WAEC SSCE - General Mathematics - 2001

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To solve the equation 2x : (x + 1) = 3 : 2, we need to cross-multiply the ratios. That means we multiply the numerator of the left-hand side ratio by the denominator of the right-hand side ratio, and we multiply the denominator of the left-hand side ratio by the numerator of the right-hand side ratio. So we have: 2x * 2 = 3 * (x + 1) Simplifying the right-hand side: 4x = 3x + 3 Subtracting 3x from both sides: x = 3 Therefore, the value of x is 3. To check, we can substitute x = 3 back into the original equation and see if both sides are equal: 2(3) : (3 + 1) = 6 : 4 = 3 : 2 So the equation is true when x = 3, and that is our final answer.

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The mode is the most frequently occurring value in a set of data. In this case, the data set is {22, 23, 24, 25, 27, 30}. The value that appears most frequently in this set is 23 years, which appears twice, making it the mode. Therefore, the answer is 23.0 years.

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To evaluate \(log_{10}5 + log_{10}20\), we can use the logarithmic rule that states: $log_{a}x + log_{a}y = log_{a}(xy)$ Applying this rule, we get: $log_{10}5 + log_{10}20 = log_{10}(5 \times 20)$ Simplifying the right-hand side: $log_{10}(5 \times 20) = log_{10}100$ Finally, we know that \(log_{10}100 = 2\), so: \(log_{10}5 + log_{10}20 = log_{10}100 = 2\) Therefore, the answer is 2.

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The standard deviation is the square root of the variance. Therefore, to find the standard deviation when given the variance, we simply take the square root of the variance. In this case, the variance is given as 25. Taking the square root of 25 gives us: sqrt(25) = 5 So the standard deviation of the distribution is 5. To summarize, the formula for finding the standard deviation from the variance is to take the square root of the variance. In this case, the variance is 25, so the standard deviation is 5.

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Substituting x = -1, we get: \[\frac{(-1)^2 + (-1) - 2}{2(-1)^2 + (-1) - 3}\] \[=\frac{1 - 1 - 2}{2 - 1 - 3}\] \[=\frac{-2}{-2}\] \[=1\] Therefore, the value of the expression is 1 when x = -1. Hence the correct option is (d) 1.

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Akọwa Nkọwa

We can solve this problem using a Venn diagram. Let's draw two circles, one for Economics and one for Geography, with some overlap between them. We know that 65 students study Economics, so we'll put 65 in the circle for Economics. Similarly, we know that 50 students study Geography, so we'll put 50 in the circle for Geography. Now we need to figure out how many students study both subjects. We can call this number "x" for now. We'll put "x" in the overlap between the two circles. We also know that every student in the class studies either Economics or Geography or both. So we need to account for all 80 students in the class. To do this, we can add up the numbers we've put in the circles and subtract the overlap once (since the students in the overlap were counted twice). In other words, Number of students studying Economics + Number of students studying Geography - Number of students studying both = Total number of students in the class Or, 65 + 50 - x = 80 Simplifying, 115 - x = 80 x = 35 So 35 students study both Economics and Geography. Therefore, the answer is option (C) 35.

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A solid cylinder has two circular bases and a curved lateral surface. To find its total surface area, we need to find the area of both circular bases and the area of the lateral surface, and then add them together. The area of one circular base is \(\pi r^2\), where r is the radius of the cylinder. Since the radius is given as 7 cm, the area of one circular base is: \begin{align*} \text{Area of one circular base} &= \pi \times (7 \text{ cm})^2 \\ &= 49 \pi \text{ cm}^2 \\ \end{align*} Since there are two circular bases, the total area of both circular bases is: \begin{align*} \text{Total area of both circular bases} &= 2 \times \text{Area of one circular base} \\ &= 2 \times 49 \pi \text{ cm}^2 \\ &= 98 \pi \text{ cm}^2 \\ \end{align*} The area of the lateral surface is the curved surface area of the cylinder, which is given by \(2\pi rh\), where r is the radius of the cylinder and h is the height (or length) of the cylinder. Since the radius is given as 7 cm and the length is given as 10 cm, the area of the lateral surface is: \begin{align*} \text{Area of lateral surface} &= 2\pi rh \\ &= 2\pi \times (7 \text{ cm}) \times (10 \text{ cm}) \\ &= 140 \pi \text{ cm}^2 \\ \end{align*} Therefore, the total surface area of the cylinder is the sum of the area of both circular bases and the area of the lateral surface: \begin{align*} \text{Total surface area} &= \text{Area of both circular bases} + \text{Area of lateral surface} \\ &= 98 \pi \text{ cm}^2 + 140 \pi \text{ cm}^2 \\ &= 238 \pi \text{ cm}^2 \\ \end{align*} Therefore, the total surface area of the cylinder is 238\(\pi\) cm\(^2\).

