WAEC SSCE - Further Mathematics - 2016
Ajụjụ 1 Ripọtì
Solve for x in the equation \(5^{x} \times 5^{x + 1} = 25\).
Akọwa Nkọwa
To solve the equation \(5^{x} \times 5^{x + 1} = 25\), we can use the rules of exponents. Since the bases are the same (both are 5), we can add the exponents to get: $$5^{x} \times 5^{x + 1} = 5^{2}$$ $$5^{2x + 1} = 5^{2}$$ Now we can solve for x by equating the exponents: $$2x + 1 = 2$$ $$2x = 1$$ $$x = \frac{1}{2}$$ Therefore, the solution is \(\frac{1}{2}\).
Ajụjụ 2 Ripọtì
Find the minimum value of \(y = 3x^{2} - x - 6\).
Akọwa Nkọwa
To find the minimum value of the quadratic function, we first need to find the vertex of the parabola. We can do this by using the formula: \(-\frac{b}{2a}\) where \(a\) is the coefficient of the squared term, and \(b\) is the coefficient of the linear term. In this case, \(a = 3\) and \(b = -1\), so: \(-\frac{b}{2a} = -\frac{-1}{2(3)} = \frac{1}{6}\) This tells us that the vertex of the parabola occurs at \(x = \frac{1}{6}\). To find the minimum value of the function, we substitute this value of \(x\) into the function: \(y = 3\left(\frac{1}{6}\right)^{2} - \frac{1}{6} - 6 = -6\frac{1}{12} = -\frac{25}{4}\) Therefore, the minimum value of the function is \(-6\frac{1}{12}\), which is equivalent to \(-\frac{25}{4}\). So the answer is option B.
Ajụjụ 3 Ripọtì
The remainder when \(x^{3} - 2x + m\) is divided by \(x - 1\) is equal to the remainder when \(2x^{3} + x - m\) is divided by \(2x + 1\). Find the value of m.
Akọwa Nkọwa
Ajụjụ 4 Ripọtì
A force 10N acts in the direction 060° and another force 6N acts in the direction 330°. Find the y component of their resultant force.
Akọwa Nkọwa
We can use vector addition to find the resultant force, then take the y component of it. First, we need to convert the forces into their x and y components. For the force of 10N at 060°, the x component is 10 cos(60°) = 5 N and the y component is 10 sin(60°) = 5√3 N. For the force of 6N at 330°, the x component is 6 cos(330°) = 3√3 N and the y component is 6 sin(330°) = -3 N. To find the resultant force, we add the x components and the y components separately: x component: 5 N + 3√3 N = 5 N + 5.2 N = 10.2 N y component: 5√3 N - 3 N = 8.66 N So the y component of the resultant force is 8.66 N. Therefore, the answer is (B) \((-3 + 5\sqrt{3})N\).
Ajụjụ 5 Ripọtì
If \(y = 4x - 1\), list the range of the domain \({-2 \leq x \leq 2}\), where x is an integer.
Akọwa Nkọwa
The given equation is \(y = 4x - 1\), which represents a linear function where the output value of \(y\) depends on the input value of \(x\). The domain of the function is defined as the set of all possible input values of \(x\) that the function can take. Here, the domain of the function is given as \({-2 \leq x \leq 2}\), which means that \(x\) can take integer values between -2 and 2 (both inclusive). To find the range of the function, we need to substitute the given values of \(x\) in the equation and calculate the corresponding values of \(y\). When we substitute \(x = -2\), we get \(y = 4(-2) - 1 = -9\). When we substitute \(x = -1\), we get \(y = 4(-1) - 1 = -5\). When we substitute \(x = 0\), we get \(y = 4(0) - 1 = -1\). When we substitute \(x = 1\), we get \(y = 4(1) - 1 = 3\). When we substitute \(x = 2\), we get \(y = 4(2) - 1 = 7\). Therefore, the range of the function for the given domain is {-9, -5, -1, 3, 7}.
Ajụjụ 6 Ripọtì
The radius of a circle increases at a rate of 0.5\(cms^{-1}\). Find the rate of change in the area of the circle with radius 7cm. \([\pi = \frac{22}{7}]\)
Akọwa Nkọwa
We know that the area of a circle is given by the formula: $$A = \pi r^2$$ We need to find the rate of change of the area of the circle, i.e., \(\frac{dA}{dt}\). We can use differentiation to find the rate of change of the area with respect to time: $$\frac{dA}{dt} = \frac{d}{dt} (\pi r^2)$$ Since the radius is increasing at a rate of 0.5\(cms^{-1}\), we have: $$\frac{dr}{dt} = 0.5$$ Using the chain rule of differentiation, we have: $$\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} = 2 \pi r \cdot 0.5 = \pi r$$ Substituting the given value of the radius, we get: $$\frac{dA}{dt} = \pi \cdot 7 = 22\(cm^{2}s^{-1}\)$$ Therefore, the rate of change in the area of the circle with radius 7cm is 22\(cm^{2}s^{-1}\). Hence, the correct option is: 22\(cm^{2}s^{-1}\).
Ajụjụ 7 Ripọtì
Given that \(-6, -2\frac{1}{2}, ..., 71\) is a linear sequence , calculate the number of terms in the sequence.
Akọwa Nkọwa
Ajụjụ 8 Ripọtì
The lines \(2y + 3x - 16 = 0\) and \(7y - 2x - 6 = 0\) intersect at point P. Find the coordinates of P.
Akọwa Nkọwa
Ajụjụ 9 Ripọtì
If \(P = {x : -2 < x < 5}\) and \(Q = {x : -5 < x < 2}\) are subsets of \(\mu = {x : -5 \leq x \leq 5}\), where x is a real number, find \((P \cup Q)\).
Akọwa Nkọwa
Ajụjụ 10 Ripọtì
Given that \(\tan x = \frac{5}{12}\), and \(\tan y = \frac{3}{4}\), Find \(\tan (x + y)\).
Akọwa Nkọwa
Ajụjụ 11 Ripọtì
Find the fourth term in the expansion of \((3x - y)^{6}\).
Akọwa Nkọwa
To find the fourth term in the expansion of \((3x - y)^6\), we can use the binomial theorem, which states that the \(r\)th term in the expansion of \((a+b)^n\) is given by: \[\binom{n}{r}a^{n-r}b^r\] In this case, \(a = 3x\) and \(b = -y\) (note the negative sign). So, using the binomial theorem, the fourth term in the expansion of \((3x - y)^6\) is: \[\binom{6}{3}(3x)^3(-y)^3 = \frac{6!}{3!3!}(27x^3)(-y^3) = -540x^3y^3\] Therefore, the answer is \(-540x^{3}y^{3}\).
Ajụjụ 13 Ripọtì
A box contains 14 white balls and 6 black balls. Find the probability of first drawing a black ball and then a white ball without replacement.
Akọwa Nkọwa
Ajụjụ 14 Ripọtì
A body of mass 10kg moving with a velocity of 5\(ms^{-1}\) collides with another body of mass 15kg moving in the same direction as the first with a velocity of 2\(ms^{-1}\). After collision, the two bodies move together with a common velocity v\(ms^{-1}\).
