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**Question 1**
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H varies directly as p and inversely as the square of y. If H = 1, p = 8 and y = 2, find H in terms of p and y.

**Question 2**
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The total surface area of a solid cylinder 165cm\(^2\). Of the base diameter is 7cm, calculate its height. [Take \(\pi = \frac{22}{7}\)]

**Question 3**
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If m : n = 2 : 1, evaluate \(\frac{3m^2 - 2n^2}{m^2 + mn}\)

**Question 4**
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If 2\(^{a}\) = \(\sqrt{64}\) and \(\frac{b}{a}\) = 3, evaluate a\(^2 + b^{2}\)

**Answer Details**

First, let's solve for a using the given equation 2\(^{a}\) = \(\sqrt{64}\). We know that \(\sqrt{64}\) is equal to 8, so we can substitute this value in the equation: 2\(^{a}\) = 8 To solve for a, we need to find the exponent that 2 is raised to in order to get 8. This exponent is 3, so a = 3. Next, we need to find the value of b. We are given that \(\frac{b}{a}\) = 3, so we can rearrange this equation to solve for b: b = 3a Substituting the value we found for a, we get: b = 3 x 3 = 9 Finally, we can evaluate a\(^2\) + b\(^2\) using the values we found for a and b: a\(^2\) + b\(^2\) = 3\(^2\) + 9\(^2\) = 9 + 81 = 90 Therefore, the answer is 90.

**Question 5**
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In XYZ, |YZ| = 32cm, < YXZ 53\(^o\) and XZY = 90\(^o\). Find, correct to the nearest centimetre, |XZ|

**Question 6**
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The following are scores obtained by some students in a test. Find the median score

8 | 18 | 10 | 14 | 18 | 11 | 13 |

14 | 13 | 17 | 15 | 8 | 16 | 13 |

**Question 7**
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8, 18, 10,14, 18, 11, 13, 14, 13, 17, 15, 8, 16, and 13

**Question 8**
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If 7 + y = 4 (mod 8), find the least value of y, 10 \(\leq y \leq 30\)

**Question 9**
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In the diagram, PQ is parallel to RS, < QFG = 105\(^o\) and < FEG = 50\(^o\). Find the value of n

**Question 10**
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The fourth term of an Arithmetic Progression (A.P) is 37 and the first term is -20. Find the common difference.

**Answer Details**

In an arithmetic progression (AP), the terms are equally spaced. Let's call the common difference "d". If we have the first term "a" and the nth term "an", we can find the common difference using the formula: d = (an - a) / (n - 1) In this case, the fourth term is 37 and the first term is -20, so we can plug in the values into the formula: d = (37 - (-20)) / (4 - 1) = 57 / 3 = 19 So the common difference is 19.

**Question 11**
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Evaluate: \(\frac{0.42 + 2.5}{0.5 \times 2.95}\), leaving the answer in the standard form.

**Answer Details**

To solve this expression, we need to follow the order of operations which is parentheses, exponents, multiplication and division (from left to right), and addition and subtraction (from left to right). We do not have any parentheses or exponents, so we can proceed to the multiplication and division step. We have: \[\frac{0.42 + 2.5}{0.5 \times 2.95} = \frac{2.92}{1.475}\] Next, we can simplify the fraction by dividing both the numerator and denominator by the greatest common factor which is 0.05. \[\frac{2.92}{1.475} = \frac{292}{147.5}\] Finally, we can express this fraction in standard form by moving the decimal point in the numerator to the left by two places and simultaneously moving the decimal point in the denominator to the left by two places. \[\frac{292}{147.5} = 1.98\] Therefore, the answer is 1.639 x 10\(^1\) in standard form.

**Question 12**
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Factorize completely; (2x + 2y)(x - y) + (2x - 2y)(x + y)

**Answer Details**

To factorize the given expression, we can use the distributive property of multiplication over addition. This means that we need to multiply each term in the first set of parentheses by each term in the second set of parentheses, and then simplify: (2x + 2y)(x - y) + (2x - 2y)(x + y) = 2x(x - y) + 2y(x - y) + 2x(x + y) - 2y(x + y) (using distributive property) = 2x^2 - 2xy + 2xy + 2y^2 + 2x^2 + 2xy - 2xy - 2y^2 (combining like terms) = 4x^2 - 4y^2 Therefore, the answer is 4(x - y)(x + y). Explanation: We can factorize the given expression by grouping the terms into two sets, each set containing two terms with a common factor. We can then factor out that common factor from each set, and simplify. In this case, we can factor out 2 from the first set of parentheses, and -2 from the second set of parentheses, giving us: 2(x + y)(x - y) - 2(x + y)(x - y) We can then combine the two sets of parentheses, giving us: (2(x + y) - 2(x - y))(x - y) Simplifying the expression inside the parentheses gives us: (2x + 2y - 2x + 2y)(x - y) Which further simplifies to: 4y(x - y) However, this is not the fully factored form, since we can factor out 4 from the expression to get: 4(x - y)(y) So the answer is 4(x - y)(y), which is equivalent to option A.

**Question 14**
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In the diagram, RT is a tangent to the circle at R, < PQR = 70\(^o\), < QRT = 52\(^o\), < QSR and < PRQ = x. Calculate the value of x.

**Question 15**
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The interior angles of a polygon are 3x\(^o\), 2x\(^o\), 4x\(^o\), 3x\(^o\) and 6\(^o\). Find the size of the smallest angle of the polygon.

**Question 16**
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Evaluate: (0.064) - \(\frac{1}{3}\)

**Question 17**
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Simplify: \(\frac{x^2 - 5x - 14}{x^2 - 9x + 14}\)

**Answer Details**

To simplify the fraction, we need to find a common denominator for the numerator and denominator. The denominator, \(x^2 - 9x + 14\), can be factored into the product of two binomials: \((x - 7)(x - 2)\). Now, to simplify the numerator, we can factor it as well: \(x^2 - 5x - 14 = (x - 7)(x + 2)\). With this factoring, we can cancel out the common factor of \(x - 7\) in the numerator and denominator, leaving us with the simplified fraction: \(\frac{x^2 - 5x - 14}{x^2 - 9x + 14} = \frac{(x - 7)(x + 2)}{(x - 7)(x - 2)} = \frac{x + 2}{x - 2}\). So the simplified fraction is.

**Question 18**
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Evaluate: 2\(\sqrt{28} - 3\sqrt{50} + \sqrt{72}\)

**Answer Details**

To evaluate the expression 2\(\sqrt{28} - 3\sqrt{50} + \sqrt{72}\), we need to simplify the square roots first. \(\sqrt{28} = 2\sqrt{7}\) \(\sqrt{50} = 5\sqrt{2}\) \(\sqrt{72} = 2\sqrt{18} = 2\sqrt{9}\sqrt{2} = 6\sqrt{2}\) Now that we have simplified the square roots, we can substitute the values back into the expression: 2\(\sqrt{28} - 3\sqrt{50} + \sqrt{72}\) = 2 * 2\(\sqrt{7}\) - 3 * 5\(\sqrt{2}\) + 6\(\sqrt{2}\) = 4\(\sqrt{7}\) - 3 * 5\(\sqrt{2}\) + 6\(\sqrt{2}\) = 4\(\sqrt{7}\) - 15\(\sqrt{2}\) + 6\(\sqrt{2}\) = 4\(\sqrt{7}\) - 9\(\sqrt{2}\) So the expression evaluates to 4\(\sqrt{7}\) - 9\(\sqrt{2}\).