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The area of a triangle is given by the formula: $$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $$ In triangle PQR, the base is |RQ| = 3cm and the height is |PS|, where S is the foot of the perpendicular from point P to line QR. Since the cross section is a right-angled triangle, we can use Pythagoras theorem to find |PS| as follows: $$ \begin{align*} |PS|^2 &= |PR|^2 - |RS|^2 \\ &= 5^2 - 3^2 \\ &= 16 \\ \end{align*} $$ Therefore, |PS| = 4cm. Substituting the values for base and height into the area formula, we get: $$ \text{Area} = \frac{1}{2} \times 3\text{cm} \times 4\text{cm} = 6\text{cm}^2 $$ Hence, the area of the cross-section is 6 cm².

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The perimeter of a sector is given by the sum of the length of the two radii and the arc length between them. In this question, we are given the radius of the sector to be 14cm and the central angle to be 135°. The arc length can be found using the formula for the circumference of a circle, C = 2πr, where r is the radius of the circle. The central angle of 135° is equivalent to \(\frac{135}{360}\) of the full circle, so the arc length of the sector is: \(\frac{135}{360} \times 2\pi \times 14 \approx 32.91cm\) The two radii have the same length and are equal to 14cm each. Therefore, the perimeter of the sector is: 14cm + 14cm + 32.91cm ≈ 60.91cm Rounding this to the nearest whole number gives us 61cm. Hence, the correct answer is 61cm.

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To find which term of the sequence is \(2^9\), we need to solve for n in the equation \(2^{2n-1}=2^9\). First, we can simplify the equation by dividing both sides by \(2^9\), giving us \(2^{2n-1-9}=2^{2n-10}=1\). Next, we can solve for n by taking the logarithm of both sides of the equation. Since any logarithm base can be used, we can use the natural logarithm, denoted as ln: \begin{align*} 2n-10 &= \ln 1 \\ 2n-10 &= 0 \\ 2n &= 10 \\ n &= 5 \end{align*} Therefore, the fifth term of the sequence is \(2^9\).

Akọwa Nkọwa

A quadratic expression is a polynomial of degree two. This means that the highest exponent in the expression is 2. Therefore, the expression that is not quadratic is y = 5(x-1) which is a linear expression of degree 1.

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To convert 25° 45' to decimal, we first note that there are 60 minutes in a degree. So, to convert minutes to degrees, we divide by 60. Thus, 25° 45' = 25 + 45/60 degrees Simplifying, 25° 45' = 25 + 0.75 degrees Combining the whole and fractional parts, we get: 25° 45' = 25.75 degrees Therefore, the answer is: 25.75°.

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We know that \(\sin p = \dfrac{5}{13}\) and $p$ is acute. Using Pythagorean identity, we can find $\cos p$: \begin{align*} \cos^2 p &= 1 - \sin^2 p \\ \cos^2 p &= 1 - \left(\dfrac{5}{13}\right)^2 \\ \cos^2 p &= \dfrac{144}{169} \\ \cos p &= \dfrac{12}{13} \end{align*} Using the definition of tangent, we can find $\tan p$: \begin{align*} \tan p &= \dfrac{\sin p}{\cos p} \\ \tan p &= \dfrac{\frac{5}{13}}{\frac{12}{13}} \\ \tan p &= \dfrac{5}{12} \end{align*} Therefore, \begin{align*} \cos p - \tan p &= \dfrac{12}{13} - \dfrac{5}{12} \\ &= \dfrac{144}{156} - \dfrac{65}{156} \\ &= \dfrac{79}{156} \end{align*} Hence, the answer is \(\frac{79}{156}\).

