WAEC SSCE - General Mathematics - 1994

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The ratio of the volumes of two spheres is equal to the cube of the ratio of their radii. Therefore, (volume of first sphere) / (volume of second sphere) = (radius of first sphere)³ / (radius of second sphere)³ Given the volumes of the spheres as 250 cm³ and 128 cm³, we have: 250 / 128 = (radius of first sphere)³ / (radius of second sphere)³ Simplifying this equation, we get: (radius of first sphere) / (radius of second sphere) = (250 / 128)^1/3 Using a calculator, we can evaluate the cube root of 250/128 to be approximately 1.25. Therefore, the ratio of the radii of the two spheres is: (radius of first sphere) / (radius of second sphere) = 1.25 Simplifying this ratio, we can express it as 5:4. Therefore, the correct answer is 5:4.

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The logarithm of a number to base 10 is the power to which 10 must be raised to obtain that number. In this case, we are given the logarithm of a number to base 10, which is \(\bar{3}.4771\). This means that the number is equal to 10 raised to the power of \(\bar{3}.4771\). Using a calculator, we can evaluate this expression to get: 10^{\(\bar{3}.4771\)} = 0.0003441 Therefore, the number whose logarithm to base 10 is \(\bar{3}.4771\) is 0.0003441. So the answer is option (D) 0.003.

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We can use the logarithmic identities to simplify the given equation: log₈x - 4log₈x = log₈(x) - log₈(x^4) = log₈(x / x^4) = log₈(1/x^3) Substituting this into the original equation, we have: log₈(1/x^3) = 2 Rewriting in exponential form, we have: 8^2 = 1/x^3 Simplifying this expression, we get: 64 = 1/x^3 x^3 = 1/64 Taking the cube root of both sides, we get: x = 1/4 Therefore, the solution to the equation is x = ¹/₄. Answer: ¹/₄.

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The area of a sector of a circle can be found using the formula A = (θ/360)πr², where θ is the angle of the sector in degrees and r is the radius of the circle. In this case, the radius of the circle is given as x cm and the angle of the sector is 135°. Substituting these values in the formula, we get: A = (135/360)πx² Simplifying this expression, we get: A = (3/8)πx² Therefore, the area of the sector is (3/8)πx², which is.

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The property that is not true for all parallelograms is: "the diagonal cut at right angles". Explanation: A parallelogram is a four-sided figure with opposite sides parallel. Its diagonals (lines connecting opposite corners) bisect each other, meaning they divide each other into two equal parts. Also, each diagonal bisects the area of the parallelogram, and the opposite sides and angles are equal. However, the diagonals do not always cut at right angles. In fact, they only cut at right angles if and only if the parallelogram is a rectangle (a special type of parallelogram with all angles equal to 90 degrees). Therefore, the correct answer is "the diagonal cut at right angles".

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We can use logarithmic rules to simplify the expression: 2log\(_{3}\) 6 + log\(_{3}\) 12 - log\(_{3}\) 16 = log\(_{3}\) (6\(^{2}\)) + log\(_{3}\) 12 - log\(_{3}\) 16 (using the rule log\(_{b}\) (x) + log\(_{b}\) (y) = log\(_{b}\) (xy)) = log\(_{3}\) (36) + log\(_{3}\) 12 - log\(_{3}\) 16 = log\(_{3}\) (36 x 12) - log\(_{3}\) 16 (using the rule log\(_{b}\) (x) - log\(_{b}\) (y) = log\(_{b}\) (x/y)) = log\(_{3}\) (432/16) = log\(_{3}\) 27 = 3 Therefore, the answer is 3.

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The loss is calculated by finding the difference between the selling price (N80,000) and the cost price (N100,000): Loss = Cost price - Selling price = N100,000 - N80,000 = N20,000 To find the loss percentage, we divide the loss by the cost price and then multiply by 100: Loss % = (Loss/Cost price) x 100% = (N20,000/N100,000) x 100% = 20% Therefore, the loss percentage is 20%. Option A is the correct answer.

