WAEC SSCE - General Mathematics - 2006

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To make a perfect square, we need to add the square of half of the coefficient of the x-term. In this case, the coefficient of x is -3, so we need to add \(\left(\frac{-3}{2}\right)^2 = \frac{9}{4}\) to make it a perfect square. Thus, the answer is \(\frac{9}{4}\).

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In the number 3.0492, 9 is in the thousandth place. The places in a decimal number are counted from left to right starting from the tenths place, which is immediately to the right of the decimal point. Each place value to the right of the tenths place is ten times smaller than the one to its left. So the first digit to the right of the decimal point is in the tenths place, the second digit is in the hundredths place, and the third digit is in the thousandths place, and so on. Therefore, the place value of 9 in the number 3.0492 is \(\frac{9}{1000}\).

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When a fair coin is tossed, there are two possible outcomes, heads (H) or tails (T), each with a probability of 1/2. Since the coin is tossed three times, the total number of possible outcomes is 2 × 2 × 2 = 8. We need to find the probability of getting two heads and one tail, which can happen in three different ways: HHT, HTH, or THH. Each of these outcomes has a probability of (1/2) × (1/2) × (1/2) = 1/8. Therefore, the probability of getting two heads and one tail is the sum of the probabilities of the three possible outcomes: P(HHT) + P(HTH) + P(THH) = 1/8 + 1/8 + 1/8 = 3/8. Hence, the answer is 3/8.

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Let the radius of the smaller circle be r, and the radius of the larger circle be R. The area of a circle is given by the formula A = \(\pi r^2\). Therefore, the radius of the smaller circle is \(\sqrt{\frac{49\pi}{\pi}} = 7\). Similarly, the radius of the larger circle is \(\sqrt{\frac{100\pi}{\pi}} = 10\). Since O is the centre of both circles, the distance between O and the point where the two circles intersect is R - r = 10 - 7 = 3. From the diagram, we can see that x is also equal to this distance, so x = 3. Therefore, the answer is 3.

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The circumference of the wheel is the distance it covers in one revolution. Circumference of the wheel = π × diameter Substituting the given values, Circumference of the wheel = π × 1.4m = 4.4m (rounded to one decimal place) Therefore, the distance covered in 100 revolutions is: Distance covered = Circumference of the wheel × Number of revolutions Substituting the given values, Distance covered = 4.4m × 100 = 440m Therefore, the wheel covers a distance of 440m in 100 complete revolutions. Hence, the correct option is (D) 440m.

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The formula for simple interest is: I = PRT Where: I = Interest P = Principal R = Rate of Interest T = Time From the question: R = 3% per annum T = 2.5 years I = N123 We can substitute the values given into the formula and solve for P: 123 = P x 0.03 x 2.5 123 = 0.075P P = 123 ÷ 0.075 P = N1,640 Therefore, the principal is N1,640. Answer option (C).

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When three angles at a point are added together, they sum up to 360 degrees. Therefore, we have: (3y - 45) + (y + 25) + y = 360 Simplifying the equation, we get: 5y - 20 = 360 Adding 20 to both sides, we get: 5y = 380 Dividing both sides by 5, we get: y = 76 Hence, the value of y is 76 degrees, and the correct option is (d) 76^o.

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Let's assume that the rectangular floor has a length of L meters and a width of W meters. We know that the rectangular carpet covers 5% of the rectangular floor, which means that the area of the carpet is 5% of the area of the floor. We can write this as an equation: 2.5 × 2.4 = 0.05 × L × W Multiplying the left-hand side, we get: 6 = 0.05LW Dividing both sides by 0.05, we get: 120 = LW Therefore, the area of the rectangular floor is LW, which is 120 square meters. Hence, the area of the floor is 120 m², which is the third option.

