mathematics knowledge. This knowledge is on skills and step-by-step procedures without explicit reference to mathematical ideas. Students committing procedural errors do not have an understanding of why or how procedure works; and do not recognize the importance of applying, and computing the procedure correctly (Elbrink, 2008). Conceptual errors result from lack or poor understanding of conceptual mathematical knowledge. Hope (2006) refers conceptual mathematics understanding as a knowledge that involves clear and systematic understanding of fundamental basic points behind the algorithms performed in mathematics. As Hull and Miles (2010) argue, students demonstrate conceptual understanding in mathematics when they provide evidence that they can generate examples of concepts; use and interrelate varied representations of concepts; identify and apply principles; know and apply facts and definitions; and apply the signs, symbols, and terms used to represent concepts. Conceptual understanding also allows a student to apply and possibly adapt some acquired mathematical ideas to new situations. Thus, to avoid procedural and conceptual errors in mathematics, both procedural and conceptual learning are necessary with a consideration that, procedural learning needs to be coupled to conceptual learning and to real-life applications. When students make some generalizations which are not correct, they create misconceptions. A misconception is not wrong thinking. Instead, it stands as a premature concept or a local generalization that the student has made. Thus, it is very difficult to avoid misconceptions during learning since are formed when students systematically use incorrect rules, or use correct rules beyond their proper
discussion, as well as some responses obtained through questionnaire. This combination of instruments contributed to record what was intended to record as well as to measure what implied to measure. 3.8 Ethical Dimensions et al. (2000) note that, ethical issues are likely to be raised by social scientists from the kinds of problems investigated with regards the methods used to obtain data. They explain further that, each stage in the research sequence may be a potential source of ethical problems. Such ethical issues depend on the extent to which the willingness and collaboration of the participants are guaranteed, promise of confidentiality, and thorough discussion on potential consequences of their involvement in a particular study. A research clearance letter by the Vice Chancellor, The Open University of Tanzania (Appendix E) was used to introduce the researcher to the Regional and Districts authorities in the three regions (Dar es Salaam, Morogoro and Coast). Responsible officers in the said regions and districts were requested for permission to allow the researcher to collect data from their schools. Copies of the questionnaire and FGD guide prepared for research purpose were attached with the letter for showing the nature of data required from the schools. Also, the purpose of this study was made clear to them that, it was only for fulfillment of the requirement for the degree of Doctor of Philosophy (PhD) - Education at The Open University of Tanzania. The researcher discussed with the responsible officers in the regional and districts offices, heads of schools, as well as respondents about the intention and benefits of the study. This was aimed at informing them that, as education stakeholders, they
