WAEC SSCE - General Mathematics - 2019

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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.

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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.

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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.

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Detalhes da Resposta

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Detalhes da Resposta

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.

Detalhes da Resposta

Detalhes da Resposta

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: \(\frac{2}{3}\).

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Detalhes da Resposta

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.

Detalhes da Resposta

Detalhes da Resposta

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.

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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.

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Detalhes da Resposta

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}.

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Detalhes da Resposta

Detalhes da Resposta

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\(^2\).

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Detalhes da Resposta

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).

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Detalhes da Resposta

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.

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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: \(\frac{2}{3}\).

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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.

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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 \(\frac{1}{3}\).

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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.

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Detalhes da Resposta

Detalhes da Resposta

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\(\sqrt{2}\).

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Detalhes da Resposta

Detalhes da Resposta

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.

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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.

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Detalhes da Resposta

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 = -\(\frac{5}{2}\).

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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 -\(\frac{1}{2}\).

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Detalhes da Resposta

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%.

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Detalhes da Resposta

Detalhes da Resposta

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\(^o\).

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Detalhes da Resposta

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Detalhes da Resposta

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.

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(a) Value of m. The same quantity of rice is fixed, so boy-days are equal:

\(m\times 10 = (m+4)\times 8 \Rightarrow 10m = 8m+32 \Rightarrow 2m=32 \Rightarrow m=\mathbf{16}\).

(b) The number. Let the number be \(n\):

\(\dfrac{5}{6}n = \dfrac{1}{3}n + 10 \Rightarrow \left(\dfrac{5}{6}-\dfrac{2}{6}\right)n = 10 \Rightarrow \dfrac{1}{2}n = 10 \Rightarrow n=\mathbf{20}\).

(c) Equation of the line through \(\left(2,\tfrac{1}{2}\right)\) and \(\left(-1,-\tfrac{1}{2}\right)\).

Gradient \(m=\dfrac{\frac{1}{2}-(-\frac{1}{2})}{2-(-1)}=\dfrac{1}{3}\).

\(y-\dfrac{1}{2}=\dfrac{1}{3}(x-2)\Rightarrow y=\dfrac{1}{3}x-\dfrac{2}{3}+\dfrac{1}{2}=\dfrac{1}{3}x-\dfrac{1}{6}\).

\(y=\dfrac{1}{3}x-\dfrac{1}{6}\), i.e. \(2x-6y-1=0\).

Detalhes da Resposta

Detalhes da Resposta

(a) A right-angled triangle: the wall is vertical, the ground horizontal, and the \(11\text{ m}\) ladder is the hypotenuse making \(75^{\circ}\) with the ground. The top rests on the wall; pushing it up raises the top by \(0.2\text{ m}\) and moves the foot nearer the wall.

Original position. Height on wall \(=11\sin 75^{\circ}=11(0.9659)=10.625\text{ m}\).

Foot from wall \(=11\cos 75^{\circ}=11(0.2588)=2.847\text{ m}\).

(b)(i) New angle. New height \(=10.625+0.2=10.825\text{ m}\). The ladder length is unchanged:

\(\sin\theta=\dfrac{10.825}{11}=0.9841 \Rightarrow \theta=79.8^{\circ}\approx \mathbf{80^{\circ}}\).

(b)(ii) Distance the foot moved. New foot from wall \(=11\cos 79.8^{\circ}=11(0.1776)=1.953\text{ m}\).

Distance moved \(=2.847-1.953=0.894\text{ m}\approx \mathbf{1\ m}\).

Detalhes da Resposta

(a)

We know that log₁₀a = 1.3010. Therefore, we can use the definition of logarithms to write:

10^1.3010 = a
a = 19.9526

We also know that log₁₀b = 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

Detalhes da Resposta

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.

Detalhes da Resposta

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 = log₈(32^(x-1)) / log₈(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) log₈(32) / log₈(2) + log₈(1/8).

Solving for x, we find that x = 3.

