WAEC SSCE - General Mathematics - 1996

Bayanin Amsa

Bayanin Amsa

To find the value of q, we need to solve the given system of equations:

3p - q = 6 ...(1)

2p + 3q = 4 ...(2)

One way to solve this system is to use the method of elimination, where we eliminate one of the variables by adding or subtracting the equations.

Multiplying equation (1) by 3, we get:

9p - 3q = 18 ...(3)

Now, we can eliminate q by adding equations (2) and (3):

2p + 3q = 4

9p - 3q = 18

-----------

11p = 22

Dividing both sides by 11, we get:

p = 2

Substituting this value of p in equation (1), we get:

3(2) - q = 6

Simplifying this, we get:

6 - q = 6

Subtracting 6 from both sides, we get:

-q = 0

Dividing both sides by -1, we get:

q = 0

Therefore, q = 0 is the solution of the given system of equations.

In summary, to find the value of q, we used the method of elimination by multiplying equation (1) by 3 and adding it to equation (2) to eliminate q. This resulted in finding the value of p as 2. Substituting this value of p in equation (1), we found the value of q as 0.

Bayanin Amsa

A nonagon has nine sides and nine interior angles. In a regular nonagon, all the sides and angles are equal. To find the measure of each interior angle of a regular nonagon, we can use the formula: Interior angle = (n - 2) x 180° / n where n is the number of sides of the polygon. Substituting n = 9 in the formula, we get: Interior angle = (9 - 2) x 180° / 9 = 7 x 20° = 140° Therefore, each interior angle of a regular nonagon is equal to 140°. So the correct answer is option (C) 140^o.

Bayanin Amsa

We are given that "M varies directly as n and inversely as the square of p." This means that M is directly proportional to n and inversely proportional to the square of p. We can represent this relationship mathematically as: M ∝ n/p^2 where the symbol ∝ means "is proportional to". We are also given that M = 3 when n = 2 and p = 1. We can use this information to find the constant of proportionality k: M ∝ n/p^2 3 ∝ 2/1^2 3 ∝ 2 To find k, we can write: M = k(n/p^2) Substituting the values we know: 3 = k(2/1^2) k = 3/2 Now we can use k to find M in terms of n and p: M = (3/2)(n/p^2) Simplifying, we get: M = (3n)/(2p^2) Therefore, the answer is option D: M = 3n/2p^2.

Bayanin Amsa

The interquartile range (IQR) is the range between the first quartile (25th percentile) and the third quartile (75th percentile) of a distribution. Half of the observations lie within this range, which means that the other half lie outside of it. Therefore, the percentage of observations that lie outside the IQR is 100% - 50% = 50%. So, the answer is 50%.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

To expand the expression (2x - 5)(x - 3), we need to apply the distributive property, which states that the product of a sum (or difference) and a term is equal to the sum (or difference) of the products of each term with the given term. So, we start by multiplying the first term of the first factor, 2x, by each term of the second factor, x and -3, and then do the same for the second term of the first factor, -5: (2x)(x) + (2x)(-3) + (-5)(x) + (-5)(-3) Simplifying this expression by multiplying and adding the terms, we get: 2x² - 6x - 5x + 15 Combining like terms, we have: 2x² - 11x + 15 Therefore, the correct answer is 2x² - 11x + 15.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

The length of an arc is given by the formula L = 2πr(θ/360), where r is the radius of the circle and θ is the central angle of the arc in degrees. In this case, the radius of the circle is 3.5cm and the central angle of the arc PXQ is 60 degrees. Substituting these values into the formula, we have: L = 2 x (22/7) x 3.5 x (60/360) L = 11cm Therefore, the length of the arc PXQ is 11cm. Answer (C)

Bayanin Amsa

Using the logarithmic identity Log_b(MN) = Log_bM + Log_bN, we can simplify Log₁₀4 + Log₁₀25 as follows: Log₁₀4 + Log₁₀25 = Log₁₀(4 x 25) = Log₁₀100 Since 100 is 10 raised to the power of 2, we can rewrite this as: Log₁₀100 = Log₁₀(10²) = 2 Therefore, the simplified expression is 2.

Bayanin Amsa

To find where the graph of the equation 2y = 5x² - 3x² - 2 intersects the y-axis, we substitute x = 0 and solve for y: 2y = 5(0)^2 - 3(0)^2 - 2 2y = -2 y = -1 Therefore, the graph of the equation intersects the y-axis at the point (0, -1). Hence, the correct option is (c) (0, -1).