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The length of the exercise book is given as 20cm correct to the nearest centimeter. This means the actual length of the exercise book could be 0.5cm more or less than the given measurement. Therefore, the actual length of the exercise book can lie within the range of 19.5 - 20.5cm, which is option A: 19.5 - 20.4cm. Option B: 19.1 - 20.9cm and option D: 19.95 - 20.05cm are too wide and do not take into account the fact that the measurement is correct to the nearest centimeter. Option C: 19.9 - 20.1cm is too narrow and does not allow for the possible error in measurement. Hence, the correct option is A: 19.5 - 20.4cm.

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The volume of the cuboid is given by the product of its length, width, and height. Let's call the height of the cuboid "h". Then, we have: Volume of cuboid = length x width x height 1 liter = 1000 cm^3 Substituting the given values, we have: 1000 cm^3 = 12.5 cm x 20 cm x h Simplifying this expression, we get: 1000 cm^3 = 250 cm x h Dividing both sides by 250 cm, we get: h = 1000 cm^3 / 250 cm h = 4 cm Therefore, the height of the cuboid is 4cm. Answer: 4cm.

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We can solve for the values of p and m by adding the two given equations:

2p - m = 6
2p + 4m = 1
-----------
4p + 3m = 7

So the value of (4p + 3m) is 7. Therefore, the correct option is (7).

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In a circle, the angle subtended by a chord at the centre of the circle is twice the angle subtended by the chord at any point on the circumference. Therefore, if the angle subtended by a chord at the centre is 120°, then the angle subtended by the chord at any point on the circumference is 60°.

Consider the triangle formed by joining the endpoints of the chord to the centre of the circle. We know that the angle at the centre of the circle is 120°, and we know that the radius of the circle is 6 cm.

Now we can use trigonometry to find the length of the chord. Let the length of the chord be 2x. Then, in the triangle we have:

\[\sin 60^\circ = \frac{x}{6}\]

\[\Rightarrow x = 6\sin 60^\circ = 6\cdot\frac{\sqrt{3}}{2} = 3\sqrt{3}\]

Therefore, the length of the chord is 2x, which is equal to:

\[2\cdot3\sqrt{3} = 6\sqrt{3}\]

Hence, the answer is (E) \(6\sqrt{3}\) cm.

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The formula for the length of an arc of a circle is given by:

L = (θ/360) x 2πr

where θ is the angle subtended at the centre of the circle in degrees, r is the radius of the circle, and π is pi.

Using the given values, we have:

θ = 75°
r = 3.5cm
π = 22/7

Substituting these values into the formula, we get:

L = (75/360) x (2 x 22/7 x 3.5)
L = (5/24) x (44/7)
L = 110/24
L ≈ 4.58cm (correct to 2 significant figures)

Therefore, the length of the arc of the circle is approximately 4.58cm. The correct option is (b) 4.6cm.

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To find the roots of the given equation, we need to solve for x when x\(^2\) + 6x = 0. Factorizing the equation, we get x(x+6) = 0 Hence, the roots are x = 0 and x = -6. So, the correct answer is 0 or -6.

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To determine which of the options is not a factor of 2p\(^2\) - 2, we can factorize the expression first. We can start by taking out a common factor of 2: 2p\(^2\) - 2 = 2(p\(^2\) - 1) We can then further factorize the expression inside the parentheses using the difference of squares: 2p\(^2\) - 2 = 2(p - 1)(p + 1) Now we can check each of the options to see if they are factors of the expression. - 2 is a factor since it is a common factor of the original expression. - p - 1 is a factor since it is one of the factors obtained from the factorization. - p + 1 is a factor since it is the other factor obtained from the factorization. - 2p - 2 is a factor since it is equivalent to 2(p - 1). - 2p + 1 is not a factor of the expression since it cannot be obtained by any combination of the factors we found earlier. Therefore, the answer is 2p + 1.

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To factorize the expression 6x\(^2\) + 7xy - 5y\(^2\), we need to find two binomials whose product equals the expression. First, we need to find two numbers that multiply to 6, the coefficient of the x\(^2\) term, and two numbers that multiply to -5, the coefficient of the y\(^2\) term. We also need to find the signs of the two numbers that add up to 7, the coefficient of the xy term. After some trial and error, we can determine that the factors are (3x - y) and (2x + 5y). Multiplying these two binomials using FOIL method, we get: (3x - y)(2x + 5y) = 6x\(^2\) + 7xy - 5y\(^2\) Therefore, the correct answer is option (D): (3x + 5y)(2x - y).