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The bearing S40°E is equivalent to an angle that is 40 degrees east of the southern direction. The direction opposite to south is north, so if we draw a line going north and another line going 40 degrees to the east of south, the angle formed between these two lines is the direction that corresponds to the bearing S40°E. This angle is equal to 180° - 40° = 140°. Therefore, the correct answer is 140^o.

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There are 12 balls in total in the box, out of which 5 are red. When the boy takes out the first ball, there are 11 balls left, out of which 4 are red. Therefore, the probability of the first ball being red is 5/12, and the probability of the second ball being red is 4/11. To find the probability of both events happening together, we multiply the probabilities: \(\frac{5}{12}\times\frac{4}{11}=\frac{5}{33}\) Therefore, the probability that the two balls taken out are both red is 5/33. Answer: \(\frac{5}{33}\)

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Let the age of the son be x. Then, the age of the man will be 4x (as he is four times as old as his son). The difference in their ages is 36 years, so we have: 4x - x = 36 Simplifying this, we get: 3x = 36 x = 12 So, the age of the son is 12 and the age of the man is 4 times that, which is 48. Therefore, the sum of their ages is: 12 + 48 = 60 years. Hence, the correct option is (c) 60 years.

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The ratio of good to bad oranges is 5:4, which means that for every 5 good oranges, there are 4 bad ones. If we know there are 36 bad oranges, we can use this ratio to find the number of good oranges as follows: 5/4 = x/36 where x is the number of good oranges. We can solve for x by cross-multiplying: 4x = 5*36 x = 45 So there are 45 good oranges in the bag. To find the total number of oranges, we add the number of good and bad ones: 45 + 36 = 81 Therefore, there are 81 oranges in the bag altogether.

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To express 48 as a product of prime factors, we need to keep dividing it by prime numbers until we get to 1. Let's start with 2, which is the smallest prime number that can divide 48: 48 ÷ 2 = 24 We can continue dividing by 2 until we can't divide by 2 anymore: 24 ÷ 2 = 12 12 ÷ 2 = 6 6 ÷ 2 = 3 Now, we can see that we have reached a prime number, 3, so we stop dividing. Therefore: 48 = 2 x 2 x 2 x 2 x 3 So the correct option is 2 x 2 x 2 x 2 x 3 or.

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The error in the boy's estimate is the difference between what he estimated and the actual cost of the journey, which is N200 - N190 = N10. To find the percentage error, we divide the error by the actual cost and then multiply by 100%. Percentage error = (error / actual cost) x 100% = (10 / 200) x 100% = 5% Therefore, the percentage error in the boy's estimate is 5%. Option (d) is the correct answer.

Akọwa Nkọwa

To find the value of k, we need to expand the right-hand side of the equation, which is equal to the left-hand side: \begin{align*} (x+2)(2x-7) &= 2x^2 - 3x - 14 \\ \end{align*} Now we can compare the coefficients of the terms on both sides of the equation: \begin{align*} \text{Coefficient of } x^2: \quad &2 = 2 \\ \text{Coefficient of } x: \quad &-3 = k \\ \text{Coefficient of the constant term:} \quad &-14 = -14 \\ \end{align*} Therefore, we have found that the value of k is -3.

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To evaluate the given expression, we follow the order of operations or BODMAS (Brackets, Order, Division, Multiplication, Addition, and Subtraction) and simplify the expression step by step. First, we simplify the fraction in the middle of the expression, \(\frac{3}{4}of\frac{2}{5} = \frac{3}{4} \times \frac{2}{5} = \frac{3\times 2}{4\times 5} = \frac{3}{10}\) Now, we can rewrite the expression as \(\frac{1}{2}+\frac{3}{10}\div 1\frac{3}{5}\) Next, we simplify the mixed number in the denominator by converting it to an improper fraction, \(1\frac{3}{5} = \frac{8}{5}\) Now, we can rewrite the expression as \(\frac{1}{2}+\frac{3}{10}\div \frac{8}{5}\) To divide by a fraction, we multiply by its reciprocal, \(\frac{1}{2}+\frac{3}{10}\times \frac{5}{8} = \frac{1}{2}+\frac{3\times 5}{10\times 8} = \frac{1}{2}+\frac{3}{16} = \frac{8}{16}+\frac{3}{16} = \frac{11}{16}\) Therefore, the value of the given expression is \(\frac{11}{16}\). Thus, option (B) is the correct answer.