Bayanin Amsa

The correct perimeter of the rectangular lawn can be calculated by adding twice the length and twice the breadth, i.e., 2 × length + 2 × breadth = 2 × 60m + 2 × 40m = 120m + 80m = 200m The perimeter calculated by the boy would be: 2 × 59.6m + 2 × 40.3m = 119.2m + 80.6m = 199.8m The difference between the correct perimeter and the calculated perimeter is: 200m - 199.8m = 0.2m To find the percentage error, we divide the difference by the correct perimeter and multiply by 100: (0.2m / 200m) × 100% = 0.1% Therefore, the percentage error in the boy's calculation of the perimeter of the lawn is 0.1%. The correct option is: 0.1%.

Bayanin Amsa

Bayanin Amsa

The mid values of the grouped marks can be found by taking the average of the upper and lower limits of each class interval. For example, the mid value of the first class interval (0 - 4) is (0+4)/2 = 2. Similarly, the mid values of the other class intervals can be calculated as 7, 12, and 17. The class interval is the difference between the upper limit of a class interval and the lower limit of the previous class interval. For example, the class interval between 0-4 and 5-9 is 5-4 = 1. The class interval in this case is not 4, but rather it is 5-0 = 5. The class boundaries are the values that separate one class interval from another. They are found by adding and subtracting half of the class interval from the upper and lower limits of each class interval. For example, the lower class boundary of the first class interval (0 - 4) is 0 - 0.5 = -0.5, and the upper class boundary of the second class interval (5 - 9) is 9 + 0.5 = 9.5. Therefore, the class boundaries for the given grouped marks are -0.5, 4.5, 9.5, 14.5, and 19.5. Therefore, the statement that is not true is II only, since the class interval is not 4. The correct class interval is 5, as explained above. So the answer is (B) II only.

Bayanin Amsa

The modal score is the score that occurs most frequently. Looking at the table, we can see that the score of 2 occurs the most frequently, appearing 6 times. Therefore, the modal goal scored is 2.

Bayanin Amsa

Bayanin Amsa

We can use trigonometry to solve this problem. Let's draw a diagram to help us visualize the situation:

             X
             |
             |
             |
             |
             |
             |
             |
             |
             |
             |
             P

We know that the angle of elevation of X from P is 60 degrees, which means that the angle XPX' is also 60 degrees, where X' is the foot of the pole. We also know that the distance from P to X' is 55m.

Let H be the height of the pole. Then we have:

tan 60 = H / PX'

We can simplify this expression using trigonometric ratios:

√3 = H / 55

Solving for H, we get:

H = 55√3

Therefore, the height of the pole is 55√3 meters. So the correct option is (c).

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

We know that if a quadratic equation has roots α and β, then the equation can be written as: (x - α)(x - β) = 0 Expanding the above expression, we get: x² - (α + β)x + αβ = 0 Here, the roots of the quadratic equation are -¹/₄ and 3. Therefore, α = -¹/₄ and β = 3. Substituting these values in the above equation, we get: x² - (α + β)x + αβ = 0 x² - (-¹/₄ + 3)x + (-¹/₄ × 3) = 0 x² - ¹¹/₄ x - ³/₄ = 0 Hence, the quadratic equation is 4x² - 11x - 3 = 0. Therefore, the correct option is: 4x² - 11x - 3 = 0

Bayanin Amsa

Bayanin Amsa

We know that the diagonal of a rectangle divides it into two right triangles with the diagonal as the hypotenuse, and the sides of the rectangle as the legs of the right triangles. Let's call the other side of the rectangle "x". Using the Pythagorean theorem, we have: 10² = 8² + x² Simplifying and solving for x, we get: x = √(10² - 8²) = √36 = 6 Therefore, the area of the rectangle is: 8 x 6 = 48 cm² Hence, the answer is 48cm².

