y is proportional to \(\frac{1}{x^2}\), then we can write: y = k\(\frac{1}{x^2}\) where k is a constant of proportionality. To find k, we can use the fact that when x = 3, y = 4: 4 = k\(\frac{1}{3^2}\) Simplifying: k = 36 Now we can use k to find y when x = 2: y = k\(\frac{1}{2^2}\) y = 9 Therefore, the answer is y = 9.

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To solve this problem, we need to first add up the parts of the ratio, which gives us 6+7+8 = 21. Next, we divide the total amount shared (N10.50) by the sum of the ratio parts (21) to find the value of one part: N10.50 ÷ 21 = N0.50 Now that we know the value of one part, we can find the value of each part of the ratio by multiplying the value of one part by the corresponding number in the ratio: - 6 parts: 6 × N0.50 = N3.00 - 7 parts: 7 × N0.50 = N3.50 - 8 parts: 8 × N0.50 = N4.00 So the largest share is N4.00, which corresponds to the 8 parts of the ratio. Therefore, the correct answer is option (C) N4.00.

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(a) For each value of x, evaluate \(y=2\cos 2x-1\).

\(x\)	\(0^\circ\)	\(30^\circ\)	\(60^\circ\)	\(90^\circ\)	\(120^\circ\)	\(150^\circ\)	\(180^\circ\)
\(y=2\cos2x-1\)	1.0	0	-2.0	-3.0	-2.0	0	1.0

(b) and (c) The graphs of \(y=2\cos2x-1\) and \(y=\dfrac{1}{180}(x-360)\), drawn on the same axes using the stated scales, are shown below.

For the straight line, the plotting values are:

\(x\)	\(0^\circ\)	\(30^\circ\)	\(60^\circ\)	\(90^\circ\)	\(120^\circ\)	\(150^\circ\)	\(180^\circ\)
\(y=\dfrac{x-360}{180}\)	-2.00	-1.83	-1.67	-1.50	-1.33	-1.17	-1.00

(d)(i) \[2\cos2x+\frac12=0\] \[2\cos2x-1=-\frac32=-1.5\] Hence, reading the points where the cosine curve has \(y=-1.5\), \[\boxed{x\approx51^\circ\text{ or }129^\circ.}\]

(d)(ii) \[2\cos2x-\frac{x}{180}+1=0\] \[2\cos2x-1=\frac{x}{180}-2=\frac{1}{180}(x-360).\] Thus, the required roots are the \(x\)-coordinates of the intersections of the curve and the straight line: \[\boxed{x\approx54^\circ\text{ or }132^\circ.}\]

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(a) Finding /PZ/

The diagram shows the parallelograms PQSR and SRYZ sharing the common side SR, with the points P, Q, Y, Z lying on one straight line in that order.

In a parallelogram opposite sides are equal, so:

• In parallelogram PQSR, side \(PQ\) is opposite side \(SR\), hence \(PQ = SR = 3\text{ cm}\).
• In parallelogram SRYZ, side \(ZY\) is opposite side \(SR\), hence \(ZY = SR = 3\text{ cm}\).

Since \(P, Q, Y, Z\) are collinear in that order:

\[ PZ = PQ + QY + YZ \]

\[ PZ = 3 + 2 + 3 = 8\text{ cm} \]

/PZ/ = 8 cm.

(b) Distance PQ along the parallel of latitude 56\(^{\circ}\)N

Both towns lie on latitude \(56^{\circ}\)N. Their longitudes are \(25^{\circ}\)E and \(95^{\circ}\)E, so the difference in longitude is:

\[ \theta = 95^{\circ} - 25^{\circ} = 70^{\circ} \]

The radius of the parallel of latitude is \(r = R\cos 56^{\circ}\), and the required distance is an arc of that parallel:

\[ PQ = \frac{\theta}{360^{\circ}} \times 2\pi R\cos 56^{\circ} \]

Substituting \(R = 6400\text{ km}\), \(\pi = \frac{22}{7}\) and \(\cos 56^{\circ} = 0.5592\):

\[ PQ = \frac{70}{360} \times 2 \times \frac{22}{7} \times 6400 \times 0.5592 \]

\[ PQ = \frac{70}{360} \times 40228.57 \times 0.5592 \]

\[ PQ = 0.19444 \times 40228.57 \times 0.5592 \approx 4374\text{ km} \]

Distance PQ \(\approx\) 4374 km (to the nearest km).

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(a) "Partly constant and partly varies inversely as the square of \(x\)" means \[y = a + \frac{b}{x^2},\] where \(a\) and \(b\) are constants.

