WAEC SSCE - General Mathematics - 1996
Ajụjụ 1 Ripọtì
Simplify: \(\frac{5}{x - y} - \frac{4}{y - x}\)
Akọwa Nkọwa
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 \(\boxed{\frac{9}{x-y}}\).
Ajụjụ 2 Ripọtì
In the diagram above, PQ is parallel to RST. /RS/ = /SQ/ = /TQ/ and ?PQR = 35o.Calculate ?SQT
Ajụjụ 4 Ripọtì
Convert the decimal number 89 to a binary number
Akọwa Nkọwa
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.
Ajụjụ 5 Ripọtì
The 6th term of a G.P is -2 and its first term is 18. What is the common ratio?
Akọwa Nkọwa
Ajụjụ 6 Ripọtì
In the diagram, ?PMN = ?PRQ and ?PNM = ?PQR. If /Pm/ = 3cm, /MQ/ = 7cm and /PN/ = 5cm, find /NR/
Akọwa Nkọwa
Ajụjụ 7 Ripọtì
Two angles (a + 31) and (b - 49o) are adjacent angles on a straight line. Which of the following is true
Akọwa Nkọwa
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°.
Ajụjụ 8 Ripọtì
Simplufy Log104 + Log1025
Akọwa Nkọwa
Using the logarithmic identity Logb(MN) = LogbM + LogbN, 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.
Ajụjụ 9 Ripọtì
The angle of elevation of the top X of a vertical pole from a point P on a level ground is 60o. The distance from P to the foot of the pole is 55m. Without using tables, find the height of the pole.
Akọwa Nkọwa
We can use trigonometry to solve this problem. Let's draw a diagram to help us visualize the situation:
X
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|
|
|
|
|
|
|
|
|
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).
Ajụjụ 10 Ripọtì
In the diagram above, PQ is a tangent to the circle MTN at T. What is the size of ?MTN?
Akọwa Nkọwa
Ajụjụ 11 Ripọtì
A bag contains 3 red, 4 black and 5 green identical balls. Two balls are picked at random, one after the other without replacement. Find the probability that one is red and the other is green
Ajụjụ 12 Ripọtì
The angle of a sector of a circle of radius 35cm is 288o. Find the perimeter of the sector. [Take π = 22/7]
Akọwa Nkọwa
Ajụjụ 14 Ripọtì
What percentage of observation lie outside interquartile range of any distribution?
Akọwa Nkọwa
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%.
Ajụjụ 15 Ripọtì
A town P is 150km from a town Q in the direction 050o. What is the bearing of Q from P?
Akọwa Nkọwa
Ajụjụ 16 Ripọtì
The cumulative frequency curve may be used to find the
Akọwa Nkọwa
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.
Ajụjụ 17 Ripọtì
Expand (2x - 5)(x - 3)
Akọwa Nkọwa
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.
Ajụjụ 18 Ripọtì
The number of goals scored by a football team in 20 matches is shown in the table above
What is the modal goal scored?
Akọwa Nkọwa
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.
Ajụjụ 19 Ripọtì
The graph of 2y = 5x2 - 3x2 - 2 cuts the y axis at the point
Akọwa Nkọwa
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).
Ajụjụ 20 Ripọtì
Factorize 32x3 - 8xy2
Akọwa Nkọwa
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)
Ajụjụ 21 Ripọtì
In the diagram above, /PQ/ = /PS/ and /QR/ = /SR/. Which of the following is/are true? i. the line PR bisects ?QRS ii. The line PR is the perpendicular bisector of the line segment QS iii. Every point on PR is equidistant from SP and QP
Akọwa Nkọwa
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.
Ajụjụ 22 Ripọtì
Ajụjụ 23 Ripọtì
In the diagram above, O is the center of the circle of radius 3.5cm, ?POQ = 60o
Use the information to answer the question below [Take ? = 22/7]
What is the length of the arc PXQ?
Akọwa Nkọwa
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)
Ajụjụ 24 Ripọtì
In the diagram above, TRQ is a straight line. Find p, if p = 1/3(a + b + c)
Akọwa Nkọwa
Ajụjụ 27 Ripọtì
If tan x = \(2\frac{2}{5}\), find the value of sin x; 0 \(\leq\) x \(\leq\) 90o
Akọwa Nkọwa
Ajụjụ 28 Ripọtì
The marks obtained by pupils of class are grouped as shown below; 0 - 4, 5 - 9,10 -14, 15 -19. Which of the following is/are not true? l. The mid values of the grouped marks are 2, 7, 12 and 17. II The class interval is 4. III The class boundaries are 0.5, 4.5, 9.5, 14.5 and 19.5
Akọwa Nkọwa
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.
