WAEC SSCE - General Mathematics - 1991

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To understand this problem, let's first define a half-plane. A half-plane is a part of the plane that lies on one side of a straight line and extends infinitely far in that direction. In this problem, we have an inequality that defines a half-plane.

The inequality is: 4x + 3y ≥ 6

To graph this inequality, we can first plot the line 4x + 3y = 6. To plot this line, we can find two points on the line by setting x = 0 and y = 0 and solving for the other variable.

When x = 0, we get: 3y = 6, y = 2

When y = 0, we get: 4x = 6, x = 3/2

So the two points on the line are (0, 2) and (3/2, 0). We can plot these points and draw a straight line passing through them.

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Now we need to determine which side of the line represents the half-plane defined by the inequality 4x + 3y ≥ 6.

To do this, we can choose a test point not on the line, such as the origin (0,0), and substitute its coordinates into the inequality:

4(0) + 3(0) ≥ 6

0 ≥ 6

Since this is false, the point (0,0) is not in the half-plane defined by the inequality. Therefore, we shade the half-plane that does not include the origin:

      |xxxxxxx
      |xxxxxxx
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  xxxxx|
  xxxxx|
  xxxxx|

The shaded portion shows the outer boundary of the half-plane defined by the inequality 4x + 3y ≥ 6.

Answer: 4x + 3y ≥ 6.

Detalhes da Resposta

Detalhes da Resposta

Detalhes da Resposta

To solve the given equation, we need to bring all the terms to one side and then factorize it to find the values of 'a'. We start by subtracting '3a' from both sides of the equation: 3a + 10 - 3a = a^2 - 3a 10 = a^2 - 3a Now, we can rearrange the terms and factorize the equation as: a^2 - 3a - 10 = 0 (a - 5)(a + 2) = 0 Using the zero product property, we get: a - 5 = 0 or a + 2 = 0 a = 5 or a = -2 Therefore, the values of 'a' that satisfy the given equation are a = 5 or a = -2. Hence, the answer is a = 5 or a = -2.

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Let the first term of the G.P. be a. The formula for the nth term of a G.P. with first term a and common ratio r is given by: an = ar^(n-1) Since the common ratio is 2, the 5th term will be a x 2^(5-1) = 16a. It is given that the 5th term is greater than the 1st term by 45. Therefore, 16a - a = 45 Simplifying this equation gives: 15a = 45 Dividing both sides by 15 gives: a = 3 Therefore, the 5th term is 16a = 16 x 3 = 48. Hence, the answer is 48.

Detalhes da Resposta

Detalhes da Resposta

Detalhes da Resposta

The expression \(\frac{y + 2}{y^{2} - 3y - 10}\) is undefined when the denominator is equal to zero, since division by zero is undefined. So we can set the denominator equal to zero and solve for y: y^2 - 3y - 10 = 0 We can factor this quadratic equation as: (y - 5)(y + 2) = 0 And using the zero product property, we know that this equation is only true when either (y - 5) = 0 or (y + 2) = 0. Therefore, the expression is undefined when either y = 5 or y = -2. Therefore, the answer to the question is y = 5 or y = -2.

Detalhes da Resposta

Detalhes da Resposta

Detalhes da Resposta

Detalhes da Resposta

When expressing a number to three significant figures, we only consider the first three digits of the number and round it up or down depending on the value of the fourth digit. In this case, the number is already given as 5846. Since the fourth digit is 6, which is greater than or equal to 5, we round up the last significant digit (the third digit) by adding 1 to it. Therefore, the answer rounded to three significant figures is 5850. So the correct option is: 5850.

Detalhes da Resposta

Detalhes da Resposta

If events X and Y are mutually exclusive, it means they cannot occur at the same time. In other words, if X happens, Y cannot happen, and vice versa. This also means that the probability of both events happening at the same time (P(X∩Y)) is equal to zero. We can use the formula for the probability of the union of two events: P(X∪Y) = P(X) + P(Y) - P(X∩Y) Since X and Y are mutually exclusive, we know that P(X∩Y) is equal to zero: P(X∪Y) = P(X) + P(Y) - 0 We can substitute the given probabilities into this formula: P(X∪Y) = 1/3 + 2/5 To add these fractions, we need a common denominator. The smallest common multiple of 3 and 5 is 15, so we can rewrite the fractions with this denominator: P(X∪Y) = 5/15 + 6/15 P(X∪Y) = 11/15 Therefore, the answer is 11/15.