Akọwa Nkọwa
We can use the law of conservation of momentum to solve this problem. According to the law, the total momentum of the system before the collision is equal to the total momentum after the collision. Before collision: - Body 1 mass = 10kg, velocity = 5\(ms^{-1}\), momentum = 50 kg\(ms^{-1}\) - Body 2 mass = 15kg, velocity = 2\(ms^{-1}\), momentum = 30 kg\(ms^{-1}\) - Total momentum before collision = 50 + 30 = 80 kg\(ms^{-1}\) After collision: - Total mass = 10 + 15 = 25kg - Let v be the common velocity of the two bodies after the collision - Total momentum after collision = 25v Equating the two momenta, we get: 80 = 25v v = 80/25 = 3.2 Therefore, the common velocity of the two bodies after the collision is 3.2\(ms^{-1}\). Answer: (a) 3.2
Ajụjụ 15 Ripọtì
A ball falls from a height of 18m above the ground. Find the speed with which the ball hits the ground. \([g = 10ms^{-2}]\)
Akọwa Nkọwa
We can use the formula for the final velocity of an object falling from a height under the influence of gravity: $v^2 = u^2 + 2gh$ where $v$ is the final velocity, $u$ is the initial velocity (in this case, 0), $g$ is the acceleration due to gravity and $h$ is the height from which the object is falling. Substituting the given values, we have: $v^2 = 0 + 2\times10\times18 = 360$ Taking the square root of both sides, we get: $v = \sqrt{360} = 6\sqrt{10} \approx 18.97ms^{-1}$ Therefore, the speed with which the ball hits the ground is approximately 18.97\(ms^{-1}\). Hence, the correct option is (d) 18.97\(ms^{-1}\).
Ajụjụ 16 Ripọtì
If the solution set of \(x^{2} + kx - 5 = 0\) is (-1, 5), find the value of k.
Akọwa Nkọwa
To solve this problem, we can use the fact that the sum and product of the roots of a quadratic equation are related to its coefficients. Specifically, for the equation \(ax^2+bx+c=0\), the sum of the roots is given by \(-b/a\) and the product of the roots is given by \(c/a\). In this problem, we are given that the solution set of \(x^2+kx-5=0\) is (-1, 5). This means that the quadratic equation has roots -1 and 5. Therefore, we can write the equation in factored form as \((x+1)(x-5)=0\). Expanding this expression, we get: $$x^2-4x-5=0$$ Comparing this to the original equation \(x^2+kx-5=0\), we can see that \(k=-4\). Therefore, the answer is (B) -4. Alternatively, we can use the sum and product of roots formula to solve for k. Since the roots of the equation are -1 and 5, we know that: $$-b/a=-1+5=4$$ and $$c/a=-5$$ Using the fact that \(a=1\), we can solve for \(b\) and \(c\): $$b=-a(-b/a)=4(1)=4$$ $$c=a(c/a)=1(-5)=-5$$ Therefore, the original equation is \(x^2+4x-5=0\), and the answer is (B) -4.
Ajụjụ 17 Ripọtì
An operation * is defined on the set, R, of real numbers by \(p * q = p + q + 2pq\). If the identity element is 0, find the value of p for which the operation has no inverse.
Akọwa Nkọwa
To find the value of p for which the operation has no inverse, we need to solve for the value of p that makes the equation \(p * q = q * p = 0\) have no solution for q other than q = 0. Let's set up the equation: \[p * q = q * p = p + q + 2pq\] We want to find the value of p that makes the equation have no solution for q other than q = 0. So let's substitute q = 0: \[p * 0 = 0 * p = p + 0 + 2p(0)\] \[p = 0\] Now, let's check if there is any other value of p that would make the equation have no solution for q other than q = 0. If p is not 0, we can rewrite the equation as: \[q = \frac{-p}{2p + 1}\] Now, if p is any value other than \(-\frac{1}{2}\), the denominator 2p+1 will not be zero, which means that there will always be a real number q that satisfies the equation. However, if p = \(-\frac{1}{2}\), the denominator becomes zero, which means that the equation has no solution for q other than q = 0. Therefore, the value of p for which the operation has no inverse is \(-\frac{1}{2}\), and the answer is option A: \(\frac{-1}{2}\).
Ajụjụ 18 Ripọtì
If \(\log_{10}y + 3\log_{10}x \geq \log_{10}x\), express y in terms of x.
Akọwa Nkọwa
We can start by using the logarithmic rule that states: \(\log_{a}b + \log_{a}c = \log_{a}(bc)\). So, we can rewrite the given inequality as follows: \[\log_{10}y + 3\log_{10}x \geq \log_{10}x\] \[\Rightarrow \log_{10}(yx^{3}) \geq \log_{10}x\] \[\Rightarrow yx^{3} \geq x\] Dividing both sides by \(x\) (since \(x > 0\)), we get: \[yx^{2} \geq 1\] \[\Rightarrow y \geq \frac{1}{x^{2}}\] Therefore, the answer is \(y \geq \frac{1}{x^{2}}\).
Ajụjụ 19 Ripọtì
If \(\begin{vmatrix} m-2 & m+1 \\ m+4 & m-2 \end{vmatrix} = -27\), find the value of m.
Akọwa Nkọwa
The determinant of a 2x2 matrix is given by the formula: \(\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc\) Using this formula, we can find the determinant of the given matrix: \(\begin{vmatrix} m-2 & m+1 \\ m+4 & m-2 \end{vmatrix} = (m-2)(m-2) - (m+1)(m+4)\) Simplifying this expression, we get: \((m-2)^2 - (m+1)(m+4) = m^2 - 4m + 4 - (m^2 + 5m + 4) = -9m\) Setting this equal to the given determinant of -27, we have: \(-9m = -27\) Solving for m, we get: \(m = 3\) Therefore, the answer is: \(3\).
Ajụjụ 20 Ripọtì
Find an expression for y given that \(\frac{\mathrm d y}{\mathrm d x} = x^{2}\sqrt{x}\)
Akọwa Nkọwa
Ajụjụ 21 Ripọtì
A man of mass 80kg stands in a lift. If the lift moves upwards with acceleration 0.5\(ms^{-2}\), calculate the reaction from the floor of the lift on the man. \([g = 10ms^{-2}]\)
Ajụjụ 22 Ripọtì
The 3rd and 6th terms of a geometric progression (G.P.) are \(\frac{8}{3}\) and \(\frac{64}{81}\) respectively, find the common ratio.
Akọwa Nkọwa
Ajụjụ 24 Ripọtì
Find the variance of 11, 12, 13, 14 and 15.
Akọwa Nkọwa
To find the variance, first we need to calculate the mean of the given data set: Mean = (11+12+13+14+15)/5 = 13 Next, we subtract the mean from each data point and square the differences: (11-13)^2 = 4 (12-13)^2 = 1 (13-13)^2 = 0 (14-13)^2 = 1 (15-13)^2 = 4 Then, we find the average of the squared differences: (4+1+0+1+4)/5 = 2 Therefore, the variance of the given data set is 2. Answer is correct.
Ajụjụ 25 Ripọtì
Find the unit vector in the direction of \(-2i + 5j\).
Akọwa Nkọwa
To find the unit vector in the direction of \(-2i + 5j\), we need to first find the magnitude of the vector and then divide the vector by its magnitude. The magnitude of the vector \(-2i + 5j\) is given by: \begin{align*} \sqrt{(-2)^2 + (5)^2} &= \sqrt{4+25} \\ &= \sqrt{29} \end{align*} Therefore, the unit vector in the direction of \(-2i + 5j\) is: \begin{align*} \frac{1}{\sqrt{29}}(-2i + 5j) &= \frac{1}{\sqrt{29}}\begin{pmatrix}-2 \\ 5\end{pmatrix} \\ &= \frac{1}{\sqrt{29}}\begin{pmatrix}-2/1 \\ 5/1\end{pmatrix} \\ &= \boxed{\frac{1}{\sqrt{29}}(-2i + 5j)} \end{align*} Therefore, option (B) is the correct answer.
Ajụjụ 26 Ripọtì
Given that \(r = 3i + 4j\) and \(t = -5i + 12j\), find the acute angle between them.