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The expression \(\frac{x^2 + 15x + 50}{x - 5}\) is not defined when the denominator, which is \(x - 5\), equals zero. In other words, the expression is undefined when \(x = 5\). Therefore, the value of x that makes the expression undefined is 5.

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To find P' ∩ Q, we first need to find the complement of P, which consists of all the elements in E that are not in P. Since P is the set of multiples of 3, P' is the set of all integers in E that are not multiples of 3. Therefore, P' = {1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20} Next, we need to find the elements that are common to both P' and Q. Since Q is the set of multiples of 4, we can simply list the multiples of 4 that are in P', which are: 4, 8, 16, 20 Therefore, P' ∩ Q = {4, 8, 16, 20}. So the correct option is (2) (4, 8, 16, 20).

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Given sin(\(\theta\)) = K, we need to find tan(\(\theta\)). We know that sin(\(\theta\)) = opposite/hypotenuse and cos(\(\theta\)) = adjacent/hypotenuse in a right-angled triangle with angle \(\theta\). Using Pythagoras theorem, we also know that hypotenuse² = opposite² + adjacent². So, let's assume a right-angled triangle with angle \(\theta\) and opposite side as K. We can find the adjacent side using Pythagoras theorem as hypotenuse² = opposite² + adjacent², which gives us adjacent = \(\sqrt{1 - K^2}\). Now, we can use the definition of tangent, which is tan(\(\theta\)) = opposite/adjacent. Thus, tan(\(\theta\)) = K/\(\sqrt{1 - K^2}\). Therefore, the correct option is \( \frac{k}{\sqrt{1 - k^2}} \).

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To simplify this expression, we need to use the rules of exponents: - 16\(^{\frac{5}{4}}\) can be written as (2\(^4\))\(^{\frac{5}{4}}\) since 16 is equal to 2\(^4\). - Using the power of a power rule, we can multiply the exponents 4 and 5/4 to get 2\(^5\). - 2\(^{-3}\) is the same as 1/2\(^3\) or 1/8 since we have a negative exponent. - 3\(^0\) is equal to 1 since any number raised to the power of 0 is 1. Therefore, 16\(^{\frac{5}{4}}\) x 2\(^{-3}\) x 3\(^0\) simplifies to: (2\(^4\))\(^{\frac{5}{4}}\) x 1/8 x 1 = 2\(^5\) x 1/8 = 32/8 = 4 So the answer is 4.

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The sum of the interior angles of any polygon can be found using the formula: (n-2) x 180, where n is the number of sides of the polygon. For a regular polygon, all interior angles are equal, so we can divide the sum of the angles by the number of angles to find the measure of each interior angle. In this case, the sum of the angles is 2520 degrees. Let's plug this value into the formula: (n-2) x 180 = 2520 Simplifying this equation: n - 2 = 14 n = 16 So the polygon has 16 sides.

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We can begin solving this equation by isolating x on one side of the equation. We do this by adding 10 to both sides and subtracting 6x from both sides: 8x - 4 = 6x - 10 8x - 6x = -10 + 4 2x = -6 x = -3 Now that we have found the value of x, we can substitute it into the expression 5x to find its value: 5x = 5(-3) = -15 Therefore, the value of 5x is -15. So, the correct option is (b) -15.

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The distance between the two ships along the equator is equal to the length of the arc that connects the two longitudes. We can start by finding the total length of the equator, which is the circumference of the earth, using the formula: C = 2πr where C is the circumference, r is the radius, and π is the mathematical constant pi. Substituting the given values, we get: C = 2 x (22/7) x 6400 C ≈ 40,320 km Since the two ships are on longitudes that are 90 degrees apart, the arc that connects them along the equator is one-fourth of the total circumference of the earth. Therefore, the length of the arc is: (1/4) x C = (1/4) x 40,320 ≈ 10,080 km Rounding this to two significant figures gives us: ≈ 10,000 km Therefore, the answer is, 10,000km.

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We know that prime numbers are numbers that are divisible only by 1 and themselves. To solve this problem, we can simply list the prime numbers between 30 and 50, and count how many there are. The prime numbers between 30 and 50 are: 31, 37, 41, 43, and 47. There are 5 prime numbers in this range.