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The measure of dispersion gives information on how spread out or clustered a set of data is. It quantifies the variability or diversity of the data values. Out of the options provided, the measure that is not a measure of dispersion is the mode. The mode represents the value that occurs most frequently in a dataset. It is not a measure of dispersion because it does not provide any information about how much the data values deviate or spread out from the mode. The other options provided are all measures of dispersion. - Mean deviation: measures the average distance between each data point and the mean of the dataset. - Range: measures the difference between the highest and lowest value in the dataset. - Interquartile range: measures the difference between the upper and lower quartiles of the dataset. - Standard deviation: measures how much the data values deviate from the mean of the dataset. Therefore, the mode is not a measure of dispersion, while the others are.

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The logarithm to base 10 of a number is the power to which 10 must be raised to give the number. So, if the logarithm to base 10 of a number is 2.6025, we can write it as: 10^2.6025 = x Using a calculator, we get: 10^2.6025 ≈ 400.401 Therefore, the number whose logarithm to base 10 is 2.6025 is approximately 400.401. Thus, the correct option is (a) 400.4.

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To find the distance between towns P and Q along their line of latitude, we need to calculate the length of the arc formed by the angle between the two towns and the center of the Earth. The distance between P and Q can be calculated using the formula: Distance = angle (in radians) x radius of the Earth First, we need to find the angle between P and Q. Since both towns are on the same line of latitude, the angle between them is simply the difference between their longitudes, which is: 20^oE - 40^oW = 60^o Next, we need to convert this angle to radians by multiplying it by π/180, since there are π radians in 180 degrees. 60^o x π/180 = π/3 radians Finally, we can plug this value into the formula for distance: Distance = angle (in radians) x radius of the Earth = (π/3) x R = (πR)/3 Therefore, the correct answer is:

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The mean of the 30 observations is given as 5. If we remove the largest observation of 34, we will be left with 29 observations. To find the mean of the remaining observations, we need to sum them up and divide by the number of observations. The sum of the 30 observations is 30 x 5 = 150. If we remove the observation of 34, the sum of the remaining 29 observations will be 150 - 34 = 116. Therefore, the mean of the remaining 29 observations is 116/29 = 4. Hence, the answer is 4.

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To determine the probability of having a blue-eyed boy, we need to consider the probability of two independent events happening together: the baby being a boy and having blue eyes. The probability of having a boy is 1/2, and the probability of having blue eyes is 1/4. Since these events are independent, we can multiply their probabilities to find the probability of both events happening together: 1/2 * 1/4 = 1/8 Therefore, the probability that Mrs. Jones will have a blue-eyed boy is 1/8.

Detalhes da Resposta

Detalhes da Resposta

We are given that every student had to study Economics or Geography or both. Therefore, the total number of students in the class is 80.

Let's denote the number of students who studied only Economics by 'E', the number of students who studied only Geography by 'G', and the number of students who studied both subjects by 'B'. Then we can use a Venn diagram to represent the information given in the problem:

   -----------------------------------------
  |                   |                     |
  |     Economics      |      Geography     |
  |                   |                     |
   -----------------------------------------
             |                     |
             |                     |
             |                     |
             B                    G
             |                     |
             |                     |
             |                     |
   -----------------------------------------
  |                   |                     |
  |  Only Economics   | Only Geography     |
  |                   |                     |
   -----------------------------------------

We are given that 65 students studied Economics, which includes both those who studied only Economics and those who studied both subjects. So we can write:

E + B = 65

Similarly, we are given that 50 students studied Geography, which includes both those who studied only Geography and those who studied both subjects. So we can write:

G + B = 50

We want to find the value of B, the number of students who studied both subjects. To do this, we can add the two equations above:

E + B = 65

G + B = 50

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E + G + 2B = 115

We know that the total number of students in the class is 80, so we can write:

E + G + B = 80

Substituting the expression for E + G + 2B into this equation, we get:

115 - B = 80

Solving for B, we get:

B = 35

Therefore, 35 students studied both Economics and Geography.