**Question 19**
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If log\(_x\) 2 = 0.3, evaluate log\(_x\) 8.

**Answer Details**

We can use the property of logarithms that states: $$\log_{a}(b^c) = c \log_{a}(b)$$ Using this property, we can rewrite the expression for log\(_x\) 8 as: $$\log_{x}(8) = \log_{x}(2^3) = 3\log_{x}(2)$$ We are given that log\(_x\) 2 = 0.3, so we can substitute this value: $$3\log_{x}(2) = 3(0.3) = 0.9$$ Therefore, the value of log\(_x\) 8 is 0.9. Therefore, option (C) is the correct answer: 0.9.

**Question 20**
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Find the angle at which an arc of length 22 cm subtends at the centre of a circle of radius 15cm. [Take \(\pi = \frac{22}{7}\)]

**Question 21**
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A box contains 3 white and 3 blue identical balls. If two balls are picked at random from the box, one after the other with replacement, what is the probability that they are of different colours?

**Question 22**
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In the diagram, RT is a tangent to the circle at R, < PQR = 70\(^o\), < QRT = 52\(^o\), < QSR and < PRQ = x. Find the value of y.

**Question 23**
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The foot of a ladder is 6m from the base of an electric pole. The top of the ladder rest against the pole at a point 8m above the ground. How long is the ladder?

**Answer Details**

To find the length of the ladder, we can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides. In this case, the ladder is the hypotenuse, and the distance from the foot of the ladder to the pole and the height of the pole where the ladder is resting form the other two sides of the right triangle. So we have: Length of ladder^2 = (distance from foot of ladder to pole)^2 + (height of pole where ladder is resting)^2 Length of ladder^2 = 6^2 + 8^2 Length of ladder^2 = 36 + 64 Length of ladder^2 = 100 Length of ladder = √100 = 10 Therefore, the length of the ladder is 10 meters. Answer option C (10m) is correct.

**Question 24**
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Evaluate; \(\frac{\log_3 9 - \log_2 8}{\log_3 9}\)

**Answer Details**

We can use the laws of logarithms to simplify the expression: \(\frac{\log_3 9 - \log_2 8}{\log_3 9} = \frac{\log_3 (3^2) - \log_2 (2^3)}{\log_3 (3^2)}\) Using the laws of logarithms, we can simplify the numerator: \(\frac{\log_3 (3^2) - \log_2 (2^3)}{\log_3 (3^2)} = \frac{2\log_3 3 - 3\log_2 2}{2\log_3 3}\) We know that \(\log_3 3 = 1\) and \(\log_2 2 = 1\), so we can substitute these values: \(\frac{2\log_3 3 - 3\log_2 2}{2\log_3 3} = \frac{2 - 3}{2} = -\frac{1}{2}\) Therefore, the value of the expression is -\(\frac{1}{2}\).

**Question 25**
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Find the equation of a straight line passing through the point (1, -5) and having gradient of \(\frac{3}{4}\)

**Question 26**
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In the diagram, POS and ROT re straight lines. OPOR is a parallelogram, |OS| = |OT| and

**Question 27**
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The pie chart represents fruits on display in a grocery shop. If there are 60 oranges on display, how many apples are there?

**Question 28**
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Make b the subject of the relation lb = \(\frac{1}{2}\) (a + b)h

**Question 29**
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In the diagram, RT is a tangent to the circle at R, < PQR = 70\(^o\), < QRT = 52\(^o\), < QSR and < PRQ = x. Calculate the variance of 2, 4, 7, 8 and 9

**Question 30**
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If T = {prime numbers} and M = {odd numbers} are subsets of \(\mu\) = {x : 0 < x **≤ **10} and x is an integer, find (T\(^{\prime}\) n M\(^{\prime}\)).

**Answer Details**

The set T is the set of prime numbers less than or equal to 10, which are {2, 3, 5, 7}. The prime complement (T\(\prime\)) is the set of non-prime numbers less than or equal to 10, which are {1, 4, 6, 8, 10}. The set M is the set of odd numbers less than or equal to 10, which are {1, 3, 5, 7, 9}. The odd complement (M\(\prime\)) is the set of even numbers less than or equal to 10, which are {2, 4, 6, 8, 10}. Finally, the intersection of T\(\prime\) and M\(\prime\) is the set of numbers that are both non-prime and even, which is {4, 6, 8, 10}. So (T\(\prime\) n M\(\prime\)) = {4, 6, 8, 10}.

**Question 31**
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If tan x = \(\frac{3}{4}\), 0 < x < 90\(^o\), evaluate \(\frac{\cos x}{2 sin x}\)

**Answer Details**

We are given that \(\tan x = \frac{3}{4}\) and \(0 < x < 90^o\). Since \(\tan x = \frac{\sin x}{\cos x}\), we can rewrite the given equation as: \[\frac{\sin x}{\cos x} = \frac{3}{4}\] Multiplying both sides by \(\cos x\) gives us: \[\sin x = \frac{3}{4}\cos x\] Dividing both sides by \(2\sin x\) gives us: \[\frac{\cos x}{2\sin x} = \frac{\frac{4}{3}\sin x}{2\sin x} = \frac{4}{3} \cdot \frac{1}{2} = \frac{2}{3}\] Therefore, the value of \(\frac{\cos x}{2\sin x}\) is \(\frac{2}{3}\). Hence, the answer is: \(\frac{2}{3}\).

**Question 32**
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If 6, P, and 14 are consecutive terms in an Arithmetic Progression (AP), find the value of P.

**Answer Details**

In an arithmetic progression (AP), the difference between any two consecutive terms is constant. So, we can find the common difference (d) between any two consecutive terms in the given AP by subtracting one term from the preceding term. Let's use the second term (P) and the first term (6) to find the common difference: d = P - 6 Now, to check whether 14 is the third term of the AP, we can use the formula for the nth term of an AP: an = a1 + (n-1)d where an = nth term a1 = first term d = common difference If 14 is the third term, then n = 3, so we have: 14 = 6 + (3-1)d 14 = 6 + 2d 2d = 8 d = 4 Now, we can substitute the value of d in the expression we got earlier: d = P - 6 4 = P - 6 P = 10 Therefore, the value of P is 10.

**Question 33**
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If (0.25)\(^y\) = 32, find the value of y.

**Answer Details**

We can solve the given equation by taking the logarithm of both sides. Any base of the logarithm can be used, but we will use the common logarithm (base 10). log\((0.25)^y\) = log 32 Using the logarithmic identity log\((a^b)\) = b log\((a)\), we get: y log\((0.25)\) = log 32 Now, we can evaluate log\((0.25)\) using the logarithmic identity log\((a^n)\) = n log\((a)\), as follows: log\((0.25)\) = log\((\frac{1}{4})\) = log\((4^{-1})\) = -log\((4)\) We know that log\((10^x)\) = x, so log\((4)\) = log\((10^{0.602})\) \(\approx\) 0.602 Therefore, y log\((0.25)\) = log 32 y (-log\((4)\)) = log 32 y (-0.602) = log 32 y = \(\frac{\text{log } 32}{-0.602}\) Using a calculator, we get: y \(\approx\) -2.5 Therefore, the value of y is approximately -2.5. Hence, the answer is: y = -\(\frac{5}{2}\).

**Question 34**
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Simplify: \(\log_{10}\) 6 - 3 log\(_{10}\) 3 + \(\frac{2}{3} \log_{10} 27\)

**Question 35**
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Which of these values would make \(\frac{3p^-}{p^{2-}}\) undefined?