Bayanin Amsa

We are given that the angles (a + 31) and (b - 49) are adjacent angles on a straight line. When two adjacent angles form a straight line, they add up to 180 degrees. This is known as the straight angle theorem. Therefore, we can write: (a + 31) + (b - 49) = 180 Simplifying the expression, we get: a + b - 18 = 180 Adding 18 to both sides, we get: a + b = 198 Therefore, the sum of the angles a and b is 198 degrees. Hence, the answer is: a + b = 198°.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

In a right-angle isosceles triangle, the two legs are congruent to each other. Let x be the length of each leg. By the Pythagorean theorem, we know that: x² + x² = 2² Simplifying the equation gives: 2x² = 4 Dividing both sides by 2, we have: x² = 2 Taking the square root of both sides gives: x = √2 Therefore, the length of each leg is √2, which is approximately 1.41. So, the correct answer is not in the options given, but it is approximately equal to, √2.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

We can factorize the given expression by first finding the greatest common factor (GCF) of the two terms, which is 8x. We can factor out the GCF from the given expression as: 32x³ - 8xy² = 8x(4x² - y²) We can then use the identity a² - b² = (a + b)(a - b) to factorize the expression further: 8x(4x² - y²) = 8x(2x + y)(2x - y) Therefore, the fully factorized form of 32x³ - 8xy² is 8x(2x + y)(2x - y). So, the correct option is: - 8x(2x + y)(2x - y)

Bayanin Amsa

Given that /PQ/ = /PS/ and /QR/ = /SR/, we can see that triangle QPR and SPR are congruent by the Side-Side-Side (SSS) criterion. Hence, we have ∠QPR = ∠SPR and ∠PQR = ∠PSR. i. From the above statement, we can say that the line PR bisects ∠QRS and ∠QSR, but it does not necessarily bisect the whole angle ∠QSRQ. Hence, statement (i) is true. ii. As triangle QPR and SPR are congruent, we can say that /QP/ = /SP/ and /QR/ = /SR/. Hence, the line PR is the perpendicular bisector of the line segment QS. Therefore, statement (ii) is also true. iii. Every point on the perpendicular bisector of a line segment is equidistant from the endpoints of the line segment. As we have already proved that line PR is the perpendicular bisector of the line segment QS, every point on PR is equidistant from SP and QP. Hence, statement (iii) is true. Therefore, all the three statements are true, and the correct answer is (v) I, II and III.

Bayanin Amsa

To convert a decimal number to binary, we need to continuously divide the decimal number by 2, until the quotient becomes 0. The binary number is obtained by writing the remainders (0 or 1) obtained in reverse order. Here's the step-by-step process: - Divide 89 by 2. Quotient is 44 and remainder is 1. - Divide 44 by 2. Quotient is 22 and remainder is 0. - Divide 22 by 2. Quotient is 11 and remainder is 0. - Divide 11 by 2. Quotient is 5 and remainder is 1. - Divide 5 by 2. Quotient is 2 and remainder is 1. - Divide 2 by 2. Quotient is 1 and remainder is 0. - Divide 1 by 2. Quotient is 0 and remainder is 1. So the remainders obtained in reverse order are: 1 0 1 1 0 0 0 Therefore, the binary representation of 89 is 1011001.

Bayanin Amsa

The formula for the volume of a cylinder is V = πr²h, where r is the radius of the base of the cylinder and h is the height of the cylinder. Substituting the given values in the formula, we have: V = π(3cm)²(8cm) V = 72π cm³ V ≈ 226.1cm³ (rounded to 1 decimal place, using π ≈ 3.142) Therefore, the correct option is (D) 226.2cm³.

Bayanin Amsa

To find the number of sides of a polygon given the sum of its interior angles, we can use the formula: Sum of interior angles = (n - 2) × 180° where "n" is the number of sides of the polygon. We are given that the sum of the interior angles of the polygon is 1440°. Substituting this value into the formula, we get: 1440° = (n - 2) × 180° Simplifying this equation, we can divide both sides by 180: 8 = n - 2 Adding 2 to both sides, we get: n = 10 Therefore, the polygon has 10 sides.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

To simplify the expression, we can use the rule of dividing powers with the same base. We divide the coefficients and subtract the exponents: 56x^-4 ÷ 14x^-8 = (56/14)x^(-4)-(-8) = 4x^-4+8 = 4x⁴ Therefore, the simplified expression is 4x⁴.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

To simplify \(\frac{5}{x-y}-\frac{4}{y-x}\), we first notice that \(y-x=-(x-y)\). Thus, we can rewrite the expression as \(\frac{5}{x-y}+\frac{4}{x-y}\). Now, we can combine the fractions by finding a common denominator, which is \(x-y\). Thus, we have \[\frac{5}{x-y}+\frac{4}{x-y}=\frac{5+4}{x-y}=\frac{9}{x-y}.\] Therefore, the simplified expression is \(\boxed{\frac{9}{x-y}}\).