Detalhes da Resposta

(a) Finding tan(y) using sin(y):

We cannot directly find the tangent (tan) of an angle using only the sine (sin). However, we can use the trigonometric identity:

tan^2(x) + 1 = sec^2(x)

where x is the angle and sec(x) is the secant (1/cos(x)).

Here's how to solve for tan(y):

1. We are given sin(y) = 8/17. We cannot find cos(y) directly from this information.
2. Since we cannot isolate tan(y) from the given identity, we cannot directly calculate its value.

Therefore, it's impossible to determine the value of

tan(y) + 2tan(y)/(1 + 2tan(y)) with the given information.

(b) Sharing Money:

Step 1: Calculate the remaining amount after Otobo's share.

Total money - Otobo's share

= N300,000.00 - N60,000.00 = N240,000.00

Step 2: Find Ada's share.

Ada receives 5/10 (which is the same as 1/2) of the remaining amount.

Ada's share = N240,000.00 * (1/2) = N120,000.00

Step 3: Calculate Adeola's share.

The remaining amount goes to Adeola.

Adeola's share = Total remaining - Ada's share

= N240,000.00 - N120,000.00 = N120,000.00

Step 4: Determine the ratio of shares.

Otobo : Ada : Adeola

= N60,000.00 : N120,000.00 : N120,000.00

Ratio: 3 : 6 : 6 (simplifying, 1 : 2 : 2)

Therefore, the money was shared in a ratio of 1 : 2 : 2 among Otobo, Ada, and Adeola respectively.

Detalhes da Resposta

(a) Finding x

From the diagram, \(RT\) and \(RS\) are tangents drawn from the external point R, touching the circle (centre O) at T and S. U lies on the circle, with the inscribed angle \(\angle TUS = 68^\circ\), and \(\angle SRT = x\).

The inscribed angle \(\angle TUS\) stands on the arc \(TS\) that faces R (the arc not containing U). By the inscribed-angle theorem, the central angle on that same arc is

In quadrilateral \(OTRS\), the radii meet the tangents at right angles, so \(\angle OTR = \angle OSR = 90^\circ\). The angles of the quadrilateral sum to \(360^\circ\):

Therefore \(x = 44^\circ\).

(b) Capacity of the two diesel tanks

Let the capacity of tank B be \(b\) litres. Then tank A holds \(b + 600\) litres.

After pumping out 100 litres from each, tank A contains 3 times as much as tank B:

So tank B holds \(400\) litres and tank A holds \(400 + 600 = 1000\) litres.

Check: after pumping, A = \(900\), B = \(300\), and \(900 = 3\times 300\). Correct.

Tank A has a capacity of \(1000\) litres and tank B has a capacity of \(400\) litres.

Detalhes da Resposta

(a) Angle MON

From the diagram, MN is a chord of the circle with centre O, and \(|MN| = 22.42\ \text{cm}\). OM and ON are radii, so \(|OM| = |ON| = r\).

The perimeter of triangle MON is 55.6 cm:

\[|OM| + |ON| + |MN| = 55.6\]

\[r + r + 22.42 = 55.6\]

\[2r = 33.18 \quad\Rightarrow\quad r = 16.59\ \text{cm}\]

Triangle MON is isosceles with \(OM = ON = 16.59\). Drop the perpendicular from O to the midpoint of MN; it bisects angle MON. Then

\[\sin\!\left(\frac{\angle MON}{2}\right) = \frac{\tfrac{1}{2}|MN|}{r} = \frac{11.21}{16.59} = 0.6757\]

\[\frac{\angle MON}{2} = \sin^{-1}(0.6757) = 42.5^\circ\]

\[\angle MON = 85.0^\circ \approx 85^\circ\]

(b) Bearings

(i) Illustration. T is equidistant from P and Q, so \(|TP| = |TQ|\). From T, the bearing of P is \(060^\circ\) and the bearing of Q is \(130^\circ\). Measuring both angles clockwise from the north line at T, P lies to the north-east and Q further round to the south-east.