**Question 36**
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Eric sold his house through an agent who charged 8% commission on the selling price. If Eric received $117,760.00 after the sale, what was the selling price of the house?

**Answer Details**

The commission charged by the agent is 8% of the selling price. This means that the remaining amount that Eric received after the sale is 92% of the selling price (100% - 8% = 92%). We can set up an equation to solve for the selling price: 92% of selling price = $117,760.00 To solve for the selling price, we can divide both sides by 92% (or multiply by the reciprocal of 92%, which is 100/92): selling price = $117,760.00 ÷ 0.92 Using a calculator, we get: selling price = $128,000.00 Therefore, the selling price of the house was $128,000.00.

**Question 37**
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A rectangular board has a length of 15cm and width x cm. If its sides are doubled, find its new area.

**Answer Details**

Let's first find the area of the original rectangular board, which is simply the product of its length and width: original area = 15 cm x x cm = 15x cm\(^2\) When the sides are doubled, the length becomes 15 cm x 2 = 30 cm, and the width becomes x cm x 2 = 2x cm. So, the new area is: new area = 30 cm x 2x cm = 60x cm\(^2\) So, the new area of the rectangular board is 60x cm\(^2\).

**Question 38**
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There are 8 boys and 4 girls in a lift. What is the probability that the first person who steps out of the lift will be a boy?

**Answer Details**

The probability of an event happening is the number of ways it can happen divided by the total number of outcomes. In this case, there are 8 boys and 4 girls in the lift. If the first person who steps out of the lift is a boy, it means that any one of the 8 boys can be the first person to step out of the lift. Therefore, the number of ways the event can happen is 8. The total number of outcomes is the total number of people who can step out of the lift first. Since there are 12 people in the lift, any one of the 12 people can be the first person to step out of the lift. Therefore, the total number of outcomes is 12. Therefore, the probability that the first person who steps out of the lift will be a boy is: probability = number of ways the event can happen / total number of outcomes probability = 8 / 12 probability = 2 / 3 Hence, the answer is: \(\frac{2}{3}\).

**Question 39**
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Solve: \(\frac{y + 1}{2} - \frac{2y - 1}{3}\) = 4

**Answer Details**

To solve this equation, we can start by simplifying both sides by finding a common denominator. \(\frac{y + 1}{2} - \frac{2y - 1}{3} = 4\) Multiplying the first term by 3 and the second term by 2 (the least common multiple of 2 and 3), we get: \(\frac{3(y+1)}{6} - \frac{2(2y-1)}{6} = 4\) Simplifying this gives: \(\frac{3y + 3 - 4y + 2}{6} = 4\) Combining like terms: \(\frac{-y + 5}{6} = 4\) Multiplying both sides by 6: \(-y + 5 = 24\) Subtracting 5 from both sides: \(-y = 19\) Finally, multiplying both sides by -1 to solve for y: \(y = -19\) Therefore, the solution is y = -19.

**Question 40**
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From the top of a vehicle cliff 20m high, a boat at sea can be sighted 75m away and on the same horizontal position as the cliff. Calculate, correct to the nearest degree, the angle of depression of the boat from the top of the cliff.

**Answer Details**

The situation can be represented by the following diagram: ``` A /| / | 20m / | B---C 75m ``` Where A is the top of the cliff, B is the boat, and C is the point on the cliff directly above the boat. We want to find the angle of depression, which is the angle ACB. We can use trigonometry to find this angle. First, we need to find the length of AC. This is simply the height of the cliff, which is given as 20m. Next, we need to find the length of BC. This is the horizontal distance from the base of the cliff to the boat, which is given as 75m. Now we can use the tangent function to find the angle of depression: tan(ACB) = opposite / adjacent = 20 / 75 Taking the inverse tangent (or arctan) of both sides gives: ACB = arctan(20 / 75) = 15.02 degrees (to two decimal places) Therefore, the angle of depression of the boat from the top of the cliff is approximately 15 degrees (to the nearest degree). So the answer is option D: 15\(^o\).

**Question 41**
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An arc subtends an angle of 72\(^o\) at the centre of a circle. Find the length of the arc if the radius of the circle is 3.5 cm. [Take \(\pi = \frac{22}{7}\)]

**Answer Details**

To find the length of the arc, we need to use the formula: Length of arc = \(\frac{\text{angle subtended by the arc}}{360^\circ} \times 2\pi r\) Here, the angle subtended by the arc is 72\(^o\) and the radius of the circle is 3.5 cm. We can substitute these values in the formula to get: Length of arc = \(\frac{72^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 3.5\) cm Simplifying this expression, we get: Length of arc = \(\frac{1}{5} \times \frac{44}{7} \times 3.5\) cm Length of arc = \(\frac{22}{5}\) cm Length of arc = 4.4 cm (approx.) Therefore, the length of the arc is 4.4 cm (option C).

**Question 42**
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The following are scores obtained by some students in a test, Find the mode of the distribution

8 | 18 | 10 | 14 | 18 | 11 | 13 |

14 | 13 | 17 | 15 | 8 | 16 | 13 |

**Answer Details**

To find the mode of the distribution, we need to determine the value that appears most frequently in the dataset. Looking at the table, we can see that the value "13" appears three times, which is more than any other value. Therefore, the mode of this distribution is 13. Alternatively, we could create a frequency table or a histogram to visualize the distribution and identify the mode more easily.

**Question 43**
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Simplify, correct to three significant figures, (27.63)\(^2\) - (12.37)\(^2\)

**Answer Details**

We can start by squaring each of the numbers in the expression: (27.63)\(^2\) = 762.9869 (12.37)\(^2\) = 153.4369 Now we can subtract the two squares: 762.9869 - 153.4369 = 609.55 Rounding the result to three significant figures, we get: 609.55 = 610 So (27.63)\(^2\) - (12.37)\(^2\) = 610.

**Question 44**
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Bala sold an article for #6,900.00 and made a profit of 15%. Calculate his percentage profit if he had sold it for N6,600.00.

**Answer Details**

If Bala sold an article for #6,900.00 and made a profit of 15%, this means that his cost price for the article was lower than the selling price by 15%. We can calculate his cost price as follows: Cost Price = Selling Price / (1 + Profit%) Cost Price = 6,900 / (1 + 0.15) Cost Price = 6,000 Therefore, Bala's cost price was #6,000.00, and he sold the article for #6,900.00, making a profit of #900.00. Now, we need to calculate the percentage profit he would have made if he sold the article for #6,600.00 instead of #6,900.00. To do this, we can use the following formula: Percentage Profit = (Profit / Cost Price) x 100% Percentage Profit = (6,600 - 6,000) / 6,000 x 100% Percentage Profit = 10% Therefore, if Bala had sold the article for #6,600.00 instead of #6,900.00, he would have made a profit of 10%. The correct option is 10%.

**Question 45**
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If 3p = 4q and 9p = 8q - 12, find the value of pq.

**Question 46**
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In the diagram, O is the centre of the circle of radius 18cm. If < zxy = 70\(^o\), calculate the length of arc ZY. [Take \(\pi = \frac{22}{7}\)]

**Question 47**
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In the diagram, PQ is parallel to RS, < QFG = 105\(^o\) and < FEG = 50\(^o\). Find the value of m.

**Question 48**
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A box contains 5 red, 6 green and 7 yellow pencils of the same size. What is the probability of picking a green pencil at random?

**Answer Details**

There are 5 + 6 + 7 = 18 pencils in the box. Since there are 6 green pencils, the probability of picking a green pencil at random is 6/18, which simplifies to 1/3. Therefore, the answer is \(\frac{1}{3}\).