Bayanin Amsa

The cumulative frequency curve can be used to find the median. A cumulative frequency curve represents the cumulative frequencies of a dataset in ascending order, with the class boundaries on the x-axis and the cumulative frequency on the y-axis. To find the median from a cumulative frequency curve, we need to locate the point on the curve where the cumulative frequency is half of the total frequency. This point will correspond to the median class, and we can then use the formula for finding the median of grouped data to calculate the exact value of the median. Therefore, the answer is median.

Bayanin Amsa

(a) Make d the subject of \(S=\frac{n}{2}[2a+(n-1)d]\).

(b)(i) Prove that \(\angle AOB=2\angle APB\).

Given: circle with centre O; A, B and P on the circumference.
To prove: \(\angle AOB=2\angle APB\).
Construction: Join P to O and produce it to a point Q.
Proof: \(OA=OP\) (radii), so triangle OAP is isosceles and its base angles are equal, \(x_1=x_2\). The exterior angle of a triangle equals the sum of the two interior opposite angles, so

Similarly, \(OB=OP\) (radii), giving \(\angle BOQ=2y_2\) where \(y_2=\angle OPB\). Adding the two central angles:

(b)(ii) Find x, y and z. The inscribed angle standing on the same arc is \(126^\circ\). The reflex angle at the centre is twice the angle at the circumference:

The angle at the centre going the other way completes the full turn:

The angle at the circumference on this arc is half the central angle:

Bayanin Amsa

(a) Multiply numerator and denominator by \(10^4\) to remove decimals:

Since \(0.45 \times 0.91 = 0.4095\) and \(0.0117 \times 35 = 0.4095\),

(b) The number halfway between two values is their average. Write both as improper fractions over 28:

(c) Each interior angle \(= 5 \times\) exterior angle, and interior \(+\) exterior \(= 180^\circ\):

(d) Pipe length 20 cm, internal radius \(r = 3\) cm, external radius \(R = 4\) cm, \(\pi = \tfrac{22}{7}\). Volume of material (hollow cylinder):

Bayanin Amsa

None

Bayanin Amsa

(a) P and Q lie on the parallel \(68.7^\circ\)S, with longitudes \(124^\circ\)W and \(56^\circ\)E. The longitude difference is

The radius of the parallel of latitude is \(R\cos 68.7^\circ\), so the distance along the parallel is

With \(\cos 68.7^\circ = 0.3633\):

(b) Bag: 4 red, 3 white, 5 green, total 12.

(i) \(P(\text{not green}) = 1 - P(\text{green}) = 1 - \dfrac{5}{12} = \dfrac{7}{12}.\)

(ii) Two balls without replacement, one red and one white (red-then-white or white-then-red):

Bayanin Amsa

(a) Let the number be \(N\).

\[\frac{N}{3}+8=18-\frac{N}{2}\]

\[\frac{N}{3}+\frac{N}{2}=18-8\]

\[\frac{5N}{6}=10\]

\[5N=60\]

\[\boxed{N=12}\]

---

(b) Construction of trapezium \(ABCD\)

1. Draw \(AB=8\text{ cm}\).
2. At \(A\), construct \(\angle DAB=60^\circ\), and draw the ray \(AD\).
3. Construct a perpendicular to \(AB\) at \(A\). Mark a point \(4\text{ cm}\) from \(A\) on this perpendicular and draw through it a line parallel to \(AB\). This line meets the \(60^\circ\) ray at \(D\).
4. With centre \(B\) and radius \(5\text{ cm}\), draw an arc to cut the parallel line at \(C\), taking the intersection to the left of \(B\).
5. Join \(BC\) and \(DC\).

From the completed construction,

\[\boxed{DC\approx 2.7\text{ cm}}\]

Bayanin Amsa

(a) Histogram

Use continuous class boundaries. Since all class widths are 10 marks, the bar heights are the frequencies.

