WAEC SSCE - General Mathematics - 2005

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The formula for the total surface area of a cylinder is given by: `2πrh + 2πr^2` Where `r` is the radius of the circular base of the cylinder and `h` is its height. In this case, the diameter of the cylinder is given as 7cm. We need to find the radius of the cylinder first. We know that the diameter is 7cm, which means the radius is half of the diameter. `radius = 7/2 = 3.5cm` The height of the cylinder is given as 10cm. Now, we can substitute the values in the formula to find the total surface area of the cylinder. `2πrh + 2πr^2` `= 2 × (22/7) × 3.5 × 10 + 2 × (22/7) × 3.5^2` `= 220 + 77` `= 297cm^2` Therefore, the total surface area of the cylinder is 297cm^2. The answer is.

Akọwa Nkọwa

Let's use the Pythagorean Theorem which states that in a right-angled triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides. Therefore, for this triangle, we have: \begin{align*} (\text{hypotenuse})^2 &= (\text{one side})^2 + (\text{other side})^2 \\ (\sqrt{13})^2 &= x^2 + (x-1)^2 \\ 13 &= x^2 + (x-1)^2 \\ 13 &= x^2 + x^2 - 2x + 1 \\ 0 &= 2x^2 - 2x - 12 \\ 0 &= x^2 - x - 6 \\ 0 &= (x-3)(x+2) \end{align*} We get two values: x=3 and x=-2. However, x must be positive, so we choose x=3 as the correct value. Therefore, the value of x is 3.

Akọwa Nkọwa

Akọwa Nkọwa

We can use the formula:

distance = speed x time

to solve the problem.

We are given that the train covers 36 km in 21 minutes. We want to find how long it will take to cover 60 km.

Let's first convert 21 minutes to hours:

21 minutes = 21/60 hours = 0.35 hours

Now we can use the formula to find the speed of the train:

36 km = speed x 0.35 hours

Solving for speed, we get:

speed = 36/0.35 km/h

Using a calculator, we can evaluate this expression to get:

speed = 102.86 km/h

Now we can use the speed to find the time it takes to cover 60 km:

60 km = 102.86 km/h x time

Solving for time, we get:

time = 60/102.86 hours

Converting this to minutes, we get:

time = (60/102.86) x 60 minutes

Simplifying this expression, we get:

time = 35 minutes (approx)

Therefore, the train will take approximately 35 minutes to cover 60 km.

So the correct option is 1.

Akọwa Nkọwa

Akọwa Nkọwa

We can use the formula for percentage increase: percentage increase = (new value - old value) / old value x 100% Here, the old salary was N2,700 and the new salary is N3,200. So, the percentage increase in salary would be: = (3200 - 2700) / 2700 x 100% = 500 / 2700 x 100% = 0.185 x 100% = 18.5% Therefore, the percentage increase in the man's salary is 18.5%. Hence, the answer is: 18.5%.

Akọwa Nkọwa

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Akọwa Nkọwa

To solve the system of equations 2x - 3y = 22 and 3x + 2y = 7, we can use the method of elimination. Multiplying the first equation by 2 and the second equation by 3, we get: 4x - 6y = 44 (1) 9x + 6y = 21 (2) Adding equations (1) and (2), we get: 13x = 65 Dividing both sides by 13, we get: x = 5 Substituting x = 5 into the second equation, we get: 3(5) + 2y = 7 Simplifying: 15 + 2y = 7 Subtracting 15 from both sides, we get: 2y = -8 Dividing both sides by 2, we get: y = -4 Therefore, the solution to the system of equations is x = 5 and y = -4, which corresponds to option (D).

Akọwa Nkọwa

Since PQRS is a quadrilateral with equal sides and opposite sides parallel, it must be a parallelogram. Since PR and QS are diagonals of this parallelogram and intersect perpendicularly at O, O is the midpoint of both PR and QS. Therefore, |PO|=|RO|. If PQR is an equilateral triangle, then PQ=QR=RP, which means that PQRS would be a rhombus (a special case of a parallelogram with all sides equal), not just a parallelogram. This contradicts the given information, so option (B) cannot be correct. If PQRX is a parallelogram, then PQ and RX are parallel and PQ=RX because PQRS is a parallelogram with opposite sides parallel and equal in length. This would mean that the diagonals PR and QS intersect at the midpoint of PQ and RX, not perpendicularly. This contradicts the given information that PR and QS intersect perpendicularly at O, so option (C) cannot be correct. There are no conditions given about the symmetry of the quadrilateral PQRS, so option (D) may or may not be correct. The presence or absence of lines of symmetry does not contradict any of the given information or the other options. Therefore, the statement that cannot be correct is option (B), "PQR is an equilateral triangle," since it contradicts the given information that PQRS is a parallelogram with equal sides and opposite sides parallel.

Akọwa Nkọwa

To find the area of the sector, we first need to find the length of the arc XOY.