The total number of integers between 30 and 50 (inclusive) is 21 (50 - 30 + 1 = 21). Therefore, the probability of selecting a prime number from this range is:

    5 prime numbers
    --------------- = 5/21
         21 numbers

So the answer is 5/21.

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To evaluate the expression, we can use the laws of exponents, which state that \(a^{m/n} = (a^m)^{1/n}\) and \(a^{-n} = \frac{1}{a^n}\). Applying these laws, we have: \(\frac{27^{\frac{1}{3}}}{16^{-\frac{1}{4}}} = \frac{(3^3)^{\frac{1}{3}}}{(2^4)^{-\frac{1}{4}}} = \frac{3^{3\cdot\frac{1}{3}}}{2^{4\cdot(-\frac{1}{4})}} = \frac{3}{2^{-1}} = 3 \cdot 2 = 6\) Therefore, the answer is 6.

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Let us denote the first term of the G.P. by a and the common ratio by r. Then the 2nd term is ar, the 3rd term is ar^2, and so on. Therefore, the 5th term is ar^4. We have two pieces of information, namely that ar = 6 and ar^4 = 48. Dividing the second equation by the first, we obtain (ar^4)/(ar) = r^3 = 48/6 = 8. Therefore, r = 2. Using this value of r, we can solve for a by using the equation ar = 6. Substituting r = 2, we obtain 2a = 6, so a = 3. Therefore, the first four terms of the G.P. are 3, 6, 12, 24, and their sum is 45. Thus, the correct answer is 45.

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If the bearing of Q from P is 122^o, then the bearing of P from Q is 180^o + 122^o = 302^o. This is because bearings are measured clockwise from north and the bearing of Q from P is 122^o which is to the south-west of P. Therefore, the bearing of P from Q will be in the opposite direction, which is to the north-east, and we add 180^o to 122^o to get the bearing of P from Q, which is 302^o.

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Let's start by finding the common difference of the arithmetic progression using the 1st and 2nd terms. Let the first term be "a" and the common difference be "d". The first term is "a", the second term is "a + d", and the 10th term is "a + 9d". Given that the sum of the 1st and 2nd terms is 4, we have: a + (a + d) = 4 2a + d = 4 -----(1) Also, the 10th term is 19, so: a + 9d = 19 -----(2) We now have two equations with two variables, which we can solve to find the values of "a" and "d". From equation (1), we get: d = 4 - 2a Substituting into equation (2), we get: a + 9(4 - 2a) = 19 Solving for "a", we get: a = 1 Substituting this value of "a" into equation (1), we get: 2 + d = 4 d = 2 So the common difference is 2. Now, we can find the 5th and 6th terms of the arithmetic progression: The 5th term = a + 4d = 1 + 4(2) = 9 The 6th term = a + 5d = 1 + 5(2) = 11 Therefore, the sum of the 5th and 6th terms is: 9 + 11 = 20 So the answer is 20.

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To simplify \(\frac{72}{125}\), we can factorize both numbers. The prime factorization of 72 is \(2^3 \times 3^2\) and the prime factorization of 125 is \(5^3\). Therefore, we can write: \[\frac{72}{125} = \frac{2^3 \times 3^2}{5^3}\] Hence, the answer is \( \frac{2^3 \times 3^2}{5^3} \).

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When a card is drawn from a well-shuffled deck of 52 cards, the probability of getting an Ace is 4/52 since there are 4 Aces in the deck. Since we're drawing two cards with replacement, the probability of getting two Aces is the product of the probability of getting an Ace on the first draw and the probability of getting an Ace on the second draw. So: Probability of getting two Aces = (4/52) x (4/52) = 16/2704 = 1/169 Therefore, the probability that the two cards drawn are both Aces is 1/169.

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(a) Divide \(11111111_{two}\) by \(101_{two}\)

We use binary long division. Divisor is \(101_{two}\).

         1 1 0 0 1 1
       ___________________
1 0 1 ) 1 1 1 1 1 1 1 1
        1 0 1
        -----
          1 0 1
          1 0 1
          -----
            0 0 1 1
                1 1 1
                1 0 1
                -----
                  1 0 1
                  1 0 1
                  -----
                    0 0

The quotient is \(11111111_{two} \div 101_{two} = 110011_{two}\), remainder \(0\).

Check in base ten: \(11111111_{two}=255\), \(101_{two}=5\), and \(255 \div 5 = 51 = 110011_{two}\). Correct.