Akọwa Nkọwa

To subtract (-y + 3x + 5z) from (4y - x - 2z), we can just subtract the corresponding coefficients of each variable. So, - the coefficient of y is 4 - (-1) = 5 - the coefficient of x is -1 - 3 = -4 - the coefficient of z is -2 - 5 = -7 Therefore, the result is: 5y - 4x - 7z. So the correct option is (a) 5y - 4x - 7z.

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To subtract two fractions, they need to have a common denominator. The common denominator of 3 and 2 is 6, so we can write: \begin{align*} \frac{2x-1}{3}-\frac{x+3}{2} &= \frac{2(2x-1)}{2 \cdot 3}-\frac{3(x+3)}{3 \cdot 2} \\ &= \frac{4x-2}{6}-\frac{3x+9}{6} \\ &= \frac{4x-3x-2-9}{6} \\ &= \frac{x-11}{6} \end{align*} Therefore, the answer is \(\frac{x-11}{6}\).

Akọwa Nkọwa

The sum of the probabilities of an event occurring and not occurring is always equal to 1. So, if p is the probability that an event will occur, then the probability that it will not occur is 1-p (since the total probability is 1). Therefore, we have: p + (1-p) = 1 which simplifies to: p - p + 1 = 1-p + p 1 = 1 So, the correct option is: p + q = 1.

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To solve this inequality, we can start by simplifying both sides. First, we can multiply both sides by 6 to eliminate the denominators: 3(y - 3) < 4(2y - 1) Expanding the right side gives: 3(y - 3) < 8y - 4 Simplifying and collecting like terms, 3y - 9 < 8y - 4 Subtracting 3y from both sides, -9 < 5y - 4 Adding 4 to both sides, -5 < 5y Dividing both sides by 5, -1 < y So the inequality is true for all values of y greater than -1. Therefore, the correct answer is: y > -7.

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To find the values of p and q, we need to use the information given in the problem. We are given that: y = px + q When x = 3, y = 5 5 = 3p + q --- (Equation 1) When x = 2, y = 4 4 = 2p + q --- (Equation 2) We can solve these two equations simultaneously to get the values of p and q. We can do this by subtracting equation 2 from equation 1: 5 = 3p + q - (4 = 2p + q) 1 = p Substituting p = 1 into equation 2, we get: 4 = 2(1) + q 4 = 2 + q q = 2 Therefore, the values of p and q are p = 1 and q = 2. Therefore, the answer is (B) p = 1, q = 2.

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When two variables are said to vary directly with each other, it means that as one increases or decreases, the other also increases or decreases by the same factor or proportion. This relationship can be expressed as: N = kM where k is the constant of proportionality. To find the value of k, we can use the initial values given: N = 8 when M = 20. 8 = k(20) Solving for k, we get: k = 8/20 = 0.4 Now that we know the value of k, we can use it to find M when N = 7. 7 = 0.4M Solving for M, we get: M = 7/0.4 = 17.5 Therefore, M is equal to 17.5 when N is equal to 7. The answer is \(17\frac{1}{2}\).

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To factorize \(6x^2 + 7x - 20\), we need to find two numbers that multiply to -120 (the product of the leading coefficient, 6, and the constant, -20) and add up to the coefficient of the middle term, 7. These numbers are 15 and -8. Therefore, we can write: \begin{align*} 6x^2 + 7x - 20 &= 6x^2 + 15x - 8x - 20 \\ &= 3x(2x+5) - 4(2x+5) \\ &= (3x-4)(2x+5) \end{align*} So the factorization of \(6x^2 + 7x - 20\) is \((3x-4)(2x+5)\).

Akọwa Nkọwa

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(a) Pass probabilities \(\tfrac{2}{3},\tfrac{5}{8},\tfrac{3}{4}\); fail probabilities \(\tfrac{1}{3},\tfrac{3}{8},\tfrac{1}{4}\).

(i) All three pass \(=\dfrac{2}{3}\times\dfrac{5}{8}\times\dfrac{3}{4}=\dfrac{30}{96}=\dfrac{5}{16}\).