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To simplify the given expression, we first need to find a common denominator for the two fractions. In this case, the least common multiple of 4 and 6 is 12. So, we can write: \[\frac{x - 4}{4} - \frac{x - 3}{6} = \frac{3(x - 4)}{12} - \frac{2(x - 3)}{12}\] Now, we can combine the two fractions by subtracting the numerators: \[\frac{3(x - 4)}{12} - \frac{2(x - 3)}{12} = \frac{3x - 12 - 2x + 6}{12}\] Simplifying the numerator, we get: \[\frac{3x - 2x - 12 + 6}{12} = \frac{x - 6}{12}\] Therefore, the simplified expression is \(\frac{x - 6}{12}\). So, the answer is option (B).

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The volume of a cone can be calculated using the formula: V = (1/3)πr²h where V is the volume of the cone, r is the radius of the circular base, h is the height of the cone, and π is a mathematical constant approximately equal to 22/7. Given that the diameter of the circular base is 7cm, the radius r can be calculated as half of the diameter: r = 7/2 = 3.5cm. The height of the cone is given as 6cm. Substituting these values into the formula for the volume of a cone, we get: V = (1/3) x (22/7) x (3.5cm)² x 6cm Simplifying this expression, we get: V = (1/3) x (22/7) x 12.25cm² x 6cm V = (22/21) x 74.25cm³ V = 77cm³ Therefore, the volume of water the conical water-jug can hold is 77cm³. Answer option (C) is correct.

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To find the value of x for which y is undefined, we need to look for values of x that make the denominator of the expression for y equal to zero, since division by zero is undefined. The denominator of the expression for y is 4x - 3. Therefore, we need to find the value of x that makes 4x - 3 equal to zero. Solving the equation 4x - 3 = 0, we get: 4x = 3 x = 3/4 Therefore, the value of x for which y is undefined is x = 3/4. Explanation: To understand why we need to look for values of x that make the denominator of the expression for y equal to zero, we need to remember that division by zero is undefined. When we divide a number by zero, there is no answer because it is impossible to divide any number into equal parts of zero size. In the given expression for y, the denominator is 4x - 3. If we plug in x = 3/4, the denominator becomes: 4x - 3 = 4(3/4) - 3 = 3 - 3 = 0 This means that when x = 3/4, the denominator becomes zero, and therefore division by zero occurs, which makes the value of y undefined. Hence, the value of x for which y is undefined is x = 3/4.

Akọwa Nkọwa

If the seller allows a 20% discount for cash payment, then the customer will only have to pay 80% of the marked price. Let MP be the marked price and CP be the cash payment. CP = 80% of MP CP/MP = 80/100 CP/MP = 4/5 Therefore, the ratio of the cash payment to the marked price is 4:5. So the correct answer is option (D) 4:5.

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The temperature in the chemical plant was -5^oC at 2.00 am. The temperature fell by 6^oC, so the temperature became (-5 - 6) = -11^oC. The temperature then rose by 7^oC, which means the temperature increased from -11^oC to (-11 + 7) = -4^oC. Therefore, the final temperature was -4^oC. Hence, the correct option is (-4^oC).

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At the end of the first year, the value of the machine will be 90% of N20,000 = N18,000. At the end of the second year, the value of the machine will be 90% of N18,000 = N16,200. Therefore, the value of the machine at the end of two years will be N16,200. So the correct option is: - N16,200

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In a triangle, the sum of the angles is always 180 degrees. In the given diagram, we can see that the angles x, y, and z form a triangle. Therefore, we have: x + y + z = 180^o Rearranging this equation, we get: x + y = 180^o - z Hence, the value of x + y in terms of z is 180^o - z. Therefore, the answer is (a) 180^o - z.

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The ratio of boys to girls in the class is 5:3. This means that for every 5 boys, there are 3 girls in the class. Let B be the number of boys in the class and G be the number of girls in the class. Then, we can write: B:G = 5:3 We can simplify this ratio by dividing both sides by the greatest common factor of 5 and 3, which is 1. B:G = 5/1 : 3/1 B:G = 5 : 3 This means that there are 5 parts for boys and 3 parts for girls in the class. The total number of parts is 5+3 = 8. To find the probability of selecting a girl at random, we need to find the number of parts that represent the girls and divide it by the total number of parts. The number of parts that represent the girls is 3, since there are 3 parts for girls out of a total of 8 parts. Therefore, the probability of selecting a girl at random is: P(selecting a girl) = number of parts that represent the girls / total number of parts P(selecting a girl) = 3/8 Therefore, the probability of selecting a girl at random from the class is 3/8.