**Question 49**
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If 23\(_y\) = 1111\(_{\text{two}}\), find the value of y

**Question 50**
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Express, correct to three significant figures, 0.003597.

**Answer Details**

To express 0.003597 correct to three significant figures, we need to round off the number to the third digit after the decimal point. The third digit after the decimal point is 7. Since 7 is greater than or equal to 5, we need to round up the second digit after the decimal point. Therefore, the third digit becomes zero. Hence, the rounded value of 0.003597 correct to three significant figures is 0.00360. Therefore, the answer is: 0.00360.

**Question 51**
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(a) Copy and complete the table of values for y = 2 cos x + 3 sin x for 0\(^o\) \(\geq\) x \(\geq\) 360\(^o\)

x | 0\(^o\) | 60\(^0\) | 120\(^o\) | 180\(^o\) | 240\(^o\) | 300\(^o\) | 360\(^o\) |

y | 2.0 | - 3.6 |

(b) Using a scale of 2cm to 60\(^o\) on the x-axis and 2cm to 1 unit in the y-axis, draw the graph of y = 2 cos x + 3 sin x for 0\(^o\) \(\geq\) 360\(^o\)

(c) Using the graph,

(i) Solve 2 cos x + 3 sin x = -1

(ii) Find, correct to one decimal place, the value of y when x = 342\(^o\)

**a) To complete the table of values for y = 2 cos x + 3 sin x, we need to substitute the given x values into the equation and solve for y. Here are the complete values:**

x | y |
---|---|

0° | 2.0 |

60° | 4.6 |

120° | 3.0 |

180° | -3.6 |

240° | -2.0 |

300° | -0.6 |

360° | 2.0 |

**b) To draw the graph of y = 2 cos x + 3 sin x, we can start by plotting the points from the table of values onto the x-y plane. Then, we can connect the points to form the graph of the equation.**

**c) (i) To solve 2 cos x + 3 sin x = -1, we can use the inverse trigonometric functions to find the angle x that satisfies this equation. For example, we can use the identity cos^2 x + sin^2 x = 1 to solve for either cos x or sin x in terms of the other, then use the inverse cosine or sine function to find the angle x.**

(ii) To find the value of y when x = 342°, we substitute this value of x into the equation y = 2 cos x + 3 sin x and solve for y. For example, using a calculator, we can find that y = 2 cos 342° + 3 sin 342° = 1.6.

**Answer Details**

**a) To complete the table of values for y = 2 cos x + 3 sin x, we need to substitute the given x values into the equation and solve for y. Here are the complete values:**

x | y |
---|---|

0° | 2.0 |

60° | 4.6 |

120° | 3.0 |

180° | -3.6 |

240° | -2.0 |

300° | -0.6 |

360° | 2.0 |

**b) To draw the graph of y = 2 cos x + 3 sin x, we can start by plotting the points from the table of values onto the x-y plane. Then, we can connect the points to form the graph of the equation.**

**c) (i) To solve 2 cos x + 3 sin x = -1, we can use the inverse trigonometric functions to find the angle x that satisfies this equation. For example, we can use the identity cos^2 x + sin^2 x = 1 to solve for either cos x or sin x in terms of the other, then use the inverse cosine or sine function to find the angle x.**

(ii) To find the value of y when x = 342°, we substitute this value of x into the equation y = 2 cos x + 3 sin x and solve for y. For example, using a calculator, we can find that y = 2 cos 342° + 3 sin 342° = 1.6.

**Question 52**
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In the diagram, \(\overline{RT}\) and \(\overline{RT}\) are tangent to the circle with centre O. < TUS = 68 °, < SRT = x, and < UTO = y. Find the value of x.

(b) Two tanks A and B am filled to capacity with diesel. Tank A holds 600 litres of diesel more than tank B. If 100 litres of diesel was pumped out of each tank, tank A would then contain 3 times as much diesel as tank B. Find the capacity of each tank.

(a) Since *\(\overline{RT}\)* and *\(\overline{RU}\)* are tangent to the circle, *\(\angle TRU\)* is equal to the angle between the radii drawn to the points of tangency. Therefore, *\(\angle TRU = \angle TRO = 90^{\circ}\)*.

Also, since *\(\overline{RT}\)* is tangent to the circle, *\(\angle SRT\)* is equal to the angle between the tangent and the chord *\(\overline{ST}\)*. Therefore, *\(\angle SRT = \angle OTU\)*.

Finally, since the sum of the angles in a triangle is *180^{\circ}*, we have:

\[\angle TUS + \angle SRT + \angle UTO = 180^{\circ}\]

\[68^{\circ} + x + y = 180^{\circ}\]

\[x = 112^{\circ} - y\]

Therefore, the value of *x* is *112^{\circ} - y*.

(b) Let the capacity of tank B be *x* litres. Then, the capacity of tank A is *x+600* litres.

After 100 litres of diesel is pumped out of each tank, tank B has *(x-100)* litres of diesel and tank A has *(x+500)* litres of diesel. Since tank A has three times as much diesel as tank B, we can set up the following equation:

\[(x+500) = 3(x-100)\]

Expanding the right-hand side gives:

\[x + 500 = 3x - 300\]

Simplifying and solving for *x* gives:

\[x = 400\]

Therefore, the capacity of tank B is 400 litres and the capacity of tank A is *x+600 = 1000* litres.

**Answer Details**

(a) Since *\(\overline{RT}\)* and *\(\overline{RU}\)* are tangent to the circle, *\(\angle TRU\)* is equal to the angle between the radii drawn to the points of tangency. Therefore, *\(\angle TRU = \angle TRO = 90^{\circ}\)*.

Also, since *\(\overline{RT}\)* is tangent to the circle, *\(\angle SRT\)* is equal to the angle between the tangent and the chord *\(\overline{ST}\)*. Therefore, *\(\angle SRT = \angle OTU\)*.

Finally, since the sum of the angles in a triangle is *180^{\circ}*, we have:

\[\angle TUS + \angle SRT + \angle UTO = 180^{\circ}\]

\[68^{\circ} + x + y = 180^{\circ}\]

\[x = 112^{\circ} - y\]

Therefore, the value of *x* is *112^{\circ} - y*.

(b) Let the capacity of tank B be *x* litres. Then, the capacity of tank A is *x+600* litres.

After 100 litres of diesel is pumped out of each tank, tank B has *(x-100)* litres of diesel and tank A has *(x+500)* litres of diesel. Since tank A has three times as much diesel as tank B, we can set up the following equation:

\[(x+500) = 3(x-100)\]

Expanding the right-hand side gives:

\[x + 500 = 3x - 300\]

Simplifying and solving for *x* gives:

\[x = 400\]

Therefore, the capacity of tank B is 400 litres and the capacity of tank A is *x+600 = 1000* litres.

**Question 53**
**Report**

NOT DRAWN TO SCALE

In the diagram, is a chord of a circle with centre 0. 22.42 cm and the perimeter of triangle MON is 55.6 cm. Calculate, correct to the nearest degree. < MON.

(b) T is equidistant from P and Q. The bearing of P from T is 60\(^o\) and the bearing of Q from T is 130\(^o\).

(i) Illustrate the information on a diagram.

(ii) Find the bearing of Q from P.