(ii) None passes \(=\dfrac{1}{3}\times\dfrac{3}{8}\times\dfrac{1}{4}=\dfrac{3}{96}=\dfrac{1}{32}\).

(iii) Exactly two pass:

• 1&2 pass, 3 fails: \(\tfrac{2}{3}\cdot\tfrac{5}{8}\cdot\tfrac{1}{4}=\tfrac{10}{96}\)
• 1&3 pass, 2 fails: \(\tfrac{2}{3}\cdot\tfrac{3}{8}\cdot\tfrac{3}{4}=\tfrac{18}{96}\)
• 2&3 pass, 1 fails: \(\tfrac{1}{3}\cdot\tfrac{5}{8}\cdot\tfrac{3}{4}=\tfrac{15}{96}\)

Total \(=\dfrac{10+18+15}{96}=\dfrac{43}{96}\).

(b) Marked price \(=\text{N}36{,}000\) gives \(20\%\) profit, so cost price \(=\dfrac{36000}{1.20}=\text{N}30{,}000\).

Discount of \(5\%\): selling price \(=36000\times 0.95=\text{N}34{,}200\).

Profit \(=34200-30000=\text{N}4{,}200\); actual percentage profit \(=\dfrac{4200}{30000}\times 100=\mathbf{14\%}\).

Akọwa Nkọwa

(a) Table of values for \(y=x^2\)

\(x\)	\(-3\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)
\(y=x^2\)	9	4	1	0	1	4	9	16

(b) Values for the straight line \(y=2x+3\)

\(x\)	\(-3\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)
\(y=2x+3\)	\(-3\)	\(-1\)	1	3	5	7	9	11

Using the stated scales, plot the points and draw smooth curve \(y=x^2\) and the straight line \(y=2x+3\). A tangent to the curve at \((-2,4)\) is also shown for part (c)(ii).

(c)(i) The graphs intersect at \((-1,1)\) and \((3,9)\). Hence, the roots of \(x^2=2x+3\) are

\[\boxed{x=-1\text{ and }x=3}.\]

(c)(ii) From the tangent at \(x=-2\), take the points \((-3,8)\) and \((-1,0)\):

\[\text{gradient}=\frac{0-8}{-1-(-3)}=\frac{-8}{2}=\boxed{-4}.\]

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Given \(\log_{10}2=0.3010,\ \log_{10}3=0.4771\).

(a) \(54=2\times 27=2\times 3^{3}\).

\(\log_{10}54=\log_{10}2+3\log_{10}3=0.3010+3(0.4771)=0.3010+1.4313=\mathbf{1.7323}\).

(b) \(0.24=\dfrac{24}{100}=\dfrac{2^{3}\times 3}{100}\).

\(\log_{10}24=3\log_{10}2+\log_{10}3=0.9030+0.4771=1.3801\).

\(\log_{10}0.24=\log_{10}24-\log_{10}100=1.3801-2=\mathbf{-0.6199}\) (i.e. \(\bar{1}.3801\)).

Akọwa Nkọwa

(a) Volume of the balloon. Diameter \(=15\text{ cm}\Rightarrow r=7.5\text{ cm}\).

\(V=\dfrac{4}{3}\pi r^{3}=\dfrac{4}{3}\times\dfrac{22}{7}\times 7.5^{3}=\dfrac{88}{21}\times 421.875=\dfrac{37125}{21}=1767.86\).

Volume \(\approx \mathbf{1768\ cm^{3}}\).

(b) X: \(28^{\circ}\)N, \(105^{\circ}\)W; Y: \(28^{\circ}\)N, \(35^{\circ}\)E. Difference in longitude \(=105+35=140^{\circ}\).

(i) Distance X to Y along latitude \(28^{\circ}\)N:

\(d=\dfrac{140}{360}\times 2\pi R\cos 28^{\circ}=\dfrac{140}{360}\times 2\times\dfrac{22}{7}\times 6400\times\cos 28^{\circ}\).

\(=\dfrac{140}{360}\times 40228.57\times 0.8829\approx 13{,}800\text{ km}\ (\mathbf{1.38\times 10^{4}\ km})\).