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

length of arc = (angle/360) x 2 x ϖ x radius

Substituting the given values, we get:

length of arc = (144/360) x 2 x ϖ x 3.5

Simplifying this expression, we get:

length of arc = 1.4ϖ

So the length of arc XOY is 1.4ϖ cm.

To find the area of the sector, we use the formula:

area of sector = (angle/360) x ϖ x radius^2

Substituting the given values, we get:

area of sector = (144/360) x ϖ x 3.5^2

Simplifying this expression, we get:

area of sector = 4.9ϖ

Therefore, the area of the sector is 4.9ϖ cm².

So the correct option is 4.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

To expand (2x-3y)(x-5y), we use the distributive property of multiplication, which states that a(b+c) = ab + ac. Applying this property, we get: (2x-3y)(x-5y) = 2x(x-5y) - 3y(x-5y) = 2x^2 - 10xy - 3xy + 15y^2 = 2x^2 - 13xy + 15y^2 Therefore, the answer is (2x^2 - 13xy + 15y^2).

Akọwa Nkọwa

The locus of a point R which moves in a plane such that it is equidistant from two intersecting lines is a bisector of the angle formed by the lines. This is because a point equidistant from two intersecting lines is located at the same distance from each line, and the only points that satisfy this condition lie on the bisector of the angle formed by the lines. Therefore, the correct option is: the bisector of the angle formed by the lines.

Akọwa Nkọwa

To find the distance between points P and R, we can use the Pythagorean theorem which states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In this case, we can consider P, R, and the point where the southward line from P intersects with the westward line from R, forming a right-angled triangle as shown in the diagram below:

                   R
                   *
                   |
                   |
                   |12 km
                   |
                   |
            *------P
          5 km

Let's call the point where the two lines intersect Q. Then we can see that the distance between P and Q is 5km, and the distance between Q and R is 12km. Therefore, we can use the Pythagorean theorem to find the distance between P and R as follows:

distance between P and R = sqrt(5^2 + 12^2)
                         = sqrt(169)
                         = 13 km

Therefore, the correct answer is 13km.

Akọwa Nkọwa

To make `f` the subject of the relation `v = u + ft`, we need to isolate `f` on one side of the equation. Starting with `v = u + ft`, we can begin by subtracting `u` from both sides to get: `v - u = ft` Then, dividing both sides by `t`, we get: `f = (v - u)/t` Therefore, the correct answer is: - `(v-u)/t`

Akọwa Nkọwa

A fair die has six equally likely outcomes, which are the numbers 1, 2, 3, 4, 5, and 6. To find the probability of obtaining neither 5 nor 2, we first need to find the number of outcomes that satisfy this condition. There are four outcomes that are not 5 or 2, which are 1, 3, 4, and 6. Therefore, the probability of obtaining neither 5 nor 2 is the number of outcomes that satisfy this condition divided by the total number of possible outcomes, which is 4/6 or 2/3. Thus, the answer is option (B), \(\frac{2}{3}\). To summarize, the probability of obtaining neither 5 nor 2 when a fair die is tossed once is 2/3, since there are four outcomes that satisfy this condition out of a total of six possible outcomes.

Akọwa Nkọwa

The equation of the curve is: \(y=x^2-5x+4\) This is a quadratic equation in standard form, where the coefficient of the \(x^2\) term is 1, the coefficient of the \(x\) term is -5, and the constant term is 4. The graph of a quadratic equation is a parabola. The coefficient of the \(x^2\) term determines whether the parabola opens upwards or downwards. In this case, since the coefficient is positive, the parabola opens upwards. To find the vertex of the parabola, we can use the formula: \(-\frac{b}{2a}\), where \(a\) is the coefficient of the \(x^2\) term and \(b\) is the coefficient of the \(x\) term. Substituting the values, we get: \(-\frac{b}{2a}=-\frac{-5}{2(1)}=\frac{5}{2}\) So the vertex is at \(\left(\frac{5}{2},\frac{1}{4}\right)\). Therefore, the equation of the curve is \(y=x^2-5x+4\).

Akọwa Nkọwa

When rounding a number to a certain number of significant figures, we look at the digit to the right of the last significant figure we want to keep. If that digit is 5 or greater, we round up the last significant figure. If it is less than 5, we simply drop all the digits to the right of the last significant figure. In this case, we want to round 0.04945 to two significant figures. The last significant figure we want to keep is the 9, so we look at the digit to its right, which is 4. Since 4 is less than 5, we simply drop all the digits to the right of the last significant figure, and the rounded value is 0.049. Therefore, the correct answer is option (ii) 0.049.

Akọwa Nkọwa

If 2 is a root of the equation \(2x^2 + (k+2)x+k=0\), then we can substitute x = 2 into the equation and get: \[2(2)^2 + (k+2)(2) + k = 0\] Simplifying this gives: \[8 + 2k + 4 + k = 0\] \[3k = -12\] \[k = -4\] Therefore, the value of k is -4.