The angle between the two bearings at T is

\[\angle PTQ = 130^\circ - 60^\circ = 70^\circ\]

(ii) Bearing of Q from P. Since \(|TP| = |TQ|\), triangle PTQ is isosceles, so the base angles are equal:

\[\angle TPQ = \angle TQP = \frac{180^\circ - 70^\circ}{2} = 55^\circ\]

The bearing of T from P is the reverse of the bearing of P from T:

\[060^\circ + 180^\circ = 240^\circ\]

Q lies \(55^\circ\) from PT, measured back towards the south. Hence the bearing of Q from P is

\[240^\circ - 55^\circ = 185^\circ\]

The bearing of Q from P is \(\mathbf{185^\circ}\).

Detalhes da Resposta

(a) Solve \(\;\dfrac{1}{3}x-\dfrac{1}{4}(x+2)\ge 3x-1\tfrac{1}{3}.\)

Write \(1\tfrac{1}{3}=\dfrac{4}{3}\) and multiply every term by the LCM of the denominators, \(12\):

\[12\cdot\tfrac{1}{3}x-12\cdot\tfrac{1}{4}(x+2)\ge 12\cdot 3x-12\cdot\tfrac{4}{3}.\]

\[4x-3(x+2)\ge 36x-16.\]

\[4x-3x-6\ge 36x-16.\]

\[x-6\ge 36x-16.\]

Collect like terms:

\[-6+16\ge 36x-x\;\Rightarrow\;10\ge 35x.\]

\[x\le\frac{10}{35}=\frac{2}{7}.\]

\(\therefore x\le\dfrac{2}{7}.\)

(b) In right-angled triangle \(ABC\) on horizontal ground, \(\angle BAC=90^{\circ}\), \(\angle ACB=35^{\circ}\) and \(|BC|=66\text{ m}\). \(AD\) is a vertical tower and \(\angle ABD=52^{\circ}\) is the angle of elevation of the top \(D\) from \(B\).

(i) Height of the tower \(|AD|\).

In triangle \(ABC\), \(BC\) is the hypotenuse and \(AB\) is opposite the \(35^{\circ}\) angle at \(C\):

\[\sin 35^{\circ}=\frac{|AB|}{|BC|}\;\Rightarrow\;|AB|=66\sin 35^{\circ}=66\times0.5736=37.86\text{ m}.\]

Also \(|AC|=66\cos 35^{\circ}=66\times0.8192=54.06\text{ m}.\)

The tower \(AD\) is vertical and \(AB\) horizontal, so triangle \(ABD\) is right-angled at \(A\):

\[\tan 52^{\circ}=\frac{|AD|}{|AB|}\;\Rightarrow\;|AD|=37.86\times\tan 52^{\circ}=37.86\times1.2799=48.45\text{ m}.\]

\(\therefore\) the height of the tower \(|AD|\approx 48.45\text{ m}\) (2 d.p.).

(ii) Angle of elevation of the tower from \(C\).

In right-angled triangle \(ACD\):

\[\tan(\angle ACD)=\frac{|AD|}{|AC|}=\frac{48.45}{54.06}=0.8962,\]

\[\angle ACD=\tan^{-1}(0.8962)=41.87^{\circ}.\]

Adding the horizontal angle \(\angle ACB=35^{\circ}\), the angle of elevation of the tower measured from \(C\) is

\[35^{\circ}+41.87^{\circ}=76.87^{\circ}\text{ (2 d.p.).}\]

Detalhes da Resposta

(a) Value of x. \(110_x = 40_{five}\).

Right side: \(40_{five}=4\times 5 + 0 = 20\).

Left side: \(110_x = 1\cdot x^{2} + 1\cdot x + 0 = x^{2}+x\).

\(x^{2}+x=20 \Rightarrow x^{2}+x-20=0 \Rightarrow (x+5)(x-4)=0\).

Since a base is positive, \(x=\mathbf{4}\).

(b) Simplify \(\dfrac{15}{\sqrt{75}}+\sqrt{108}+\sqrt{432}\).

\(\dfrac{15}{\sqrt{75}}=\dfrac{15}{5\sqrt{3}}=\dfrac{3}{\sqrt{3}}=\sqrt{3}\).

