WAEC SSCE - General Mathematics - 2007

If x =3, y = 2 and z = 4 what is the value of \(3x^{2}-2y+z\)

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Substituting x = 3, y = 2 and z = 4 into the expression \(3x^{2}-2y+z\), we get: $$3(3)^2 - 2(2) + 4 = 27 - 4 + 4 = 27$$ Therefore, the value of \(3x^{2}-2y+z\) when x = 3, y = 2 and z = 4 is 27. The correct option is (b) 27.

For what range of values of x is 4x - 3(2x - 1) > 1?

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To solve the inequality 4x - 3(2x - 1) > 1, we can simplify it as follows: 4x - 6x + 3 > 1 -2x + 3 > 1 -2x > -2 x < 1 Therefore, the range of values of x that satisfy the inequality is x < 1. So, the correct option is "x < 1".

The volume of a cylinder is 1200cm3 and the area of its base is 150cm2. Find the height of the cylinder.

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We can use the formula for the volume of a cylinder to solve this problem, which is: V = πr²h where V is the volume of the cylinder, r is the radius of the base, and h is the height of the cylinder. We are given that the volume of the cylinder is 1200cm³, so we can substitute this value into the formula: 1200 = πr²h We are also given that the area of the base is 150cm², and the formula for the area of a circle is: A = πr² where A is the area of the circle and r is the radius. Since the base of the cylinder is a circle, we can use this formula to find the radius: 150 = πr² r² = 150/π r ≈ 6.12 cm (rounded to two decimal places) Now we can substitute this value for r in the first formula and solve for h: 1200 = π(6.12)²h h ≈ 8 cm (rounded to two decimal places) Therefore, the height of the cylinder is approximately 8cm. So the correct option is (b) 8.00cm.

The area of a square field is 110.25m². Find the cost of fencing it round at N75.00 per metre

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The area of the square field is given as 110.25m². We know that the area of a square is given by the formula A = s², where A is the area and s is the side length. Therefore, we can find the side length of the square as follows: s² = A s² = 110.25 s = √110.25 = 10.5m The perimeter of the square is given by the formula P = 4s, where P is the perimeter and s is the side length. Therefore, we can find the perimeter of the square as follows: P = 4s = 4(10.5m) = 42m To find the cost of fencing it round at N75.00 per metre, we need to multiply the perimeter by N75.00: Cost = P × N75.00 = 42m × N75.00 = N3,150.00 Therefore, the cost of fencing the square field at N75.00 per metre is N3,150.00. So the correct option is (B) N3,150.00.

Simplify\(\frac{3x^{3}}{(3x)^{3}}\)

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To simplify the expression, we need to first expand the denominator of the fraction, which is \((3x)^3\). This can be written as \(3^3 x^3\) or \(27x^3\). Substituting this in the original fraction gives: $$\frac{3x^3}{(3x)^3} = \frac{3x^3}{27x^3} = \frac{1}{9}$$ Therefore, the simplified form of the expression is \(\frac{1}{9}\). Hence, the correct option is \(\frac{1}{9}\).

In the diagram, |SP| = |SR| and < PRS = 50^o. Calculate < PQR

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Find the product of 0.0409 and 0.0021 leaving your answer in the standard form

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I am x years old and my brother is 3 years older. How old was my brother last year?

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If you are currently x years old, and your brother is 3 years older than you, then your brother's current age is x + 3. To find out how old your brother was last year, you need to subtract 1 from his current age: (x + 3) - 1 = x + 2 Therefore, your brother was x + 2 years old last year. So, the correct option is (b) (x + 2) years.

A chord of length 6cm is drawn in a circle of radius 5cm. Find the distance of the chord from the centre of the circle.