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Using the logarithmic property that log_a(b) - log_a(c) = log_a(b/c), we can simplify the given expression as: log_a(270/10) + log_a(1/3) = 2 Simplifying further, we get: log_a(27) + log_a(1/3) = 2 Using the logarithmic property that log_a(b) + log_a(c) = log_a(bc), we can write: log_a(27 x 1/3) = 2 log_a(9) = 2 Now, using the definition of logarithm, we can write: a^2 = 9 Taking the square root on both sides, we get: a = 3 or -3 However, a cannot be negative, as the base of a logarithm must be positive. Therefore, the value of a is 3. Hence, the value of a is 3.

Akọwa Nkọwa

Range is not a measure of central tendency. Measures of central tendency are statistical measures that represent the typical or central values of a data set. They include the mean, median, and mode. The mean is the arithmetic average of a data set, the median is the middle value when the data is arranged in order, and the mode is the most frequently occurring value in the data set. On the other hand, range is a measure of dispersion, which describes how spread out the data is. It is the difference between the highest and lowest values in the data set.

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The baker used 40% of a 50kg bag of flour, which is equivalent to: $$ 40\% \times 50kg = \frac{40}{100} \times 50kg = 20kg $$ If \(\frac{1}{8}\) of the amount used was for the cake, then the amount of flour used for the cake is: $$ \frac{1}{8} \times 20kg = \frac{20}{8} kg = 2\frac{1}{2} kg $$ Therefore, the answer is \boxed{2\frac{1}{2}}.

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To evaluate \((0.13)^3\) correct to three significant figures, we need to first cube the number and then round the answer to three significant figures. \((0.13)^3 = 0.002197\) Rounding to three significant figures, we get 0.00220. Therefore, the correct answer is 0.00220.

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We are given two equations as follows: x + y = 12 .... (Equation 1) 3x - y = 20 .... (Equation 2) We need to find the value of 2x - y. To solve the problem, we can use the method of substitution or elimination. Here, we will use the method of substitution. From Equation 1, we can write: y = 12 - x Substituting this value of y in Equation 2, we get: 3x - (12 - x) = 20 Simplifying the equation, we get: 4x - 12 = 20 Adding 12 to both sides, we get: 4x = 32 Dividing both sides by 4, we get: x = 8 Now, substituting the value of x in Equation 1, we get: y = 12 - 8 y = 4 Therefore, we have found the values of x and y as 8 and 4, respectively. Finally, substituting these values in the expression 2x - y, we get: 2x - y = 2(8) - 4 = 12 Hence, the value of 2x - y is 12.

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We can use trigonometry to solve this problem. Let x be the distance from the bottom of the ladder to the electric pole. Then, we can create a right triangle with the ladder as the hypotenuse, x as one leg, and the distance we are trying to find as the other leg. Using trigonometric functions, we know that: sin(65^o) = opposite/hypotenuse sin(65^o) = (distance up the pole)/16 (distance up the pole) = 16*sin(65^o) (distance up the pole) ≈ 14.5m Therefore, the answer is 14.5m.

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The equation 2x + y = 10 can be rearranged as 2x = 10 - y. Dividing both sides by 2 gives x = 5 - (y/2). Therefore, the value of x will depend on the value of y. Now, if we substitute each of the given values of x into the equation 2x + y = 10, we can determine which ones are not possible. For x = 4, we get 2(4) + y = 10, which simplifies to y = 2. This is a possible value of y, so x = 4 is a possible value of x. For x = 5, we get 2(5) + y = 10, which simplifies to y = 0. But the question states that y ≠ 0, so x = 5 is not a possible value of x. For x = 8, we get 2(8) + y = 10, which simplifies to y = -6. This is a possible value of y, so x = 8 is a possible value of x. For x = 10, we get 2(10) + y = 10, which simplifies to y = -10. This is also a possible value of y, so x = 10 is a possible value of x. Therefore, the answer is x = 5, which is not a possible value of x.