Akọwa Nkọwa

We can rewrite the equation in exponential form as follows: $$9^{1.5} = x$$ Using the exponent rules, we can simplify $9^{1.5}$ as: $$9^{1.5} = \sqrt{9^3} = 3^3 = 27$$ Therefore, $x = 27$. So the correct option is (B) 27.

Akọwa Nkọwa

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To evaluate the expression, we must follow the order of operations, which is also known as PEMDAS (parentheses, exponents, multiplication and division, addition and subtraction). Starting with the parentheses, we have: $$7\frac{1}{2}-\left(2\frac{1}{2}+3\right)\div\frac{33}{2}$$ $$=7\frac{1}{2}-\left(5+\frac{3}{1}\right)\div\frac{33}{2}$$ $$=7\frac{1}{2}-\left(5+\frac{3}{1}\right)\times\frac{2}{33}$$ $$=7\frac{1}{2}-\left(5\times\frac{2}{33}+\frac{3}{1}\times\frac{2}{33}\right)$$ $$=7\frac{1}{2}-\left(\frac{10}{33}+\frac{6}{33}\right)$$ $$=7\frac{1}{2}-\frac{16}{33}$$ Now, we need to find a common denominator to subtract the fractions. We can convert the mixed number to an improper fraction and multiply by $\frac{33}{33}$ to get: $$=7\frac{1}{2}\times\frac{33}{33}-\frac{16}{33}$$ $$=\frac{15}{2}\times\frac{33}{33}-\frac{16}{33}$$ $$=\frac{495}{66}-\frac{16}{33}$$ $$=\frac{495-32}{66}$$ $$=\frac{463}{66}$$ To round to the nearest whole number, we divide 463 by 66 and round to the nearest whole number: $$\frac{463}{66}\approx7$$ Therefore, the correct answer is 7.

Akọwa Nkọwa

Akọwa Nkọwa

The probability that John passes the exam is 3/4, so the probability that John fails the exam is 1-3/4=1/4. Similarly, the probability that James fails the exam is 1-3/5=2/5. Now, we need to find the probability that both John and James fail the exam. We can use the multiplication rule of probability which states that the probability of two independent events occurring together is the product of their individual probabilities. Therefore, the probability of both John and James failing the exam is: (1/4) x (2/5) = 1/10 So, the answer is \(\frac{1}{10}\).

Akọwa Nkọwa

To find the perimeter of the sector, we need to find the length of the arc and the length of the two radii that form the sector. The circumference of a circle with radius 14cm is given by: C = 2πr = 2π(14) = 28π Therefore, the length of the arc subtending an angle of 300 degrees is: L = (300/360) × 28π = 7π/3 × 28 ≈ 73.33cm The length of the two radii that form the sector is: 2r = 2 × 14 = 28cm Therefore, the perimeter of the sector is: P = L + 2r ≈ 73.33 + 28 = 101.33cm Hence, the correct option is (c) 101.33cm.

Akọwa Nkọwa

Akọwa Nkọwa

To determine the values of y for which the expression \(\frac{6y-1}{y^2 - y-6}\) is defined, we need to find the values of y that make the denominator of the expression nonzero. The denominator of the expression is \(y^2 - y-6\), which can be factored as \((y-3)(y+2)\). Therefore, the expression is undefined when \(y=3\) or \(y=-2\), because in those cases the denominator becomes zero. So, the values of y for which the expression is defined are all the real numbers except 3 and -2. That means, y can be any number other than 3 and -2. Therefore, the correct answer is: y can be any number other than 3 and -2.

Akọwa Nkọwa

To simplify the expression \(\frac{2}{3xy} - \frac{3}{4yz}\), we need to find a common denominator and combine the two fractions.

The common denominator of the two fractions is 12xyz, which is the lowest multiple of 3xy and 4yz. We can convert each fraction to an equivalent fraction with the common denominator of 12xyz as follows:

 2     3     4
--  - --  = --
3xy   4yz   12xyz

To find the numerators of the equivalent fractions, we multiply each numerator by the missing factor in the denominator of the other fraction. For example, we multiply the numerator of the first fraction by 4z to get a denominator of 12xyz, and we multiply the numerator of the second fraction by 3x to get a denominator of 12xyz:

 2(4z)     3(3x)     8z - 9x
------  - ------  = --------
3xy(4z)   4yz(3x)   12xyz

Now we can simplify the expression by combining the two fractions:

 8z - 9x
--------
12xyz

Therefore, the simplified expression is \(\frac{8z-9x}{12xyz}\), which corresponds to.

Therefore, the correct answer is: \(\frac{8z-9x}{12xyz}\).

Akọwa Nkọwa

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The sum of the exterior angles of any polygon is 360 degrees. In a regular polygon, all the exterior angles have the same measure, so to find the measure of one exterior angle, we divide 360 by the number of sides. For a regular polygon with 9 sides, each exterior angle has a measure of 360/9 = 40 degrees. Therefore, the answer is option (B), 40^o. To summarize, the measure of one exterior angle of a regular polygon with 9 sides is 40 degrees, since the sum of the exterior angles is 360 degrees and they are all equal in measure.