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When a chord is drawn in a circle, it divides the circle into two equal parts. The line connecting the midpoint of the chord and the center of the circle is perpendicular to the chord. Therefore, to find the distance of the chord from the center of the circle, we need to draw a perpendicular bisector to the chord and measure the distance from the center of the circle to the perpendicular bisector. Let O be the center of the circle, AB be the chord of length 6cm and M be the midpoint of AB. Then OM is the perpendicular bisector of AB, and AM = MB = 3cm. Using the Pythagorean theorem in triangle OAM, we have: $$OA^2 = OM^2 + AM^2 = 5^2 - 3^2 = 16$$ Taking the square root of both sides, we have OA = 4cm. Therefore, the distance of the chord from the center of the circle is 4cm. So the correct option is (d) 4.0cm.

A sector of a circle of radius 14cm containing an angle 60^o is folded to form a cone. Calculate the radius of the base of the cone.

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The sector of a circle can be folded to form a cone if and only if the arc length of the sector is equal to the circumference of the base of the cone. Given that the radius of the sector is 14 cm and the angle is 60^o, the arc length of the sector is \(\frac{60}{360}\times2\pi(14)=\frac{14\pi}{3}\) cm. Let's denote the radius of the base of the cone by r. Then, the circumference of the base of the cone is \(2\pi r\) cm. Since the arc length of the sector is equal to the circumference of the base of the cone, we have: \[\frac{14\pi}{3}=2\pi r\] Dividing both sides by 2π gives: \[r=\frac{14}{3}\div2=\frac{7}{3}=2\frac{1}{3}\,\text{cm}\] Therefore, the radius of the base of the cone is \(2\frac{1}{3}\,\text{cm}\). Answer:.

Evaluate(111_two)² and leave your answer in base 2.

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To evaluate (111_two)², we need to convert the binary number 111_two to its decimal equivalent, then square the result, and finally convert the answer back to binary. 111_two = 1 × 2² + 1 × 2¹ + 1 × 2⁰ = 7_ten (111_two)² = (7_ten)² = 49_ten To convert 49_ten back to binary, we repeatedly divide by 2 and write down the remainders in reverse order: 49 ÷ 2 = 24 remainder 1 24 ÷ 2 = 12 remainder 0 12 ÷ 2 = 6 remainder 0 6 ÷ 2 = 3 remainder 0 3 ÷ 2 = 1 remainder 1 1 ÷ 2 = 0 remainder 1 Reading the remainders in reverse order gives us: 49_ten = 110001_two Therefore, the answer is (B) 110001_two.

Simplify \(3\sqrt{45}-12\sqrt{5}+16\sqrt{20}\)leaving your answer in surd form

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First, we need to simplify each term under the square root sign. We can write: \begin{align*} 3\sqrt{45} &= 3\sqrt{9\times 5} = 3\times 3\sqrt{5} = 9\sqrt{5}\\ 12\sqrt{5} &= 2\times 2\times 3\sqrt{5} = 4\sqrt{5}\times 3\\ 16\sqrt{20} &= 16\sqrt{4\times 5} = 16\times 2\sqrt{5} = 32\sqrt{5} \end{align*} Now we can substitute these simplified expressions back into the original expression and simplify: \begin{align*} 3\sqrt{45}-12\sqrt{5}+16\sqrt{20} &= 9\sqrt{5}-4\sqrt{5}\times 3+32\sqrt{5}\\ &= (9-12+32)\sqrt{5}\\ &= 29\sqrt{5} \end{align*} Therefore, the answer is \(29\sqrt{5}\).

Each of the interior angle of a regular polygon is 162^o. How many sides has the polygon?

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The sum of the interior angles of a polygon can be found by using the formula (n-2) x 180, where n is the number of sides of the polygon. For a regular polygon, all interior angles are equal. Let's call the measure of each interior angle x. Since the polygon is regular, we can use the fact that the sum of the measures of the interior angles of a polygon with n sides is (n-2) x 180, and that each interior angle of this polygon has measure x. Thus, we can set up the equation: n x x = (n-2) x 180 Simplifying this equation, we get: nx = 180n - 360 nx - 180n = -360 n(x - 180) = -360 n = -360 / (x - 180) We know that each interior angle of the polygon has a measure of 162 degrees, so x = 162. Substituting this value into the equation we just derived, we get: n = -360 / (162 - 180) n = -360 / (-18) n = 20 Therefore, the polygon has 20 sides. Answer: 20.

