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**Question 1**
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An open cone with base radius 28cm and perpendicular height 96cm was stretched to form sector of a circle. calculate the arc of the sector (Take \(\pi = \frac{22}{7}\))

**Question 3**
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Which of the following is not a probability of Mary scoring 85% in a mathematics test?

**Answer Details**

The probability of an event happening can never be greater than 1. Therefore, 1.01 cannot be a probability of Mary scoring 85% in a mathematics test.

**Question 4**
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A pyramid has a rectangular base with dimensions 12m by 8m. If its height is 14m, calculate the volume

**Answer Details**

The formula for the volume of a pyramid is given by V = 1/3 * B * h, where B is the area of the base and h is the height of the pyramid. In this case, the base of the pyramid is a rectangle with dimensions 12m by 8m, so its area is A = 12m * 8m = 96m^{2}. The height of the pyramid is given as 14m. Substituting these values into the formula, we have V = 1/3 * 96m^{2} * 14m = 448m^{3}. Therefore, the volume of the pyramid is 448m^{3}.

**Question 6**
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If x = 64 and y = 27, evaluate: \(\frac{x^{\frac{1}{2}} - y^{\frac{1}{3}}}{y - x^{\frac{2}{3}}}\)

**Answer Details**

We can start by substituting the given values: \begin{align*} \frac{x^{\frac{1}{2}} - y^{\frac{1}{3}}}{y - x^{\frac{2}{3}}} &= \frac{64^{\frac{1}{2}} - 27^{\frac{1}{3}}}{27 - 64^{\frac{2}{3}}} \\ &= \frac{8 - 3}{27 - 16} \\ &= \frac{5}{11} \end{align*} Therefore, the answer is $\frac{5}{11}$.

**Question 7**
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The pie chart shows the distribution of 600 mathematics textbooks for Arts, Business, Science and Technical Classes. What percentage of the total number of textbooks belongs to science?

**Question 8**
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In the diagram, PQ is a straight line. Calculate the value of the angle labelled 2y

**Question 9**
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In the diagram, PQRST is a regular polygon with sides QR and TS produced to meet at V. Find the size of < RVS

**Question 10**
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The bearing of Y from X is 060^{o} and the bearing of Z from Y = 060^{o}. Find the bearing of X from Z

**Question 11**
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In the diagram, PQRS is a rhombus and < PSQ = 35^{o}. Calculate the size of < PRO

**Question 12**
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If \(\sqrt{50} - K\sqrt{8} = \frac{2}{\sqrt{2}}\), find K

**Answer Details**

We can start by simplifying the left-hand side of the equation using the laws of square roots: \begin{align*} \sqrt{50} - K\sqrt{8} &= \sqrt{25\cdot 2} - K\sqrt{4\cdot 2} \\ &= 5\sqrt{2} - 2K\sqrt{2} \\ &= \sqrt{2}(5 - 2K). \end{align*} Now, we can rewrite the given equation as: \begin{align*} \sqrt{2}(5 - 2K) &= \frac{2}{\sqrt{2}} \\ 5 - 2K &= \frac{2}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \\ 5 - 2K &= \frac{2}{2} \\ 5 - 2K &= 1 \\ -2K &= -4 \\ K &= 2. \end{align*} Therefore, the value of K that satisfies the equation is 2.

**Question 13**
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The distance between two towns is 50km. It is represented on a map by 5cm. Find the scale used

**Question 14**
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If sin x = \(\frac{5}{13}\) and 0^{o} \(\leq\) x \(\leq\) 90^{o}, find the value of (cos x - tan x)

**Answer Details**

We know that sin x = \(\frac{5}{13}\) and 0^{o} \(\leq\) x \(\leq\) 90^{o}. First, we can find the value of cos x using the identity: sin^{2} x + cos^{2} x = 1 sin^{2} x + cos^{2} x = 1 \(\frac{25}{169}\) + cos^{2} x = 1 cos^{2} x = \(\frac{144}{169}\) cos x = \(\pm\)\(\frac{12}{13}\) Since 0^{o} \(\leq\) x \(\leq\) 90^{o}, we know that cos x is positive. Therefore, cos x = \(\frac{12}{13}\). Next, we can find the value of tan x using the identity: tan x = \(\frac{sin x}{cos x}\) tan x = \(\frac{sin x}{cos x}\) = \(\frac{\frac{5}{13}}{\frac{12}{13}}\) = \(\frac{5}{12}\) Finally, we can find the value of (cos x - tan x) as: cos x - tan x = \(\frac{12}{13}\) - \(\frac{5}{12}\) = \(\frac{79}{156}\) Therefore, the answer is (cos x - tan x) = \(\frac{79}{156}\).