Akọwa Nkọwa

When Ladi sold the car, he made a loss of 4%. This means that he sold the car for 96% of its original value (100% - 4% = 96%). We can use this information to find the original price of the car. Let's assume that the original price of the car was x Naira. Then, the selling price of the car (i.e., the price at which Ladi sold the car) would be 96% of x, or 0.96x Naira. We know that Ladi sold the car for N84,000, so we can set up an equation: 0.96x = 84,000 Solving for x, we get: x = 84,000 / 0.96 x = 87,500 Therefore, the original price of the car was N87,500. So the correct option is 4.

Akọwa Nkọwa

We know that $8 = 2^3$, so $8^{5-x} = (2^3)^{5-x} = 2^{3(5-x)} = 2^{15-3x}$. Therefore, the equation becomes: $$2^7 = 2^{15-3x}$$ Since the bases are equal, we can equate the exponents: $$7 = 15-3x$$ Solving for $x$ gives: $$x = \frac{15-7}{3} = \frac{8}{3}$$ Therefore, the solution is $\frac{8}{3}$.

Akọwa Nkọwa

Akọwa Nkọwa

To find the mean of a set of numbers, we add up all the numbers and then divide the sum by the total number of numbers. So, in this case, we add up the numbers: $$1+3+4+8+8+4+7=35$$ There are 7 numbers in the set, so we divide the sum by 7 to get: $$\frac{1+3+4+8+8+4+7}{7}=\frac{35}{7}=5$$ Therefore, the mean (also known as the average) of the numbers 1, 3, 4, 8, 8, 4, and 7 is 5. So, the correct answer is 5.

Akọwa Nkọwa

Akọwa Nkọwa

To calculate the percentage increase in the salary of a man, we need to find the difference between the new and old salary and then express it as a percentage of the old salary. Let's assume the old salary was 40x and the new salary is 47x (where x is a constant). Then, the increase in salary = 47x - 40x = 7x. Now, the percentage increase in salary = (increase in salary / old salary) x 100% = (7x / 40x) x 100% = 17.5% Therefore, the percentage increase in the salary of the man is 17.5%. Answer is correct.

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First, let's find P^I, which is the complement of P in ξ: P^I = ξ \ P = {1, 7, 15} Similarly, Q^I = ξ \ Q = {1, 11, 13} The intersection of P^I and Q^I is the set of elements that are in both sets. In this case, the intersection is: P^I ∩ Q^I = {1} Therefore, the answer is option (C) {1}.

Akọwa Nkọwa

To solve this problem, we need to calculate the volume of the rectangular packet and the volume of each cube of sugar, and then divide the volume of the packet by the volume of each cube. The volume of the rectangular packet is: $$V_{packet} = 16 \text{cm} \times 12 \text{cm} \times 6 \text{cm} = 1152 \text{cm}^3$$ The volume of each cube of sugar is: $$V_{cube} = 2 \text{cm} \times 2 \text{cm} \times 2 \text{cm} = 8 \text{cm}^3$$ To calculate how many cubes of sugar can be packed into the packet, we divide the volume of the packet by the volume of each cube: $$\frac{V_{packet}}{V_{cube}} = \frac{1152 \text{cm}^3}{8 \text{cm}^3} = 144$$ Therefore, 144 cubes of sugar of side 2cm can be neatly packed into the rectangular packet. Hence, the correct answer is 144.

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To convert a binary number to a decimal number, we need to multiply each digit of the binary number by its corresponding power of 2, starting from the rightmost digit with the power of 0, and then add the products.

Let's apply this method to convert the binary number 101101 to a decimal number:

1    0    1    1    0    1    (binary digits)
25  24  23  22  21  20  (corresponding powers of 2)
32   0    8    4    0    1     (products)

Now we can add the products to get the decimal equivalent of the binary number:

32 + 0 + 8 + 4 + 0 + 1 = 45

Therefore, the binary number 101101_two is equal to the decimal number 45, which corresponds to.

Therefore, the correct answer is: 45.

Akọwa Nkọwa

The given sequence is an arithmetic sequence, because the difference between consecutive terms is constant. To find the common difference d, we subtract the second term from the first term:

5 - 2.5 = 2.5

So the common difference is 2.5. We can use this common difference to find any term of the sequence, using the formula:

an = a1 + (n - 1)d

where an is the nth term, a1 is the first term, n is the term number, and d is the common difference.

We know that the nth term is 25, so we can set an = 25 and solve for n:

25 = 2.5 + (n - 1)2.5

Simplifying the equation, we get:

22.5 = 2.5n - 2.5
25 = 2.5n
n = 10

Therefore, the term number n for which the nth term of the sequence is 25 is 10, which corresponds to.

Therefore, the correct answer is: 10.

Akọwa Nkọwa

(a) The regular polygon

(i) Value of n. At any vertex the interior and exterior angles are supplementary, and here the interior angle is 120° greater than the exterior angle. Let the exterior angle be \(x\):

For any polygon the exterior angles sum to \(360^\circ\), so

The polygon has 12 sides.