As shown in Table 4.3, the highest failure rates are found in questions 47, 48, 46, 44, 45, 36 and 43. These were questions on Units of area, Volume, Capacity, Length, Time and Weight, respectively. The failure rates in these items, in their respective order, range from 83.9% to 70.7%. The next category in the order of difficulty, were questions with failure rates ranging between 50% and 70% which were on: Percentages, Angles, Number patterns, Algebra and Arithmetic operations involving negative numbers. 4.3: Percentages of Students’ Failure on Competence Tested in the Task No Task/Question Class level taught Competence or skill required Percentage failures 47 Badili m2 100 kuwa sm2 Convert 100 m2 into cm2 IV, V, VI, Conversion of units of area 83.9 48 Badili m3 10 kuwa sm3 Convert 10m3 into cm3 V, VI, VII Conversion of of volume 82.9 46 Badili m3 1000 kuwa lita Convert 1000m3 into litres V, VI, VII Conversion of units of volume 81.9 44 Kadiria urefu wa mlango wa chumba cha darasa lako kwa kutumia sentimita. Estimate the length of your classroom door in cm. III, IV, V, VI Estimation of length 76.9 45 Kadiria muda utakaotumia kutembea kutoka mlango wa chumba cha darasa lako hadi kwenye mlango wa ofisi ya Mkuu wa Shule yako. How much time will you take to walk from your classroom to the School head’s office Estimation of time 74.7 36 Tafuta wastani wa muda ufuatao: saa 1:40, saa 2:55, saa 8:05, na saa 3:20. Find the average of time durations 1:40, 2:55, 8:05 and 3:20 Addition of units of time 72.7 43 Kadiria uzito wa mtoto mchanga katika gramu Estimate the weight of a born baby in grammes Estimation of weight 70.7
27 Tafuta jibu la/ Evaluate: – 6 – ( + 3 ) VI,VII Subtraction involving positive and negative numbers 51.4 8 Toa/ Subtract: 97 – 971 IV,V,VI Subtraction of a big integer from the small integer. 49.6 25 Kokotoa jibu/calculate: 12 – 10 × 7 VII Use of BODMAS rule 47.9 13 Toa/subtract: 0.89 – (– 8.9 ) VI Subtraction of a negative decimal number from a positive decimal number. 47.6 15 Gawanya/Divide: 7.65 ÷ 34 VI Division of decimal numbers by a whole number. 47.4 22 Zidisha/Multiply : 3.5 × V,VI Multiplication of a decimal number with a number 46.2 23 Zidisha/Multiply 2.14 × 0.029 V,VI Multiplication of decimal numbers 45.2 19 Zidisha/Multiply V,VI Multiplication of a mixed number by a fraction 41.4 7 Toa/subtract 0 – 8 IV,V Subtraction of a positive number from Zero 40.2 29 Kokotoa jibu la/Calculate: – 3 – (–3) V,VI,VII Subtraction of two negative numbers 40.2 31 Badili kuwa sehemu rahisi. Convert into fraction in its lowest form. V,VI,VII Converting percentages to fractions 40.0 17 Gawanya/Divide V,VI Division of fractions 38.0 32 Badili namba ya desimali 2.6 kuwa asilimia Change 2.6 into percentages V,VI Converting decimals to percentages 37.0 28 Gawanya/Divide (+ 78 ) ÷ (– 13 ) Division of positive and negative numbers 37.0 18 Gawanya/Divide: 0.12 ÷ 0.003 V,VI Division of decimal numbers 35.5 20 Zidisha: 0.29 × 1000 V,VI Multiplication of a decimal number with multiples of 10 35.5 21 Zidisha/ multiply : 58 × 347 V,VI Multiplication of Large integers 31.8 24 Kokotoa jibu la/ Calculate 6 – 8 ÷ 2 VII Application of BODMAS rule. 31.0 16 Gawanya/Divide: V,VI Division of numbers 30.8
14 Gawanya/ Divide: 9113 ÷ 13 V,VI Division of integers using the long method 30.3 11 Toa/Subtract: V,VI Subtraction of numbers 26.6 6 Toa/Subtract: 2.11 – 1.173 V,VI Subtraction of a number with three decimal places from one with two decimal places. 26.3 12 Toa/Subtract: 7.03 – 1.174 V,VI Subtraction of a number with three decimal places from one with two decimal points 24.3 4 Jumlisha/Add: V,VI Addition of mixed numbers 23.1 26 Gawanya/Divide: V,VI Division with fractions 18.1 3 Jumlisha/Add : 43.003 + 14.01 + 58.507 V,VI Addition of decimal numbers(two stages) 17.1 2 Jumlisha/Add: 0.021 + 9.98 V,VI Addition of decimal numbers(one stage) 12.2 1 Jumlisha/Add: 856 + 9134 IV,V,VI Addition of whole numbers 10.9 5 Jumlisha/Add: 4012 + 679 IV,V,VI Addition of whole numbers 6.5 The third category includes questions of failure rates ranging between 25% and 50%. The questions were on a wide spectrum of competences falling into four groups: Subtraction of whole numbers, decimals, integers and mixed numbers Use of the BODMAS rule Multiplication of decimals, mixed numbers and decimals or fractions, decimals by multiples of 10, three digits numbers, Division of decimal numbers, fractions, positive and negative integers, numbers, integers by long division method. The lowest range of failure rate (below 25%) was in questions involving addition of