At what rate per cent per annum will N520.00 yield a simple interest of N39.00 in three years?

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We can use the formula for simple interest: Simple Interest = (P * r * t) / 100 where P is the principal, r is the rate of interest per annum, and t is the time period in years. We are given that P = N520.00 and the simple interest, SI = N39.00. We are also given that the time period is 3 years. Substituting these values into the formula, we get: N39.00 = (N520.00 * r * 3) / 100 Simplifying this equation, we get: r = (N39.00 * 100) / (N520.00 * 3) = 2.5% Therefore, the rate of interest per annum is 2.5%. So, the correct option is: - 2\(\frac{1}{2}\)%

The table shows the ages(in years) of twenty children chosen at random from a community. What is the median of the distribution? \(\begin{array}{c|c} Age(years) & 1 & 2 & 3 & 4 & 5 \\ \hline {\text {Number of children}} & 2 & 3 & 5 & 6 & 4 \end{array}\)

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A messenger was paid N2.50 an hour during the normal working hours and 4.00 n hour during overtime. If he received N31.00 for 10 hours work, how many hours are for overtime?

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Let's call the number of normal working hours the messenger did as "x" and the number of overtime hours as "y". We can create two equations based on the given information: Equation 1: The messenger worked a total of 10 hours x + y = 10 Equation 2: The total amount paid is N31.00 2.5x + 4y = 31 We can use the first equation to solve for x in terms of y: x = 10 - y Now we can substitute this expression for x into the second equation: 2.5(10 - y) + 4y = 31 Simplifying and solving for y, we get: 25 - 2.5y + 4y = 31 1.5y = 6 y = 4 Therefore, the messenger worked 4 hours of overtime. Answer: (c) 4.

Which of the following is a factor of 2 - x - x²?

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A sector of a circle of radius 14cm containing an angle 60^o is folded to form a cone. Calculate the radius of the base of the cone

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The venn diagram shows the choice of food of a number of visitors to a canteen. If there were 35 visitors in all, find the value of x

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In the diagram, |XR| = 4cm

|RZ| = 12cm, |SR| = n, |XZ| = m and SR||YZ. Find m in terms of n

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Expand the expression(3a - xy)(3a + xy)

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To expand the expression (3a - xy)(3a + xy), we can use the formula for multiplying two binomials: (a + b)(a - b) = a² - b² If we let a = 3a and b = xy, then we have: (3a - xy)(3a + xy) = (3a)² - (xy)² Expanding the terms on the right side, we get: (3a)² - (xy)² = 9a² - x²y² Therefore, the answer is 9a² - x²y².

I am x years old and my brother is 3 years older how old was my brother last year

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A chord of length 6cm is drawn in a circle of radius 5cm. Find the distance of the chord from the center of the circle

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In a , < PQR = < PRQ = 45^o. which of the following statements is/are correct? i. \(\bigtriangleup\)PQR is an equalateral triangle ii. \(\bigtriangleup\)PQR is an isosceles triangle iii. \(\bigtriangleup\)PQR is a right-angled triangle

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From the diagram, find the value of x in the diagram.

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The venn diagram shows the choice of food of a number of visitors to a canteen. How many people took at least two kinds of food?  If there were 35 visitors in all

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Find the LCM of 2³ x 3 x 5², 2 x 3² x 5 and 2³ x 3³ x 5

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For what value of x is the expression \(\frac{2x - 1}{x + 3}\) not defined?

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The expression \(\frac{2x-1}{x+3}\) is undefined if the denominator is equal to zero, since division by zero is not allowed in mathematics. Therefore, we need to solve the equation \(x+3=0\) to find the value of x for which the expression is undefined. Solving for x, we have: \begin{align*} x+3 &= 0 \\ x &= -3 \end{align*} Therefore, the expression is not defined for x = -3. Thus, the correct answer is (-3).