**Question 15**
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If Un = n(n^{2} + 1), evaluate U5 - U4

**Question 16**
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Using the histogram, estimate the mode of distribution

**Answer Details**

In a histogram, the mode is the value with the highest frequency, or the tallest bar. Looking at the given histogram, it appears that the tallest bar is the one corresponding to the interval between 52 and 54, which means that the mode lies somewhere in that interval. Since the interval width is 2, we can estimate the mode to be around the middle of the interval, which is (52 + 54)/2 = 53. Therefore, the estimated mode of the distribution is 53.5. So, the answer is (c) 53.5.

**Question 17**
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A sales boy gave a change of N68 instead of N72. Calculate his percentage error

**Answer Details**

The sales boy gave a change of N68 instead of N72, which means he gave N4 less than he should have. To find the percentage error, we use the formula: Percentage Error = (Error / True Value) × 100% In this case, the True Value is N72 and the Error is N4. Substituting these values into the formula, we get: Percentage Error = (4 / 72) × 100% Percentage Error = 0.055555... × 100% Percentage Error = 5.5555...% Rounding off to the nearest whole number, we get: Percentage Error ≈ 6% Therefore, the sales boy's percentage error is approximately 6%. The closest option to this answer is 5\(\frac{5}{9}\)%, so the answer would be: - 5\(\frac{5}{9}\)%

**Question 18**
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Given that (x + 2)(x^{2} - 3x + 2) + 2(x + 2)(x - 1) = (x + 2) M, find M

**Answer Details**

We can begin by factoring out (x + 2) from both terms on the left side of the equation: (x + 2)(x^{2} - 3x + 2) + 2(x + 2)(x - 1) = (x + 2) M (x + 2)[(x^{2} - 3x + 2) + 2(x - 1)] = (x + 2) M Simplifying the expression in the brackets, we get: (x + 2)(x^{2} - x) = (x + 2) M Now, we can cancel out (x + 2) from both sides of the equation: x^{2} - x = M Therefore, the value of M is simply x^{2} - x.

**Question 19**
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In the diagrams, |XZ| = |MN|, |ZY| = |MO| and |XY| = |NO|. Which of the following statements is true?

**Question 20**
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In what number base is the addition 465 + 24 + 225 = 1050?

**Answer Details**

To solve this problem, we need to find out in what base the addition statement is true. Let's assume the base is "b". Then, in the units column, we have: - 5 + 4 = 10, so we write down 0 and carry-over 1. - In the "b" column, we have: 6 + 2 + 2 + 1 = 11, so we write down 1 and carry-over 1. - In the "b^2" column, we have: 4 + 2 + 5 + 1 = 12, so we write down 2 and carry-over 1. - In the "b^3" column, we have: 4 + 2 + 2 = 8. Putting these digits together, we get the number 8201 in base "b". Now, we need to check if this number is equal to 1050 in base 10. 8201 in base "b" means: 8 x b^3 + 2 x b^2 + 0 x b + 1 x 1 = 1050 Rearranging the terms, we get: 8 b^3 + 2 b^2 + 1 = 1050 Subtracting 1 from both sides: 8 b^3 + 2 b^2 = 1049 Since b is a positive integer, we can see that b must be greater than 5. Trying b = 6, we get: 8 x 6^3 + 2 x 6^2 = 1048 This is not equal to 1049, so we need to try a larger base. Trying b = 7, we get: 8 x 7^3 + 2 x 7^2 = 1049 This is equal to 1049, so the base is 7. Therefore, the answer is seven.

**Question 21**
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Given that P = x^{2} + 4x - 2, Q = 2x - 1 and Q - p = 2, find x

**Question 22**
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In the diagram, O is the centre of the circle. OM||XZ and < ZOM = 25^{o}

**Question 23**
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Express \(\frac{2}{x + 3} - \frac{1}{x - 2}\) as a simple fraction

**Answer Details**

To add fractions, we need to have a common denominator. In this case, the common denominator is \((x+3)(x-2)\). Therefore, we need to convert each fraction to have this denominator. \[\frac{2}{x+3} - \frac{1}{x-2} = \frac{2(x-2)}{(x+3)(x-2)} - \frac{(x+3)}{(x+3)(x-2)}\] Simplifying the above expression, we have: \[\frac{2(x-2) - (x+3)}{(x+3)(x-2)} = \frac{2x - 4 - x - 3}{(x+3)(x-2)} = \frac{x-7}{(x+3)(x-2)}\] Therefore, \(\frac{2}{x + 3} - \frac{1}{x - 2} = \boxed{\frac{x-7}{(x+3)(x-2)}}\). The correct option is (a).