Detalhes da Resposta

Detalhes da Resposta

To factorize 3a\(^2\) - 11a + 6, we need to find two binomials that, when multiplied together, result in the original expression. To do this, we can use a technique called "factoring by grouping." First, we need to identify two numbers that multiply to 3 x 6 = 18 and add up to -11. These numbers are -2 and -9. Next, we can split the middle term -11a into -2a - 9a, and then group the terms as follows: 3a\(^2\) - 2a - 9a + 6 Now, we can factor out the greatest common factor from the first two terms, and the greatest common factor from the last two terms: a(3a - 2) - 3(3a - 2) Notice that we have a common binomial factor of (3a - 2), which we can factor out: (3a - 2)(a - 3) Therefore, the correct option is (3a - 2)(a - 3).

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To find P∪Q∪R, we need to combine all the elements in P, Q and R, without repeating any element. The elements in P are {2, 1, 3, 9, 1/2}. The elements in Q are {1, 2^(1/2), 3, 7}. The elements in R are {5, 4, 2^(1/2)}. So, P∪Q∪R = {1/2, 1, 2, 2^(1/2), 3, 4, 5, 7, 9}. Therefore, the correct answer is (¹/₂, 1, 2, 2¹/₂, 3, 4, 5, 7, 9).

Detalhes da Resposta

Detalhes da Resposta

Detalhes da Resposta

To find the sum of their annual contribution to the fund, we need to calculate the amount that each of them contributes and then add them up. The headmaster contributes 7% of his income, which is 7/100 * N5,500 = N385 per annum. The wife contributes 4% of her income, which is 4/100 * N4,000 = N160 per annum. Therefore, the sum of their annual contribution to the fund is N385 + N160 = N545. So, the correct answer is N545.

Detalhes da Resposta

Detalhes da Resposta

Detalhes da Resposta

Detalhes da Resposta

In any polygon, the sum of the exterior angles is always 360 degrees. Therefore, the measure of each exterior angle of a regular polygon with n sides is 360/n degrees. The ratio of the exterior angle to the interior angle is 1:3. This means that the measure of the exterior angle is three times the measure of the interior angle. Let x be the measure of the interior angle. Then, the measure of the exterior angle is 3x. We know that the sum of the interior angles of any polygon is (n-2) times 180 degrees. Therefore, the measure of each interior angle of a regular polygon with n sides is [(n-2) x 180]/n degrees. Now we can set up an equation to solve for n. 3x = 360/n x = (n-2) x 180/n Substituting 3x for the exterior angle in the first equation: 3x = 360/n x = 120/n Substituting 120/n for x in the second equation: 120/n = (n-2) x 180/n Simplifying: 120 = 180(n-2) 120 = 180n - 360 540 = 180n n = 3 However, a polygon with 3 sides is not possible since it would be a triangle. Therefore, the answer is not 3. We can try other answer options by substituting each value of n in the formula for the measure of each exterior angle (360/n) and checking if the ratio of the exterior angle to the interior angle is 1:3. For n=4, the measure of each exterior angle is 90 degrees, and the measure of each interior angle is 90 degrees. Therefore, the ratio of the exterior angle to the interior angle is 1:1, which is not 1:3. For n=5, the measure of each exterior angle is 72 degrees, and the measure of each interior angle is 108 degrees. Therefore, the ratio of the exterior angle to the interior angle is 2:3, which is not 1:3. For n=6, the measure of each exterior angle is 60 degrees, and the measure of each interior angle is 120 degrees. Therefore, the ratio of the exterior angle to the interior angle is 1:2, which is not 1:3. For n=8, the measure of each exterior angle is 45 degrees, and the measure of each interior angle is 135 degrees. Therefore, the ratio of the exterior angle to the interior angle is 1:3, which satisfies the given condition. Therefore, the polygon has 8 sides, and the answer is 8.

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We can use the logarithmic rule that states log_a(b) + log_a(c) = log_a(bc) to simplify the expression. Applying this rule, we have: log₆(6) + log₆(2) - log₆(12) = 1 + log₆(2) - log₆(2\*2\*3) = 1 + log₆(2) - (log₆(2) + log₆(2\*3)) = 1 + log₆(2) - (log₆(2) + log₆(6)) = 1 + log₆(2) - (log₆(2) + 1) = log₆(2) - log₆(2) - 1 + 1 = 0 Therefore, the simplified expression is 0. The correct option is (C) 0.