Ajụjụ 29 Ripọtì
In the diagram above, KLMN is a cyclic quadrilateral. /KL/ = /KN/, ?NKM = 55o and ?KML = 40o. Find ?LKM
Akọwa Nkọwa
Ajụjụ 30 Ripọtì
Each interior angle of a regular nonagon(nine sided polygon) is equal to
Akọwa Nkọwa
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) 140o.
Ajụjụ 31 Ripọtì
Find the 9th term of the arithmetic progression 18, 12, 6, 0, -6 .........
Akọwa Nkọwa
Ajụjụ 32 Ripọtì
Find, correct to 1 decimal place, the volume of cylinder of height 8cm and base radius 3cm. [Take π = 3.142]
Akọwa Nkọwa
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.
Ajụjụ 33 Ripọtì
A boy measured the length and breath of a rectangular lawn as 59.6m and 40.3m respectively instead of 60m and 40m. What is the percentage error in his calculation of the perimeter of the lawn?
Akọwa Nkọwa
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%.
Ajụjụ 34 Ripọtì
M varies directly as n and inversely as the square of p. If M = 3, when n = 2 and p = 1, find M in terms of n and p.
Akọwa Nkọwa
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.
Ajụjụ 35 Ripọtì
The number of goals scored by a football team in 20 matches is shown in the table above
What is the mean goal scored?
Akọwa Nkọwa
Ajụjụ 36 Ripọtì
One side of a rectangle is 8cm and the diagonal is 10cm. What is the area of the rectangle?
Akọwa Nkọwa
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.
Ajụjụ 37 Ripọtì
If 3p - q = 6 and 2p + 3q = 4, find q
Akọwa Nkọwa
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.
Ajụjụ 38 Ripọtì
The table gives the distribution of outcomes obtained when a die was rolled 100 times.
What is the experimental probability that it shows at most 4 when rolled again?Akọwa Nkọwa
Ajụjụ 40 Ripọtì
In the diagram above, O is the center of the circle of radius 3.5cm, ∠POQ = 60°. What is the area of the minor sector POQ?
[Take π = 22/7].
Akọwa Nkọwa
Ajụjụ 41 Ripọtì
In the diagram above, find x correct to the nearest degree
Ajụjụ 42 Ripọtì
Simplify 56x\(^{-4}\) \(\div\) 14x\(^{-8}\)
Akọwa Nkọwa
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.
Ajụjụ 43 Ripọtì
If sin\(\theta\) cos\(\theta\), for 0o \(\leq\) θ \(\leq\) 360o, find the value of \(\theta\)
Ajụjụ 44 Ripọtì
two towns X and Y both on latitude 60oS have longitude 27oE and 33oW respectively. Find to the nearest kilometers, the distance between X and Y measured along the parallel of latitude. [Take 2πR = 4 x 104km, where R is the Radius of the earth]
Akọwa Nkọwa
Ajụjụ 45 Ripọtì
If the hypotenuse of a right-angle isosceles triangle is 2, what is the length of each of the other side?
Akọwa Nkọwa
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.
Ajụjụ 47 Ripọtì
How many sides has a polygon if the sum of its interior angle is 1440o?
Akọwa Nkọwa
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.
Ajụjụ 48 Ripọtì
The roots of a quadratic equation are -1/4 and 3. The quadratic equation is
Akọwa Nkọwa
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
Ajụjụ 49 Ripọtì
If x is a real number which of the following is more illustrated on the number line?
Ajụjụ 51 Ripọtì
(a) ABCD is a trapezium with AB parallel to DC and /AD/ = /AB/. If < BAD = 106°, find < BDC.
(b) The table below shows the distribution of 20 cards labelled A - E.
| Card | A | B | C | D | E |
| Frequency | 3 | 4 | 7 | 5 | 1 |
(i) If a card is selected at random from the pack, what is the probability that the card is E? (ii) If two cards are selected at random one after the other without replacement from the pack, what is the probability that one of the two cards is B?
Ajụjụ 52 Ripọtì
(a) Copy and complete the table for \(y = 3x^{2} - 5x - 7\)
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| \(y = 3x^{2} - 5x - 7\) | 35 | -7 | -9 | 5 |
(b) Using a scale of 2cm = 1 unit along the x- axis and 2cm = 5 units along the y- axis, draw the graph of \(y = 3x^{2} - 5x - 7\).
(c) On the same axis, draw the graph of \(y + 3x + 2 = 0\).