If \(8^{x+1}=\frac{1}{4}\), find x

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We can solve the given equation by taking logarithms to the base 2 on both sides. This gives: \begin{align*} 8^{x+1} &= \frac{1}{4}\\ \Rightarrow \quad 2^{3(x+1)} &= 2^{-2}\\ \Rightarrow \quad 3(x+1) &= -2\\ \Rightarrow \quad 3x+3 &= -2\\ \Rightarrow \quad 3x &= -5\\ \Rightarrow \quad x &= -\frac{5}{3} \end{align*} Therefore, the value of $x$ is $-\frac{5}{3}$. So, the correct option is \(-\frac{5}{3}\).

The bearing of a point P from another point Q is 310^o. If |PQ| = 200m, how far west of Q is P?

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make w the subject of the relation \(\frac{a + bc}{wd + f}\) = g

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To make w the subject of the relation \(\frac{a + bc}{wd + f} = g\), we can follow these steps: 1. Multiply both sides by \((wd + f)\) 2. Divide both sides by g 3. Divide both sides by (a + bc) This gives: \begin{align*} \frac{a + bc}{wd + f} &= g \\ (a + bc) &= g(wd + f) \\ a + bc &= gwd + gf \\ gwd &= a + bc - gf \\ w &= \frac{a + bc - gf}{gd} \end{align*} Therefore, the solution is: \begin{equation*} w = \frac{a + bc - gf}{gd} \end{equation*} Hence, the option that matches this answer is: - \(\frac{a + bc - fg}{dg}\)

Simplify(0.3 x 10⁵) \(\div\) (0.4 \(\times\) 10⁷), leaving you answer in the standard form.

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To simplify this expression, we need to perform division and then express the answer in standard form. (0.3 x 10⁵) \(\div\) (0.4 x 10⁷) = (0.3/0.4) x 10⁵−7 = 0.75 x 10^-2 = 7.5 x 10^-3 Therefore, the answer is 7.5 x 10^-3.

PQR is an equilateral triangle with sides 2\(\sqrt{3cm}\). calculate its height

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In the diagram, O is the centre of the circle and < PQR = 106^o, find the value of y

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If the probability of an event occurring is x, what is the probability of the event not occurring?

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The probability of an event occurring is the measure of the likelihood that the event will happen, given as a number between 0 and 1, inclusive. The probability of an event not occurring is the measure of the likelihood that the event will not happen. Since the sum of the probabilities of an event occurring and not occurring is 1, we can find the probability of an event not occurring by subtracting the probability of the event occurring from 1. Therefore, the probability of an event not occurring is equal to 1 minus the probability of the event occurring. Mathematically, if the probability of an event occurring is x, then the probability of the event not occurring is 1 - x. So, option A, 1 - x, is the correct answer.

In the diagram, O is the centre of the circle and PQ is a diameter. Triangle RSO is an equilateral triangle of side 4cm. Find the area of the shaded region

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N140,000 is shared between ABU, Kayode and Uche. Abu has twice as much as Kayode, and Kayode has twice as much as Uche. What is Kayode's share?

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Let's start by using variables to represent the amount of money each person gets. Let Uche's share be x. Then, we know that Kayode's share is twice that amount, so his share is 2x. And we also know that Abu's share is twice Kayode's share, so his share is 2(2x) = 4x. The sum of their shares is given as N140,000, so we can set up an equation: x + 2x + 4x = 140,000 Simplifying the left side of the equation, we get: 7x = 140,000 Dividing both sides by 7, we get: x = 20,000 So Uche's share is N20,000. We can then find Kayode's share, which is twice that amount: 2x = 2(20,000) = 40,000 Therefore, Kayode's share is N40,000. So the answer is, N40,000.

Simplify \((0.3\times 10^{5})\div (0.4\times 10^{7})\)leaving your answer in standard form

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Find the volume of a solid cylinder with base radius 10cm and height 14cm.

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The volume of a cylinder is given by the formula V = πr²h, where r is the radius of the base and h is the height of the cylinder. In this case, the base radius is 10cm and the height is 14cm. Substituting these values into the formula, we have: V = π(10cm)²(14cm) = π(100cm²)(14cm) = 1400πcm³ Using the value of π ≈ 3.14, we can approximate the answer to: V ≈ 1400(3.14) ≈ 4400cm³ Therefore, the volume of the cylinder is approximately 4400cm³. The correct option is (D) 4400cm³.