81.9 If 1m = 100cm 1,000m3 = 100,000cm3 Since 1,000cm3 = 1l 100,000cm3 = x By cross multiplication 1,000 × cm3 = 100,000cm3 × litres x = =100 litres Incorrect use of the relationship 1m = 100cm to deduce that 1m3= 100m3 Cubing only the unit- prefix instead of both the prefix and quantity expressed in centimeters. Defined 1m3 as 1m3= 100cm3 instead of 1m3 = (1m)3 = 1m × 1m × 1m = = 1,000,000cm3 There were two types of errors noted in the respondents’ work of converting square metres into square centimeters, cubic metres into cubic centimeters and cubic metres into litres. The first is a conceptual error. Over 81% of respondents applied incorrectly the relationships of units. This error occurred because they did not have a good understanding of the relationship between metre and centimetres. They knew that 1m = 100cm, but did not understand the definitions or rather the expanded forms of ‘a square metre’ and ‘a cubic metre’. These respondents had a misconception that since the relationship of one metre and centimetres was that 1m = 100cm, then 1m2 = 100cm2. Likewise they deduced that 1m3 = 100cm3. Clearly, the respondents squared or cubed only the prefixes, forgetting to do the same for the numerical quantities as well. Therefore, respondents lacked correct conceptual understanding, that 1m2 = 12m2 = (1m)2= 1m × 1m. Similarly, they did not understand that 1m3 = (1m)3 = 13m3 = 1m ×1m ×1m.
numeration of minutes and hours. Table 4.6: Student Failure Rates (%) by Work, Errors, Causes of Errors and Misconceptions on Computation of Units of Time, and Length % rate Student’s work Errors Associated misconceptions Noted errors Causes 72.7 2:55 1:40 8:05 + 3:20 15:20 addition of minutes is done in the same way as for base 10 numeration. Unable to differentiate the hours/minutes numeration from the base ten decimal numeration. Did not distinguish decimal number notation from hour/minute notation. 57.3 km m cm 3 25 75 × 5 16 125 375 km m cm 3 25 75 × 5 16 28 75 Did not split cm into m Created a ‘carrying figure’ from the product of m added it to the product of km Did not know the relationship between the given units of length and hence unable to make proper conversion. Does not know the meaning of the Prefix ‘centi’ in the word ‘centimeter’ Does not know the meaning of the prefix ‘kilo’ in the term ‘kilometre’. Thinking that 100m = 1km Each of the metric units of length is related to the metre. The relationship between kilometre is that, 1000metres = 1 kilometre (Kilo means 1000) and, the relationship between a metre and centimeters is that 100cm = 1m (Cent means 100). In item 38,
respondents were expected to decompose the product of 75m × 5 into kilometres and metres. As shown in Table 4.6, respondents committed an error called decomposition of units by leaving 375cm in the column of centimetres. Remediation of these errors on converting units of time and length can be done by familiarizing students with relationships of “an hour and minutes”, “a metre and centmetres”, and “a kilometer and metres”. This familiarization should be followed by extensive exercises on changing one unit into another. 