**Question 24**
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If y varies directly s the square root of (x + 1) and y = 6 when x = 3, find x when y = 9

**Answer Details**

We are given that y varies directly with the square root of (x+1), which can be written as y=k√(x+1), where k is the constant of variation. To find k, we use the given values of x and y: y=k√(x+1) 6=k√(3+1) 6=k√4 6=2k k=3 So the equation for y in terms of x is y=3√(x+1). To find x when y=9, we substitute these values into the equation and solve for x: 9=3√(x+1) 3=√(x+1) 9=x+1 x=8 Therefore, x = 8 when y = 9.

**Question 25**
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Using the venn diagram, find n(x \(\cap\) y^{1})

**Answer Details**

The intersection of two sets x and y1 is represented by the overlapping region of the two circles. From the diagram, we can see that the number of elements in the intersection is 2. Therefore, n(x \(\cap\) y1) = 2.

**Question 26**
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Simplify \(\frac{1\frac{7}{8} \times 2\frac{2}{5}}{6\frac{3}{4} \div \frac{3}{4}}\)

**Question 27**
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in the diagram, the height of a flagpole |TF| and the length of its shadow |FL| re in the ratio 6:8. Using k as a constant of proportionality, find the shortest distance between T and L

**Question 28**
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If \(\frac{1}{2}\)x + 2y = 3 and \(\frac{3}{2}\)x and \(\frac{3}{2}\)x - 2y = 1, find (x + y)

**Answer Details**

Given the equations: \begin{align*} \frac{1}{2}x + 2y &= 3 \\ \frac{3}{2}x - 2y &= 1 \\ \end{align*} We can solve for x and y using simultaneous equations: First, multiply the first equation by 2: \begin{align*} x + 4y &= 6 \\ \frac{3}{2}x - 2y &= 1 \\ \end{align*} Then, multiply the second equation by 2: \begin{align*} x + 4y &= 6 \\ 3x - 4y &= 2 \\ \end{align*} Add the equations together: \begin{align*} 4x &= 8 \\ x &= 2 \\ \end{align*} Substitute x = 2 back into the first equation: \begin{align*} \frac{1}{2}(2) + 2y &= 3 \\ 1 + 2y &= 3 \\ 2y &= 2 \\ y &= 1 \\ \end{align*} Finally, substitute x = 2 and y = 1 into (x + y): \begin{align*} x + y &= 2 + 1 \\ &= 3 \\ \end{align*} Therefore, (x + y) = 3, and the answer is.

**Question 29**
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When a number is subtracted from 2, the result equals 4 less than one-fifth of the number. Find the number

**Answer Details**

Let's call the number we are looking for "x". According to the problem, when we subtract x from 2, the result is equal to 4 less than one-fifth of the number. In mathematical terms, this can be written as: 2 - x = (1/5)x - 4 To solve for x, we can start by simplifying the equation by adding x to both sides: 2 = (6/5)x - 4 Next, we can add 4 to both sides: 6 = (6/5)x Finally, we can multiply both sides by 5/6 to isolate x: x = 5 Therefore, the answer is 5.

**Question 30**
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Given that p^{\(\frac{1}{3}\)} = \(\frac{3\sqrt{q}}{r}\), make q the subject of the equation

**Answer Details**

We want to solve for q, so we need to isolate it on one side of the equation. First, we can isolate the cube root of p by cubing both sides of the equation: p^{(1/3)} = (3√q)/r (p^{(1/3)})³ = (3√q)³/r³ p = (27q)/r³ Next, we can isolate q by multiplying both sides by r³/27: (p/27)r³ = q Therefore, the solution is q = (p/27)r³, which is equivalent to q = pr^{(1/3)}.

**Question 31**
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Multiply 2.7 x 10^{-4} by 6.3 x 10^{6} and leave your answers in standard form

**Answer Details**

To multiply two numbers in scientific notation, we simply multiply their coefficients and add their exponents. (2.7 x 10^{-4}) x (6.3 x 10^{6}) = (2.7 x 6.3) x 10^{-4+6} = 17.01 x 10^{2} We can express 17.01 x 10^{2} in standard form by moving the decimal point to the left two places, which gives us: 1.701 x 10^{3} Therefore, the correct answer is (c) 1.701 x 10^{3}.