Convert 42₅ to base three numeral

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To convert a number from base 5 to base 3, we need to first convert it to base 10, and then convert it to base 3. 42₅ can be written as: $42_{5} = 4\cdot5^{1} + 2\cdot5^{0} = 20 + 2 = 22_{10}$ Now we convert 22₁₀ to base 3. We can do this by repeatedly dividing 22 by 3 and taking the remainders, until the quotient is zero. The remainders, read from bottom to top, give us the base 3 numeral. \begin{array}{c|c} \text{Dividend} & \text{Remainder}\\ \hline 22 & 1 \\ 7 & 2 \\ 2 & 2 \\ 0 & \\ \end{array} So, 22₁₀ is equal to 212₃. Therefore, the correct option is (3) 211₃.

Q is 32 km away from P on a bearing 042^o and R is 25km from P on a bearing of 132^o. Calculate the bearing of R from Q.

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Divide the sum of 8, 6, 7, 2, 0, 4, 7, 2, 3, by their mean

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To divide the sum of the given numbers by their mean, we need to first find their sum and mean. Sum = 8 + 6 + 7 + 2 + 0 + 4 + 7 + 2 + 3 = 39 Mean = (8 + 6 + 7 + 2 + 0 + 4 + 7 + 2 + 3) / 9 = 4.33 (rounded to two decimal places) Now, to divide the sum by the mean, we simply divide the sum by the mean: 39 / 4.33 = 9 (rounded to the nearest whole number) Therefore, the answer is 9.

The area of a square field is 110.25m². Find the cost of fencing it round at N75.00 per meter square

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The table above gives the distribution of the marks of a number of students in a test.
\(\begin{array}{c|c} Mark &1 & 2 & 3 & 4 & 5 & 6 \\ \hline Frequency & 5 & 3 & 6 & 4 & 2 & 5\end{array}\), find the mode of the distribution.

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In the diagram |LN| = 4cm, LNM = 90^o and tan y = \(\frac{2}{3}\). What is the area of the \(\bigtriangleup\)LMN?

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If p \(\alpha \frac{I}{Q}\) which of the following is true?

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Given that p \(\alpha \frac{I}{Q}\), where \(\alpha\) means 'is proportional to'. To determine the relationship between q and p, we need to manipulate the equation so that q is isolated on one side. We can write p = k\(\frac{I}{Q}\), where k is the constant of proportionality. Multiplying both sides by Q, we get pQ = kI. Dividing both sides by p, we get Q = \(\frac{k}{p}\)I. Since k is a constant of proportionality, we can write it as k = cp for some other constant c. Therefore, Q = \(\frac{c}{p}\)I. This means that q is inversely proportional to p. So, the correct option is q \(\alpha \frac{1}{p}\).

If x = 3, Y = 2 and z = 4, what is the value of 3x² - 2y + z?

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Substituting x = 3, y = 2, and z = 4 into the expression 3x² - 2y + z gives: 3x² - 2y + z = 3(3)² - 2(2) + 4 = 27 - 4 + 4 = 27 Therefore, the value of 3x² - 2y + z is 27.

Two circles have radii 16cm and 23cm. What is the difference between their circumference? take \(\pi = \frac{22}{7}\)

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The circumference of a circle is given by the formula 2πr, where r is the radius of the circle and π is a constant value approximately equal to 22/7 or 3.14. Using the given formula, the circumference of the circle with radius 16cm is 2 x (22/7) x 16cm = 100.57cm (to 2 decimal places). Similarly, the circumference of the circle with radius 23cm is 2 x (22/7) x 23cm = 144.57cm (to 2 decimal places). Therefore, the difference between the circumferences of the two circles is 144.57cm - 100.57cm = 44.00cm (to 2 decimal places). Hence, the correct option is (C) 44.00cm.