(ii) Sum of all the interior angles.

The interior angles sum to 1800°.

(b) The boy's journey

(i) Sketch of the movement. He walks 6 km from \(P\) to \(Q\) on a bearing of 065°, then 13 km from \(Q\) to \(R\) on a bearing of 146°. A North line is drawn at each point so the bearings can be measured clockwise from North.

(ii) Distance PR. First find the interior angle of the triangle at \(Q\). The bearing of \(P\) from \(Q\) is \(065^\circ + 180^\circ = 245^\circ\), and the bearing of \(R\) from \(Q\) is \(146^\circ\). Working from the North line at \(Q\), the angle between \(QP\) and the North line is \(180^\circ - 65^\circ = 115^\circ\) on the western side, giving the angle inside the triangle:

Apply the cosine rule to triangle \(PQR\):

Since \(\cos 99^\circ = -0.1564\):

Correct to the nearest kilometre, \(|PR| \approx \mathbf{15\text{ km}}\).

Akọwa Nkọwa

(a) Smallest integer satisfying the inequality

\[x + 8 < 4x - 15\]

Collect like terms:

\[8 + 15 < 4x - x\]

\[23 < 3x\]

\[x > \frac{23}{3} = 7\tfrac{2}{3} \approx 7.67\]

The smallest integer greater than \(7.67\) is \(\mathbf{8}\).

(b) Monthly income of the sales girl

Fixed salary \(= \text{N}2{,}500\).

Commission \(= 5\text{ kobo in the naira} = \dfrac{5}{100}\text{ naira per naira} = \text{N}0.05\) per naira of sales.

Commission on N200,000 sales:

\[0.05 \times 200{,}000 = \text{N}10{,}000\]

Total income:

\[2{,}500 + 10{,}000 = \text{N}12{,}500\]

Her income for the month is \(\mathbf{\text{N}12{,}500.00}\).

(c) Area of the window

The window is a rectangle with a semicircle on top. From the diagram, the semicircular part has radius \(35\text{ cm}\), so the width of the rectangle equals the diameter:

\[\text{width} = 2 \times 35 = 70\text{ cm}, \qquad \text{height} = 50\text{ cm}\]

Area of rectangle:

\[A_1 = 70 \times 50 = 3{,}500\text{ cm}^2\]

Area of semicircle: (with \(\pi = \tfrac{22}{7}\))

\[A_2 = \frac{1}{2}\pi r^2 = \frac{1}{2}\times \frac{22}{7} \times 35^2 = \frac{1}{2}\times \frac{22}{7} \times 1225\]

\[A_2 = \frac{1}{2}\times 22 \times 175 = \frac{1}{2}\times 3{,}850 = 1{,}925\text{ cm}^2\]

Total area of window:

\[A = A_1 + A_2 = 3{,}500 + 1{,}925 = \mathbf{5{,}425\text{ cm}^2}\]

Akọwa Nkọwa

(a) Circle through A, B, C, D with tangent XY at A.

From the diagram: \(\angle ADB=20^\circ\) (at D), and the tangent-secant angle at Y (between tangent \(YA\) and the secant through \(B\) and \(C\)) is \(\angle AYC=69^\circ\), with \(C,\,B,\,Y\) in a straight line.

(i) \(\angle CAX\).

\(\angle ADB=20^\circ\) is an inscribed angle on chord \(AB\), so

\[\text{arc } AB=2\times20^\circ=40^\circ.\]

The tangent-chord angle \(\angle BAY\) equals the angle in the alternate segment \(\angle ADB\):

\[\angle BAY=20^\circ.\]

In triangle \(ABY\), \(\angle AYB=69^\circ\), so

\[\angle ABY=180^\circ-69^\circ-20^\circ=91^\circ,\]

and since \(C,B,Y\) are collinear,

\[\angle ABC=180^\circ-\angle ABY=180^\circ-91^\circ=89^\circ.\]

By the alternate segment theorem, the tangent-chord angle \(\angle CAX\) equals the inscribed angle \(\angle ABC\) in the alternate segment:

\[\angle CAX=\angle ABC=\boxed{89^\circ}.\]

(ii) \(\angle ABY\).

From the working above,

\[\angle ABY=180^\circ-69^\circ-20^\circ=\boxed{91^\circ}.\]

(b) Factors of \(m^{2}-km+c\).

If \((m+1)\) and \((m-3)\) are factors, then

\[m^{2}-km+c=(m+1)(m-3)=m^{2}-3m+m-3=m^{2}-2m-3.\]

Comparing coefficients:

\[-k=-2\ \Rightarrow\ \boxed{k=2},\qquad c=\boxed{-3}.\]

Akọwa Nkọwa

(a) Value of x

In the diagram \(\triangle ABD\) is right-angled at \(B\), with \(|AB| = 3\ \text{cm}\) and hypotenuse \(|AD| = 5\ \text{cm}\). Point \(C\) lies on \(BD\), \(\angle ACB = 61^\circ\) and \(\angle DAC = x^\circ\).