(d) From your graph, find the : (i) range of values of x for which \(3x^{2} - 5x - 7 < 0\) ; (ii) roots of the equation \(3x^{2} - 2x - 5 = 0\).
Akọwa Nkọwa
None
Ajụjụ 53 Ripọtì
(a) Make d the subject of the formula \(S = \frac{n}{2}[2a + (n - 1) d]\).
(b) (i) 
In the diagram, O is the centre of the circle, A, B and P are points on the circumference. Prove that < AOB = 2 < APB.
(ii) 
Find the angles x, v and z in the diagram.
Akọwa Nkọwa
None
Ajụjụ 54 Ripọtì
The table below shows the marks obtained by forty pupils in a Mathematics test.
| Marks | 0 - 9 | 10 - 19 | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 |
| No of pupils | 4 | 5 | 6 | 12 | 8 | 5 |
(a) Draw a histogram for the mark distribution ;
(b) Use your histogram to estimate the mode ;
(c) Calculate the median of the distribution.
Akọwa Nkọwa
None
Ajụjụ 55 Ripọtì
(a) P and Q are points on the parallel of latitude 68.7°S, their longitudes being 124°W and 56°E respectively. What is their distance apart measured along the parallel of latitude? [Take R = 6400km, \(\pi = 3.142\)]. (Give your answers to 3 significant figures).
(b) A bag contains four red, three white and five green balls. (i) If one ball is picked at random, what is the probability that it is not green? (ii) if two balls are picked at random without replacement, what is the probability that one is red and the other white?
Ajụjụ 56 Ripọtì
(a) Find the number N such that when \(\frac{1}{3}\) of it is added to 8, the result is the same as when \(\frac{1}{2}\) of it is subtracted from 18.
(b) Using a ruler and a pair of compasses only, construct a trapezium ABCD, in which the parallel sides AB and DC are 4 cm apart. < DAB = 60°, /AB/ = 8 cm and /BC/ = 5 cm. Measure /DC/.
Akọwa Nkọwa
None
Ajụjụ 57 Ripọtì
Two men P and Q set off from a base camp R, prospecting for oil. P moves 20km on a bearing of 205° and Q moves 15km on a bearing of 060°. Calculate the:
(a) distance of Q from P ;
(b) bearing of Q from P.
(Give your answer in each case to the nearest whole number)
Ajụjụ 58 Ripọtì
(a) If \(\log_{10} 2 = 0.3010\) and \(\log_{10} 3 = 0.4771\), calculate without using tables, the value of \(\log_{10} 0.72\).
(b) A hawk on top of a tree, 20 metres high views a chick on the ground at an angle of depression of 39°. Find, correct to 2 significant figures, the distance of the chick from the bottom of the tree.
Akọwa Nkọwa
None
Ajụjụ 59 Ripọtì
A shopkeeper buys 40 kg of fruits for N120.00. He sells 20 kg at N5.00 per kg, 10 kg at N3.00 per kg, 5 kg at N2.00 per kg and the remaining 5 kg at 50k per kg. Calculate the :
(a) amount he realises from the sales ;
(b) total profit / loss ;
(c) percentage profit/ loss on his outlay of N120.00.
Akọwa Nkọwa
None
Ajụjụ 60 Ripọtì
(a) The universal set U is the set of integers, P, Q and R are subsets of U defined as follows:
\(P = x : x \leq 2 \) ; \(Q = x : -7 < x < 15\) ; \(R = x : -2 \leq x < 19\).
Find (i) \(P \cap Q\) ; (ii) \(P \cap (Q \cup R')\), where R' is the complement of R with respect to U.
(b) The following data shows the marks of 40 students in a History examination.
41 52 37 56 63 48 65 46 54 32 51 66 74 23 35 61 58 44 49 53 45 57 56 38 59 28 50 49 67 56 36 45 79 68 43 56 26 47 55 71.
(i) Form a grouped frequency table with the class intervals 20 - 29, 30 - 39, 40 - 49 etc; (ii) Find the mean of the distribution.
Ajụjụ 61 Ripọtì
(a) Without using tables, find the value of \(\frac{0.45 \times 0.91}{0.0117}\)
(b) Find the number which is exactly halfway between \(1\frac{6}{7}\) and \(2\frac{11}{28}\).
(c) If each interior angle of a regular polygon is five times the exterior angle, how many sides has the polygon?
(d) Calculate the volume of the material used in making a pipe 20cm long, with an internal diameter 6cm and external diameter 8cm. [Take \(pi = \frac{22}{7}\)].