Akọwa Nkọwa
Ajụjụ 27 Ripọtì
If \(f(x) = \frac{4}{x} - 1, x \neq 0\), find \(f^{-1}(7)\).
Akọwa Nkọwa
To find \(f^{-1}(7)\), we need to find the value of \(x\) for which \(f(x) = 7\). So, we start by setting \(f(x) = 7\) and solving for \(x\): \begin{align*} f(x) &= 7 \\ \frac{4}{x} - 1 &= 7 \\ \frac{4}{x} &= 8 \\ x &= \frac{4}{8} \\ x &= \frac{1}{2} \end{align*} Therefore, \(f^{-1}(7) = \frac{1}{2}\). So, the correct option is: - \(\frac{1}{2}\)
Ajụjụ 29 Ripọtì
If \(P = \begin{pmatrix} 1 & 2 \\ 5 & 1 \end{pmatrix}\) and \(Q = \begin{pmatrix} 0 & 1 \\ 1 & 3 \end{pmatrix}\), find PQ.
Akọwa Nkọwa
To find the product PQ of matrices P and Q, we multiply the elements of each row of matrix P by the corresponding elements of each column of matrix Q, and add up the products. Thus, we have: \begin{align*} PQ &= \begin{pmatrix} 1 & 2 \\ 5 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 3 \end{pmatrix} \\ &= \begin{pmatrix} (1)(0)+(2)(1) & (1)(1)+(2)(3) \\ (5)(0)+(1)(1) & (5)(1)+(1)(3) \end{pmatrix} \\ &= \begin{pmatrix} 2 & 7 \\ 1 & 8 \end{pmatrix} \end{align*} Therefore, the answer is option (C): \(\begin{pmatrix} 2 & 7 \\ 1 & 8 \end{pmatrix}\).
Ajụjụ 30 Ripọtì
Find the coefficient of \(x^{3}\) in the expansion of \([\frac{1}{3}(2 + x)]^{6}\).
Akọwa Nkọwa
The binomial theorem states that the term containing \(x^r\) in the expansion of \((a+b)^n\) is given by: $$\frac{n!}{r!(n-r)!}a^{n-r}b^{r}$$ In this case, we have \([\frac{1}{3}(2 + x)]^{6}\), so \(a = \frac{2}{3}\) and \(b = \frac{1}{3}x\). We want to find the coefficient of \(x^3\), so \(r = 3\). Using the formula above, the term containing \(x^3\) is: $$\frac{6!}{3!(6-3)!}\left(\frac{2}{3}\right)^3\left(\frac{1}{3}x\right)^3 = 20\times\frac{8}{27}\times\frac{1}{27}x^3 = \frac{160}{2187}x^3$$ Therefore, the coefficient of \(x^3\) in the expansion is \(\boxed{\frac{160}{729}}\).
Ajụjụ 31 Ripọtì
Express \(\frac{8 - 3\sqrt{6}}{2\sqrt{3} + 3\sqrt{2}}\) in the form \(p\sqrt{3} + q\sqrt{2}\).
Akọwa Nkọwa
Ajụjụ 32 Ripọtì
Given n = 3, evaluate \(\frac{1}{(n-1)!} - \frac{1}{(n+1)!}\)
Akọwa Nkọwa
We are given that n=3. Substituting n=3, we have: \(\frac{1}{(n-1)!} - \frac{1}{(n+1)!} = \frac{1}{(3-1)!} - \frac{1}{(3+1)!} = \frac{1}{2!} - \frac{1}{4!}\) Evaluating the factorials, we get: \(\frac{1}{2!} - \frac{1}{4!} = \frac{1}{2} - \frac{1}{24} = \frac{12}{24} - \frac{1}{24} = \frac{11}{24}\) Therefore, the value of \(\frac{1}{(n-1)!} - \frac{1}{(n+1)!}\) when n=3 is \(\frac{11}{24}\). Hence, the answer is, \(\frac{11}{24}\).
Ajụjụ 33 Ripọtì
Find the equation of the line which passes through (-4, 3) and parallel to line y = 2x + 5.
Akọwa Nkọwa
To find the equation of a line that is parallel to a given line, we need to use the fact that parallel lines have the same slope. The given line has a slope of 2 (since it is in the form y = mx + b, where m is the slope and b is the y-intercept). Therefore, any line parallel to it must also have a slope of 2. We also know that the line passes through the point (-4, 3). We can use the point-slope form of the equation of a line to find the equation of the line: y - y1 = m(x - x1) where (x1, y1) is the point the line passes through, and m is the slope. Substituting in the values we know, we get: y - 3 = 2(x - (-4)) Simplifying: y - 3 = 2(x + 4) y - 3 = 2x + 8 y = 2x + 11 Therefore, the equation of the line which passes through (-4, 3) and parallel to line y = 2x + 5 is y = 2x + 11. So the answer is: y = 2x + 11.
Ajụjụ 34 Ripọtì
A fair coin is tossed 3 times. Find the probability of obtaining exactly 2 heads.
Akọwa Nkọwa
When a fair coin is tossed once, there are two possible outcomes, either heads or tails. Therefore, the probability of obtaining heads is \(\frac{1}{2}\) and the probability of obtaining tails is also \(\frac{1}{2}\). Now, when a fair coin is tossed 3 times, there are \(2^3 = 8\) possible outcomes as each toss has 2 possible outcomes (heads or tails). These 8 possible outcomes are: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT Out of these 8 outcomes, we need to find the probability of obtaining exactly 2 heads. We can count the number of outcomes in which exactly 2 heads are obtained as follows: - HHT - HTH - THH Therefore, there are 3 outcomes in which exactly 2 heads are obtained. So, the probability of obtaining exactly 2 heads is: \[\frac{\text{number of outcomes in which exactly 2 heads are obtained}}{\text{total number of possible outcomes}} = \frac{3}{8}\] Hence, the answer is \(\frac{3}{8}\).
Ajụjụ 36 Ripọtì
Evaluate \(\cos 75°\), leaving the answer in surd form.
Akọwa Nkọwa
We can use the trigonometric identity \(\cos 75^\circ = \cos(45^\circ+30^\circ) = \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ\). We know that \(\cos 45^\circ = \frac{\sqrt{2}}{2}\) and \(\sin 45^\circ = \frac{\sqrt{2}}{2}\), and we can find \(\cos 30^\circ\) and \(\sin 30^\circ\) using a 30-60-90 triangle or the unit circle. In either case, we get \(\cos 30^\circ = \frac{\sqrt{3}}{2}\) and \(\sin 30^\circ = \frac{1}{2}\). Substituting these values, we have: \begin{align*} \cos 75^\circ &= \cos(45^\circ+30^\circ) \\ &= \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ \\ &= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} \\ &= \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} \\ &= \frac{\sqrt{6} - \sqrt{2}}{4} \end{align*} Therefore, the answer is \(\frac{\sqrt{2}}{4}(\sqrt{3} - 1)\).
Ajụjụ 37 Ripọtì
Points E(-2, -1) and F(3, 2) are the ends of the diameter of a circle. Find the equation of the circle.
Akọwa Nkọwa
Ajụjụ 38 Ripọtì
Find the gradient to the normal of the curve \(y = x^{3} - x^{2}\) at the point where x = 2.
Akọwa Nkọwa
To find the gradient to the normal, we need to find the gradient of the tangent and then take the negative reciprocal of that value. To find the gradient of the tangent, we need to differentiate the equation of the curve. \[\frac{dy}{dx} = 3x^{2} - 2x\] At the point where x = 2, the gradient of the tangent is: \[\frac{dy}{dx}\Bigr|_{x=2} = 3(2)^{2} - 2(2) = 8\] So the gradient to the normal is the negative reciprocal of 8: \[\frac{-1}{8}\] Therefore, the answer is (a) \(\frac{-1}{8}\).