4.4.2 Estimating Length/Height, Weight and Time It is very useful to be able to estimate length, mass or weight, volume and time, because it may always not be easy to measure them. Estimation of these quantities is taught in Standard IV, and VI. In items 43, 44 and 45, respondents were requested to estimate the length of a classroom door in centimeters, the time it takes to walk from the classroom to the head teacher’s office, and to estimate the weight of a newly born baby. Knowledge and Competences/Skills tested Estimating heights and lengths Estimating the time on an activity Estimating the weight of a substance Sample questions Item 44: Estimate the length of your classroom door in centimeters Item 45: How much time will you take to walk from your classroom to the head teacher’s office? Item: 43: Estimate the weight of a born baby
Table 4.7 presents errors noted in solutions of items 44, 45 and 43, associated misconceptions and causes. Table 4.7: Student Failure Rates (%) By Work, Errors, Causes of Errors and Misconceptions on Estimating Length, Time and Weight % Failure rate Student’s work Errors Associated misconceptions Noted errors Causes 76.9 Sample answers: of answers given i. 2cm to 7cm ii ii. 12cm to 18cm iii. 30cm to 38 cm iv.100cm to 120cm v. 1000cm Too low estimates Too high estimates They were used to: i. instruments (scale) for measuring length ii. instruments with standard lengths like a metre rule and the ‘30 cm ruler’ Did not know the range or width of a ‘unit centimeter’ 74.7 Sample answers 25sec 48sec 5min 10min 50min Very low estimates Very high estimates They were not used to measure time duration when doing something. Did not know the duration of a second or a minute. 70.7 Sample answers 1.5gm, 2gms to 10gms, 100gms to 600gms, 5kg to 9kg, 10kg to 25kg Very low estimates Very high estimates Were not used to describe how much is there in a substance. Had unrealistic heavier a unit is. Respondents were physically sitting in the classroom looking at the classroom door. The lengths of doors in their classrooms were normal, slightly more than the height of an adult person. This means that, by knowing their physical heights, which is
2.6 = Wrote 2.6 = Instead of 2 Multiplied by 100% as for the case of expressing number into percentages Thought that a decimal point in 2.6 was separating the numerator and denominator as for the short line in. . Did not know: i. the meaning of a decimal point in the decimal number as well as the purpose of the short line in the fraction ii. the rule for changing a decimal number into fractional form
Solving simple algebraic word problem Sample questions: Item 37: If and n = 3, evaluate the expression Item 40: Simplify the expression Item 41: Find the of if Item 42: I am thinking of a number when it is multiplied by 3 and then adding to the result, the answer is 1. Find the number I am thinking of. Table 4.11 shows errors noted in solutions of the four items as well as the associated misconceptions and causes. In item 37 respondents were required to evaluate an algebraic expression m2 – mn – 2 by substituting m= –2 and n = 3. The substitution error as noted from respondents work, was in substituting m = –2 into mn. They wrote either (–2)2 + (–2)3 – 2 or –22 + (–2 × 3) –2 instead of (–2)2 + – (–2)3 – 2. The misconception was that the two signs in – (–2)3 make a positive sign outside brackets and hence obtained + (–2 × 3) instead of +2(3). When subtracting a positive number from a negative as in –2– 2, some respondents obtained 0 as the answer instead of – 4. The respondents were confusing this with the fact that, one number minus the same number, the answer is 0, (a – a = 0). With this misconception they regarded –2 – 2 as subtraction of the same numbers and hence committed an error of confusing numbers with opposite signs. The respondents in the Focused-group discussion also gave 0 as the answer for the two expressions a – a and – a – a. They thought that the two expressions were the same.