(b) Sector: radius 6 cm, angle 105°, \(\pi=\tfrac{22}{7}\)

(i) Perimeter. First find the arc length.

\[\text{Arc} = \frac{\theta}{360}\times 2\pi r = \frac{105}{360}\times 2\times\frac{22}{7}\times 6 = \frac{7}{24}\times\frac{264}{7} = 11\text{ cm}\]

Perimeter = arc + the two straight radii:

\[P = 11 + 2(6) = 23\text{ cm}\]

(ii) Area.

\[A = \frac{\theta}{360}\times \pi r^{2} = \frac{105}{360}\times\frac{22}{7}\times 6^{2} = \frac{7}{24}\times\frac{792}{7} = 33\text{ cm}^{2}\]

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This is a discrete frequency distribution, so each score \(x\) already has its frequency \(f\). Build an \(fx\) column, add the columns, then apply the mean formula.

(a) Mean mark.

Correct to 3 significant figures, the mean mark is \(\mathbf{47.9}\).

(b)(i) Mode. The mode is the score with the highest frequency. The largest frequency is \(24\), which belongs to the score \(45\), so the mode is \(\mathbf{45}\).

(b)(ii) Range.

Examination note: multiply every score by its own frequency before adding, and divide by \(\sum f = 70\), not by the number of different scores (7). Dividing by 7 is the usual slip that gives a wrong mean near \(46\text{-}48\) instead of the correct \(47.9\).

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(a) Express everything as powers of 3: \(9 = 3^2\), \(27 = 3^3\).

\(3\times 9^{1+x} = 3^1 \times 3^{2(1+x)} = 3^{1 + 2 + 2x} = 3^{3 + 2x}\)

\(27^{-x} = 3^{-3x}\)

Equating indices: \(3 + 2x = -3x \ \Rightarrow\ 5x = -3 \ \Rightarrow\ \mathbf{x = -\dfrac{3}{5}}\).

(b) Use \(\log\sqrt{a} = \tfrac{1}{2}\log a\) and combine:

\(\log_{10}\sqrt{35} + \log_{10}\sqrt{2} - \log_{10}\sqrt{7} = \log_{10}\!\left(\dfrac{\sqrt{35}\times\sqrt{2}}{\sqrt{7}}\right)\)

\(= \log_{10}\sqrt{\dfrac{35\times 2}{7}} = \log_{10}\sqrt{10} = \log_{10} 10^{1/2} = \dfrac{1}{2}\)

Value \(= \mathbf{\dfrac{1}{2}}\).

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Let the horizontal distance between the building and the foot \(F\) of the tower be \(x\), and the tower height be \(H\). \(P\) is at the foot of the building, \(Q\) is its top, \(20\text{ m}\) directly above \(P\).

From \(P\): \(\tan 65° = \dfrac{H}{x}\), so \(H = x\tan 65°\).

From \(Q\) (height 20 m, same horizontal distance \(x\)): \(\tan 25° = \dfrac{H - 20}{x}\), so \(H - 20 = x\tan 25°\).

Subtracting: \(20 = x(\tan 65° - \tan 25°) = x(2.1445 - 0.4663) = 1.6782x\).

\(x = \dfrac{20}{1.6782} = 11.92\text{ m}\).

(a) Distance PT: \(PT\) is the line of sight from \(P\) to \(T\), at elevation \(65°\):

\(PT = \dfrac{x}{\cos 65°} = \dfrac{11.92}{0.4226} = 28.20 \approx \mathbf{28.2\text{ m}}\) (3 s.f.).

(b) Height of tower: \(H = x\tan 65° = 11.92 \times 2.1445 = 25.56 \approx \mathbf{25.6\text{ m}}\) (3 s.f.).

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(a) Speed along the parallel of latitude

Both points are on latitude \(32^\circ\)N. The difference in longitude is

\[47^\circ + 25^\circ = 72^\circ\]

Distance along a parallel of latitude \(\phi\):

\[d = \frac{\theta}{360}\times 2\pi R\cos\phi\]

\[d = \frac{72}{360}\times 2\times\frac{22}{7}\times 6400\times\cos32^\circ\]

Using \(\cos32^\circ = 0.8480\):

\[d = 0.2\times \frac{44}{7}\times 6400\times 0.8480 \approx 6823\text{ km}\]

Speed \(= \dfrac{\text{distance}}{\text{time}} = \dfrac{6823}{11} \approx 620\) km/h.