Akọwa Nkọwa

(a) Simultaneous equations.

\[2x-y=\frac{9}{2}\quad(1),\qquad x+4y=0\quad(2).\]

From (2), \(x=-4y\). Substitute into (1):

\[2(-4y)-y=\frac{9}{2}\]

\[-8y-y=\frac{9}{2}\]

\[-9y=\frac{9}{2}\ \Rightarrow\ y=-\frac{1}{2}.\]

Then \(x=-4y=-4\left(-\dfrac{1}{2}\right)=2\).

\[\boxed{x=2,\quad y=-\tfrac{1}{2}}.\]

(b) Tangent TA to the circle at A.

From the diagram, \(B, C, D, A\) lie on the circle, \(TA\) is the tangent at \(A\), \(\angle BCA=40^\circ\) (at \(C\)) and \(\angle DAT=52^\circ\) (between the tangent \(AT\) and chord \(AD\)).

Chord AB. By the alternate segment theorem, the tangent-chord angle between the tangent (on the side towards \(X\), opposite \(T\)) and chord \(AB\) equals the angle in the alternate segment \(\angle BCA\):

\[\angle BAX=\angle BCA=40^\circ.\]

Straight tangent line. The points \(X, A, T\) lie on the straight tangent line, so the three angles at \(A\) on one side sum to \(180^\circ\):

\[\angle BAX+\angle BAD+\angle DAT=180^\circ.\]

\[40^\circ+\angle BAD+52^\circ=180^\circ\]

\[\angle BAD=180^\circ-40^\circ-52^\circ=\boxed{88^\circ}.\]

Akọwa Nkọwa

(a)

Construct a line segment \(XY=8\text{ cm}\). At \(X\) and \(Y\), construct angles of \(45^\circ\) on the same side of \(XY\). Let the two rays meet at \(Z\). Bisect \(\angle YXZ\) and \(\angle XYZ\); their intersection is \(P\), the point equidistant from the three sides of the triangle.

With centre \(P\), draw a perpendicular to any side, say \(XY\), meeting it at \(T\). Using \(PT\) as radius, draw the circle. It touches all three sides of \(\triangle XYZ\).

Since \(\angle XZY=90^\circ\),

\[XZ=YZ=\frac{8}{\sqrt2}=4\sqrt2\text{ cm}.\]

Hence the radius of the inscribed circle is

\[r=\frac{XZ+YZ-XY}{2}=\frac{4\sqrt2+4\sqrt2-8}{2}=4\sqrt2-4\approx1.66\text{ cm}.\]

Radius of the circle \(\approx 1.7\text{ cm}\).

(b)

\[\begin{aligned}(x-5)+2x+2x+(2x+7)+2x+(2x-1)&=144\\9x+1&=144\\9x&=143\\x&=\frac{143}{9}\\x&=15\frac{8}{9}\text{ cm}.\end{aligned}\]

Akọwa Nkọwa

(a) Sample space

Let the first number in each ordered pair be the score on the first die and the second number be the score on the second die.

Hence, there are \(6\times6=36\) equally likely outcomes.

(b)(i) Probability of a total of 7 or 8

A total of \(7\) occurs in 6 outcomes:

\((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\).

A total of \(8\) occurs in 5 outcomes:

\((2,6),(3,5),(4,4),(5,3),(6,2)\).

Therefore,

\[P(7\text{ or }8)=\frac{6+5}{36}=\frac{11}{36}.\]

(b)(ii) Probability of a total less than 4

The possible totals are \(2\) and \(3\), obtained from:

\((1,1),(1,2),(2,1)\).