Step 1 - the whole angle at A. In right triangle \(ABD\), \(AB\) is adjacent to \(\angle BAD\) and \(AD\) is the hypotenuse, so

\[\cos(\angle BAD)=\frac{|AB|}{|AD|}=\frac{3}{5}=0.6,\qquad \angle BAD=\cos^{-1}(0.6)=53.13^\circ.\]

Step 2 - the lower part of that angle. Triangle \(ACB\) is also right-angled at \(B\), so its three angles give

\[\angle BAC=180^\circ-90^\circ-61^\circ=29^\circ.\]

Step 3 - subtract. Since \(C\) lies on \(BD\), the angle \(\angle DAC\) is the difference between the whole angle at \(A\) and \(\angle BAC\):

\[x=\angle BAD-\angle BAC=53.13^\circ-29^\circ=24.13^\circ.\]

Correct to one decimal place, \(x = \mathbf{24.1}\).

(b) The pyramid

\(OABCD\) is a pyramid on a square base of side \(2\ \text{cm}\) with a slant height of \(4\ \text{cm}\). In a pyramid the slant height is the distance from the apex \(O\) down the middle of a triangular face to the midpoint of a base edge (call that midpoint \(M\)). It is not the sloping corner edge \(OA\); mixing the two up is the usual error here. The foot \(N\) of the vertical height is the centre of the square base.

(i) Vertical height

\(N\) is the centre of the base and \(M\) is the midpoint of a base edge, so \(NM\) is half the side:

\[NM=\tfrac{1}{2}\times 2=1\ \text{cm}.\]

The vertical height \(h=ON\), the slant height \(l=OM=4\ \text{cm}\) and \(NM=1\ \text{cm}\) form a right triangle at \(N\). By Pythagoras' theorem,

\[l^{2}=h^{2}+NM^{2}\ \Rightarrow\ h^{2}=4^{2}-1^{2}=16-1=15,\]

\[h=\sqrt{15}=3.873\ \text{cm}.\]

Correct to three significant figures, \(h = \mathbf{3.87\ \text{cm}}\).

(ii) Volume

\[V=\frac{1}{3}\times(\text{base area})\times h=\frac{1}{3}\times(2\times 2)\times\sqrt{15}=\frac{4\sqrt{15}}{3}.\]

\[V=\frac{4\times 3.873}{3}=\frac{15.49}{3}=5.164\ \text{cm}^{3}.\]

Correct to three significant figures, \(V = \mathbf{5.16\ \text{cm}^{3}}\).

Akọwa Nkọwa

(a) A negative index inverts the base:

\[\left(\tfrac{64}{125}\right)^{-\frac{2}{3}} = \left(\tfrac{125}{64}\right)^{\frac{2}{3}} = \left(\sqrt[3]{\tfrac{125}{64}}\right)^{2} = \left(\tfrac{5}{4}\right)^{2} = \tfrac{25}{16}.\]

\[2 \div \tfrac{25}{16} = 2 \times \tfrac{16}{25} = \tfrac{32}{25} = 1\tfrac{7}{25}.\]

(b) At the point of intersection k the two y-values are equal:

\[3x + 5 = -4x - 1 \Rightarrow 7x = -6 \Rightarrow x = -\tfrac{6}{7}.\]

\[y = 3\left(-\tfrac{6}{7}\right) + 5 = -\tfrac{18}{7} + \tfrac{35}{7} = \tfrac{17}{7}.\]

Coordinates of k: \(\left(-\tfrac{6}{7},\ \tfrac{17}{7}\right)\).

Akọwa Nkọwa

(a) Simplify inside the root first:

\[\frac{0.18\times12.5}{0.05\times0.2} = \frac{2.25}{0.01} = 225,\qquad \sqrt{225} = 15.\]

(b) Write each surd in simplest form: \(\sqrt{18} = 3\sqrt2\) and \(\sqrt{50} = 5\sqrt2\).

\[\frac{8 - 4\sqrt{18}}{\sqrt{50}} = \frac{8 - 12\sqrt2}{5\sqrt2} = \frac{8}{5\sqrt2} - \frac{12\sqrt2}{5\sqrt2} = \frac{4\sqrt2}{5} - \frac{12}{5} = \frac{4\sqrt2 - 12}{5}.\]

(c) The relation is \(x = \dfrac{k y^3}{z^2}\). Use \(x=108,\,y=3,\,z=4\):

\[108 = \frac{k(27)}{16}\Rightarrow k = \frac{108\times16}{27} = 64.\]

Now find z when \(x=4000,\,y=10\):

\[4000 = \frac{64(1000)}{z^2}\Rightarrow z^2 = \frac{64000}{4000} = 16\Rightarrow z = 4.\]

Akọwa Nkọwa

(a) Completing the table. For each value of \(x\) we evaluate \(y = x^{2} - 2x - 2\) by building it up in rows:

The three missing entries are \(y=13\) (at \(x=-3\)), \(y=1\) (at \(x=-1\)) and \(y=-3\) (at \(x=1\)).