Simplify 3\(\sqrt{45} - 12\sqrt{5} + 16\sqrt{20}\), leaving your answer in surd form.

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First, we need to simplify each of the surds in the expression. \[\begin{aligned} 3\sqrt{45} &= 3\sqrt{9\times 5} = 3\times 3\sqrt{5} = 9\sqrt{5} \\ 16\sqrt{20} &= 16\sqrt{4\times 5} = 16\times 2\sqrt{5} = 32\sqrt{5} \end{aligned}\] Now we can substitute these simplified surds back into the original expression and simplify it further: \[\begin{aligned} 3\sqrt{45} - 12\sqrt{5} + 16\sqrt{20} &= 9\sqrt{5} - 12\sqrt{5} + 32\sqrt{5} \\ &= (9-12+32)\sqrt{5} \\ &= 29\sqrt{5} \end{aligned}\] Therefore, the simplified expression is 29\(\sqrt{5}\). The correct option is (a).

Simplify \(\frac{3x^3}{(3x)^3}\)

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We can simplify this expression by using the rule that states \(\frac{a^m}{a^n}=a^{m-n}\) for any non-zero real number a and integers m and n. Applying this rule, we get: \[\frac{3x^3}{(3x)^3} = \frac{3x^3}{27x^3} = \frac{1}{9}\] Therefore, the answer is \(\frac{1}{9}\).

Simplify \(\frac{4}{2x} - \frac{2x + x}{x}\)

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Find the smallest value of k such that 2\(^2\) x 3\(^3\) x 5 x k is a perfect square.

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If \(p\propto \frac{1}{q}\), which of the following is true?

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Find the value of a if log10 a + log10a2 = 0.9030

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Find the LCM of \(2^{3}\times 3\times 5^{2}, 2\times 3^{2}\times 5 \hspace{1mm}and \hspace{1mm}2^{2}\times 3^{2}\times 5\)

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If the ratio x:y = 3:5 and y:z = 4:7, find the ratio x:y:z

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Solve the simultaneous equations x + y = 2 and 3x - 2y = 1

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To solve the simultaneous equations x + y = 2 and 3x - 2y = 1, we can use the method of substitution or elimination. Method of Substitution: We can solve for one variable in terms of the other from the first equation and substitute it into the second equation, then solve for the other variable. Let's solve for y in terms of x from the first equation: x + y = 2 y = 2 - x Now substitute y = 2 - x into the second equation: 3x - 2y = 1 3x - 2(2 - x) = 1 3x - 4 + 2x = 1 5x = 5 x = 1 Now substitute x = 1 into either equation to find y: x + y = 2 1 + y = 2 y = 1 Therefore, the solution is x = 1 and y = 1. So, the correct answer is: x = 1 and y = 1.

N140,000 is shared between Abu, Kayode and Uche. Abu has twice as much as Kayode and Kayode has twice as much as Uche. What is Kayode'sshare?

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Let Uche's share be x. Since Kayode has twice as much as Uche, Kayode's share will be 2x. Also, since Abu has twice as much as Kayode, Abu's share will be 2(2x) = 4x. The sum of their shares is N140,000, so we have: x + 2x + 4x = N140,000 7x = N140,000 x = N20,000 Therefore, Kayode's share is 2x = 2(N20,000) = N40,000. Hence, the answer is N40,000.

Solve the simultaneous equation: x+y=2 and 3x-2y=1

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To solve this system of simultaneous equations, we need to find the values of x and y that satisfy both equations at the same time. One way to do this is to use the method of substitution. From the first equation, we have x + y = 2, which we can rearrange as x = 2 - y. We can then substitute this expression for x into the second equation, giving 3(2-y) - 2y = 1. Simplifying the left-hand side, we get 6 - 5y = 1, and solving for y, we find y = 1. We can then substitute this value for y into either equation to find x. Using the first equation, we get x + 1 = 2, so x = 1. Therefore, the solution to the simultaneous equations is x = 1 and y = 1.