\(\sqrt{108}=\sqrt{36\times 3}=6\sqrt{3}\); \(\sqrt{432}=\sqrt{144\times 3}=12\sqrt{3}\).

Sum \(=\sqrt{3}+6\sqrt{3}+12\sqrt{3}=19\sqrt{3}\).

\(a=19,\ b=3\).

Detalhes da Resposta

(Reading the garbled cost as \(\text{N}52{,}000.00\) for \(130\text{ kg}\); the figures divide exactly, supporting this.)

Cost price per kg \(=\dfrac{52000}{130}=\text{N}400\).

First half (65 kg) at 30% profit. Selling price \(=400\times 1.30=\text{N}520\) per kg.

Revenue \(=65\times 520=\text{N}33{,}800\).

(a) New selling price per kg. Reduced by 12%: \(520\times(1-0.12)=520\times 0.88=\mathbf{\text{N}457.60}\) per kg.

(ii) Percentage profit on entire sales. Of the remaining 65 kg she throws away 5 kg, selling \(60\text{ kg}\) at \(\text{N}457.60\):

Revenue \(=60\times 457.60=\text{N}27{,}456\).

Total revenue \(=33800+27456=\text{N}61{,}256\); total cost \(=\text{N}52{,}000\).

Profit \(=61256-52000=\text{N}9{,}256\).

Percentage profit \(=\dfrac{9256}{52000}\times 100\approx \mathbf{17.8\%}\).

Detalhes da Resposta

(a) Setting up the joint variation

F varies directly as \(m_1 m_2\) and inversely as \(d^2\):

\[F = \frac{k\, m_1 m_2}{d^2}\]

(i) Finding k and the expression

Substitute \(F = 20,\ m_1 = 25,\ m_2 = 10,\ d = 5\):

\[20 = \frac{k(25)(10)}{5^2} = \frac{250k}{25} = 10k\]

\[k = 2\]

Therefore the required expression is

\[F = \frac{2\,m_1 m_2}{d^2}\]

(ii) Finding d

Substitute \(F = 30,\ m_1 = 7.5,\ m_2 = 4\):

\[30 = \frac{2(7.5)(4)}{d^2} = \frac{60}{d^2}\]

\[d^2 = \frac{60}{30} = 2\]

\[d = \sqrt{2} \approx 1.41\ \text{m}\]

(b) Finding x in the diagram

The figure is a five-sided polygon (pentagon) whose interior angles, read from the diagram, are:

\[x,\quad (x+20^\circ),\quad (x+40^\circ),\quad (x+80^\circ),\quad (x+60^\circ)\]

The sum of the interior angles of a pentagon is

\[(5-2)\times 180^\circ = 540^\circ\]

So:

\[x + (x+20^\circ) + (x+40^\circ) + (x+80^\circ) + (x+60^\circ) = 540^\circ\]

\[5x + 200^\circ = 540^\circ\]

\[5x = 340^\circ\]

\[x = 68^\circ\]

Detalhes da Resposta

(a) Geometric Progression. \(T_3=ar^{2}=\dfrac{1}{4}\) and \(T_6=ar^{5}=\dfrac{1}{32}\).

Divide: \(\dfrac{ar^{5}}{ar^{2}}=r^{3}=\dfrac{1/32}{1/4}=\dfrac{1}{8}\Rightarrow r=\dfrac{1}{2}\).

Then \(a\left(\tfrac{1}{2}\right)^{2}=\tfrac{1}{4}\Rightarrow \tfrac{a}{4}=\tfrac{1}{4}\Rightarrow a=1\).

(i) First term \(a=1\), common ratio \(r=\dfrac{1}{2}\).

(ii) \(T_7=ar^{6}=1\cdot\left(\dfrac{1}{2}\right)^{6}=\dfrac{1}{64}\).

(b) If \(2\) and \(-3\) are roots of \(ax^{2}+bx+c=0\):

Sum of roots \(=-\dfrac{b}{a}=2+(-3)=-1\); product \(=\dfrac{c}{a}=2\times(-3)=-6\).