(b) Graph of \(y = x^{2} - 2x - 2\). Plotting the nine points \((x,y)\) and joining them with a smooth curve gives the parabola below. The dashed horizontal line \(y = 2.5\) is used in part (c)(ii) and the vertical dashed line \(x = 1\) is the line of symmetry from part (c)(iii).

(c) Using the graph.

(i) Roots of \(x^{2} - 2x - 2 = 0\). These are the values of \(x\) where the curve cuts the \(x\)-axis, i.e. where \(y = 0\). The curve crosses the \(x\)-axis at \[ x \approx -0.7 \quad \text{and} \quad x \approx 2.7. \]

(ii) Values of \(x\) for which \(x^{2} - 2x - 4\tfrac{1}{2} = 0\). Rearranging so that the left side becomes our plotted expression: \[ x^{2} - 2x - 4\tfrac{1}{2} = 0 \;\Rightarrow\; x^{2} - 2x - 2 = 2\tfrac{1}{2} = 2.5. \] So we draw the horizontal line \(y = 2.5\) and read off where it meets the curve: \[ x \approx -1.3 \quad \text{and} \quad x \approx 3.3. \]

(iii) Equation of the line of symmetry. The lowest point (vertex) of the curve occurs at \(x = \dfrac{-(-2)}{2(1)} = 1\), where \(y = -3\). The curve is symmetrical about the vertical line through this point, so the line of symmetry is \[ x = 1. \]

Akọwa Nkọwa

(a) Work "of" and \(\div\) before subtraction:

\[\tfrac{5}{8}\text{ of }2\tfrac{1}{2} = \tfrac{5}{8}\times\tfrac{5}{2} = \tfrac{25}{16},\qquad \tfrac{3}{4}\div\tfrac{3}{5} = \tfrac{3}{4}\times\tfrac{5}{3} = \tfrac{5}{4} = \tfrac{20}{16}.\]

\[\tfrac{25}{16} - \tfrac{20}{16} = \tfrac{5}{16}.\]

(b) The cone and the pyramid have equal heights \(h\) and equal volumes.

\[V_{\text{cone}} = \tfrac{1}{3}\pi r^2 h,\qquad V_{\text{pyramid}} = \tfrac{1}{3}(\text{base area})h.\]

Since the volumes and heights are equal, the base areas are equal:

\[\pi r^2 = 154 \Rightarrow \tfrac{22}{7}r^2 = 154 \Rightarrow r^2 = 154\times\tfrac{7}{22} = 49.\]

\[r = 7\text{ cm}.\]

Akọwa Nkọwa

Solution

The completed scale construction is shown below. Everything that follows is read directly from it.

(a) Accurate scale drawing (1 cm to 10 m). Each real length is divided by 10 to get the drawing length:

\(|AB| = 85\text{ m} = \dfrac{85}{10} = 8.5\text{ cm}, \qquad |CA| = |CB| = 111\text{ m} = \dfrac{111}{10} = 11.1\text{ cm}.\)

Points A, B and C are plotted with ruler and compasses to these lengths, giving the triangular plot above.

(b) The two loci.

(i) \(l_1\), the locus of points equidistant from the sides \(AC\) and \(BC\), is the bisector of angle \(ACB\). It is constructed from C (shown by the two equal angle marks) and drawn as the blue dashed line.

(ii) \(l_2\), the locus of points 60 m from A, is a circle centred at A with radius \(\dfrac{60}{10} = 6\text{ cm}\). The relevant part of this arc is drawn in red.

(c) Locating T and finding \(|TC|\). The tree T lies on both loci, so it is the point where \(l_1\) meets \(l_2\) inside the plot. Measuring the drawing length of \(CT\):

\(CT = 6.0\text{ cm}.\)

Converting back to the real distance using the scale (multiply by 10):

\(|TC| = 6.0 \times 10 = 60\text{ m}.\)

(d) Region for the flagpole P. P must be:

• nearer AC than BC – this is the side of the bisector \(l_1\) lying toward \(AC\) (the upper part of the plot), and
• more than 60 m from A – this is the region outside the arc \(l_2\), i.e. on the C-side of the arc.

The set of points satisfying both conditions is the green hatched region above (bounded by side \(CA\), the bisector \(l_1\) and the arc \(l_2\)). Any point P chosen inside that shaded region is a valid position for the flagpole.

Akọwa Nkọwa

(a) Evaluate \(\log_{10}\sqrt{30} - \log_{10}\sqrt{6} + \log_{10}\sqrt{2}\).

Write each surd as a power and combine using the laws of logarithms. Since \(\sqrt{n}=n^{1/2}\),

Using \(\log a - \log b + \log c = \log\dfrac{ac}{b}\):

Since \(\log_{10}10 = 1\),

(b) \(U=\{1,2,3,\dots,10\}\), \(A=\{1,2,3,4,5\}\), \(B=\{2,3,5\}\), \(C=\{6,8,10\}\).