Table 4.12: Students Failure rates (%) by Work, Errors, Causes of Errors and Misconceptions on Using BODMAS Rule % Failure rates Students work Errors Associated misconceptions Noted errors Causes 53.8 − (− 8 ÷ 4 ) = − 8 ÷ 4 = − 2 − (− 8 ÷ 4 ) = + ( 8 ÷ 4 ) = 8 ÷ 4 = 2 − (− 8 ÷ 4 ) = +8 ÷ − 4 = − 2 Neglected the negative sign outside the brackets Multiplied the two negative signs in − (−8, to obtain + as for the case of distributive law of multiplication Opened brackets as in the case of distributive law of multiplication Did not understand the two negative signs multiplied together. Thought that the negative sign outside the brackets have the same meaning as the one inside. Thought that division is associated distributive property like multiplication. 47.9 12 − 10 7 = 2 7 = 14 Performed 12 – 10 first instead of 10 × 7 Did not know BODMAS rule. Thought that the computation can be done by starting with any operation. 31.0 6 − 8 ÷ 2 = –2 ÷ 2 = –1 Performed 6 – 8 first instead of 8 ÷ 2
Errors and misconceptions committed by respondents in solving items 10, 27, 28 and 29 as well as causes of the misconceptions are given in Table 4.13. Table 4.13: Students Failure Rates (%) by Work, Errors, Cause of Errors and Misconceptions on Subtraction of Integers % Failure Students work Errors Associated misconceptions Noted errors Causes 54.1 –9 – 3 = –6 –9 – 3 = +6 Obtained – 6 as the answer instead of –12 Gave +6 as the answer instead of –12 Did not know the different ways of writing a – b, –a – b and – (+a) They were also confusing the rules of adding two consecutive positive and negative integers. Thought that, the two negative signs in‘ – 9 –3’ are multiplied to give + 3 hence –9 + 3 = –6 Thought that the two consecutive negative in ‘–9 –3’ are multiplied to give a positive sign as for the case of –a × –a = +a 51.4 − 6 − (+ 3 ) = 6 + 3 i. − 6 − = +6 ii. − (+ 3 ) = + 3 Did not know the: i. of opening brackets ii. meaning of − (+ a ), – (–a) Thought that the two signs in ‘ – 6 – ‘ are multiplied to give + 6 47.6 –8.9 –0.89 –8.19 Interchanged the terms from 0.89 – (–8.9) to Did not know the meaning of – (– a) Thought that it was not possible to subtract –8.9 from 0.89
% Failure rates Students work Errors Associated misconceptions Noted errors Causes 37.0 (+78) ÷ (−13) = = 6 Neglected the positive and negative signs in the computation Did not know the rules for dividing positive and negative numbers. Thought that the positive and negative signs are not involved when performing division as it applies for multiplication. 30.3 -91 13 -13 00 Skipped a zero between 7 and 1 in the quotient. The zero was the ‘number of times 1 goes into 13. Did not have clear understanding of using the long division method. Thought that, it was unrealistic to write 0 as the number of times 1 goes into 13. In item 28, respondents were required to divide a positive integer by a negative integer, while in item 14 they were required to divide positive integers. As shown in Table 4.15, the proportion of respondents who failed item 14, was greater than the proportion of failures in item 28. The common error noted on division of positive and negative numbers was omission of the positive and negative signs in the computation. Respondents in the Focused-group discussions from schools C3SGB and NG2SGB appeared to know only the products of two negative numbers, two positive numbers as well as products of the numbers with opposite signs. However, some of them were confusing the result of: dividing two negative numbers, two positive numbers, negative and positive numbers. As such, they conceptualized wrongly the rules governing division of positive and negative numbers by confusing
38.0 = = = Used incorrect rule: Did not understand the logic of the rule: Thought that, the rule for dividing two fractions was: ‘multiply reciprocals of both Dividend and Divisor’. 30.8 4 ÷ 2 = = = = Used incorrect rule: Did not understand the logic of the rule: Thought that, the rule for dividing fractions was: to multiply divisor by the reciprocal of dividend. 18.1 = = 12 = = Used incorrect rule: Used incorrect rule: = Did not understand the rule for dividing fractions. Thought that, the rule for dividing two fractions was: ‘multiply reciprocals of both Dividend and Divisor Thought that numerators and denominators are divided separately as for the case of fractions. Generally, as in the table, the failure rate on these items ranged from 38.0% to 18.1%. Failure to obtain correct answers was due to the incorrect use of the rule for dividing fractions. While the correct rule was to multiply the dividend by the reciprocal of the divisor that is:, some respondents either multiplied the reciprocals of dividend and divisor, or multiplied the dividend by reciprocal of the divisor, that is ×. It was also noted that, there were respondents (18.1%) who divided the