Thus,

\[P(\text{total less than }4)=\frac{3}{36}=\frac{1}{12}.\]

Akọwa Nkọwa

(a) For \(y=2\sin x-\cos 2x\), the completed table is:

\(x\)	\(0^\circ\)	\(30^\circ\)	\(60^\circ\)	\(90^\circ\)	\(120^\circ\)	\(150^\circ\)	\(180^\circ\)
\(y\)	\(-1.0\)	\(0.5\)	\(2.2\)	\(3.0\)	\(2.2\)	\(0.5\)	\(-1.0\)

For example,

\[y(60^\circ)=2\sin60^\circ-\cos120^\circ=2\left(\frac{\sqrt3}{2}\right)-\left(-\frac12\right)=2.232\ldots\approx2.2.\]

(b) The graph of \(y=1.25\) is the horizontal line shown on the same axes.

(c)

(i) The curve cuts the \(x\)-axis at approximately

\[\boxed{x=21^\circ\text{ and }159^\circ}.\]

(ii) The intersections of the curve with \(y=1.25\) give

\[\boxed{x=42^\circ\text{ and }138^\circ}.\]

Akọwa Nkọwa

(a) Finding |XD|. Since \(\angle CAX=\angle BDX\) (given) and \(\angle AXC=\angle DXB\) (vertically opposite), triangles \(AXC\) and \(DXB\) are similar (AA). Matching the equal angles gives

\[\frac{AX}{DX}=\frac{CX}{BX}\Rightarrow \frac{6}{XD}=\frac{3}{4}\]

\[XD=\frac{6\times4}{3}=8\text{ cm}.\]

(b) The angle at Y. The bearing of Y from X is \(312°\), so the bearing of X from Y is \(312°-180°=132°\). The bearing of Y from Z is \(022°\), so the bearing of Z from Y is \(022°+180°=202°\). Hence

\[\angle XYZ=202°-132°=70°.\]

(i) |XZ|. By the cosine rule,

\[XZ^2=XY^2+ZY^2-2\,XY\cdot ZY\cos70°=32^2+50^2-2(32)(50)\cos70°\]

\[XZ^2=3524-3200(0.3420)=2429.5\Rightarrow XZ=\sqrt{2429.5}=49.3\text{ km}.\]

(ii) Bearing of Z from X. By the sine rule,

\[\frac{\sin\angle YXZ}{ZY}=\frac{\sin70°}{XZ}\Rightarrow \sin\angle YXZ=\frac{50\sin70°}{49.3}=0.9532\]

\[\angle YXZ=72.4°.\]

From the diagram Z lies to the west of Y as seen from X, so the bearing of Z from X is measured anticlockwise from that of Y:

\[312°-72.4°=239.6°.\]

Akọwa Nkọwa

(a)

Since 12 students offered neither subject, place 12 outside the two circles. Let the number in the overlap be \(x\).

\[ (32-x)+x+(28-x)+12=45 \]
\[72-x=45\]
\[x=27\]

Thus, \(P\) only \(=32-27=5\), and \(G\) only \(=28-27=1\).

(i) The required Venn diagram is:

(ii) \[\boxed{n(P\cap G)=27}\] Therefore, 27 students offered both Physics and Government.

(iii) \[n(P\cup G)=45-12=\boxed{33}\]

(b)

\[p+q=\frac{2u}{1-u}+\frac{1+u}{1-u}=\frac{2u+1+u}{1-u}=\frac{3u+1}{1-u}\]

\[p-q=\frac{2u}{1-u}-\frac{1+u}{1-u}=\frac{2u-1-u}{1-u}=\frac{u-1}{1-u}=-1\]

Hence,

\[\frac{p+q}{p-q}=\frac{\frac{3u+1}{1-u}}{-1}=-\frac{3u+1}{1-u}=\boxed{\frac{3u+1}{u-1}}\]

Akọwa Nkọwa

(a) When the sector is folded into a cone, the arc length of the sector becomes the circumference of the base, and the sector radius becomes the slant height \(l = 8\text{ cm}\).