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We can solve the given equation by using the distributive property of multiplication and combining like terms:

6(x - 4) + 3(x + 7) = 3
6x - 24 + 3x + 21 = 3
9x - 3 = 3
9x = 6
x = 6/9

Simplifying the fraction 6/9, we get:

x = 2/3

Therefore, the solution to the given equation is x = 2/3. So, the correct answer is:

2/3

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To find the median of a distribution, we need to arrange the values in order from lowest to highest or highest to lowest. Then, we locate the middle value of the distribution. If the number of values is odd, the median is the middle value. In this case, we have five values, which is an odd number. Therefore, we can directly locate the middle value. If the number of values is even, the median is the average of the two middle values. In this case, we do not have an even number of values. Arranging the given values in order from lowest to highest, we have: 4, 4.5, 5, 6.5, 9 The middle value of this distribution is 5, which is the third value in the ordered set. Therefore, the median of this distribution is 5.

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Detalhes da Resposta

Detalhes da Resposta

Detalhes da Resposta

Detalhes da Resposta

Detalhes da Resposta

We know that 1 kilometer is equal to 1000 meters and 1 hour is equal to 3600 seconds. Therefore, to convert the car's speed from kilometers per hour (km/hr) to meters per second (m/s), we can use the following conversion factors: 1 km = 1000 m 1 hr = 3600 s We start by converting the speed from km/hr to meters per hour: 80 km/hr x (1000 m/km) = 80,000 m/hr Next, we convert from hours to seconds: 80,000 m/hr x (1 hr/3600 s) = 22.22 m/s (rounded to two decimal places) Therefore, the car's speed in meters per second (m/s) is approximately 22.22 m/s. Answer: 22.2m/s.

Detalhes da Resposta

Detalhes da Resposta

In a pentagon, the sum of the angles is equal to (5 - 2) times 180 degrees, which is 540 degrees. So we can set up an equation to solve for x:

x + 2x + (x + 60) + (x + 10) + (x - 10) = 540

Simplifying this equation, we get:

6x + 60 = 540

Subtracting 60 from both sides, we get:

6x = 480

Dividing both sides by 6, we get:

x = 80

Therefore, the value of x is 80. So, the correct answer is:

80

Detalhes da Resposta

There are a total of 5 marbles in the box, 2 of which are white and 3 are blue. If two marbles are picked one after the other without replacement, there are two possible scenarios: either a white marble is picked first or a blue marble is picked first. Case 1: A white marble is picked first. In this case, there are 4 marbles left, out of which 3 are blue. Therefore, the probability of picking a blue marble after picking a white marble is 3/4. Case 2: A blue marble is picked first. In this case, there are 4 marbles left, out of which 2 are white. Therefore, the probability of picking a white marble after picking a blue marble is 2/4 or 1/2. Since there are two possible scenarios and each scenario is mutually exclusive, we can add the probabilities of the two scenarios to get the probability of picking two marbles of different colors: Probability = (Probability of picking a white marble first x Probability of picking a blue marble second) + (Probability of picking a blue marble first x Probability of picking a white marble second) Probability = (2/5 x 3/4) + (3/5 x 1/2) Probability = 3/10 + 3/10 Probability = 6/10 or 3/5 Therefore, the probability of picking two marbles of different colors is 3/5 or 0.6. Answer is the correct answer.

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Detalhes da Resposta

Detalhes da Resposta

Detalhes da Resposta

Detalhes da Resposta

In an arithmetic progression (A.P), the difference between any two consecutive terms is constant. Let's call this constant difference "d". We're told that the first term "a" is equal to twice the common difference, so: a = 2d To find the 5th term of the A.P, we can use the formula: an = a + (n-1)d where "an" is the nth term of the A.P. Substituting the given values, we get: a5 = 2d + (5-1)d a5 = 2d + 4d a5 = 6d Therefore, the 5th term of the A.P is 6d. Answer: 6d.