The table shows the ages(in years) of twenty children chosen at random from a community. What is the mean age? \(\begin{array}{c|c} Age(years) & 1 & 2 & 3 & 4 & 5 \\ \hline {\text {Number of children}} & 2 & 3 & 5 & 6 & 4 \end{array}\)

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To find the mean age of the twenty children, we need to calculate the sum of the ages and divide by the total number of children. Sum of the ages = (1 x 2) + (2 x 3) + (3 x 5) + (4 x 6) + (5 x 4) = 2 + 6 + 15 + 24 + 20 = 67 Total number of children = 2 + 3 + 5 + 6 + 4 = 20 Mean age = sum of the ages/total number of children = 67/20 = 3.35 years Therefore, the mean age of the twenty children is 3.35 years. The correct option is (b) 3.35 years.

Find the value of x in the diagram

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Which of the following is true for the set \(P = \{-3.2\leq x< 5\}\) where x is an integer

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Evaluate \((111_{two})^2\) and leave your answer in base 2

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To solve this question, we need to convert the number 111 from binary to decimal, then square the result and convert the answer back to binary. Starting with the binary number 111, we can convert it to decimal using the place value system. The rightmost digit represents 1, the second digit from the right represents 2, and the leftmost digit represents 4. Adding these values together, we get: 1 + 2 + 4 = 7 So 111 in binary is equal to 7 in decimal. To square 7, we simply multiply it by itself: 7 x 7 = 49 So the decimal equivalent of (111_two)² is 49. To convert this back to binary, we use the same place value system but in reverse. Starting with the largest power of 2 that is less than or equal to 49, we subtract that value and place a 1 in the corresponding digit. We then repeat this process with the remainder until we reach 0. 49 is greater than or equal to 32, so we subtract 32 and place a 1 in the 6th digit from the right. The remainder is 17. 17 is greater than or equal to 16, so we subtract 16 and place a 1 in the 5th digit from the right. The remainder is 1. 1 is less than 2, so we place a 1 in the 1st digit from the right. The remainder is 0, so we have our final answer: 49 in binary is equal to 110001_two Therefore, the correct answer is (b) 110001_two.

For what value of x is the expression \(\frac{2x-1}{x+3}\)not defined?

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The expression \(\frac{2x-1}{x+3}\) is not defined when the denominator, \(x+3\), is equal to zero. Therefore, we can solve the equation \(x+3=0\) to find the value of x that makes the expression undefined. \(x+3=0\) \(x=-3\) So, the value of x that makes the expression undefined is -3. Therefore, the correct answer is (-3).

IN the diagram, |LN| = 4cm, LNM = 90^o and tan y = 2/3. What is the area of the ?LMN?

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Which of the following statement is true for the ste P = {-3.2 \(\leq\) x < 5} where x is an integer?

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The given set P = {-3.2 ≤ x < 5} contains all the integers that are greater than or equal to -3.2 but less than 5. However, since x has to be an integer, the least possible value of x is -3 (since -3 is the greatest integer less than or equal to -3.2) and the greatest possible value of x is 4 (since 4 is the greatest integer less than 5). Therefore, the statement "least value of x is -3" is true for the set P, and the statement "greatest value of x is 5" is false.

If 8^x-1 = \(\frac{1}{4}\), find x

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We know that 8^x-1 = \(\frac{1}{4}\). We can write 8^x-1 as (2³)^x-1 = 2^3x-3. Similarly, we can write 1/4 as 2^-2. Therefore, we can write the given equation as 2^3x-3 = 2^-2. Since the bases are equal, we can equate the exponents to get: 3x - 3 = -2 Adding 3 to both sides, we get: 3x = 1 Dividing both sides by 3, we get: x = \(\frac{1}{3}\) Therefore, the value of x is \(\frac{1}{3}\).

(a) A cylinder with radius 3.5 cm has its two ends closed, if the total surface area is \(209 cm^{2}\), calculate the height of the cylinder. [Take \(\pi = \frac{22}{7}\)].