(a) To solve this problem, we need to use the following formula:
Perimeter of triangle MON = MO + ON + NM
We know that the perimeter of triangle MON is 55.6 cm. Let's assume that MO = x, ON = y, and NM = z. Then we have:
x + y + z = 55.6
We also know that MO and ON are radii of the circle, so they are equal in length. Let's assume that MO = ON = r. Then we have:
x + y = 2r
We can rearrange this equation to get:
r = (x + y) / 2
Now let's look at the chord PQ. We know that the perpendicular bisector of a chord passes through the centre of the circle. Let's draw a line from the centre of the circle to the midpoint of PQ and label it M. We also know that MO is a radius of the circle, so MO = r. Let's label the distance from M to the midpoint of PQ as d. Then we have:
r^2 = d^2 + (PQ/2)^2
We know that PQ = 22.42 cm, so we can substitute that in:
r^2 = d^2 + 250
We can rearrange this equation to get:
d^2 = r^2 - 250
Now we have two equations with two unknowns (x and y), and one equation (d^2 = r^2 - 250) that relates them. We can substitute r = (x + y) / 2 into the equation for d^2:
d^2 = ((x + y) / 2)^2 - 250
Simplifying this equation gives:
d^2 = (x^2 + 2xy + y^2) / 4 - 250
Multiplying both sides by 4 gives:
4d^2 = x^2 + 2xy + y^2 - 1000
We also know that x + y = 2r, so we can substitute that in:
4d^2 = x^2 + 4xr + y^2 - 1000
Substituting r = (x + y) / 2 gives:
4d^2 = x^2 + 4x((x+y)/2) + y^2 - 1000
Simplifying this equation gives:
4d^2 = 5x^2 + 10xy + 5y^2 - 1000
We can now substitute x + y = 2r and simplify:
4d^2 = 5x^2 + 10xr + 5r^2 - 1000
Substituting r^2 = d^2 + 250 gives:
4d^2 = 5x^2 + 10xd^2/(x+y) + 5d^2 + 1250
Simplifying this equation gives:
4d^2 = 5x^2(x+y) + 10xd^2 + 5d^2(x+y) + 1250(x+y)
We know that x + y + z = 55.6, so we can substitute z = 55.6 - x - y into the equation for the perimeter of triangle MON:
x + y + z = 55.6
z = 55.6 - x - y
Substituting this into the equation for MO + ON + NM gives:
r + r + z = 55.6
2r + z = 55.6
2r = 55.6 - z
r

**Answer Details**

(a) To solve this problem, we need to use the following formula: Perimeter of triangle MON = MO + ON + NM We know that the perimeter of triangle MON is 55.6 cm. Let's assume that MO = x, ON = y, and NM = z. Then we have: x + y + z = 55.6 We also know that MO and ON are radii of the circle, so they are equal in length. Let's assume that MO = ON = r. Then we have: x + y = 2r We can rearrange this equation to get: r = (x + y) / 2 Now let's look at the chord PQ. We know that the perpendicular bisector of a chord passes through the centre of the circle. Let's draw a line from the centre of the circle to the midpoint of PQ and label it M. We also know that MO is a radius of the circle, so MO = r. Let's label the distance from M to the midpoint of PQ as d. Then we have: r^2 = d^2 + (PQ/2)^2 We know that PQ = 22.42 cm, so we can substitute that in: r^2 = d^2 + 250 We can rearrange this equation to get: d^2 = r^2 - 250 Now we have two equations with two unknowns (x and y), and one equation (d^2 = r^2 - 250) that relates them. We can substitute r = (x + y) / 2 into the equation for d^2: d^2 = ((x + y) / 2)^2 - 250 Simplifying this equation gives: d^2 = (x^2 + 2xy + y^2) / 4 - 250 Multiplying both sides by 4 gives: 4d^2 = x^2 + 2xy + y^2 - 1000 We also know that x + y = 2r, so we can substitute that in: 4d^2 = x^2 + 4xr + y^2 - 1000 Substituting r = (x + y) / 2 gives: 4d^2 = x^2 + 4x((x+y)/2) + y^2 - 1000 Simplifying this equation gives: 4d^2 = 5x^2 + 10xy + 5y^2 - 1000 We can now substitute x + y = 2r and simplify: 4d^2 = 5x^2 + 10xr + 5r^2 - 1000 Substituting r^2 = d^2 + 250 gives: 4d^2 = 5x^2 + 10xd^2/(x+y) + 5d^2 + 1250 Simplifying this equation gives: 4d^2 = 5x^2(x+y) + 10xd^2 + 5d^2(x+y) + 1250(x+y) We know that x + y + z = 55.6, so we can substitute z = 55.6 - x - y into the equation for the perimeter of triangle MON: x + y + z = 55.6 z = 55.6 - x - y Substituting this into the equation for MO + ON + NM gives: r + r + z = 55.6 2r + z = 55.6 2r = 55.6 - z r

**Question 54**
**Report**

Three red balls, five green balls, and a number of blue balls are put together in a sack. One ball is picked at random from the sack. If the probability of picking a red ball is \(\frac{1}{6}\) find;

(a) The number of blue balls in the sack

(b) the probability of picking a green ball

**(a)** Let the number of blue balls be x.

The probability of picking a red ball is 1/6, which means there is only one red ball in the sack.

Therefore, the total number of balls in the sack is:

1 (red ball) + 5 (green balls) + x (blue balls) = 6 + x

The probability of picking a red ball is the number of red balls in the sack divided by the total number of balls in the sack. So we can write:

1/(6 + x) = 1/6

Solving for x, we get:

6 + x = 6

x = 0

Therefore, there are **0 blue balls** in the sack.

**(b)** The probability of picking a green ball is the number of green balls in the sack divided by the total number of balls in the sack.

Since we know there are 5 green balls and 1 red ball in the sack (and 0 blue balls, as we found in part (a)), the total number of balls in the sack is:

1 (red ball) + 5 (green balls) + 0 (blue balls) = 6

So the probability of picking a green ball is:

5/6

Therefore, the probability of picking a green ball is **5/6**.

**Answer Details**

**(a)** Let the number of blue balls be x.

The probability of picking a red ball is 1/6, which means there is only one red ball in the sack.

Therefore, the total number of balls in the sack is:

1 (red ball) + 5 (green balls) + x (blue balls) = 6 + x

The probability of picking a red ball is the number of red balls in the sack divided by the total number of balls in the sack. So we can write:

1/(6 + x) = 1/6

Solving for x, we get:

6 + x = 6

x = 0

Therefore, there are **0 blue balls** in the sack.

**(b)** The probability of picking a green ball is the number of green balls in the sack divided by the total number of balls in the sack.

Since we know there are 5 green balls and 1 red ball in the sack (and 0 blue balls, as we found in part (a)), the total number of balls in the sack is:

1 (red ball) + 5 (green balls) + 0 (blue balls) = 6

So the probability of picking a green ball is:

5/6

Therefore, the probability of picking a green ball is **5/6**.

**Question 55**
**Report**

(a) Ali and Yusif shared N420.000.00 in the ratio 3.,; 5 : 8 respectively. Find the sum of Ali and Yusuf's shares

(b) Solve: 2(\(\frac{1}{8}\))\(^x\) = 32\(^{x - 1}\).

**a) To find out the share of Ali and Yusuf:**

To find the total ratio of the shares, we have 3 : 5 : 8, which simplifies to 3 : 13. We can then calculate the fraction of N420,000.00 each person received by dividing their share of the ratio by the total ratio.

For Ali, the fraction would be ^{3}/_{13} of N420,000.00. So, Ali received N420,000.00 * ^{3}/_{13} = N120,000.00.