Taking \(a=1\): \(b=1,\ c=-6\). So \(\mathbf{a=1,\ b=1,\ c=-6}\) (equation \(x^{2}+x-6=0\)).

Detalhes da Resposta

The cost \(C\) is partly constant and partly varies as the number of tourists \(n\), so

where \(a\) (the constant part) and \(b\) (the rate per tourist) are constants.

When \(n=20\), \(C=740\): \(\;740=a+20b\).
When the number increases by 10, \(n=30\), \(C=960\): \(\;960=a+30b\).

Subtract the first equation from the second:

Then \(a=740-20(22)=740-440=300\).

So \(C=300+22n\). When \(n=15\):

The dinner costs \(\$630.00\) when 15 tourists are present.

Detalhes da Resposta

(a) Solve \(\dfrac{1+4x}{2}-\dfrac{5+2x}{7}\[7(1+4x)-2(5+2x)<14(x-2)\]\[7+28x-10-4x<14x-28\]\[24x-3<14x-28\]\[24x-14x<-28+3\]\[10x<-25\Rightarrow x<-2.5\]

\(x<-2.5\)

(b) \(x:y=3:5\), so let \(x=3k,\ y=5k\).

\(\dfrac{2x^{2}-y^{2}}{y^{2}-x^{2}}=-\dfrac{7}{16}\)

Detalhes da Resposta

3 red, 5 green, \(b\) blue. Total \(=8+b\). \(P(\text{red})=\dfrac{3}{8+b}=\dfrac{1}{6}\).

(a) \(3\times 6 = 8+b \Rightarrow 18=8+b \Rightarrow b=\mathbf{10}\) blue balls.

Total balls \(=3+5+10=18\).

(b) \(P(\text{green})=\dfrac{5}{18}\).

Detalhes da Resposta

From the diagram, \(PQRS\) is split by the diagonal \(PR\) into two right-angled triangles: triangle \(PQR\) (right-angled at \(Q\)) and triangle \(PRS\) (right-angled at \(R\)). Given \(|PQ| = 3\) cm, \(|QR| = 4\) cm and \(|PS| = 13\) cm.

Step 1 - find the diagonal \(|PR|\). In right-angled triangle \(PQR\) (\(\angle PQR = 90^{\circ}\)), \(PR\) is the hypotenuse:

\[ |PR| = \sqrt{|PQ|^2 + |QR|^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ cm}. \]

Step 2 - find \(|RS|\). In right-angled triangle \(PRS\) (\(\angle PRS = 90^{\circ}\)), \(PS\) is the hypotenuse:

\[ |RS| = \sqrt{|PS|^2 - |PR|^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm}. \]

Step 3 - add the two triangle areas.

\[ \text{Area of } PQR = \tfrac{1}{2}\times |PQ| \times |QR| = \tfrac{1}{2}\times 3 \times 4 = 6\text{ cm}^2. \]

\[ \text{Area of } PRS = \tfrac{1}{2}\times |PR| \times |RS| = \tfrac{1}{2}\times 5 \times 12 = 30\text{ cm}^2. \]

\[ \text{Area of } PQRS = 6 + 30 = 36\text{ cm}^2. \]

Area of the quadrilateral \(= 36\) cm\(^2\).

Detalhes da Resposta

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.

Detalhes da Resposta

(a) For \(y=2\cos x+3\sin x\), with \(0^\circ\leq x\leq360^\circ\):

\(x\)	\(0^\circ\)	\(60^\circ\)	\(120^\circ\)	\(180^\circ\)	\(240^\circ\)	\(300^\circ\)	\(360^\circ\)
\(y\)	\(2.0\)	\(3.6\)	\(1.6\)	\(-2.0\)	\(-3.6\)	\(-1.6\)	\(2.0\)

For example,

\[y(60^\circ)=2\cos60^\circ+3\sin60^\circ=1+2.598=3.598\approx3.6.\]

(b) The completed graph is shown below. The plotted points are joined with a smooth curve.

(c)(i) Drawing the horizontal line \(y=-1\) and reading the points of intersection with the curve gives

\[\boxed{x\approx162^\circ\text{ or }310^\circ}.\]

(c)(ii) From the graph, at \(x=342^\circ\),

\[\boxed{y\approx1.0}\]

to one decimal place.