Detalhes da Resposta

Detalhes da Resposta

Detalhes da Resposta

(a) Let \(N=\dfrac{\sqrt[3]{1.376}}{\sqrt[4]{0.007}}\). Using logarithms:

\[\log N=0.04623-(-0.53873)=0.58496.\]

\[N=\text{antilog}(0.58496)=3.85\ (3\text{ s.f.}).\]

(b) Express in one base: \(81=3^{4}\) and \(\tfrac13=3^{-1}\).

\[\frac{\log81}{\log\frac13}=\frac{\log3^{4}}{\log3^{-1}}=\frac{4\log3}{-\log3}=-4.\]

Detalhes da Resposta

(i) Interior angle of the pentagon. The sum of the interior angles of an \(n\)-sided polygon is \((n - 2)\times 180^\circ\). For a regular pentagon \(n = 5\):

\[\text{Each interior angle} = \frac{(5 - 2)\times 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ.\]

(ii) < BXA. In the rectangle \(AXYE\), the side \(AE\) is a side of the pentagon and \(AX \perp AE\), so \(< XAE = 90^\circ\).

The interior angle of the pentagon at \(A\) is \(< BAE = 108^\circ\). Since \(X\) lies on \(BC\), the ray \(AX\) splits \(< BAE\):

\[< BAX = < BAE - < XAE = 108^\circ - 90^\circ = 18^\circ.\]

Also \(X\) lies on \(BC\), so in triangle \(ABX\) the angle at \(B\) equals the interior angle of the pentagon:

\[< ABX = 108^\circ.\]

The angles of triangle \(ABX\) sum to \(180^\circ\):

\[< BXA = 180^\circ - < ABX - < BAX = 180^\circ - 108^\circ - 18^\circ = 54^\circ.\]

Therefore \(< BXA = \mathbf{54^\circ}\).

Detalhes da Resposta

None

Detalhes da Resposta

(a) Construction of parallelogram \(PQRS\)

1. Draw \(PR=9\text{ cm}\).
2. Bisect \(PR\) to obtain its midpoint \(K\), so that \(KP=KR=4.5\text{ cm}\).
3. At \(K\), construct a ray making \(60^\circ\) with \(KR\). With centre \(K\) and radius \(3\text{ cm}\), mark \(Q\) on this ray.
4. Produce \(QK\) through \(K\), and with the same radius \(3\text{ cm}\), mark \(S\) on the opposite ray. Thus \(QS=6\text{ cm}\) and \(K\) is the midpoint of \(QS\).
5. Join \(P\) to \(Q\), \(Q\) to \(R\), \(R\) to \(S\), and \(S\) to \(P\). Since the diagonals bisect each other at \(K\), \(PQRS\) is a parallelogram.

(b) Construction of the equal-area rectangle

1. From \(Q\), construct \(QD\perp PS\).
2. At \(P\), construct a perpendicular to \(PS\) on the side containing \(Q\). Transfer the length \(QD\) onto this perpendicular to locate \(A\), such that \(PA=QD\).
3. Through \(A\), construct a line parallel to \(PS\). Through \(S\), construct a line parallel to \(PA\); let the two lines meet at \(B\).
4. \(PABS\) is the required rectangle, since \(PA=QD\) and both figures have base \(PS\) and the same perpendicular height.

For the measurement,

\[|PS|=\sqrt{4.5^2+3^2-2(4.5)(3)\cos60^\circ}=\sqrt{15.75}\approx3.97\text{ cm}.\]

The area of the parallelogram is

\[\text{Area}=\frac12(9)(6)\sin60^\circ\approx23.38\text{ cm}^2.\]

Hence

\[|PA|=\frac{23.38}{3.97}\approx5.89\text{ cm}.\]

Therefore, \(|PA|\approx5.9\text{ cm}\).

Detalhes da Resposta

Let the pilot be at a height \(H\) above the ground, with the horizontal distance to the tower equal to 8.5 km. The angle of depression to the base is \(33^\circ\) and to the top is \(30^\circ\) (the top is higher, so its depression is smaller).

Height of pilot above the ground, using the base:

\[\tan 33^\circ = \frac{H}{8.5} \Rightarrow H = 8.5\tan 33^\circ = 8.5(0.6494) = 5.520\text{ km}.\]

(a) Shortest distance from the pilot to the base. This is the straight line (line of sight):

\[d = \frac{8.5}{\cos 33^\circ} = \frac{8.5}{0.8387} = 10.135\text{ km} \approx \mathbf{10.1\text{ km}}.\]

(b) Height of the control tower. The height of the pilot above the top of the tower is

\[8.5\tan 30^\circ = 8.5(0.5774) = 4.908\text{ km}.\]

Therefore the tower height is

\[H - 8.5\tan 30^\circ = 5.520 - 4.908 = 0.612\text{ km}.\]

The height of the control tower \(\approx \mathbf{0.612\text{ km}}\) (about 612 m), correct to three significant figures.