Arc length \(= \dfrac{90}{360}\times 2\pi(8) = \tfrac{1}{4}\times 16\pi = 4\pi\text{ cm}\).

(i) Base circumference \(2\pi r = 4\pi \ \Rightarrow\ \mathbf{r = 2\text{ cm}}\).

(ii) Height \(h = \sqrt{l^2 - r^2} = \sqrt{8^2 - 2^2} = \sqrt{60} = \mathbf{7.75\text{ cm}}\) (2 d.p.).

(iii) Volume \(= \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\times\tfrac{22}{7}\times 4 \times 7.746 = \mathbf{32.5\text{ cm}^3}\) (3 s.f.).

(b) Scale \(1 : 20{,}000\) means \(1\text{ cm}\) on the map \(= 20{,}000\text{ cm}\) on the ground.

(i) \(4.5\text{ cm}\) represents \(4.5\times 20{,}000 = 90{,}000\text{ cm} = 900\text{ m} = \mathbf{0.9\text{ km}}\).

Akọwa Nkọwa

(a)(i) Solve both inequalities.

\(4x < 2 + 3x \ \Rightarrow\ x < 2\)

\(x - 8 < 3x \ \Rightarrow\ -8 < 2x \ \Rightarrow\ x > -4\)

Both hold when \(\mathbf{-4 < x < 2}\).

(ii) Number line: an open circle at \(-4\) and an open circle at \(2\), with the segment between them shaded (values strictly between, endpoints excluded).

(b) Let the correct bill be the two-digit number \(10a + b\) (\(a\) = tens digit, \(b\) = units digit). Interchanging gives \(10b + a\), which overcharges by 45:

\((10b + a) - (10a + b) = 45 \ \Rightarrow\ 9(b - a) = 45 \ \Rightarrow\ b - a = 5\)

Also \(a + b = 9\). Adding: \(2b = 14 \ \Rightarrow\ b = 7\), so \(a = 2\).

The bill should be \(10a + b = \mathbf{£27}\). (Check: \(72 - 27 = 45\).)

Akọwa Nkọwa

Setting up. The figure \(ABCDEO\) is two-thirds of a circle, centre \(O\), radius \(AO=7\text{ cm}\). The sector angle is

The equal chords \(|AB|=|BC|=|CD|=|DE|\) divide the \(240^\circ\) arc into \(4\) equal arcs, each

So each of \(\triangle OAB,\ \triangle OBC,\ \triangle OCD,\ \triangle ODE\) has two sides equal to the radius \(7\) with a \(60^\circ\) angle between them, making each an equilateral triangle of side \(7\text{ cm}\).

Shaded portion = area of the \(240^\circ\) sector \(-\) area of the four triangles (it is the four segments between the chords and the outer arc).

Area of the \(240^\circ\) sector with \(\pi=\tfrac{22}{7}\):

Area of one equilateral triangle (side \(7\)):

Four of them: \(4\times21.22=84.87\text{ cm}^2.\)

Shaded area:

(Equivalently, one \(60^\circ\) segment \(=\tfrac16(154)-21.22=25.67-21.22=4.45\text{ cm}^2\), and \(4\times4.45=17.8\text{ cm}^2.\))

Akọwa Nkọwa

(a) Convert mixed numbers to improper fractions.

Numerator: \(4\tfrac{2}{9} - 1\tfrac{13}{15} = \tfrac{38}{9} - \tfrac{28}{15} = \tfrac{190}{45} - \tfrac{84}{45} = \tfrac{106}{45}\).

Denominator: First \(\tfrac{4}{7}\times 2\tfrac{1}{3} = \tfrac{4}{7}\times\tfrac{7}{3} = \tfrac{4}{3}\). Then \(2\tfrac{1}{5} + \tfrac{4}{3} = \tfrac{11}{5} + \tfrac{4}{3} = \tfrac{33}{15} + \tfrac{20}{15} = \tfrac{53}{15}\).