(i) Filling the Venn diagram. Every element of \(B\) is also in \(A\) (\(2,3,5\in A\)), so the circle \(B\) lies completely inside circle \(A\), exactly as shown. \(C\) has no element in common with \(A\), so it is drawn separately. The regions are filled as:

• Inside \(B\) (which is inside \(A\)): \(2,\;3,\;5\)
• Inside \(A\) but outside \(B\): \(1,\;4\)
• Inside \(C\): \(6,\;8,\;10\)
• Outside every circle: \(7,\;9\)

(The diagram already shows \(1\) in the \(A\)-only region, \(5\) in \(B\), \(6,10\) in \(C\) and \(9\) outside; the remaining elements \(2,3\) join \(B\), \(4\) joins the \(A\)-only region, \(8\) joins \(C\) and \(7\) joins the outside.)

(ii) \(A\cap C\). \(A=\{1,2,3,4,5\}\) and \(C=\{6,8,10\}\) share no element, so

(iii) \(A\cap B'\). First \(B' = U - B = \{1,4,6,7,8,9,10\}\). Then take the elements common to \(A\):

Akọwa Nkọwa

(a) Distance A to B along the parallel of latitude 53°N, through a longitude difference of \(85^\circ - 25^\circ = 60^\circ\):

\[AB = \frac{60}{360}\times 2\pi R\cos53^\circ = \frac{60}{360}\times 2\times\frac{22}{7}\times6400\times\cos53^\circ.\]

\[AB = \frac{1}{6}\times 40228.6\times0.6018 = 4035\text{ km (to the nearest km)}.\]

(b) Time to reach B at 400 km/h:

\[t = \frac{4035}{400} = 10.09 \approx 10\text{ hours}.\]

(c) Latitude of C. Flying south along a meridian, 1° corresponds to

\[\frac{2\pi R}{360} = \frac{1}{360}\times2\times\frac{22}{7}\times6400 = 111.75\text{ km}.\]

The 2000 km southward change in latitude is

\[\frac{2000}{111.75} = 17.9^\circ.\]

\[\text{Latitude of C} = 53^\circ - 17.9^\circ = 35.1^\circ \approx 35^\circ\text{N}.\]

Akọwa Nkọwa

(a) Two dice give \(6\times6 = 36\) equally likely outcomes.

(i) Same digit: the pairs are (1,1),(2,2),...,(6,6), that is 6 outcomes.

\[P(\text{same digit}) = \tfrac{6}{36} = \tfrac{1}{6}.\]

(ii) Total greater than 5: count the totals that are 5 or less: total 2 (1 way), 3 (2), 4 (3), 5 (4), giving \(1+2+3+4 = 10\) outcomes.

\[P(\text{total} > 5) = \tfrac{36-10}{36} = \tfrac{26}{36} = \tfrac{13}{18}.\]

(b) \(x = \cos30^\circ = \tfrac{\sqrt3}{2}\Rightarrow x^2 = \tfrac{3}{4};\quad y = \sin30^\circ = \tfrac{1}{2}\Rightarrow y^2 = \tfrac{1}{4}.\)

\[\frac{x^2 + y^2}{y^2 - x^2} = \frac{\tfrac{3}{4}+\tfrac{1}{4}}{\tfrac{1}{4}-\tfrac{3}{4}} = \frac{1}{-\tfrac{1}{2}} = -2.\]

Akọwa Nkọwa

(a) Mean age

Take the class mid-value (class-mark) \(x\) of each interval and form \(fx\).

\[ \bar{x}=\frac{\sum fx}{\sum f}=\frac{1835}{80}=22.9375 \]

Mean age \( \approx \mathbf{22.9\ years}\) (to 1 d.p.).

(b)(i) Histogram

First convert each class to its continuous class boundaries, then draw bars whose heights equal the frequencies (the bars touch, since the boundaries are continuous).

(b)(ii) Modal age from the histogram

The tallest bar is the modal class \(20\text{-}24\) (boundaries \(19.5\text{-}24.5\)). To read the mode, join the top-left corner of the modal bar to the top-left corner of the bar after it, and the top-right corner of the modal bar to the top-right corner of the bar before it. From the point where these two lines cross, drop a vertical line to the age axis.

The vertical line meets the axis at approximately \(\mathbf{21.8\ years}\).

This agrees with the calculation \[ \text{Mode}=L+\frac{\Delta_1}{\Delta_1+\Delta_2}\times c =19.5+\frac{25-18}{(25-18)+(25-17)}\times 5 =19.5+\frac{7}{15}\times 5 =19.5+2.3=21.8\ \text{years}. \]

(c) Probability of being in the modal class

The modal class \(20\text{-}24\) contains \(25\) members out of a total of \(80\):

\[ P(\text{modal class})=\frac{25}{80}=\frac{5}{16}=0.3125 \]