For Yusuf, the fraction would be ^{5}/_{13} of N420,000.00. So, Yusuf received N420,000.00 * ^{5}/_{13} = N200,000.00.

The sum of Ali and Yusuf's shares is N120,000.00 + N200,000.00 = N320,000.00.

**b) To solve the equation 2((1/8)^x) = 32^(x-1):**

We need to isolate x on one side of the equation. To do this, we can take the logarithm base 8 of both sides of the equation.

The equation becomes: x = log_{8}(32^(x-1)) / log_{8}(2(1/8)^x).

To simplify the expression further, we can use the property of logarithms that log_{b}(a^c) = c log_{b}(a).

So, x = (x-1) log_{8}(32) / log_{8}(2) + log_{8}(1/8).

Solving for x, we find that x = 3.

**Answer Details**

**a) To find out the share of Ali and Yusuf:**

To find the total ratio of the shares, we have 3 : 5 : 8, which simplifies to 3 : 13. We can then calculate the fraction of N420,000.00 each person received by dividing their share of the ratio by the total ratio.

For Ali, the fraction would be ^{3}/_{13} of N420,000.00. So, Ali received N420,000.00 * ^{3}/_{13} = N120,000.00.

For Yusuf, the fraction would be ^{5}/_{13} of N420,000.00. So, Yusuf received N420,000.00 * ^{5}/_{13} = N200,000.00.

The sum of Ali and Yusuf's shares is N120,000.00 + N200,000.00 = N320,000.00.

**b) To solve the equation 2((1/8)^x) = 32^(x-1):**

We need to isolate x on one side of the equation. To do this, we can take the logarithm base 8 of both sides of the equation.

The equation becomes: x = log_{8}(32^(x-1)) / log_{8}(2(1/8)^x).

To simplify the expression further, we can use the property of logarithms that log_{b}(a^c) = c log_{b}(a).

So, x = (x-1) log_{8}(32) / log_{8}(2) + log_{8}(1/8).

Solving for x, we find that x = 3.

**Question 56**
**Report**

(a) If logo a = 1.3010 and log\(_{10}\)b - 1.4771. find the value of ab

(b)

In the diagram. O is the centre of the circle,< ACB = 39\(^o\) and < CBE = 62\(^o\). Find: (i) the interior angle AOC;

(ii) angle BAC.

**(a)**

We know that log_{10}a = 1.3010. Therefore, we can use the definition of logarithms to write:

10^1.3010 = a a = 19.9526

We also know that log_{10}b = 1.4771. Using the same process as above, we can find that:

10^1.4771 = b b = 30.059

Finally, we can find the value of ab by multiplying these two values:

ab = 19.9526 * 30.059 ab = 600.001

Therefore, ab is approximately equal to 600 (to the nearest whole number).

**(b)**

A |\ | \ e | \ c | \ | \ -------O----- | / | / d | / |/ B

We can start by using the fact that angles in a triangle add up to 180 degrees. Therefore, angle AOB is:

angle AOB = 180 - angle ACB angle AOB = 180 - 39 angle AOB = 141

Since O is the center of the circle, we know that AC and BC are radii of the circle. Therefore, they are equal in length. We can use this fact to find the length of CD, since angle CDE is a right angle:

CD = AC = BC CD = d sin(62) = CD/CE CD = CE * sin(62)

Now we can use the cosine rule to find the length of CE:

CE^2 = AC^2 + AE^2 - 2(AC)(AE)cos(39) CE^2 = d^2 + e^2 - 2de(cos(141))

We can also use the cosine rule to find the

**Answer Details**

**(a)**

We know that log_{10}a = 1.3010. Therefore, we can use the definition of logarithms to write:

10^1.3010 = a a = 19.9526

We also know that log_{10}b = 1.4771. Using the same process as above, we can find that:

10^1.4771 = b b = 30.059

Finally, we can find the value of ab by multiplying these two values:

ab = 19.9526 * 30.059 ab = 600.001

Therefore, ab is approximately equal to 600 (to the nearest whole number).

**(b)**

A |\ | \ e | \ c | \ | \ -------O----- | / | / d | / |/ B

We can start by using the fact that angles in a triangle add up to 180 degrees. Therefore, angle AOB is:

angle AOB = 180 - angle ACB angle AOB = 180 - 39 angle AOB = 141

Since O is the center of the circle, we know that AC and BC are radii of the circle. Therefore, they are equal in length. We can use this fact to find the length of CD, since angle CDE is a right angle:

CD = AC = BC CD = d sin(62) = CD/CE CD = CE * sin(62)

Now we can use the cosine rule to find the length of CE:

CE^2 = AC^2 + AE^2 - 2(AC)(AE)cos(39) CE^2 = d^2 + e^2 - 2de(cos(141))

We can also use the cosine rule to find the

**Question 57**
**Report**

A survey of 40 students showed that 23 students study Mathematics, 5 study Mathematics and Physics, 8 study Chemistry and Mathematics, 5 study Physics and Chemistry and 3 study all the three subjects. The number of students who study Physics only is twice the number who study Chemistry only.

(a) Find the number of students who study:

(i) only Physics.

(ii) only one subject

b) What is the probability that a student selected at random studies exactly 2 subjects?

a. (i) **To find the number of students who study only Physics, we need to subtract the number of students who study Mathematics and Physics, and the number of students who study Physics and Chemistry from the total number of students who study Mathematics.**

From the information given, 23 students study Mathematics, and 5 study Mathematics and Physics, so 23 - 5 = 18 students study Mathematics only.

The number of students who study Physics only is twice the number who study Chemistry only, so if x is the number of students who study Chemistry only, then 2x is the number of students who study Physics only.

(ii) **To find the number of students who study only one subject, we need to add the number of students who study each subject only.**

From the information given, 18 students study Mathematics only, x students study Chemistry only, and 2x students study Physics only, so 18 + x + 2x = 18 + 3x students study only one subject.

b. **To find the probability that a student selected at random studies exactly 2 subjects, we need to find the number of students who study 2 subjects, and divide that by the total number of students (40).**

From the information given, 8 students study Chemistry and Mathematics, and 5 students study Physics and Chemistry, so 8 + 5 = 13 students study 2 subjects.

The probability of selecting a student who studies exactly 2 subjects is 13/40.

**Answer Details**

a. (i) **To find the number of students who study only Physics, we need to subtract the number of students who study Mathematics and Physics, and the number of students who study Physics and Chemistry from the total number of students who study Mathematics.**

From the information given, 23 students study Mathematics, and 5 study Mathematics and Physics, so 23 - 5 = 18 students study Mathematics only.

The number of students who study Physics only is twice the number who study Chemistry only, so if x is the number of students who study Chemistry only, then 2x is the number of students who study Physics only.

(ii) **To find the number of students who study only one subject, we need to add the number of students who study each subject only.**

From the information given, 18 students study Mathematics only, x students study Chemistry only, and 2x students study Physics only, so 18 + x + 2x = 18 + 3x students study only one subject.

b. **To find the probability that a student selected at random studies exactly 2 subjects, we need to find the number of students who study 2 subjects, and divide that by the total number of students (40).**

From the information given, 8 students study Chemistry and Mathematics, and 5 students study Physics and Chemistry, so 8 + 5 = 13 students study 2 subjects.

The probability of selecting a student who studies exactly 2 subjects is 13/40.

**Question 58**
**Report**

(a) The third and sixth terms of a Geometric Progression (G.P) are and \(\frac{1}{4}\) and \(\frac{1}{32}\) respectively.

Find:

(i) the first term and the common ratio;

(ii) the seventh term.