\(\dfrac{106/45}{53/15} = \dfrac{106}{45}\times\dfrac{15}{53} = \dfrac{106\times 15}{45\times 53} = \dfrac{2\times 15}{45} = \dfrac{30}{45} = \mathbf{\dfrac{2}{3}}\)

(b) Rationalise \(\dfrac{7\sqrt{5}}{\sqrt{7}}\) by multiplying by \(\dfrac{\sqrt{7}}{\sqrt{7}}\):

\(\dfrac{7\sqrt{5}}{\sqrt{7}}\times\dfrac{\sqrt{7}}{\sqrt{7}} = \dfrac{7\sqrt{35}}{7} = \mathbf{\sqrt{35}}\)

Akọwa Nkọwa

(a)(i) Finding the value of \(k\)

Score, \(x\)	1	2	3	4	5	7	8	9	10	Total
Frequency, \(f\)	1	1	5	3	\(k^2+1\)	6	2	3	4	\(k^2+26\)
\(fx\)	1	2	15	12	\(5k^2+5\)	42	16	27	40	\(5k^2+160\)

Since the mean score is 6,

\[6=\frac{5k^2+160}{k^2+26}\]

\[6k^2+156=5k^2+160\]

\[k^2=4\]

\[\therefore k=2\]

Thus, the frequency for score 5 is \(k^2+1=5\), and the total number of pupils is \(4+26=30\).

(a)(ii) Median score

Score	1	2	3	4	5	7	8	9	10
Cumulative frequency	1	2	7	10	15	21	23	26	30

There are 30 scores. The 15th score is 5 and the 16th score is 7.

\[\text{Median}=\frac{5+7}{2}=6\]

(b) Bar chart of the distribution

(c) Probability of scoring less than 6

The scores less than 6 are 1, 2, 3, 4 and 5.

\[\text{Number of such pupils}=1+1+5+3+5=15\]

\[P(\text{score less than }6)=\frac{15}{30}=\boxed{\frac{1}{2}}\]

Akọwa Nkọwa

(a) Work backwards. The man sold to the woman at a 5% loss for \(N120{,}175\), so this is \(95\%\) of the man's cost price (which is the dealer's selling price).

Dealer's selling price \(= \dfrac{120{,}175}{0.95} = N126{,}500\).

The dealer made a 15% profit, so \(N126{,}500 = 115\%\) of the dealer's cost price.

Dealer's cost price \(= \dfrac{126{,}500}{1.15} = \mathbf{N110{,}000.00}\).

(b) Wheel diameter \(= 36\text{ cm}\), so circumference \(= \pi d = \dfrac{22}{7}\times 36 = \dfrac{792}{7} = 113.14\text{ cm}\).

Distance \(= 1.05\text{ km} = 105{,}000\text{ cm}\).

Number of revolutions \(= \dfrac{105{,}000}{113.14} = 928.03 \approx \mathbf{928}\) (3 s.f.).

Akọwa Nkọwa

(a) Volume of the solid (pyramid on a cuboid).

From the diagram the cuboid has base \(4\text{ m} \times 3\text{ m}\) and height \(2\text{ m}\); the pyramid stands on the top \(4\text{ m} \times 3\text{ m}\) face and has a slant edge (apex to base corner) of \(5\text{ m}\).

Volume of the cuboid:

Height of the pyramid. The apex is above the centre of the rectangular base. The distance from the centre to a base corner is half the diagonal of the \(4 \times 3\) rectangle:

Using the right triangle (height, half-diagonal, slant edge):

Volume of the pyramid:

Total volume of the shape:

Total volume \(\approx 41.32\text{ m}^3\) (to 2 d.p.).

(b) Value of x in the A.P.

The 2nd, 3rd and 4th terms are \(x-2\), \(5\) and \(x+2\). In an A.P. the difference between successive terms is constant, so the middle term equals the average of its neighbours (equivalently, consecutive differences are equal):

x = 5. (Check: terms are \(3, 5, 7\), common difference \(2\).)