**Question 32**
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Find the values of k in the equation 6k^{2} = 5k + 6

**Question 33**
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Simplify: \(\frac{x^2 - y^2}{(x + y)^2} + \frac{(x - y)^2}{(3x + 3y)}\)

**Question 36**
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Using the histogram, what is the median class?

**Answer Details**

To find the median class in a histogram, we need to identify the class interval that contains the median value. The median is the middle value in a dataset, so we need to find the midpoint of the data. To find the median class: 1. Add up the frequencies in the histogram starting from the left-hand side until the total is greater than or equal to the total number of data points divided by 2. 2. The median class is the class interval that contains the midpoint of the data. In this histogram, the total number of data points is 80. The midpoint is (80 + 1)/2 = 40.5. Starting from the left-hand side, we can add up the frequencies: 8 + 19 + 24 = 51. The median falls within the interval 40.5 - 50.5, which is the third interval. Therefore, the median class is 40.5 - 50.5.

**Question 37**
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A chord is 2cm from the centre of a circle. If the radius of the circle is 5cm, find the length of the chord

**Answer Details**

To solve this problem, we can use the Pythagorean theorem to find the length of the chord. We know that the radius of the circle is 5cm and the distance from the centre of the circle to the chord is 2cm. We can draw a perpendicular line from the centre of the circle to the chord, which will divide the chord into two equal parts. This perpendicular line will also bisect the chord and form a right triangle with the radius of the circle and half of the chord. The hypotenuse of this triangle is the radius of the circle, which is 5cm, and one leg is half of the chord, which we can call x. The other leg is the distance from the centre of the circle to the chord, which is 2cm. Using the Pythagorean theorem, we can solve for x: 5^2 = x^2 + 2^2 25 = x^2 + 4 x^2 = 21 x = √21 Therefore, the length of the chord is twice the value of x, which is 2√21cm. Hence, the correct answer is option A.

**Question 38**
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In the diagram, ST//PQ reflex angle SRQ = 198^{o} and < RQp = 72^{o}. Find the value of y

**Question 39**
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An object is 6m away from the base of a mast. The angle of depression of the object from the top pf the mast is 50^{o}, Find, correct to 2 decimal places, the height of the mast

**Question 40**
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What is the locus of the point X which moves relative to two fixed points P and M on a plane such that < PXM = 30^{o}

**Question 41**
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An interior angle of a regular polygon is 5 times each exterior angle. How many sides has the polygon?

**Answer Details**

In a regular polygon, each exterior angle is equal to 360 degrees divided by the number of sides. Let the number of sides be represented by n. Therefore, the measure of each exterior angle is 360/n. Since the interior angle and exterior angle are supplementary, we can write: Interior angle + Exterior angle = 180 degrees Let x be the measure of each interior angle. We are given that each interior angle is 5 times each exterior angle, so: x = 5(360/n - x) Simplifying and solving for x, we get: x = 150 degrees The sum of the interior angles of an n-sided polygon is (n - 2) × 180 degrees. Since each interior angle in our polygon is 150 degrees, we can write: n × 150 = (n - 2) × 180 Simplifying and solving for n, we get: n = 12 Therefore, the polygon has 12 sides. So, the correct answer is 12.

**Question 42**
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The slant height of a cone is 5cm and the radius of its base is 3cm. Find, correct to the nearest whole number, the volume of the cone. ( Take \(\pi = \frac{22}{7}\))

**Question 43**
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If p = (y : 2y \(\geq\) 6) and Q = (y : y -3 \(\geq\) 4), where y is an integer, find p\(\cap\)Q

**Answer Details**

The set p represents all values of y such that 2y is greater than or equal to 6. Simplifying the inequality 2y \(\geq\) 6 gives y \(\geq\) 3. Therefore, p can be written as p = {y: y \(\geq\) 3}. Similarly, the set Q represents all values of y such that y - 3 is greater than or equal to 4. Simplifying the inequality y - 3 \(\geq\) 4 gives y \(\geq\) 7. Therefore, Q can be written as Q = {y: y \(\geq\) 7}. The intersection of p and Q, denoted by p\(\cap\)Q, is the set of all values of y that are in both p and Q. Since p contains all values of y greater than or equal to 3 and Q contains all values of y greater than or equal to 7, the intersection of p and Q is {y: y \(\geq\) 7}. Therefore, p\(\cap\)Q = {7, 8, 9, 10, ...}. The correct option is (c) {3, 4, 5, 6, 7}.