Detalhes da Resposta

(a) Evaluate each logarithm to base 2.

\(\log_{2}8=\log_{2}2^{3}=3\), \(\log_{2}16=\log_{2}2^{4}=4\), \(4\log_{2}2=4(1)=4\).

Numerator \(=3+4-4=3\).

Denominator: \(\log_{4}16=\log_{4}4^{2}=2\).

\(=\dfrac{3}{2}\)

(b) Convert each base-five number to base ten.

Therefore \(x=222-71=151\).

\(x=151\)

Detalhes da Resposta

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

Second term: \(a+d=x-1\)
Fourth term: \(a+3d=x+1\)
Sixth term: \(a+5d=7\)

(a) Common difference. Subtract the second-term equation from the fourth-term equation:

Common difference \(d=1\).

(c) Value of x. Subtract the fourth-term equation from the sixth-term equation:

With \(d=1\): \(2=6-x\Rightarrow x=4\).

(b) First term. From \(a+d=x-1\): \(a+1=4-1=3\Rightarrow a=2\).

First term \(=2\), common difference \(=1\), \(x=4\). (Check: terms are \(2,3,4,5,6,7,\ldots\); 2nd\(=3=x-1\), 4th\(=5=x+1\), 6th\(=7\).)

Detalhes da Resposta

(a) For the missing entry between  rm x=-1.0 rm and  rm x=0.0 rm , the intended value is  rm x=-0.5 rm . The completed table is:

x	-2.0	-1.5	-1.0	-0.5	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0
y	11	0	-9	-16	-21	-24	-25	-24	-21	-16	-9	0	11

For example, when  rm x=-1.5 rm ,

\[y=4(-1.5)^2-8(-1.5)-21=9+12-21=0.\]

(b)(i) The graph of  rm y=4x^2-8x-21 rm is shown below. The plotted curve uses the stated scales: 2 cm represents 1 unit on the  rm x rm -axis and 2 cm represents 5 units on the  rm y rm -axis.

(b)(ii)( rm  rm )

\[4x^2-8x=3\]

Since  rm y=4x^2-8x-21 rm , this gives

\[y=3-21=-18.\]

From the intersections of the curve with the line  rm y=-18 rm ,

\[\boxed{x\approx -0.35\quad\text{or}\quad x\approx2.3}.\]

(b)(ii)( rm  rm )

\[4x^2-7x-21=0.\]

Writing this in terms of the plotted curve,

\[4x^2-8x-21=-x,\]

so draw the line  rm y=-x rm . From its intersections with the curve,

\[\boxed{x\approx-1.6\quad\text{or}\quad x\approx3.3}.\]

Detalhes da Resposta

(a) Height of the second cone. Curved surface area \(=\pi r l\).

Cone 1: \(\pi(5)(12)=60\pi\). Cone 2 has \(r=6\) and equal C.S.A.:

\(\pi(6)l_2 = 60\pi \Rightarrow l_2 = 10\text{ cm}\).

Height \(h=\sqrt{l_2^{2}-r^{2}}=\sqrt{10^{2}-6^{2}}=\sqrt{100-36}=\sqrt{64}=\mathbf{8\ cm}\).

(b) \(A=\begin{pmatrix}2&5\\-1&-3\end{pmatrix},\ B=\begin{pmatrix}3&-2\\4&1\end{pmatrix}\).

(i) \(BA\):

\(BA=\begin{pmatrix}(3)(2)+(-2)(-1)&(3)(5)+(-2)(-3)\\(4)(2)+(1)(-1)&(4)(5)+(1)(-3)\end{pmatrix}=\begin{pmatrix}8&21\\7&17\end{pmatrix}\).

(ii) Determinant of \(BA\): \((8)(17)-(21)(7)=136-147=\mathbf{-11}\).