Ajụjụ 39 Ripọtì
Consider the statements:
p : Musa is short
q : Musa is brilliant
Which of the following represents the statement "Musa is short but not brilliant"?
Akọwa Nkọwa
The statement "Musa is short but not brilliant" consists of two parts: "Musa is short" and "Musa is not brilliant". Using the statements p and q, we can represent "Musa is short" as p, and "Musa is not brilliant" as \(\sim q\), where \(\sim\) means "not". Now we need to find the logical operator that connects p and \(\sim q\) to form the statement "Musa is short but not brilliant". This operator is the logical AND, represented by \(\wedge\). Therefore, the statement "Musa is short but not brilliant" is represented by: \(p \wedge \sim q\).
Ajụjụ 40 Ripọtì
Given that \(n = 10\) and \(\sum d^{2} = 20\), calculate the Spearman's rank correlation coefficient.
Akọwa Nkọwa
Spearman's rank correlation coefficient is a measure of the strength of a relationship between two variables and is used when the variables are measured on an ordinal scale. The formula for calculating Spearman's rank correlation coefficient is: $$r_{s} = 1 - \frac{6\sum d^{2}}{n(n^{2} - 1)}$$ where \(d\) is the difference between the ranks of each observation in the two variables, and \(n\) is the sample size. In this case, we are given that \(n = 10\) and \(\sum d^{2} = 20\). Substituting these values into the formula, we get: $$r_{s} = 1 - \frac{6(20)}{10(10^{2} - 1)} = 1 - \frac{120}{990} = 1 - 0.121 = 0.879$$ Therefore, the Spearman's rank correlation coefficient is 0.879.
Ajụjụ 41 Ripọtì
(a) Two Mathematics books, 5 different Physics books and 3 different Chemistry are to be arranged on a shelf. How many arrangements are possible if ;
(i) books on the same subject must stand together? (ii) only the Physics books must stand together?
(b) In a certain community, 13 out of every 20 persons speak English. If 8 persons are selected at random from the community, find, correct to three significant figures, the probability that at least 3 of them speak English.
(a)(i) If books of the same subject must stand together, treat each subject as one block: 3 blocks arranged in \(3!\) ways, with the books inside each block also arranged.
\[3! \times 2! \times 5! \times 3! = 6 \times 2 \times 120 \times 6 = 8640\](ii) If only the Physics books must stand together, treat the 5 Physics books as one block. The items to arrange are: 2 Maths + 1 Physics block + 3 Chemistry \(= 6\) items.
\[6! \times 5! = 720 \times 120 = 86400\](b) \(p = \tfrac{13}{20} = 0.65,\ q = 0.35,\ n = 8\). Find \(P(\text{at least 3})\) via the complement:
\[P(0) = (0.35)^8 = 0.000225\] \[P(1) = \binom{8}{1}(0.65)(0.35)^7 = 0.003346\] \[P(2) = \binom{8}{2}(0.65)^2(0.35)^6 = 0.021747\] \[P(\ge 3) = 1 - (0.000225 + 0.003346 + 0.021747) = 1 - 0.025318 = 0.975\]The probability that at least 3 speak English is \(0.975\) (3 s.f.).
Akọwa Nkọwa
(a)(i) If books of the same subject must stand together, treat each subject as one block: 3 blocks arranged in \(3!\) ways, with the books inside each block also arranged.
\[3! \times 2! \times 5! \times 3! = 6 \times 2 \times 120 \times 6 = 8640\](ii) If only the Physics books must stand together, treat the 5 Physics books as one block. The items to arrange are: 2 Maths + 1 Physics block + 3 Chemistry \(= 6\) items.
\[6! \times 5! = 720 \times 120 = 86400\](b) \(p = \tfrac{13}{20} = 0.65,\ q = 0.35,\ n = 8\). Find \(P(\text{at least 3})\) via the complement:
\[P(0) = (0.35)^8 = 0.000225\] \[P(1) = \binom{8}{1}(0.65)(0.35)^7 = 0.003346\] \[P(2) = \binom{8}{2}(0.65)^2(0.35)^6 = 0.021747\] \[P(\ge 3) = 1 - (0.000225 + 0.003346 + 0.021747) = 1 - 0.025318 = 0.975\]The probability that at least 3 speak English is \(0.975\) (3 s.f.).
Ajụjụ 42 Ripọtì
(a) A body of mass 3kg moves with a velocity of 8ms\(^{-1}\). It collides with a second body moving in the same direction with a velocity of 5ms\(^{-1}\). After collision, the bodies move together with a velocity of 6ms\(^{-1}\). Find the mass of the second body.
(b) If the second body in (a) moves with a velocity of 5ms\(^{-1}\) in the opposite direction as that of the 3kg body with a velocity of 8ms\(^{-1}\), find, correct to two decimal places, the common velocity of the two bodies if they move together after collision.
Momentum is conserved in each collision.
(a) Let the second mass be \(m\). Both bodies move in the same direction, then coalesce at \(6\text{ ms}^{-1}\):
\[ 3(8) + m(5) = (3 + m)(6). \]
\[ 24 + 5m = 18 + 6m \Rightarrow m = 6\text{ kg}. \]
(b) Now the \(6\text{ kg}\) body moves at \(5\text{ ms}^{-1}\) opposite to the \(3\text{ kg}\) body. Take the direction of the \(3\text{ kg}\) body as positive:
\[ 3(8) + 6(-5) = (3 + 6)v. \]
\[ 24 - 30 = 9v \Rightarrow v = \frac{-6}{9} = -0.67\text{ ms}^{-1}\ (\text{2 d.p.}). \]
The common velocity is \(0.67\text{ ms}^{-1}\) in the direction of the \(6\text{ kg}\) body's original motion.
Akọwa Nkọwa
Momentum is conserved in each collision.
(a) Let the second mass be \(m\). Both bodies move in the same direction, then coalesce at \(6\text{ ms}^{-1}\):
\[ 3(8) + m(5) = (3 + m)(6). \]
\[ 24 + 5m = 18 + 6m \Rightarrow m = 6\text{ kg}. \]
(b) Now the \(6\text{ kg}\) body moves at \(5\text{ ms}^{-1}\) opposite to the \(3\text{ kg}\) body. Take the direction of the \(3\text{ kg}\) body as positive:
\[ 3(8) + 6(-5) = (3 + 6)v. \]
\[ 24 - 30 = 9v \Rightarrow v = \frac{-6}{9} = -0.67\text{ ms}^{-1}\ (\text{2 d.p.}). \]
The common velocity is \(0.67\text{ ms}^{-1}\) in the direction of the \(6\text{ kg}\) body's original motion.
Ajụjụ 43 Ripọtì
(a) Without using mathematical tables or calculator, evaluate \(\frac{\frac{3}{2}\log 27 - 3\log 5\sqrt{5}}{\log 0.6}\)
(b) Two linear transformations A and B in the \(O_{xy}\) plane, are defined by :
\(A : (x, y) (x + 2y, -x + y)\)
\(B : (x, y) (2x + 3y, x + 2y)\).
(i) Write down the matrices A and B; (ii) Find the image of the point P(-2, 2) under the linear transformation A followed by B.