(b) Given that 2 and -3 are the roots of the equation ax\(^2\) ± bx + c = 0, find the values of a, b and c.

**(a) (i) The first term and the common ratio of a Geometric Progression (G.P) can be found using the formula for the nth term of a G.P:**

a_{n} = a * r^{(n-1)}

where a is the first term, r is the common ratio, and n is the term number.

We are given that the third term is 1/4 and the sixth term is 1/32. We can use these values to solve for a and r:

a * r^{(3-1)} = 1/4

a * r^{(6-1)} = 1/32

Dividing the second equation by the first:

a * r^{(6-1)} / a * r^{(3-1)} = 1/32 / 1/4

r^{3} = 1/32 / 1/4

r^{3} = (1/32) * (4)

r^{3} = 1

Taking the cube root of both sides:

r = 1^{(1/3)} = 1

Substituting r = 1 back into the first equation:

a * 1^{(3-1)} = 1/4

a = 1/4

So, the first term is 1/4 and the common ratio is 1.

(ii) To find the seventh term, we can use the formula for the nth term of a G.P:

a_{n} = a * r^{(n-1)}

where a is the first term (1/4), r is the common ratio (1), and n is the term number (7).

a_{7} = 1/4 * 1^{(7-1)} = 1/4 * 1^{6} = 1/4 * 64 = 16

So, the seventh term is 16.

**(b) Given that 2 and -3 are the roots of the equation ax ^{2} + bx + c = 0, we can use the sum and product of the roots formula to find the values of a, b, and c:**

The sum of the roots = -b/a

The product of the roots = c/a

Since we are given that the roots are 2 and -3, we can substitute:

The sum of the roots = 2 + (-3) = -1

The product of the roots = 2 * (-3) = -6

So:

-b/a = -1

c/a = -6

Multiplying both equations by a:

-b = -a

c = -6a

Since a is non-zero, we can divide both equations by -1 to simplify:

b = a

c = -6

**Answer Details**

**(a) (i) The first term and the common ratio of a Geometric Progression (G.P) can be found using the formula for the nth term of a G.P:**

a_{n} = a * r^{(n-1)}

where a is the first term, r is the common ratio, and n is the term number.

We are given that the third term is 1/4 and the sixth term is 1/32. We can use these values to solve for a and r:

a * r^{(3-1)} = 1/4

a * r^{(6-1)} = 1/32

Dividing the second equation by the first:

a * r^{(6-1)} / a * r^{(3-1)} = 1/32 / 1/4

r^{3} = 1/32 / 1/4

r^{3} = (1/32) * (4)

r^{3} = 1

Taking the cube root of both sides:

r = 1^{(1/3)} = 1

Substituting r = 1 back into the first equation:

a * 1^{(3-1)} = 1/4

a = 1/4

So, the first term is 1/4 and the common ratio is 1.

(ii) To find the seventh term, we can use the formula for the nth term of a G.P:

a_{n} = a * r^{(n-1)}

where a is the first term (1/4), r is the common ratio (1), and n is the term number (7).

a_{7} = 1/4 * 1^{(7-1)} = 1/4 * 1^{6} = 1/4 * 64 = 16

So, the seventh term is 16.

**(b) Given that 2 and -3 are the roots of the equation ax ^{2} + bx + c = 0, we can use the sum and product of the roots formula to find the values of a, b, and c:**

The sum of the roots = -b/a

The product of the roots = c/a

Since we are given that the roots are 2 and -3, we can substitute:

The sum of the roots = 2 + (-3) = -1

The product of the roots = 2 * (-3) = -6

So:

-b/a = -1

c/a = -6

Multiplying both equations by a:

-b = -a

c = -6a

Since a is non-zero, we can divide both equations by -1 to simplify:

b = a

c = -6

**Question 59**
**Report**

(a) The curved surface areas of two cones are equal. The base radius of one is 5 cm and its slant height is 12cm. calculate the height of the second cone if its base radius is 6 cm.

(b) Given the matrices A = \(\begin{pmatrix} 2 & 5 \\ -1 & -3 \end{pmatrix}\) and B = \(\begin{pmatrix} 3 & -2 \\ 4 & 1 \end{pmatrix}\), find:

(i) BA;

(ii) the determinant of BA.

**a)** We can use the formula for the curved surface area of a cone, which is given by πrs, where r is the radius of the base and s is the slant height. Let's assume that the height of the second cone is h. Then we have:

π(5)(12) = π(6)(√(6^2 - h^2))

Simplifying the equation, we get:

60 = 6√(36 - h^2)

10 = √(36 - h^2)

Squaring both sides, we get:

100 = 36 - h^2

h^2 = 64

h = **8**

Therefore, the height of the second cone is 8 cm.

**b) (i)** To find BA, we need to multiply the matrices in the order B x A. The product is given by:

BA = ** ∡**(3 -2) (4 1) x (2 5) (-1 -3) =

`∡`

**(ii)** The determinant of a product of two matrices is equal to the product of their determinants, i.e., |AB| = |A||B|. Therefore, to find the determinant of BA, we can multiply the determinants of B and A. The determinant of a 2x2 matrix is given by the formula ad - bc, where a, b, c, and d are the elements of the matrix. Using this formula, we get:

|BA| = (8)(17) - (21)(7) = **-49**

Therefore, the determinant of BA is -49.

**Answer Details**

**a)** We can use the formula for the curved surface area of a cone, which is given by πrs, where r is the radius of the base and s is the slant height. Let's assume that the height of the second cone is h. Then we have:

π(5)(12) = π(6)(√(6^2 - h^2))

Simplifying the equation, we get:

60 = 6√(36 - h^2)

10 = √(36 - h^2)

Squaring both sides, we get:

100 = 36 - h^2

h^2 = 64

h = **8**

Therefore, the height of the second cone is 8 cm.

**b) (i)** To find BA, we need to multiply the matrices in the order B x A. The product is given by:

BA = ** ∡**(3 -2) (4 1) x (2 5) (-1 -3) =

`∡`

**(ii)** The determinant of a product of two matrices is equal to the product of their determinants, i.e., |AB| = |A||B|. Therefore, to find the determinant of BA, we can multiply the determinants of B and A. The determinant of a 2x2 matrix is given by the formula ad - bc, where a, b, c, and d are the elements of the matrix. Using this formula, we get:

|BA| = (8)(17) - (21)(7) = **-49**

Therefore, the determinant of BA is -49.

**Question 60**
**Report**

A woman bought 130 kg of tomatoes for 4452,000.00. She sold half of the tomatoes at a profit of 30%. The rest of the tomatoes began to go bad, she then reduced the selling price per kg by 12%. Calculate:

(a) the new selling price per kg;

(ii) the percentage profit on the entire sales if she threw away 5 kg of bad tomatoes.