Detalhes da Resposta

(a) Frequency distribution

Class interval	Class boundaries	Frequency, \(f\)	Cumulative frequency, \(cf\)
0 - 9	-0.5 - 9.5	1	1
10 - 19	9.5 - 19.5	3	4
20 - 29	19.5 - 29.5	3	7
30 - 39	29.5 - 39.5	6	13
40 - 49	39.5 - 49.5	8	21
50 - 59	49.5 - 59.5	12	33
60 - 69	59.5 - 69.5	6	39
70 - 79	69.5 - 79.5	3	42
80 - 89	79.5 - 89.5	4	46
90 - 99	89.5 - 99.5	4	50
Total		50

(b) Cumulative frequency curve (less-than ogive)

The cumulative frequencies are plotted against the upper class boundaries, beginning with \((-0.5,0)\), and joined by a smooth curve.

(c)(i) Median

There are \(50\) students, so the median corresponds to cumulative frequency \(\frac{50}{2}=25\). Reading from \(cf=25\) on the ogive gives a mark of approximately \(53\).

Therefore, the estimated median mark is \(53\) marks.

(c)(ii) Percentage scoring at least 66 marks

From the ogive, the cumulative frequency at \(66\) marks is approximately \(37\). Hence, the estimated number scoring at least \(66\) marks is

\[50-37=13.\]

\[\text{Percentage}=\frac{13}{50}\times100\%=26\%.\]

Therefore, the percentage is \(26\%\), correct to the nearest whole number.

Detalhes da Resposta

(a) Equation of the line through \(A(-2,7)\) and \(B(2,-3)\).

Gradient \(m=\dfrac{-3-7}{2-(-2)}=\dfrac{-10}{4}=-\dfrac{5}{2}\).

\(y-7=-\dfrac{5}{2}(x+2)\Rightarrow y=-\dfrac{5}{2}x-5+7=-\dfrac{5}{2}x+2\).

\(y=-\dfrac{5}{2}x+2\), i.e. \(5x+2y-4=0\).

(b) \(\dfrac{5b-a}{8b+3a}=\dfrac{1}{5}\).

Cross-multiply: \(5(5b-a)=8b+3a \Rightarrow 25b-5a=8b+3a\).

\(25b-8b=3a+5a \Rightarrow 17b=8a \Rightarrow \dfrac{a}{b}=\dfrac{17}{8}=2.125\).

\(\dfrac{a}{b}\approx 2.13\) (2 d.p.).

Detalhes da Resposta

(a) Finding \(x = \angle SRT\).

From the diagram, \(RT\) and \(RS\) are tangents drawn from the external point \(R\), touching the circle (centre \(O\)) at \(T\) and \(S\). \(\angle TUS = 68^\circ\) is the angle at the circumference standing on chord \(TS\).

The angle subtended at the centre by the same chord \(TS\) is twice the angle at the circumference:

\[ \angle TOS = 2 \times \angle TUS = 2 \times 68^\circ = 136^\circ \]

Since a tangent is perpendicular to the radius at the point of contact:

\[ \angle OTR = \angle OSR = 90^\circ \]

Now consider quadrilateral \(RTOS\). Its interior angles sum to \(360^\circ\):

\[ \angle SRT + \angle OTR + \angle TOS + \angle OSR = 360^\circ \]

\[ x + 90^\circ + 136^\circ + 90^\circ = 360^\circ \]

\[ x = 360^\circ - 316^\circ = 44^\circ \]

(b) Capacity of each tank. Let the capacity of tank \(B\) be \(b\) litres. Then tank \(A\) holds \((b + 600)\) litres.

After pumping out \(100\) litres from each, tank \(A\) has \(3\) times as much as tank \(B\):

\[ (b + 600) - 100 = 3\,(b - 100) \]

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

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

\[ 800 = 2b \;\Rightarrow\; b = 400 \]

So tank \(B = 400\) litres and tank \(A = 400 + 600 = 1000\) litres.

Check: \(1000 - 100 = 900\) and \(3(400 - 100) = 3 \times 300 = 900.\) Correct.

Answers: (a) \(x = 44^\circ\); (b) Tank A \(= 1000\) litres, Tank B \(= 400\) litres.