(ii) Distance of X from the equator (along its meridian):

\(d=\dfrac{28}{360}\times 2\pi R=\dfrac{28}{360}\times 40228.57\approx \mathbf{3130\ km}\).

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From the diagram: \(AB=15\text{ m}\) on a bearing of \(053^\circ\), \(BC=18\text{ m}\) on a bearing of \(161^\circ\), the pole BT is vertical at B, and \(\angle ATB=58^\circ\).

(a) Length of AC. First find \(\angle ABC\). The bearing of A from B is the back-bearing of \(053^\circ\):

The bearing of C from B is \(161^\circ\), so

Apply the cosine rule to triangle ABC:

(b) Bearing of C from A. Use the sine rule to find \(\angle BAC\):

C lies clockwise of B as seen from A, so the bearing of C from A is

(c) Height of the pole BT. The pole is vertical and BA is horizontal, so triangle ATB is right-angled at B. With \(\angle ATB=58^\circ\) and \(AB=15\text{ m}\) opposite that angle:

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(a) Frequency distribution

Number of births	Class midpoint, \(x\)	Frequency, \(f\)	\(fx\)
45–54	49.5	1	49.5
55–64	59.5	2	119.0
65–74	69.5	15	1042.5
75–84	79.5	21	1669.5
85–94	89.5	7	626.5
95–104	99.5	3	298.5
105–114	109.5	1	109.5
Total		50	3915

(b) Histogram

The continuous class boundaries are \(44.5\text{–}54.5,\ 54.5\text{–}64.5,\ldots,\ 104.5\text{–}114.5\). Since all class widths are 10, the heights of the adjacent bars are their frequencies.

(c) Estimated mode

The modal class is \(75\text{–}84\). From the histogram, or by interpolation within this class,

\[\text{Mode}=L+\frac{f_1-f_0}{(f_1-f_0)+(f_1-f_2)}\times c\]

\[=74.5+\frac{21-15}{(21-15)+(21-7)}\times10\]

\[=74.5+\frac{6}{20}\times10=\boxed{77.5\text{ births (approximately)}}.\]

(d) Mean number of births

\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{3915}{50}=\boxed{78.3\text{ births}}.\]

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(a) Find \(x\).

From the diagram, \(MN\parallel ST\) and \(NP\parallel QT\), with \(\angle STQ=70^{\circ}\) at \(T\), and \(x\) is the angle at \(N\) between \(NM\) and \(NP\).

Extend \(QT\) to cut \(MN\) at a point \(W\). Because \(MN\parallel ST\) and \(QT\) is a transversal, corresponding angles give

Now \(MN\) is a transversal of the parallel lines \(NP\) and \(QT\). The ray \(NP\) points to the same side of \(MN\) as \(P\) (above), while \(WQ\) points to the opposite side (below), so \(x\) and \(\angle MWQ\) are co-interior (allied) angles:

(b) Find \(\angle BDA\).

From the diagram \(A,B,C\) lie on the straight line \(AC\), with \(D\) above it, \(|BC|=|BD|\), \(\angle BCD=50^{\circ}\) and \(\angle BAD=55^{\circ}\).

Triangle \(BCD\) is isosceles with \(|BC|=|BD|\), so the base angles are equal:

Since \(A,B,C\) are collinear, \(\angle DBA\) and \(\angle DBC\) are angles on a straight line at \(B\):

In triangle \(ABD\):

Akọwa Nkọwa

(a) Apples \(A=40\), Oranges \(O=46\), Mangoes \(M=50\). \(A\cap O=14\), \(A\cap M=15\), all three \(=10\); every trader sells at least one, total \(=100\).

(i) Venn diagram. Three intersecting circles; centre (all three) \(=10\).

(ii) Let \(O\cap M\) (total) \(=y\). By the inclusion-exclusion principle:

\(|A|+|O|+|M|-|A\cap O|-|A\cap M|-|O\cap M|+|A\cap O\cap M|=100\).

\(40+46+50-14-15-y+10=100 \Rightarrow 117-y=100 \Rightarrow y=17\).

Oranges and mangoes only \(=y-10=17-10=\mathbf{7}\).

(b) \(312_{four}+52_x=96_{ten}\).

\(312_{four}=3(16)+1(4)+2=54\). So \(52_x=96-54=42\).