(b) Using the given values:

When \(x = 1, y = 11\): \(a + b = 11.\quad(1)\)

When \(x = 2, y = 5\): \(a + \dfrac{b}{4} = 5.\quad(2)\)

Subtract \((2)\) from \((1)\): \(b - \dfrac{b}{4} = 6 \Rightarrow \dfrac{3b}{4} = 6 \Rightarrow b = 8.\) Then \(a = 11 - 8 = 3.\)

So \(y = 3 + \dfrac{8}{x^2}.\) When \(x = 4\): \[y = 3 + \frac{8}{16} = 3 + 0.5 = \mathbf{3.5}.\]

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(a) A pack has 4 fives and 2 red nines. Drawing a five and drawing a red nine are mutually exclusive events (a five cannot also be a red nine), so the probabilities add.

\(P(\text{five or red nine}) = \frac{4}{52} + \frac{2}{52} = \frac{6}{52} = \dfrac{3}{26}\)

(b) Take P as origin and measure position as (East, North). A bearing \(\theta\) gives the direction \((\sin\theta, \cos\theta)\).

• \(Q = 5(\sin150°,\ \cos150°) = (2.500,\ -4.330)\)
• \(\vec{QR} = 3(\sin060°,\ \cos060°) = (2.598,\ 1.500)\)
• \(R = (2.500+2.598,\ -4.330+1.500) = (5.098,\ -2.830)\)

Relative to P, R lies to the East and to the South, so the bearing is between 090° and 180°.

\(\text{Bearing} = 180° - \tan^{-1}\!\left(\dfrac{5.098}{2.830}\right) = 180° - 60.96° = 119.04°\)

Bearing of R from P \(\approx \mathbf{119°}\) (to the nearest degree).

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(a) \(p = x + ym^3\). Make \(m\) the subject.

\(ym^3 = p - x \ \Rightarrow\ m^3 = \dfrac{p-x}{y} \ \Rightarrow\ \mathbf{m = \sqrt[3]{\dfrac{p-x}{y}}}\)

(b) \(x^2 - 6x + 7 = 0\) by completing the square.

\(x^2 - 6x = -7\)

Add \(\left(\tfrac{6}{2}\right)^2 = 9\) to both sides:

\((x-3)^2 = -7 + 9 = 2\)

\(x - 3 = \pm\sqrt{2} = \pm 1.414\)

\(x = 3 + 1.414 = 4.4\quad\text{or}\quad x = 3 - 1.414 = 1.6\) (1 d.p.)

(c) Let the consecutive odd numbers be \(n\) and \(n+2\).

\(n(n+2) = 195 \ \Rightarrow\ n^2 + 2n - 195 = 0\)

\((n+15)(n-13) = 0 \ \Rightarrow\ n = -15\ \text{or}\ n = 13\)

Since the numbers are positive, \(n = 13\). The numbers are \(\mathbf{13}\) and \(\mathbf{15}\).

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(a) Grouping the four terms \(px - 2qx - 4qy + 2py\) in pairs: \[(px - 2qx) + (2py - 4qy) = x(p - 2q) + 2y(p - 2q) = (p - 2q)(x + 2y).\] So the expression factorises as \(\mathbf{(p - 2q)(x + 2y)}.\)

(b) \(U = \{1,2,3,4,5,6,7,8,9,10\},\ P = \{1,2,4,6,10\},\ Q = \{2,3,6,9\}.\)

Left side: \(P \cup Q = \{1,2,3,4,6,9,10\}\), so \[(P \cup Q)' = \{5, 7, 8\}.\]

Right side: \(P' = \{3,5,7,8,9\}\) and \(Q' = \{1,4,5,7,8,10\}\), so \[P' \cap Q' = \{5, 7, 8\}.\]

Since \((P \cup Q)' = \{5,7,8\} = P' \cap Q'\), the identity is verified (De Morgan's law).

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Total number of students: \(2+3+4+4+4+4+5+3+2+1 = 32\).

(a)(i) Mean. \(\sum fx = 167\).

\[\bar{x}=\frac{167}{32}\approx 5.22.\]

(a)(ii) Median. With \(N=32\), the median is the mean of the 16th and 17th values. The cumulative frequency reaches 13 at mark 4 and 17 at mark 5, so both the 16th and 17th values are 5.

\[\text{Median}=\frac{5+5}{2}=5.\]

(a)(iii) Mode. The highest frequency (5) is at mark 7, so the mode = 7.

(b) Percentage scoring at least 8 marks. Marks 8, 9, 10 have \(3+2+1=6\) students.

\[\frac{6}{32}\times100\% = 18.75\%.\]

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Total balls \(= 5 + 3 + 2 = 10\). Since each ball is replaced, the two draws are independent and the probabilities on each draw stay the same.