Detalhes da Resposta

Table of values

\(x\)	-3	-2	-1	0	1	2	3	4
\(y=3+2x-x^2\)	-12	-5	0	3	4	3	0	-5
\(y=2x-3\)	-9	-7	-5	-3	-1	1	3	5

Using the stated scales, the required graph is:

(i) At the points of intersection,

\[3+2x-x^2=2x-3\]

\[6-x^2=0.\]

Reading the two intersection abscissae from the graph gives

\[x\approx -2.4\quad\text{or}\quad x\approx 2.4.\]

(ii) The highest point of the parabola is \((1,4)\). Hence, the maximum value of \(3+2x-x^2\) is

\[\boxed{4.0}.\]

(iii) The curve \(y=3+2x-x^2\) meets the line \(y=1\) at approximately \(x=-0.7\) and \(x=2.7\). The part of the parabola on or below \(y=1\) is therefore

\[\boxed{-3.0\leq x\leq -0.7\quad\text{or}\quad 2.7\leq x\leq4.0}.\]

Detalhes da Resposta

Total balls \(= 9\): black \(= 3\), blue \(= 2\), red \(= 9 - 3 - 2 = 4\).

(a)(i) P(not blue) \(= \dfrac{9 - 2}{9} = \dfrac{7}{9}\).

(a)(ii) P(not red) \(= \dfrac{9 - 4}{9} = \dfrac{5}{9}\).

(b)(i) Both black, no replacement. First black: \(\dfrac{3}{9}\). After removing one black, 2 black remain out of 8: \(\dfrac{2}{8}\).

\[P = \frac{3}{9} \times \frac{2}{8} = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}.\]

(b)(ii) Both blue, with replacement. The bag is restored, so each draw is \(\dfrac{2}{9}\).

\[P = \frac{2}{9} \times \frac{2}{9} = \frac{4}{81}.\]

Detalhes da Resposta

(a) Solve \(2x^2 + 7x - 11 = 0\) using the quadratic formula with \(a = 2, b = 7, c = -11\).

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{49 + 88}}{4} = \frac{-7 \pm \sqrt{137}}{4}.\]

Now \(\sqrt{137} = 11.7047\), so

\[x = \frac{-7 + 11.7047}{4} = \frac{4.7047}{4} = 1.18,\qquad x = \frac{-7 - 11.7047}{4} = \frac{-18.7047}{4} = -4.68.\]

Hence \(x = 1.18\) or \(x = -4.68\) (2 d.p.).

(b) Let \(P = \dfrac{1}{x}\) and \(Q = \dfrac{1}{y}\). The equations become:

\[2P + Q = 3 \quad(1), \qquad P - 5Q = 7 \quad(2).\]

From (1): \(Q = 3 - 2P\). Substitute into (2):

\[P - 5(3 - 2P) = 7 \Rightarrow P - 15 + 10P = 7 \Rightarrow 11P = 22 \Rightarrow P = 2.\]

Then \(Q = 3 - 2(2) = -1\).

Returning to \(x\) and \(y\): \(x = \dfrac{1}{P} = \dfrac{1}{2}\) and \(y = \dfrac{1}{Q} = -1\).

Therefore \(x = \dfrac{1}{2},\; y = -1\).

Detalhes da Resposta

(a) Class boundaries and cumulative frequencies

(b) Cumulative frequency curve

Plot cumulative frequency against the upper class boundaries, beginning with the point \((1.45,0)\).

The points to plot are:

\[ (1.45,0),\ (1.95,3),\ (2.45,13),\ (2.95,31),\ (3.45,41),\ (3.95,48),\ (4.45,50). \]

Join the points with a smooth cumulative frequency curve.

(c)(i) Interquartile range

Total number of customers, \(N=50\).

\[ Q_1=\frac{N}{4}=\frac{50}{4}=12.5 \]

\[ Q_3=\frac{3N}{4}=\frac{3(50)}{4}=37.5 \]

Detalhes da Resposta

(a)(i)

“Nobody takes Economics and Chemistry” is written as

\[n(E\cap C)=0\]

“4 pupils take Economics and Government” is written as

\[n(E\cap G)=4\]

(a)(ii) The Venn diagram is:

(b)(i) Let \(n(C\cap G)=x\). Since \(E\cap C=\varnothing\), there is no triple intersection.