(b)   In the diagram, O is the centre of the circle and ABC is a tangent at B. If \(\stackrel\frown{BDF} = 66°\) and \(\stackrel\frown{DBC} = 57°\), calculate, (i) \(\stackrel\frown{EBF}\) and (ii) \(\stackrel\frown{BGF}\).

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(a) Evaluate, without using mathematical tables or calculator, \((3.69 \times 10^{5}) \div (1.64 \times 10^{-3})\), leaving your answer in standard form.

(b) A man invested N20,000 in bank A and N25,000 in bank B at the beginning of the year. Bank A pays simple interest at a rate of y% per annum and B pays 1.5y% per annum. If his total interest at the end of the year from the two banks was N4,600, find the value of y.

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In a college, the number of absentees recorded over a period of 30 days was as shown in the frequency distribution table

Calculate the : (a) Mean

(b) Standard deviation , correct to two decimal places.

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(a) Simplify : \(\frac{3\frac{1}{12} + \frac{7}{8}}{2\frac{1}{4} - \frac{1}{6}}\)

(b) If \(p = \frac{m}{2} - \frac{n^{2}}{5m}\) ; 

(i) make n the subject of the relation ;  (ii) find, correct to three significant figures, the value of n when p = 14 and m = -8.

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Y is 60 km away from X on a bearing of 135°. Z is 80 km away from X on a bearing of 225°. Find the :

(a) distance of Z from Y ;

(b) bearing of Z from Y.

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(a) Using a ruler and a pair of compasses only, construc :

(i) a triangle PQR such that /PQ/ = 10 cm, /QR/ = 7 cm and < PQR = 90° ; (ii) the locus \(l_{1}\) of points equidistant from Q and R ; (iii) the locus \(l_{2}\) of points equidistant from P and Q.

(b) Locate the point O equidistant from P, Q and R.

(c) With O as centre, draw the circumcircle of the triangle PQR.

(d) Measure the radius of the circumcircle.

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(a) Simplify : \(\frac{x^{2} - y^{2}}{3x + 3y}\)

(b) 

In the diagram, PQRS is a rectangle. /PK/ = 15 cm, /SK/ = /KR/ and <PKS = 30°. Calculate, correct to three significant figures : (i) /PS/ ; (ii) /SK/ and (iii) the area of the shaded portion.

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(a) 

In the diagram, AOB is a straight line. < AOC = 3(x + y)°, < COB = 45°, < AOD = (5x + y)° and < DOB = y°. Find the values of x and y.

(b) From two points on opposite sides of a pole 33m high, the angles of elevation of the top of the pole are 53° and 67°. If the two points and the base are on te same horizontal level, calculate, correct to three significant figures, the distance between the two points.

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(a) Simplify : \(\frac{x^{2} - 8x + 16}{x^{2} - 7x + 12}\).

(b) If \(\frac{1}{2}, \frac{1}{x}, \frac{1}{3}\) are successive terms of an arithmetic progression (A.P), show that \(\frac{2 - x}{x - 3} = \frac{2}{3}\).

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Out of the 24 apples in a box, 6 are bad. If three apples are taken from the box at random, with replacement, find the probability that :

(a) the first two are good and the third is bad ;

(b) all three are bad ;

(c) all the three are good.

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The diagram shows the cross- section of a railway tunnel. If |AB| = 100m and the radius of the arc is 56m, calculate, correct to the nearest metre, the perimetre of the cross- section.

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(a) The 3rd and 8th terms of an arithmetic progression (A.P) are -9 and 26 respectively. Find the : (i) common difference ; (ii) first term.

(b) 

In the diagram \(\overline{PQ} || \overline{YZ}\), |XP| = 2cm, |PY| = 3 cm, |PQ| = 6 cm and the area of \(\Delta\) XPQ = 24\(cm^{2}\).Calculate the area of the trapezium PQZY.

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(a) With the aid of four- figure logarithm tables, evaluate \((0.004592)^{\frac{1}{3}}\).

(b) If \(\log_{10} y + 3\log_{10} x = 2\), express y in terms of x.

(c) Solve the equations : \(3x - 2y = 21\)

                                        \(4x + 5y = 5\).

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