(a) Let's first find the cost per kilogram of the tomatoes:

Cost per kilogram = Total cost / Total weight

Cost per kilogram = 4452,000.00 / 130 kg

Cost per kilogram = **34,246.15**

Now let's find the selling price per kilogram for the first half of the tomatoes, which were sold at a 30% profit:

Profit = Selling price - Cost price

30% of Cost price = 0.3 * 34,246.15 = **10,273.85**

Selling price = Cost price + Profit

Selling price = 34,246.15 + 10,273.85 = **44,520.00**

Selling price per kilogram = Selling price / (130 kg / 2)

Selling price per kilogram = 44,520.00 / 65 kg

Selling price per kilogram = **684.92**

Now let's find the new selling price per kilogram for the remaining tomatoes, which were sold at a 12% discount:

Discount = Selling price before discount - Selling price after discount

12% of Selling price before discount = 0.12 * Selling price before discount

Selling price after discount = Selling price before discount - 0.12 * Selling price before discount

Selling price after discount = 0.88 * Selling price before discount

Selling price per kilogram = Selling price after discount / (130 kg / 2)

Selling price per kilogram = 0.88 * Selling price before discount / (130 kg / 2)

Selling price per kilogram = 0.44 * Selling price before discount / 65

Selling price per kilogram = 0.44 * 684.92 / 65

Selling price per kilogram = **4.63**

(ii) If 5 kg of tomatoes were thrown away, then the total weight sold would be 125 kg. Let's calculate the total revenue:

Revenue = Selling price per kilogram * Total weight sold

Revenue = (684.92 * 65 / 2) + (4.63 * 60)

Revenue = **22,303.80**

The cost of the tomatoes was 4452,000.00, and since the revenue was less than that, there was a loss. Let's calculate the percentage loss:

Loss = Cost - Revenue

Loss = 4452,000.00 - 22,303.80

Loss = **4429,696.20**

Percentage loss = (Loss / Cost) * 100

Percentage loss = (4429,696.20 / 4452,000.00) * 100

Percentage loss = **99.5%**

So the percentage profit on the entire sales was **-99.5%**, which means there was a loss of 99.5%.

**Answer Details**

(a) Let's first find the cost per kilogram of the tomatoes:

Cost per kilogram = Total cost / Total weight

Cost per kilogram = 4452,000.00 / 130 kg

Cost per kilogram = **34,246.15**

Now let's find the selling price per kilogram for the first half of the tomatoes, which were sold at a 30% profit:

Profit = Selling price - Cost price

30% of Cost price = 0.3 * 34,246.15 = **10,273.85**

Selling price = Cost price + Profit

Selling price = 34,246.15 + 10,273.85 = **44,520.00**

Selling price per kilogram = Selling price / (130 kg / 2)

Selling price per kilogram = 44,520.00 / 65 kg

Selling price per kilogram = **684.92**

Now let's find the new selling price per kilogram for the remaining tomatoes, which were sold at a 12% discount:

Discount = Selling price before discount - Selling price after discount

12% of Selling price before discount = 0.12 * Selling price before discount

Selling price after discount = Selling price before discount - 0.12 * Selling price before discount

Selling price after discount = 0.88 * Selling price before discount

Selling price per kilogram = Selling price after discount / (130 kg / 2)

Selling price per kilogram = 0.88 * Selling price before discount / (130 kg / 2)

Selling price per kilogram = 0.44 * Selling price before discount / 65

Selling price per kilogram = 0.44 * 684.92 / 65

Selling price per kilogram = **4.63**

(ii) If 5 kg of tomatoes were thrown away, then the total weight sold would be 125 kg. Let's calculate the total revenue:

Revenue = Selling price per kilogram * Total weight sold

Revenue = (684.92 * 65 / 2) + (4.63 * 60)

Revenue = **22,303.80**

The cost of the tomatoes was 4452,000.00, and since the revenue was less than that, there was a loss. Let's calculate the percentage loss:

Loss = Cost - Revenue

Loss = 4452,000.00 - 22,303.80

Loss = **4429,696.20**

Percentage loss = (Loss / Cost) * 100

Percentage loss = (4429,696.20 / 4452,000.00) * 100

Percentage loss = **99.5%**

So the percentage profit on the entire sales was **-99.5%**, which means there was a loss of 99.5%.

**Question 61**
**Report**

(a) Fred bought a car for $5,600.00 and later sold it at 90% of the cost price. He spent $1,310.00 out of the amount received and invested the rest at 6% per annum simple interest. Calculate the interest earned in 3 years.

(b) Solve the equations 2\(^x\)(4\(^{-7}\)) = 2 and 3\(^{-x}\)(9\(^{2y}\)) = 3 simultaneously.

**a) First, let's calculate the amount Fred received after selling the car.** He sold the car for 90% of its cost price, so he received $5,600 * 0.9 = $5,040.00. After spending $1,310.00, he was left with $5,040 - $1,310 = $3,730.00. This is the amount he invested at 6% per annum simple interest.

The formula for simple interest is I = P * R * T, where I is the interest, P is the principal (the amount invested), R is the rate of interest (as a decimal), and T is the time (in years). Plugging in the values, we have I = $3,730 * 0.06 * 3 = $676.20.

So Fred earned $676.20 in interest over a period of 3 years.

**b) To solve the equations 2 ^{x}(4^{-7}) = 2 and 3^{-x}(9^{2y}) = 3 simultaneously,** we need to find the values of x and y that make both equations true.

Starting with the first equation, we can simplify the left-hand side to get 2^{x} * 4^{-7} = 2^{x} * (1/16) = 2. We can then divide both sides of the equation by 2^{x} to get (1/16) = 1. Since this is not true, there are no values of x that make this equation true.

Next, let's simplify the second equation to get 3^{-x} * 9^{2y} = 3^{-x} * 81^{y} = 3. Dividing both sides of the equation by 3^{-x} gives us 81^{y} = 3^{x}. Taking the log base 3 of both sides gives us y = x.

So the only solution for x and y is x = y = 1.

In conclusion, there is only one solution for the system of equations, x = y = 1.

**Answer Details**

**a) First, let's calculate the amount Fred received after selling the car.** He sold the car for 90% of its cost price, so he received $5,600 * 0.9 = $5,040.00. After spending $1,310.00, he was left with $5,040 - $1,310 = $3,730.00. This is the amount he invested at 6% per annum simple interest.

The formula for simple interest is I = P * R * T, where I is the interest, P is the principal (the amount invested), R is the rate of interest (as a decimal), and T is the time (in years). Plugging in the values, we have I = $3,730 * 0.06 * 3 = $676.20.

So Fred earned $676.20 in interest over a period of 3 years.

**b) To solve the equations 2 ^{x}(4^{-7}) = 2 and 3^{-x}(9^{2y}) = 3 simultaneously,** we need to find the values of x and y that make both equations true.

Starting with the first equation, we can simplify the left-hand side to get 2^{x} * 4^{-7} = 2^{x} * (1/16) = 2. We can then divide both sides of the equation by 2^{x} to get (1/16) = 1. Since this is not true, there are no values of x that make this equation true.

Next, let's simplify the second equation to get 3^{-x} * 9^{2y} = 3^{-x} * 81^{y} = 3. Dividing both sides of the equation by 3^{-x} gives us 81^{y} = 3^{x}. Taking the log base 3 of both sides gives us y = x.

So the only solution for x and y is x = y = 1.

In conclusion, there is only one solution for the system of equations, x = y = 1.

**Question 62**
**Report**

(a) Solve the inequality: \(\frac{1 + 4x}{2}\) -\(\frac{5 + 2x}{7}\) < x -2

(b) If x: y = 3: 5, find the value of \(\frac{2x^2 - y^2}{y^2 - x^2}\)

**(a) **

majority of the candidates multiplied through by the L.C.M which is 14 to arrive at

7(1+4x) - 2(5+2x) < 14(x - 2)

and when simplified yielded

10x < -25 x -212 $\frac{1}{2}$

**(b) **

it was expected that they express

x = 3y5 $\frac{3\mathit{y}}{5}$ or y = 5x3 $\frac{5\mathit{x}}{3}$

and thereafter, substitute into the given expression to have

2x2−y2y2−x2=2x2−(5x3)2(5x3