Speed \(\approx 620\) km/h.

(b) Height of the kite

Let the foot of the vertical from the kite \(K\) meet the ground at \(O\). \(Q\) is nearer (elevation \(45^\circ\)) and \(P\) is \(15\) m further back (elevation \(35^\circ\)). Let \(OQ = d\) and height \(= h\).

From \(Q\): \(\tan45^\circ = \dfrac{h}{d} = 1 \Rightarrow d = h\).

From \(P\): \(\tan35^\circ = \dfrac{h}{d+15}\).

Substitute \(d = h\):

\[h = (h+15)\tan35^\circ = 0.7002(h+15)\]

\[h - 0.7002h = 10.503 \;\Rightarrow\; 0.2998h = 10.503 \;\Rightarrow\; h \approx 35\text{ m}\]

Height \(\approx 35\) m.

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(a) Solve \(\log_{10}(3x-1) - \log_{10}2 = 3\)

Combine the logs using \(\log a - \log b = \log\frac{a}{b}\):

\[\log_{10}\left(\frac{3x-1}{2}\right) = 3\]

Rewrite in index form (\(\log_{10}N = 3 \Rightarrow N = 10^{3}\)):

\[\frac{3x-1}{2} = 1000\]

\[3x-1 = 2000 \;\Rightarrow\; 3x = 2001 \;\Rightarrow\; x = 667\]

(b) Evaluate \(\sqrt{\dfrac{0.897\times 3.536}{0.00249}}\) to 3 s.f. using logarithm tables

Divide (subtract the log of the denominator):

\[0.5013 - \bar{3}.3963 = 0.5013 - (-2.6037) = 3.1050\]

Take the square root (divide the log by 2):

\[\tfrac{1}{2}\times 3.1050 = 1.5525\]

Antilog of \(1.5525\) gives \(3.570\times 10^{1}\).

\[\sqrt{\frac{0.897\times 3.536}{0.00249}} \approx 35.7\]

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(a) Use the difference of two squares, \(a^2 - b^2 = (a-b)(a+b)\):

\(17.57^2 - 12.43^2 = (17.57 - 12.43)(17.57 + 12.43)\)

\(= (5.14)(30.00) = \mathbf{154.2}\)

(b) Theorem: Angles in the same segment of a circle are equal.

Given: A circle with centre \(O\); a chord \(AB\); points \(P\) and \(Q\) on the major arc (the same segment), so \(\angle APB\) and \(\angle AQB\) both stand on chord \(AB\).

To prove: \(\angle APB = \angle AQB\).

Construction: Join \(OA\) and \(OB\).

Proof: By the theorem that the angle subtended by an arc at the centre is twice the angle it subtends at the circumference (on the same arc):

\(\angle AOB = 2\,\angle APB \quad\text{and}\quad \angle AOB = 2\,\angle AQB\)

Therefore \(2\,\angle APB = 2\,\angle AQB\), giving \(\angle APB = \angle AQB\). Hence angles in the same segment are equal. \(\blacksquare\)

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(a) Cumulative frequency table

Age (years)	Number of people	Cumulative frequency
10	8	8
15	19	27
20	10	37
25	7	44
30	6	50

(b) Cumulative frequency curve (ogive)

Plot the points \((10,8), (15,27), (20,37), (25,44)\) and \((30,50)\), then join them with a smooth increasing curve.

(c)

(i) Median age

\(N=50\), so the median corresponds to cumulative frequency \(\frac{50}{2}=25\). Reading across from 25 to the ogive and down to the age axis gives:

\[\boxed{\text{Median age} \approx 14.5\text{ years}}\]

(ii) Number of people at most 15 years old

At age 15 years, the cumulative frequency is \(27\).

\[\boxed{27\text{ people}}\]

(iii) Number of people between 20 and 25 years old

\[44-37=7\]

\[\boxed{7\text{ people}}\]

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(a) Arithmetic Progression

Let the first term be \(a\) and common difference \(d\). The \(n\)th term is \(a+(n-1)d\).

Fourth term: \(a + 3d = 37\).