Using the total number of pupils,

\[22=12+8+7-4-x\]

\[22=23-x\]

\[x=1\]

Therefore, the number who take both Chemistry and Government is \(\boxed{1}\).

(b)(ii)

\[\text{Government only}=8-4-1=3\]

Therefore, the number who take Government only is \(\boxed{3}\).

Detalhes da Resposta

(a) Proof: the angle at the centre is twice the angle at the circumference

Given: A circle centre \(O\), an arc \(AB\), with \(\angle AOB\) at the centre and \(\angle ACB\) at a point \(C\) on the remaining part of the circumference, both standing on the same arc \(AB\).

To prove: \(\angle AOB = 2\,\angle ACB\).

Construction: Join \(C\) to \(O\) and produce it to a point \(D\).

Proof: In triangle \(OAC\), \(OA = OC\) (radii), so it is isosceles and \(\angle OAC = \angle OCA\). The exterior angle equals the sum of the two opposite interior angles:

\[ \angle AOD = \angle OAC + \angle OCA = 2\,\angle OCA \]

Similarly, in triangle \(OBC\), \(OB = OC\) (radii), so \(\angle OBC = \angle OCB\) and

\[ \angle BOD = \angle OBC + \angle OCB = 2\,\angle OCB \]

Adding:

\[ \angle AOD + \angle BOD = 2\,\angle OCA + 2\,\angle OCB \]

\[ \angle AOB = 2(\angle OCA + \angle OCB) = 2\,\angle ACB \]

Hence the angle subtended at the centre is twice that subtended at the circumference. (The same argument holds when \(O\) lies outside triangle \(ACB\), using subtraction instead of addition.) \(\blacksquare\)

(b) Calculations

\(O\) is the centre, \(\angle OQR = 32^{\circ}\) and \(\angle MPQ = 15^{\circ}\).

(i) \(\angle QPR\)

In triangle \(OQR\), \(OQ = OR\) (radii), so it is isosceles and \(\angle ORQ = \angle OQR = 32^{\circ}\). Then

\[ \angle QOR = 180^{\circ} - 32^{\circ} - 32^{\circ} = 116^{\circ} \]

\(\angle QOR\) is the angle at the centre and \(\angle QPR\) is the angle at the circumference standing on the same arc \(QR\). By the theorem proved in (a):

\[ \angle QPR = \tfrac{1}{2}\,\angle QOR = \tfrac{1}{2}\times 116^{\circ} = 58^{\circ} \]

(ii) \(\angle MQO\)

\(\angle MPQ = 15^{\circ}\) is the angle at the circumference standing on arc \(MQ\). The angle at the centre on the same arc is

\[ \angle MOQ = 2 \times 15^{\circ} = 30^{\circ} \]

In triangle \(OMQ\), \(OM = OQ\) (radii), so it is isosceles with equal base angles:

\[ \angle MQO = \angle OMQ = \frac{180^{\circ} - 30^{\circ}}{2} = 75^{\circ} \]

\(\angle QPR = 58^{\circ}\) and \(\angle MQO = 75^{\circ}\).

Detalhes da Resposta

Each ream is supposed to contain 480 sheets, so 5 reams should hold \(5 \times 480 = 2400\) sheets.

Actual sheets: \(435 + 420 + 405 + 415 + 440 = 2115\).

Total shortage \(= 2400 - 2115 = 285\) sheets.

(a) Percentage error.

\[\text{Percentage error} = \frac{\text{shortage}}{\text{supposed number}}\times 100 = \frac{285}{2400}\times 100 = 11.875\%.\]

To the nearest whole number, the percentage error \(= \mathbf{12\%}\).

(b) Amount cheated. A full ream of 480 sheets costs \(\text{N}35.00\), so one sheet is worth \(\dfrac{35}{480}\) naira.

The buyer paid for 2400 sheets but received only 2115, a shortage of 285 sheets:

\[\text{Amount cheated} = 285 \times \frac{35}{480} = \frac{9975}{480} = \text{N}20.78.\]

To the nearest naira, the buyer was cheated by \(\mathbf{\text{N}21}\).