\(52_x=5x+2=42 \Rightarrow 5x=40 \Rightarrow x=\mathbf{8}\).

Akọwa Nkọwa

Read the height of every bar against the vertical scale, which is marked in steps of \(5\) vehicles. Match each shading to the key: fine dots are cars, solid black is lorries, diagonal cross-hatching is buses. Reading each of the three bars in each month group gives the table below.

The same data drawn to scale:

(a) Vehicles produced in February. Add the three February bars:

\[ 40 + 25 + 30 = 95 \text{ vehicles.} \]

(b) Fraction of February vehicles that were cars. Cars in February over the February total:

\[ \frac{40}{95} = \frac{8}{19}. \]

(c) Buses produced altogether, January to March. Add the three bus readings:

\[ 20 + 30 + 50 = 100 \text{ buses.} \]

(d) Ratio of lorries in February to lorries in March. The lorry bars read \(25\) and \(35\). Divide both by their highest common factor, \(5\):

\[ 25 : 35 = 5 : 7. \]

The key point when reading a grouped bar chart is to fix the scale first (each square here is \(5\) vehicles) and to match every bar to the correct shading in the key before adding. Miscounting the scale or reading the wrong shading is what leads to a wrong total such as treating February as \(75\) instead of \(95\).

Akọwa Nkọwa

Sides \(x\) and \((x+7)\), diagonal \((x+8)\). By Pythagoras:

\(x^{2}+(x+7)^{2}=(x+8)^{2}\).

\(x^{2}+x^{2}+14x+49=x^{2}+16x+64\).

\(2x^{2}+14x+49-x^{2}-16x-64=0 \Rightarrow x^{2}-2x-15=0\).

\((x-5)(x+3)=0 \Rightarrow x=5\) (reject \(x=-3\), a length must be positive).

(a) \(x=5\ \text{m}\).

(b) Area \(=x(x+7)=5\times 12=\mathbf{60\ m^{2}}\).

Akọwa Nkọwa

(a) \(\dfrac{1}{3^{5n}}\times 9^{n-1}\times 27^{n+1}\). Convert everything to base 3:

\(9^{n-1}=3^{2(n-1)}=3^{2n-2}\); \(27^{n+1}=3^{3(n+1)}=3^{3n+3}\); \(\dfrac{1}{3^{5n}}=3^{-5n}\).

Add exponents: \(-5n+(2n-2)+(3n+3)= (-5n+2n+3n)+(-2+3)=0+1=1\).

Result \(=3^{1}=\mathbf{3}\).

(b) Let woman \(=w\), daughter \(=d\). \(w+d=46\).

In 4 years: \(\dfrac{w+4}{d+4}=\dfrac{7}{2}\Rightarrow 2(w+4)=7(d+4)\Rightarrow 2w+8=7d+28\Rightarrow 2w-7d=20\).

Substitute \(w=46-d\): \(2(46-d)-7d=20\Rightarrow 92-9d=20\Rightarrow 9d=72\Rightarrow d=8\), \(w=38\).

Woman is 38 years, daughter is 8 years. Check: in 4 years \(42:12=7:2\). \(\checkmark\)

Akọwa Nkọwa

(a) The equation is \(2x^{2}+(p+1)x+q=0\) with roots \(1\) and \(3\) (the constant term is the unknown \(q\), not a fixed number).

Sum of roots \(=1+3=4=-\dfrac{p+1}{2}\Rightarrow p+1=-8\Rightarrow p=-9\).

Product of roots \(=1\times 3=3=\dfrac{q}{2}\Rightarrow q=6\).

\(p=-9,\ q=6\).

(b) Weight \(W\propto\dfrac{1}{d^{2}}\), so \(W=\dfrac{k}{d^{2}}\). On the surface, \(d=6400\text{ km}\), \(W=80\text{ kg}\):

\(k=80\times 6400^{2}\).

At \(800\text{ km}\) above the surface, \(d=6400+800=7200\text{ km}\):

\(W=80\times\left(\dfrac{6400}{7200}\right)^{2}=80\times\dfrac{64}{81}=\dfrac{5120}{81}=63.2\).

Weight \(\approx \mathbf{63\ kg}\).