The 6th term is 12 more than the 4th term: \((a+5d) = 37 + 12 = 49\), so \(a + 5d = 49\).

Subtract the equations:

\[(a+5d)-(a+3d) = 49-37 \;\Rightarrow\; 2d = 12 \;\Rightarrow\; d = 6\]

Then \(a = 37 - 3(6) = 19\).

First term \(= 19\). Seventh term:

\[a + 6d = 19 + 6(6) = 55\]

Seventh term \(= 55\).

(b) Sets with \(P=\{1,2,3,4\}\) and \(Q=\{3,5,6\}\).

(i) \(P\cap Q = \{3\}\)

(ii) \(P\cup Q = \{1,2,3,4,5,6\}\)

(iii) \((P\cap Q)\cup Q = \{3\}\cup\{3,5,6\} = \{3,5,6\}\)

(iv) \((P\cap Q)\cup P = \{3\}\cup\{1,2,3,4\} = \{1,2,3,4\}\)

Akọwa Nkọwa

Setup. Bag A: 12 white, 8 black (20 balls). Bag B: 10 white, 15 black (25 balls). Two balls are drawn from each bag without replacement.

Useful probabilities per bag

Bag A, two black: \(\dfrac{8}{20}\times\dfrac{7}{19} = \dfrac{14}{95}\)

Bag A, exactly one white: \(2\times\dfrac{12}{20}\times\dfrac{8}{19} = \dfrac{48}{95}\)

Bag B, two black: \(\dfrac{15}{25}\times\dfrac{14}{24} = \dfrac{7}{20}\)

Bag B, exactly one white: \(2\times\dfrac{10}{25}\times\dfrac{15}{24} = \dfrac{1}{2}\)

(a) All four balls black

Both bags must give two blacks:

\[P = \frac{14}{95}\times\frac{7}{20} = \frac{98}{1900} = \frac{49}{950} \approx 0.052\]

(b) Exactly one of the four balls is white

The single white must come from one bag while the other bag yields two blacks.

White from A, two black from B: \(\dfrac{48}{95}\times\dfrac{7}{20} = \dfrac{84}{475}\)

Two black from A, white from B: \(\dfrac{14}{95}\times\dfrac{1}{2} = \dfrac{7}{95} = \dfrac{35}{475}\)

Add the mutually exclusive cases:

\[P = \frac{84}{475} + \frac{35}{475} = \frac{119}{475} \approx 0.251\]

Akọwa Nkọwa

(a)(i) Construction of triangle \(XYZ\)

1. Draw \(YZ=8\text{ cm}\).
2. At \(Y\), construct an angle of \(60^\circ\). At \(Z\), construct an angle of \(75^\circ\), using ruler and compasses only.
3. Let the two constructed rays meet at \(X\). This gives triangle \(XYZ\).

From the completed construction,

\[|XY|\approx 10.9\text{ cm}.\]

(a)(ii) The locus \(l_1\) of points equidistant from \(Y\) and \(Z\) is the perpendicular bisector of \(YZ\).

(a)(iii) The locus \(l_2\) of points equidistant from \(XY\) and \(YZ\) is the internal bisector of \(\angle XYZ\).

(b) Let \(Q\) be the intersection of \(l_1\) and \(l_2\). Measuring from the construction gives

\[\boxed{|QY|\approx 4.8\text{ cm}}\]

Akọwa Nkọwa

(a) For each value of \(x\), calculate \(y=2x^2-9x-1\).

\(x\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)
\(y\)	\(10\)	\(-1\)	\(-8\)	\(-11\)	\(-10\)	\(-5\)	\(4\)	\(17\)

(b) The plotted curve, together with the construction line \(y=3\) and the tangent at \(x=3\), is shown below.

(c)(i) Since \(2x^2-9x=4\), then

\[2x^2-9x-1=3.\]

Thus, draw or consider the horizontal line \(y=3\). From its intersections with the curve,

\[\boxed{x=-0.4\text{ or }x=4.9}\]

correct to one decimal place.

(ii) At \(x=3\), the point on the curve is \((3,-10)\). The tangent shown passes through \((3,-10)\) and \((5,-4)\). Hence,

\[\text{Gradient}=\frac{-4-(-10)}{5-3}=\frac{6}{2}=\boxed{3}.\]