Marks	Class boundaries	Frequency	Cumulative frequency
0 - 9	-0.5 - 9.5	4	4
10 - 19	9.5 - 19.5	5	9
20 - 29	19.5 - 29.5	6	15
30 - 39	29.5 - 39.5	12	27
40 - 49	39.5 - 49.5	8	35
50 - 59	49.5 - 59.5	5	40

(b) Mode

The modal class is \(30-39\), with class boundaries \(29.5-39.5\).

\[\text{Mode}=L+\frac{d_1}{d_1+d_2}\times c\]

\[=29.5+\frac{12-6}{(12-6)+(12-8)}\times10\]

\[=29.5+\frac{6}{10}\times10=\boxed{35.5\text{ marks}}\]

(c) Median

\[N=40,\qquad \frac{N}{2}=20\]

The 20th value lies in the class \(30-39\), since the cumulative frequency rises from \(15\) to \(27\) in this class.

\[\text{Median}=L+\frac{\left(\frac{N}{2}-\text{cumulative frequency before median class}\right)}{f_m}\times c\]

\[=29.5+\frac{20-15}{12}\times10\]

\[=29.5+4.1667=\boxed{33.67\text{ marks}}\]

Bayanin Amsa

Place base camp R at the origin. Convert each bearing to \((\text{East}, \text{North})\) components using East \(= r\sin\theta\), North \(= r\cos\theta\).

P: 20 km on bearing \(205^\circ\):

Q: 15 km on bearing \(060^\circ\):

(a) Distance of Q from P. The displacement is

(Check by the cosine rule with included angle \(205^\circ - 60^\circ = 145^\circ\): \(PQ^2 = 20^2 + 15^2 - 2(20)(15)\cos 145^\circ = 1116.5\), giving \(PQ = 33\) km.)

(b) Bearing of Q from P. Both components of \(Q - P\) are positive, so Q is north-east of P:

The bearing of Q from P is \(\approx 040^\circ\).

Bayanin Amsa

(a) With U the set of integers: \(P = \{x : x \le 2\}\), \(Q = \{x : -7 < x < 15\}\), \(R = \{x : -2 \le x < 19\}\).

(i) \(P \cap Q\) contains integers that are both \(\le 2\) and strictly between \(-7\) and \(15\), i.e. \(-6 \le x \le 2\):

(ii) First find \(R' = \{x : x < -2\ \text{or}\ x \ge 19\} = \{x : x \le -3\ \text{or}\ x \ge 19\}\).

Then \(Q \cup R'\): \(Q\) gives \(-6 \le x \le 14\) and \(R'\) gives all \(x \le -3\) together with \(x \ge 19\); their union is \(\{x : x \le 14\} \cup \{x : x \ge 19\}\).

Now intersect with \(P = \{x : x \le 2\}\). Every integer \(\le 2\) is already \(\le 14\), so

(b)(i) Grouped frequency table (40 marks):

(ii) Mean: \(\sum fx = 3(24.5)+5(34.5)+10(44.5)+13(54.5)+6(64.5)+3(74.5) = 2010\).

Bayanin Amsa

Cost price (outlay) \(= \text{N}120.00\) for 40 kg. Note \(50\text{k} = \text{N}0.50\).

(a) Amount realised from sales:

(b) Total profit:

Since this is positive, it is a profit.

(c) Percentage profit on outlay of N120.00:

Bayanin Amsa

(a) For \(y=3x^{2}-5x-7\), the completed table is:

\(x\)	\(-3\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)
\(y=3x^2-5x-7\)	35	15	1	−7	−9	−5	5	21

For example, when \(x=-2\),

\[ y=3(-2)^2-5(-2)-7=12+10-7=15. \]

(b) and (c) The required graphs, drawn on the same axes, are shown below. The curve is \(y=3x^2-5x-7\), and the straight line is \(y=-3x-2\).

(d)(i) \(3x^2-5x-7<0\) where the parabola lies below the \(x\)-axis. Reading the intercepts from the graph gives approximately \(-0.9\) and \(2.6\). Hence,

\[ -0.9<x<2.6. \]

(d)(ii) The roots of \(3x^2-2x-5=0\) are the \(x\)-coordinates of the points where the line and curve intersect:

\[ 3x^2-5x-7=-3x-2 \] \[ 3x^2-2x-5=0. \] From the graph,

\[ x=-1.0\quad\text{or}\quad x\approx1.7. \]

Indeed, \(3x^2-2x-5=(3x-5)(x+1)\), so the exact roots are \(x=-1\) and \(x=\frac53\).

Bayanin Amsa

(a) Express 0.72 using 2 and 3: \(0.72 = \dfrac{72}{100} = \dfrac{2^3 \times 3^2}{10^2}\).

(b) The tree is 20 m high; the chick is on the ground at angle of depression \(39^\circ\) from the top. The angle of depression equals the angle of elevation from the chick, so in the right-angled triangle (height 20 m, horizontal distance \(d\)):

The distance of the chick from the bottom of the tree is 25 m (to 2 significant figures).

Bayanin Amsa

None