Ii iii. the rule for subtracting a large integer from a small one. the successive negative signs in –14– 6 1 multiply to make +14 26.6 = 16 = 16 = 19 = = = = = = Used 1 as LCM instead of 3 Used 9 as LCM instead of 3 For each number, the numerator was subtracted from the whole number. Did not know the meaning of LCM and how to find it. Did not know how to change a mixed number into fractional form. Thought that the LCM of the two fractions is obtained by multiplying their denominators. Thought that, a mixed number is changed into fractional form by subtracting the numerator from the whole number. Respondents created a positive sign in the expression –14 – 6 as in the case of – (– 14) whose expansion form is; –1 × –4. Respondents therefore, were not able to differentiate ‘a negative sign’ from ‘a minus sign’ in the expression –14 – 6. Also,
26.3 2.110 − 1.173 937 1.173 − 2.11 -1.063 The answer has no decimal point Computed 1.173–2.11 instead of 2.11–1.173. As such they interchanged the numbers. Did not know the significance of ‘a zero’ on the left of a decimal point. Had no understanding of the rules for subtracting decimal numbers. Thought that a zero in 0.937 can be ignored as in the case of a zero in 937.0 i. Thought that .173 is larger than .11 therefore a large number can not be subtracted from a small. ii. Thought that decimal .11 and .173 were independent of integer parts 24.3 7.030 −1.174 6.956 7.03( ) −1.174 5.864 Obtained 6.956 as the answer instead of 5.856 No subtraction was done in the thousandth value. The digit (4) was simply brought down to the thousandth place value in the answer. The idea of ‘borrowing’ is not clearly understood. ------------ They had no idea of placing zero(s) to fill gap(s) in a number with digits. Thought that ‘borrowing’ is done only once Thought that if a digit is missing in the vertical alignment, then no subtraction can be done in the respective column. Numerous errors were on subtraction of decimal numbers as cited in the respondents’ work in Table 4.24. First, in the question to find 2.11– 1.173, the digits were correctly vertically aligned with a replacement of a zero in the thousandth place
= 12 Obtained as a sum of i. Did not understand the rule for adding fractions ii. Did not understand the rule for converting a number into fractional form. Thought that the sum of fractions is obtained by adding their numerators and denominators separately. i.e. = = Applied LCM before changing numbers into fractional form. Did not understand: i. the rule for changing a mixed number into fractional form ii. the conditions for using LCM. Thought that LCM is used even in the mix of whole numbers and fractions. = + Added together the denominator, whole number and numerator when changing the two mixed numbers into fractional form. Did not understand the rule for changing a mixed number into fractional form. Thought that the rule for changing a into form of was ((a + b) c instead of ((a + b) c As shown in Table 4.25, the respondents committed some errors in adding mixed numbers. Common errors noted included the error of adding numerators and denominators separately when finding the sum of proper fractions, and incorrect use of the rule for changing a mixed number into fractional form. The sum of was
Conceptual Units Estimations Making unrealistic estimates of length, weight and time. Conceptual Lacking a ‘feel’ or ‘sense’ of unit quantities or standard units of measurements Drawing unrealistic angle estimates Conceptual Addition of hours and minutes length Added units of time in the same way as for base 10 numeration Conceptual Does not differentiate base 10 numeration from the minutes and numeration Multiplication of km, m and cm by the factor. Writing a product which is not matching with a prefix Conceptual Does not understand the meaning of given prefix Algebra Simplifying algebraic expression by collecting like terms Incorrect placing of brackets when collecting like terms Incorrect opening of brackets Thinking that brackets have no effect on the signs of the grouped terms. Making direct substitution in an algebraic expression Ignoring the coefficient of the variable or term Thinking that a variable is independent of its coefficient