(a) Work on the numerator:
\[\tfrac{3}{2}\log 27 = \tfrac{3}{2}\log 3^3 = \tfrac{9}{2}\log 3,\qquad 3\log 5\sqrt{5} = 3\log 5^{3/2} = \tfrac{9}{2}\log 5\] \[\text{Numerator} = \tfrac{9}{2}\log 3 - \tfrac{9}{2}\log 5 = \tfrac{9}{2}\log\tfrac{3}{5} = \tfrac{9}{2}\log 0.6\]Dividing by \(\log 0.6\):
\[\frac{\tfrac{9}{2}\log 0.6}{\log 0.6} = \frac{9}{2} = 4.5\](b)(i) The matrices are:
\[A = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix},\qquad B = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}\](ii) Apply \(A\) to \(P(-2, 2)\):
\[A(P) = (\,-2 + 2(2),\ -(-2) + 2\,) = (2,\ 4)\]Then apply \(B\) to \((2, 4)\):
\[B(2,4) = (2(2) + 3(4),\ 2 + 2(4)) = (16,\ 10)\]The image of \(P\) under \(A\) followed by \(B\) is \((16, 10)\).
Akọwa Nkọwa
(a) Work on the numerator:
\[\tfrac{3}{2}\log 27 = \tfrac{3}{2}\log 3^3 = \tfrac{9}{2}\log 3,\qquad 3\log 5\sqrt{5} = 3\log 5^{3/2} = \tfrac{9}{2}\log 5\] \[\text{Numerator} = \tfrac{9}{2}\log 3 - \tfrac{9}{2}\log 5 = \tfrac{9}{2}\log\tfrac{3}{5} = \tfrac{9}{2}\log 0.6\]Dividing by \(\log 0.6\):
\[\frac{\tfrac{9}{2}\log 0.6}{\log 0.6} = \frac{9}{2} = 4.5\](b)(i) The matrices are:
\[A = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix},\qquad B = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}\](ii) Apply \(A\) to \(P(-2, 2)\):
\[A(P) = (\,-2 + 2(2),\ -(-2) + 2\,) = (2,\ 4)\]Then apply \(B\) to \((2, 4)\):
\[B(2,4) = (2(2) + 3(4),\ 2 + 2(4)) = (16,\ 10)\]The image of \(P\) under \(A\) followed by \(B\) is \((16, 10)\).
Ajụjụ 44 Ripọtì
Given that \(p = \begin{pmatrix} 5 \\ 3 \end{pmatrix}, q = \begin{pmatrix} -1 \\ 2 \end{pmatrix}\) and \(r = \begin{pmatrix} 17 \\ 5 \end{pmatrix}\) and \(r = \alpha r + \beta q\), where \(\alpha\) and \(\beta\) are scalars, express q in terms of r and p.
Express \(r\) as a linear combination \(r = \alpha p + \beta q\):
\[\begin{pmatrix} 17 \\ 5 \end{pmatrix} = \alpha\begin{pmatrix} 5 \\ 3 \end{pmatrix} + \beta\begin{pmatrix} -1 \\ 2 \end{pmatrix}\]This gives the equations:
\[5\alpha - \beta = 17,\qquad 3\alpha + 2\beta = 5\]From the first, \(\beta = 5\alpha - 17\). Substitute:
\[3\alpha + 2(5\alpha - 17) = 5 \ \Rightarrow\ 13\alpha = 39 \ \Rightarrow\ \alpha = 3,\quad \beta = -2\]Hence \(r = 3p - 2q\). Make \(q\) the subject:
\[2q = 3p - r \ \Rightarrow\ q = \tfrac{1}{2}(3p - r)\]Check: \(\tfrac{1}{2}\left(3\begin{pmatrix}5\\3\end{pmatrix} - \begin{pmatrix}17\\5\end{pmatrix}\right) = \tfrac{1}{2}\begin{pmatrix}-2\\4\end{pmatrix} = \begin{pmatrix}-1\\2\end{pmatrix} = q\).
Akọwa Nkọwa
Express \(r\) as a linear combination \(r = \alpha p + \beta q\):
\[\begin{pmatrix} 17 \\ 5 \end{pmatrix} = \alpha\begin{pmatrix} 5 \\ 3 \end{pmatrix} + \beta\begin{pmatrix} -1 \\ 2 \end{pmatrix}\]This gives the equations:
\[5\alpha - \beta = 17,\qquad 3\alpha + 2\beta = 5\]From the first, \(\beta = 5\alpha - 17\). Substitute:
\[3\alpha + 2(5\alpha - 17) = 5 \ \Rightarrow\ 13\alpha = 39 \ \Rightarrow\ \alpha = 3,\quad \beta = -2\]Hence \(r = 3p - 2q\). Make \(q\) the subject:
\[2q = 3p - r \ \Rightarrow\ q = \tfrac{1}{2}(3p - r)\]Check: \(\tfrac{1}{2}\left(3\begin{pmatrix}5\\3\end{pmatrix} - \begin{pmatrix}17\\5\end{pmatrix}\right) = \tfrac{1}{2}\begin{pmatrix}-2\\4\end{pmatrix} = \begin{pmatrix}-1\\2\end{pmatrix} = q\).
Ajụjụ 45 Ripọtì
When \(f(x) = 2x^{3} + mx^{2} + nx + 11\) is divided by \(x^{2} + 5x + 1\), the quotient is \(2x - 5\) and the remainder is \(30x + 16\). Find the values of m and n.
By the division identity, dividend = divisor \(\times\) quotient + remainder:
\[ f(x) = (x^2 + 5x + 1)(2x - 5) + (30x + 16). \]
Expand the product:
\[ (x^2+5x+1)(2x-5) = 2x^3 - 5x^2 + 10x^2 - 25x + 2x - 5 = 2x^3 + 5x^2 - 23x - 5. \]
Add the remainder:
\[ f(x) = 2x^3 + 5x^2 - 23x - 5 + 30x + 16 = 2x^3 + 5x^2 + 7x + 11. \]
Compare with \( f(x) = 2x^3 + mx^2 + nx + 11 \):
\[ m = 5, \qquad n = 7. \]
Akọwa Nkọwa
By the division identity, dividend = divisor \(\times\) quotient + remainder:
\[ f(x) = (x^2 + 5x + 1)(2x - 5) + (30x + 16). \]
Expand the product:
\[ (x^2+5x+1)(2x-5) = 2x^3 - 5x^2 + 10x^2 - 25x + 2x - 5 = 2x^3 + 5x^2 - 23x - 5. \]
Add the remainder:
\[ f(x) = 2x^3 + 5x^2 - 23x - 5 + 30x + 16 = 2x^3 + 5x^2 + 7x + 11. \]
Compare with \( f(x) = 2x^3 + mx^2 + nx + 11 \):
\[ m = 5, \qquad n = 7. \]
Ajụjụ 46 Ripọtì
Four vectors \(r = \alpha i + \beta j\), where \(\alpha \text{ and } \beta\) are constants, \(s = 2i -j, m = 3i + 2j\) and \(n = i + j\) are such that the magnitude of r is three times as s and is parallel to the vactor (m - n).
(a) Find the values of \(\alpha\) and \(\beta\).
(b) Calculate the magnitude and direction of (r - s).
(a) Since \(r\) is parallel to \(m - n\):
\[m - n = (3i + 2j) - (i + j) = 2i + j\]So \(r = \lambda(2i + j)\), giving \(\alpha = 2\lambda,\ \beta = \lambda\).
The magnitude of \(r\) is three times that of \(s = 2i - j\), where \(|s| = \sqrt{4 + 1} = \sqrt{5}\):
\[|r| = \sqrt{(2\lambda)^2 + \lambda^2} = |\lambda|\sqrt{5} = 3\sqrt{5} \ \Rightarrow\ |\lambda| = 3\]Taking \(\lambda = 3\): \(\alpha = 6,\ \beta = 3\) (the other solution \(\lambda = -3\) gives \(\alpha = -6,\ \beta = -3\)).