**Question 44**
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In the diagram, PRST is a square. If |PQ| = 24cm. |QR| = 10cm and < PQR = 90^{o}, find the perimeter of the polygon PQRST.

**Question 45**
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Four oranges sell for Nx and three mangoes sell for Ny. Olu bought 24 oranges and 12 mangoes. How much did he pay in terms of x and y?

**Answer Details**

Four oranges sell for Nx, so one orange costs \(\frac{N}{4}x\). Three mangoes sell for Ny, so one mango costs \(\frac{N}{3}y\). Olu bought 24 oranges and 12 mangoes, so he paid: \[24 \cdot \frac{N}{4}x + 12 \cdot \frac{N}{3}y = 6Nx + 4Ny\] Therefore, Olu paid N(6x + 4y) in terms of x and y. The answer is option (B).

**Question 46**
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A cube and cuboid have the same base area. The volume of the cube is 64cm\(^3\) while that of the cuboid is 80cm\(^3\). Find the height of the cuboid

**Answer Details**

Let the base area of the cube be x, then the length of one side of the cube is \(\sqrt[3]{64}\) = 4 cm. Since the base area of the cube and cuboid are equal, the base of the cuboid must also have an area of x. The volume of the cuboid is given as 80cm\(^3\) which can be expressed as: 80 = x × h, where h is the height of the cuboid We know that the length of the base of the cuboid is equal to the length of the side of the cube. Therefore, the dimensions of the cuboid are 4 cm by 4 cm by h cm. Using the formula for the volume of a cuboid, we get: Volume of cuboid = length × width × height = 4 × 4 × h = 16h Substituting 80 for the volume of the cuboid, we get: 16h = 80 Solving for h, we get: h = 5cm Therefore, the height of the cuboid is 5cm. Hence, the correct answer is 5cm.

**Question 47**
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The graph of the relation y = x2 + 2x + k passes through the point (2, 0). Find the values of k

**Answer Details**

We are given the equation y = x2 + 2x + k and we know that it passes through the point (2,0). We can substitute x = 2 and y = 0 into the equation to find k. Substituting, we get: 0 = 22 + 2(2) + k 0 = 4 + 4 + k 0 = 8 + k k = -8 Therefore, the value of k is -8, which is the fourth option.

**Question 48**
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The pie chart shows the distribution of 600 mathematics textbooks for Arts, Business, Science and Technical Classes. How many textbooks are for the technical class?

**Question 49**
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The frequency distribution table shows the marks obtained by 100 students in a Mathematics test.

Marks (%) |
1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 | 81-90 | 91-100 |

Frequency | 2 | 3 | 5 | 13 | 19 | 31 | 13 | 9 | 4 | 1 |

(a) Draw the cumulative curve for the distribution.

(b) Use the graph to find the : (i) 60th percentile ; (ii) probability that a student passed the test if the pass mark was fixed at 35%.

None

**Answer Details**

None

**Question 50**
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A boy 1.2m tall, stands 6m away from the foot of a vertical lamp pole 4.2m long. If the lamp is at the tip of the pole,

(a) represent this information in a diagram ;

(b) calculate the (i) length of the shadow of the boy cast by the lamp ; (ii) angle of elevation of the lamp from the boy, correct to the nearest degree.

None

**Answer Details**

None

**Question 51**
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When one end of a ladder, LM, is placed against a vertical wall at a point 5 metres above the ground, the ladder makes an angle of 37° with the horizontal ground.

(a) Represent this information in a diagram ;

(b) Calculate, correct to 3 significant figures, the length of the ladder ;

(c) If the foot of the ladder is pushed towards the wall by 2 metres, calculate,correct to the nearest degree, the angle which the ladder nows makes with the ground.

None

**Answer Details**

None

**Question 52**
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(a)

A segment of a circle is cut off from a rectangular board as shown in the diagram. If the radius of the circle is \(1\frac{1}{2}\) times the length of the chord; calculate, correct to 2 decimal places, the perimeter of the remaining portion. [Take \(\pi = \frac{22}{7}\)]

(b) Evaluate without using calculators or tables, \(\frac{3}{\sqrt{3}}(\frac{2}{\sqrt{3}} - \frac{\sqrt{12}}{6})\).

None

**Answer Details**

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