\(P(\text{blue}) = \tfrac{5}{10} = \tfrac{1}{2},\quad P(\text{black}) = \tfrac{3}{10},\quad P(\text{red}) = \tfrac{2}{10} = \tfrac{1}{5}\)

(a) Two red balls:

\(P = \tfrac{1}{5}\times\tfrac{1}{5} = \dfrac{1}{25}\)

(b) Two blue OR two black (mutually exclusive, so add):

\(P = \left(\tfrac{1}{2}\right)^2 + \left(\tfrac{3}{10}\right)^2 = \tfrac{1}{4} + \tfrac{9}{100} = \tfrac{25}{100} + \tfrac{9}{100} = \dfrac{34}{100} = \dfrac{17}{50}\)

(c) One black and one red in any order (black-then-red or red-then-black):

\(P = 2\times\tfrac{3}{10}\times\tfrac{2}{10} = 2\times\tfrac{6}{100} = \dfrac{12}{100} = \dfrac{3}{25}\)

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(a) Cumulative frequency table

Marks (%)	Number of pupils, \(f\)	Cumulative frequency
11 - 20	68	68
21 - 30	184	252
31 - 40	294	546
41 - 50	402	948
51 - 60	480	1428
61 - 70	310	1738
71 - 80	164	1902
81 - 90	98	2000

Using upper class boundaries, plot the cumulative frequencies and join the points with a smooth increasing curve.

(b)(i) Cut-off mark

The 300 candidates admitted are the highest scorers. Hence the number below the cut-off mark is

\[2000-300=1700.\]

From the ogive, at cumulative frequency \(1700\), the corresponding mark is approximately \(68\).

Therefore, the cut-off mark is 68 marks (68%).

(b)(ii) Probability of scoring at least \(45\%\)

From the ogive, the cumulative frequency at \(45\%\) is approximately \(720\).

\[\text{Number scoring at least }45\%=2000-720=1280.\]

\[P(\text{score at least }45\%)=\frac{1280}{2000}=0.64.\]

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(a)(i) From tables, \(\sin 63.35° = 0.8937\).

\(2\sin 63.35° = 2 \times 0.8937 = \mathbf{1.787}\)

(ii) \(\cos 44.74° = 0.7103\).

\(\log \cos 44.74° = \log 0.7103 = \bar{1}.8515 \;(= -0.1485)\)

(b) \(\log K - \log(K-2) = \log 5\)

\(\log\!\left(\dfrac{K}{K-2}\right) = \log 5 \ \Rightarrow\ \dfrac{K}{K-2} = 5\)

\(K = 5(K-2) = 5K - 10 \ \Rightarrow\ 4K = 10 \ \Rightarrow\ \mathbf{K = 2.5}\)

(c) Evaluate \(\dfrac{(3.68)^2 \times 6.705}{\sqrt{0.3581}}\) using logarithms.

\(\text{Log of answer} = 1.9580 - \bar{1}.7770 = 1.9580 + 0.2230 = 2.1810\)

Antilog \(2.1810 = \mathbf{151.7}\) (4 s.f.).

Antwoorddetails

(a) Take X as origin with (East, North) components; a bearing \(\theta\) gives direction \((\sin\theta,\cos\theta)\).

\(Y = 20(\sin060°,\cos060°) = (17.32,\ 10.00)\)

From Y on bearing \(195°\), distance \(YZ = d\): \(Z = Y + d(\sin195°,\cos195°) = (17.32 - 0.2588d,\ 10.00 - 0.9659d)\).

Z is due East of X, so its North component is \(0\):

\(10.00 - 0.9659d = 0 \ \Rightarrow\ d = \dfrac{10.00}{0.9659} = 10.35\)

(i) \(|YZ| \approx \mathbf{10.4\text{ km}}\).

East component of Z: \(17.32 - 0.2588(10.35) = 17.32 - 2.68 = 14.64\).

(ii) \(|ZX| \approx \mathbf{14.6\text{ km}}\).

(b) The route is along the same meridian (138°E) from 36°N to 36°S, an angular change of \(36° + 36° = 72°\).

(i) \(\text{Distance} = \dfrac{72}{360}\times 2\pi R = \dfrac{72}{360}\times 2\times\dfrac{22}{7}\times 6400\)

\(= 0.2 \times 40228.57 = 8045.7 \approx \mathbf{8050\text{ km}}\) (3 s.f.).

(ii) \(\text{Time} = \dfrac{8045.7}{800} = 10.06 \approx \mathbf{10\text{ hours}}\).

Antwoorddetails

Construction and measurement

1. Draw a straight line and mark a point B on it. With centre B and radius 8.1 cm, mark C on the line.
2. At B, construct angles of 90° and 120° with BC. Bisect the angle between them to obtain a ray making 105° with BC.
3. With centre B and radius 7.5 cm, cut this ray at A. Join A to C to complete triangle ABC.
4. To divide BC in the ratio \(3:2\), draw any auxiliary ray from B. Mark five equal intervals on it, join the fifth mark to C, and draw through the third mark a parallel to this joining line. Its intersection with BC is D. Thus \(BD=4.86\text{ cm}\) and \(DC=3.24\text{ cm}\).
5. Construct the perpendicular to BC through D. Let it meet AC at P.

The completed construction is shown below.

Measuring the constructed length gives:

\[\boxed{BP \approx 5.4\text{ cm}}\]

More accurately, \(BP\approx 5.39\text{ cm}\).