(b) Using \(r = 6i + 3j\):
\[r - s = (6 - 2)i + (3 - (-1))j = 4i + 4j\] \[|r - s| = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2} \approx 5.7\]Direction: \(\tan\theta = \dfrac{4}{4} = 1\), so \(\theta = 45^\circ\) above the positive \(x\)-axis (bearing \(045^\circ\)).
Akọwa Nkọwa
(a) Since \(r\) is parallel to \(m - n\):
\[m - n = (3i + 2j) - (i + j) = 2i + j\]So \(r = \lambda(2i + j)\), giving \(\alpha = 2\lambda,\ \beta = \lambda\).
The magnitude of \(r\) is three times that of \(s = 2i - j\), where \(|s| = \sqrt{4 + 1} = \sqrt{5}\):
\[|r| = \sqrt{(2\lambda)^2 + \lambda^2} = |\lambda|\sqrt{5} = 3\sqrt{5} \ \Rightarrow\ |\lambda| = 3\]Taking \(\lambda = 3\): \(\alpha = 6,\ \beta = 3\) (the other solution \(\lambda = -3\) gives \(\alpha = -6,\ \beta = -3\)).
(b) Using \(r = 6i + 3j\):
\[r - s = (6 - 2)i + (3 - (-1))j = 4i + 4j\] \[|r - s| = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2} \approx 5.7\]Direction: \(\tan\theta = \dfrac{4}{4} = 1\), so \(\theta = 45^\circ\) above the positive \(x\)-axis (bearing \(045^\circ\)).
Ajụjụ 47 Ripọtì
(a)(i) Write down the expansion of \((1 + x)^{7}\) in ascending powers of x.
(ii) If the coefficients of the fifth, sixth and seventh terms in the expansion in (a)(i) above form a linear sequence(A.P), find the common difference of the A.P.
(b) Using the trapezium rule with ordinates at 1, 2, 3, 4 and 5, calculate, correct to two decimal places,
\(\int_{1}^{5} \sqrt{(2x + 8x^{2})} \mathrm {d} x\).
(a)(i) By the binomial theorem:
\[(1 + x)^7 = 1 + 7x + 21x^2 + 35x^3 + 35x^4 + 21x^5 + 7x^6 + x^7\](ii) The 5th, 6th and 7th coefficients are \(35,\ 21,\ 7\). They form an A.P. with common difference:
\[d = 21 - 35 = -14 \quad(\text{and } 7 - 21 = -14)\](b) Let \(f(x) = \sqrt{2x + 8x^2}\), with \(h = 1\). Compute the ordinates:
- \(f(1) = \sqrt{10} = 3.1623\)
- \(f(2) = \sqrt{36} = 6.0000\)
- \(f(3) = \sqrt{78} = 8.8318\)
- \(f(4) = \sqrt{136} = 11.6619\)
- \(f(5) = \sqrt{210} = 14.4914\)
Trapezium rule:
\[\int_1^5 f\,dx \approx \tfrac{h}{2}\big[(f_0 + f_4) + 2(f_1 + f_2 + f_3)\big]\] \[= \tfrac{1}{2}\big[(3.1623 + 14.4914) + 2(6 + 8.8318 + 11.6619)\big]\] \[= \tfrac{1}{2}\big[17.6537 + 52.9874\big] = \tfrac{1}{2}(70.6411) = 35.32\]So the integral is approximately \(35.32\) (to 2 decimal places).
Akọwa Nkọwa
(a)(i) By the binomial theorem:
\[(1 + x)^7 = 1 + 7x + 21x^2 + 35x^3 + 35x^4 + 21x^5 + 7x^6 + x^7\](ii) The 5th, 6th and 7th coefficients are \(35,\ 21,\ 7\). They form an A.P. with common difference:
\[d = 21 - 35 = -14 \quad(\text{and } 7 - 21 = -14)\](b) Let \(f(x) = \sqrt{2x + 8x^2}\), with \(h = 1\). Compute the ordinates:
- \(f(1) = \sqrt{10} = 3.1623\)
- \(f(2) = \sqrt{36} = 6.0000\)
- \(f(3) = \sqrt{78} = 8.8318\)
- \(f(4) = \sqrt{136} = 11.6619\)
- \(f(5) = \sqrt{210} = 14.4914\)
Trapezium rule:
\[\int_1^5 f\,dx \approx \tfrac{h}{2}\big[(f_0 + f_4) + 2(f_1 + f_2 + f_3)\big]\] \[= \tfrac{1}{2}\big[(3.1623 + 14.4914) + 2(6 + 8.8318 + 11.6619)\big]\] \[= \tfrac{1}{2}\big[17.6537 + 52.9874\big] = \tfrac{1}{2}(70.6411) = 35.32\]So the integral is approximately \(35.32\) (to 2 decimal places).
Ajụjụ 48 Ripọtì
The mean of the numbers 1, 4, k, (k + 4) and 11 is (k + 1). Calculate the :
(a) value of k ;
(b) standard deviation.
(a) Value of k. The mean of \(1, 4, k, (k+4), 11\) is \((k+1)\):
\[ \frac{1 + 4 + k + (k+4) + 11}{5} = k + 1 \Rightarrow \frac{20 + 2k}{5} = k + 1. \]
\[ 20 + 2k = 5k + 5 \Rightarrow 15 = 3k \Rightarrow k = 5. \]
(b) Standard deviation. The numbers become \(1, 4, 5, 9, 11\) with mean \(k+1 = 6\).
Deviations from the mean: \(-5, -2, -1, 3, 5\); their squares: \(25, 4, 1, 9, 25\), summing to \(64\).
\[ \text{Variance} = \frac{64}{5} = 12.8, \qquad \text{S.D.} = \sqrt{12.8} \approx 3.58. \]
Akọwa Nkọwa
(a) Value of k. The mean of \(1, 4, k, (k+4), 11\) is \((k+1)\):
\[ \frac{1 + 4 + k + (k+4) + 11}{5} = k + 1 \Rightarrow \frac{20 + 2k}{5} = k + 1. \]
\[ 20 + 2k = 5k + 5 \Rightarrow 15 = 3k \Rightarrow k = 5. \]
(b) Standard deviation. The numbers become \(1, 4, 5, 9, 11\) with mean \(k+1 = 6\).
Deviations from the mean: \(-5, -2, -1, 3, 5\); their squares: \(25, 4, 1, 9, 25\), summing to \(64\).
\[ \text{Variance} = \frac{64}{5} = 12.8, \qquad \text{S.D.} = \sqrt{12.8} \approx 3.58. \]
Ajụjụ 49 Ripọtì
Given that (5, 2), (-4, k) and (2, 1) lie on a straight line, find the value of k.
Three points are collinear when the slope between any two pairs is the same.
Using \((5,2)\) and \((2,1)\): \[ m = \frac{2-1}{5-2} = \frac{1}{3}. \]
Using \((2,1)\) and \((-4,k)\) and equating the slope: \[ \frac{k-1}{-4-2} = \frac{1}{3} \Rightarrow \frac{k-1}{-6} = \frac{1}{3}. \]
\[ k - 1 = -2 \Rightarrow k = -1. \]
Akọwa Nkọwa
Three points are collinear when the slope between any two pairs is the same.
Using \((5,2)\) and \((2,1)\): \[ m = \frac{2-1}{5-2} = \frac{1}{3}. \]
Using \((2,1)\) and \((-4,k)\) and equating the slope: \[ \frac{k-1}{-4-2} = \frac{1}{3} \Rightarrow \frac{k-1}{-6} = \frac{1}{3}. \]
\[ k - 1 = -2 \Rightarrow k = -1. \]
Ajụjụ 50 Ripọtì
A binary operation \(*\) is defined on the set, R, of real numbers by \(m * n = m + n + 2\). Find the :
(a) identity element under the operation ;
(b) inverse of n under the operation .