instead of counting from the right. Evaluating the quotient of decimal numbers by the method of removing decimal points. Using two different factors (multiples of 10) to multiply the numerator and denominator. Conceptual Thinking that the decimals of numerator and denominator are independent numbers Adding fractions with different denominators Choosing the smallest multiple of denominators to be the LCM Conceptual Associating the term ‘Lowest’ as it appears in the abbreviation LCM to mean just a ‘small’ multiple. Dividing fractions reciprocals of both multiplicand and multiplied Adding or subtracting numerators and denominators separately. Procedural i. Confusing the rules of multiplication of fractions to the rules of division. ii. Treating numerators and denominators as independent numbers Percentages Changing Percentages into decimal or fractional form Multiplied by 100 instead of dividing by 100 Conceptual Considering the term ‘Percent’ to mean 100 parts instead of th part
Numbers Adding and subtracting of whole numbers using the vertical method Making incorrect alignment of digits Skipping ‘carrying figures’ the arrangement of numbers for easier computation Procedural i. Does not know the importance of aligning digits as per respective place values ii. Carelessness mistake iii. Confusing the rule of commutative in addition to apply it on subtraction Multiplication and Division of whole numbers by using long methods. Obtaining incorrect products of digits in top and bottom numbers Skipping carrying figures. twice when required to do it once Bringing down two digits at once in the algorithm of Procedural i. Had no clear understanding of multiplication tables ii. Carelessness iii. Thought that borrowing is done continuously iv. Did not know the idea of a number going ‘zero times’ into another number.
long division method. Mixed Operations Computation of mathematical expression involving more than one of the arithmetic operations. Computing from left to right irrespective of the BODMAS rule. Procedural Had no understanding of the BODMAS rule. Subtraction of integers Computation involving integers and ‘a minus sign’ both ‘a negative sign’ and a sign’ to mean the same thing Omission of ‘a negative sign’ Making incorrect generalizations of the rules for addition and Multiplication of integers. Conceptual i. Confusing a negative sign from a minus sign ii. Confusing the rules of multiplication to the rules of addition. 4.7 Remedial Approach In objective five the sought to develop an appropriate remedial approach for addressing the noted errors and misconceptions committed by Form I entrants in
Marker pens Flip chart Computer graphical displays Conceptual knowledge, rules and procedures All numbers that are less than zero are called negative numbers. On the number line, negative numbers are to the left of zero. Numbers to the right of zero are positive numbers. Negative numbers are denoted by a symbol (–) written just in front of a digit. It is also important to note that the (–) sign placed between two numbers is a minus sign. When this sign is placed just in front of a number it is a negative sign. Therefore; 2 + (–7) means 2 plus negative 7 –2 – 1 means negative 2 minus 1 5 – (–3) means 5 minus negative 3 Real-life examples can be of much help to illustrate the concept of negative numbers. Negative temperature: When water is very cold, the temperature reading on the thermometer is close to zero. When water freezes to form ice, its temperature is below zero, hence negative reading of temperature. Comparison of quantities: Brian scored 5 points less than Shedrack (–5), Meshack scored 2 points more than Fred (+2) Business: In business, capital is a zero point, profit money is positive, loss money is negative.
A certain number m is the opposite of negative nine. Express m in symbolic form. 6. Estimations What angle is made when a door is opened half way? What is the approximate sum of 6 + 11/12 ? Is the width of your thumb approximating 5mm? 7. Algebraic expressions and equations Is it true that 2y – y = 2. Give reason Is it true that 2y – y = 2y – 1y. Give reason If y = –2, find the of 7 – y Is it correct to say that if 3k – 3 , then 3k = –2 – 3? Give reason. Appendix D: Documentary for Respondents’ Performance in the PSLE or School- based Entrance Examination. Name of School: ……………......School Category…………………District… Number of Respondents…….. Number of Mathematics teachers at school………… Entrance Criterion: PSLE………School Based Examination………(Tick the appropriate)