The operation is \( m * n = m + n + 2 \).
(a) Identity element. Let \(e\) be the identity, so \( m * e = m \) for all \(m\):
\[ m + e + 2 = m \Rightarrow e = -2. \] The identity element is \( e = -2 \).
(b) Inverse of n. Let \(n^{-1}\) be the inverse, so \( n * n^{-1} = e = -2 \):
\[ n + n^{-1} + 2 = -2 \Rightarrow n^{-1} = -4 - n. \] The inverse of \(n\) is \( n^{-1} = -(n+4) \).
Akọwa Nkọwa
The operation is \( m * n = m + n + 2 \).
(a) Identity element. Let \(e\) be the identity, so \( m * e = m \) for all \(m\):
\[ m + e + 2 = m \Rightarrow e = -2. \] The identity element is \( e = -2 \).
(b) Inverse of n. Let \(n^{-1}\) be the inverse, so \( n * n^{-1} = e = -2 \):
\[ n + n^{-1} + 2 = -2 \Rightarrow n^{-1} = -4 - n. \] The inverse of \(n\) is \( n^{-1} = -(n+4) \).
Ajụjụ 51 Ripọtì
(a) If \(f(x + 2) = 6x^{2} + 5x - 8\), find \(f(5)\).
(b) Express \(\frac{7\sqrt{2} + 3\sqrt{3}}{4\sqrt{2} - 2\sqrt{3}}\) in the form \(p + q\sqrt{r}\), where p, q and r are rational numbers.
(a) To find \(f(5)\) from \( f(x+2) = 6x^2 + 5x - 8 \), set \( x + 2 = 5 \), i.e. \( x = 3 \):
\[ f(5) = 6(3)^2 + 5(3) - 8 = 54 + 15 - 8 = 61. \]
(b) Rationalise by multiplying numerator and denominator by \( 4\sqrt2 + 2\sqrt3 \):
Denominator: \( (4\sqrt2 - 2\sqrt3)(4\sqrt2 + 2\sqrt3) = (4\sqrt2)^2 - (2\sqrt3)^2 = 32 - 12 = 20. \)
Numerator: \( (7\sqrt2 + 3\sqrt3)(4\sqrt2 + 2\sqrt3) = 56 + 14\sqrt6 + 12\sqrt6 + 18 = 74 + 26\sqrt6. \)
\[ \frac{74 + 26\sqrt6}{20} = \frac{37}{10} + \frac{13}{10}\sqrt6. \]
So \( p = \dfrac{37}{10},\ q = \dfrac{13}{10},\ r = 6 \).
Akọwa Nkọwa
(a) To find \(f(5)\) from \( f(x+2) = 6x^2 + 5x - 8 \), set \( x + 2 = 5 \), i.e. \( x = 3 \):
\[ f(5) = 6(3)^2 + 5(3) - 8 = 54 + 15 - 8 = 61. \]
(b) Rationalise by multiplying numerator and denominator by \( 4\sqrt2 + 2\sqrt3 \):
Denominator: \( (4\sqrt2 - 2\sqrt3)(4\sqrt2 + 2\sqrt3) = (4\sqrt2)^2 - (2\sqrt3)^2 = 32 - 12 = 20. \)
Numerator: \( (7\sqrt2 + 3\sqrt3)(4\sqrt2 + 2\sqrt3) = 56 + 14\sqrt6 + 12\sqrt6 + 18 = 74 + 26\sqrt6. \)
\[ \frac{74 + 26\sqrt6}{20} = \frac{37}{10} + \frac{13}{10}\sqrt6. \]
So \( p = \dfrac{37}{10},\ q = \dfrac{13}{10},\ r = 6 \).
Ajụjụ 52 Ripọtì
The table shows the distribution of marks scored by some students in a test.
| Marks | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 | 81-90 | 91-100 |
| No. of students | 3 | 17 | 41 | 85 | 97 | 115 | 101 | 64 | 21 | 6 |
(a)(i) Construct a cumulative frequency table for the distribution ; (ii) Draw a cumulative frequency curve for the distribution.
(b) Use the curve to estimate the :
(i) number of students who scored marks between 32 and 74 ; (ii) pass mark, if 18% of the students failed ; (iii) lowest mark for distinction, if 8% of the students passed with distinction.
(a)(i) Cumulative frequency table
The cumulative frequency is obtained by adding each class frequency to the running total. The curve is plotted against the upper class boundary of each class, so these are tabulated as well. The total number of students is \(N = 550\).
| Marks | Class boundaries | Frequency \(f\) | Cumulative frequency \(\sum f\) |
|---|---|---|---|
| 1 - 10 | 0.5 - 10.5 | 3 | 3 |
| 11 - 20 | 10.5 - 20.5 | 17 | 20 |
| 21 - 30 | 20.5 - 30.5 | 41 | 61 |
| 31 - 40 | 30.5 - 40.5 | 85 | 146 |
| 41 - 50 | 40.5 - 50.5 | 97 | 243 |
| 51 - 60 | 50.5 - 60.5 | 115 | 358 |
| 61 - 70 | 60.5 - 70.5 | 101 | 459 |
| 71 - 80 | 70.5 - 80.5 | 64 | 523 |
| 81 - 90 | 80.5 - 90.5 | 21 | 544 |
| 91 - 100 | 90.5 - 100.5 | 6 | 550 |
(a)(ii) Cumulative frequency curve (ogive)
Plot the cumulative frequency (vertical axis) against the upper class boundary (horizontal axis), starting the curve from \((0.5,\,0)\), and join the points with a smooth S-shaped curve.
(b) Estimates read from the curve
(i) Number of students scoring between 32 and 74. Read the cumulative frequency at each mark from the ogive (dashed guide lines):
\[\text{CF}(32)\approx 61+\frac{32-30.5}{10}\times 85 \approx 74\]\[\text{CF}(74)\approx 459+\frac{74-70.5}{10}\times 64 \approx 481\]Number scoring between 32 and 74:
\[481-74 = 407\ \text{students (approximately)}.\](ii) Pass mark, if 18% failed. The number who failed is
\[18\%\ \text{of}\ 550 = \frac{18}{100}\times 550 = 99.\]The pass mark is the mark below which 99 students lie, i.e. the mark corresponding to \(\text{CF}=99\). Reading horizontally from \(99\) on the vertical axis to the curve and down to the mark axis:
\[\text{pass mark} \approx 30.5+\frac{99-61}{85}\times 10 \approx 35.\](iii) Lowest mark for distinction, if 8% passed with distinction. The distinction group is the top 8%, so the number below the distinction mark is
\[550-\Big(8\%\ \text{of}\ 550\Big)=550-44=506.\]Read the mark corresponding to \(\text{CF}=506\):
\[\text{lowest distinction mark} \approx 70.5+\frac{506-459}{64}\times 10 \approx 78.\]Akọwa Nkọwa
(a)(i) Cumulative frequency table
The cumulative frequency is obtained by adding each class frequency to the running total. The curve is plotted against the upper class boundary of each class, so these are tabulated as well. The total number of students is \(N = 550\).
| Marks | Class boundaries | Frequency \(f\) | Cumulative frequency \(\sum f\) |
|---|---|---|---|
| 1 - 10 | 0.5 - 10.5 | 3 | 3 |
| 11 - 20 | 10.5 - 20.5 | 17 | 20 |
| 21 - 30 | 20.5 - 30.5 | 41 | 61 |
| 31 - 40 | 30.5 - 40.5 | 85 | 146 |
| 41 - 50 | 40.5 - 50.5 | 97 | 243 |
| 51 - 60 | 50.5 - 60.5 | 115 | 358 |
| 61 - 70 | 60.5 - 70.5 | 101 | 459 |
| 71 - 80 | 70.5 - 80.5 | 64 | 523 |
| 81 - 90 | 80.5 - 90.5 | 21 | 544 |
| 91 - 100 | 90.5 - 100.5 | 6 | 550 |
(a)(ii) Cumulative frequency curve (ogive)
Plot the cumulative frequency (vertical axis) against the upper class boundary (horizontal axis), starting the curve from \((0.5,\,0)\), and join the points with a smooth S-shaped curve.
(b) Estimates read from the curve
(i) Number of students scoring between 32 and 74. Read the cumulative frequency at each mark from the ogive (dashed guide lines):
\[\text{CF}(32)\approx 61+\frac{32-30.5}{10}\times 85 \approx 74\]\[\text{CF}(74)\approx 459+\frac{74-70.5}{10}\times 64 \approx 481\]Number scoring between 32 and 74:
\[481-74 = 407\ \text{students (approximately)}.\](ii) Pass mark, if 18% failed. The number who failed is
\[18\%\ \text{of}\ 550 = \frac{18}{100}\times 550 = 99.\]The pass mark is the mark below which 99 students lie, i.e. the mark corresponding to \(\text{CF}=99\). Reading horizontally from \(99\) on the vertical axis to the curve and down to the mark axis:
\[\text{pass mark} \approx 30.5+\frac{99-61}{85}\times 10 \approx 35.\](iii) Lowest mark for distinction, if 8% passed with distinction. The distinction group is the top 8%, so the number below the distinction mark is
\[550-\Big(8\%\ \text{of}\ 550\Big)=550-44=506.\]Read the mark corresponding to \(\text{CF}=506\):
\[\text{lowest distinction mark} \approx 70.5+\frac{506-459}{64}\times 10 \approx 78.\]Ajụjụ 53 Ripọtì
The probabilities that Ago, Sulley and Musa will gain admission to a certain university are \(\frac{4}{5}, \frac{3}{4}\) and \(\frac{2}{3}\) respectively. Find the probability that :
(a) none of them will gain admission ;
(b) only Ago and Sulley will gain admission.
Probabilities of gaining admission: \( P(A)=\tfrac45,\ P(S)=\tfrac34,\ P(M)=\tfrac23 \). The events are independent.
Failure probabilities: \( P(A')=\tfrac15,\ P(S')=\tfrac14,\ P(M')=\tfrac13 \).
(a) None gains admission: \[ P(A')\,P(S')\,P(M') = \frac15 \cdot \frac14 \cdot \frac13 = \frac{1}{60}. \]
(b) Only Ago and Sulley gain admission (Musa fails): \[ P(A)\,P(S)\,P(M') = \frac45 \cdot \frac34 \cdot \frac13 = \frac{12}{60} = \frac{1}{5}. \]
Akọwa Nkọwa
Probabilities of gaining admission: \( P(A)=\tfrac45,\ P(S)=\tfrac34,\ P(M)=\tfrac23 \). The events are independent.
Failure probabilities: \( P(A')=\tfrac15,\ P(S')=\tfrac14,\ P(M')=\tfrac13 \).
(a) None gains admission: \[ P(A')\,P(S')\,P(M') = \frac15 \cdot \frac14 \cdot \frac13 = \frac{1}{60}. \]
(b) Only Ago and Sulley gain admission (Musa fails): \[ P(A)\,P(S)\,P(M') = \frac45 \cdot \frac34 \cdot \frac13 = \frac{12}{60} = \frac{1}{5}. \]
Ajụjụ 54 Ripọtì
A particle is projected vertically upwards from the ground with speed \(30ms^{-1}\). Calculate the :
(a) maximum height reached by the particle;
(b) time taken by the particle to return to the ground;
(c) time(s) taken for the particle to attain a height of 40m above the ground. [Take \(g = 10ms^{-2}\)]
Given \(u = 30\,\text{ms}^{-1},\ g = 10\,\text{ms}^{-2}\), upward taken positive.
(a) At maximum height \(v = 0\). Use \(v^2 = u^2 - 2gh\):
\[0 = 30^2 - 2(10)h \ \Rightarrow\ 20h = 900 \ \Rightarrow\ h = 45\,\text{m}\](b) Time to return to ground:
\[t = \frac{2u}{g} = \frac{2(30)}{10} = 6\,\text{s}\](c) Height \(40\,\text{m}\): use \(h = ut - \tfrac{1}{2}gt^2\):
\[40 = 30t - 5t^2 \ \Rightarrow\ 5t^2 - 30t + 40 = 0 \ \Rightarrow\ t^2 - 6t + 8 = 0\] \[(t - 2)(t - 4) = 0 \ \Rightarrow\ t = 2\,\text{s (going up)},\ t = 4\,\text{s (coming down)}\]Akọwa Nkọwa
Given \(u = 30\,\text{ms}^{-1},\ g = 10\,\text{ms}^{-2}\), upward taken positive.
(a) At maximum height \(v = 0\). Use \(v^2 = u^2 - 2gh\):
\[0 = 30^2 - 2(10)h \ \Rightarrow\ 20h = 900 \ \Rightarrow\ h = 45\,\text{m}\](b) Time to return to ground:
\[t = \frac{2u}{g} = \frac{2(30)}{10} = 6\,\text{s}\](c) Height \(40\,\text{m}\): use \(h = ut - \tfrac{1}{2}gt^2\):
\[40 = 30t - 5t^2 \ \Rightarrow\ 5t^2 - 30t + 40 = 0 \ \Rightarrow\ t^2 - 6t + 8 = 0\] \[(t - 2)(t - 4) = 0 \ \Rightarrow\ t = 2\,\text{s (going up)},\ t = 4\,\text{s (coming down)}\]Ajụjụ 55 Ripọtì
(a) If \(^{k}P_{2} = 72\), find the value of k.
(b) Solve the equation : \(2\cos^{2} \theta - 5\cos \theta = 3; 0° \leq \theta \leq 360°\)
(a) \(^{k}P_2 = k(k - 1) = 72\):
\[k^2 - k - 72 = 0 \ \Rightarrow\ (k - 9)(k + 8) = 0\]Since \(k > 0\), \(k = 9\).
(b) \(2\cos^2\theta - 5\cos\theta = 3\). Let \(c = \cos\theta\):
\[2c^2 - 5c - 3 = 0 \ \Rightarrow\ (2c + 1)(c - 3) = 0\] \[c = -\tfrac{1}{2} \quad\text{or}\quad c = 3\ (\text{rejected, since } |\cos\theta| \le 1)\]So \(\cos\theta = -\tfrac{1}{2}\). In \(0^\circ \le \theta \le 360^\circ\):
\[\theta = 120^\circ \quad\text{and}\quad \theta = 240^\circ\]Akọwa Nkọwa
(a) \(^{k}P_2 = k(k - 1) = 72\):
\[k^2 - k - 72 = 0 \ \Rightarrow\ (k - 9)(k + 8) = 0\]Since \(k > 0\), \(k = 9\).
(b) \(2\cos^2\theta - 5\cos\theta = 3\). Let \(c = \cos\theta\):
\[2c^2 - 5c - 3 = 0 \ \Rightarrow\ (2c + 1)(c - 3) = 0\] \[c = -\tfrac{1}{2} \quad\text{or}\quad c = 3\ (\text{rejected, since } |\cos\theta| \le 1)\]So \(\cos\theta = -\tfrac{1}{2}\). In \(0^\circ \le \theta \le 360^\circ\):
\[\theta = 120^\circ \quad\text{and}\quad \theta = 240^\circ\]
