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SECONDARYMATHEMATICSWORKSHOP

ENGLISH LANGUAGE

ENGLISH LANGUAGELEARNERSIN THEMATHEMATICSCLASSROOMENGLISH LANGUAGE

SAMPLE QUESTION 1Questions to help students rely on their own understanding,ask the following :DO YOU THINK THAT IS TRUE? WHY?DOES THAT MAKE SENSE TO YOU?HOW DID YOU GET YOUR ANSWER?DO YOU AGREE WITH THE EXPLANATION?

SAMPLE QUESTION : 2To promote problem solving, ask the following :

WHAT DO YOU NEED TO FIND OUT?WHAT INFORMATION DO YOU HAVE?WILL A DIAGRAM OR NUMBER LINE HELP YOU?WHAT TECHNIQUE COULD YOU USE?WHAT DO YOU THINK THE ANSWER WILL BE

SAMPLE QUESTION : 3Questions to encourage students to speak out, ask the followingWhat do you think about what said?Do you agree what I have said?Why?Or why not?Does anyone have the same answer but a different way to explain it?Do you understand what ?Are you confuse?

SAMPLE QUESTION : 4Question to check the students progress, ask the following:What have you found out so far?What do you notice about?What other things that you need to do?What other information you need to find out?Have you though of another way to solve the questions?

SAMPLE QUESTION : 5Question to help students when they get stuck,ask the followingWhat have you done so far?What do you need to figure out next?How would you say the questions in your own words?Could you try it the other way round?Have you compared your work with anyone else?

SAMPLE QUESTION : 6Question to make connection among ideas and application,Ask the following:What other problem does this remind you of?Can you give me an example of ?Can you write down the objective or aim?Can you write down the formulae?

EXAMPLE TO COMMUNICATECAN YOU REPEAT THAT PLEASE?HOW DO YOU SPELL________?WHAT DOES ____MEAN?CAN YOU GIVE ME AN EXAMPLE?Teacher : I am reading a book about amphibiansStudents : Can you repeat that please?Teacher : I said : Im reading a book on amphibiansStudents : How do you spell amphibians?Teacher : A-M-P-H-I-B-I-A-N-SStudents : What does amphibians mean?Teacher : It is an animal that is born in water but can live on landStudent : Can you give me an example?Teacher : A frog

KNOW YOUR KEY WORDSMORE THANLESS THANALTOGETHERAT FIRSTSUMDIFFERENTCOMPAREDIGITSFIND THE LENGTH /MASSPLACE VALUEWHOLE NUMBER

KNOW YOUR KEY WORDSORDINAL NUMBERSUBTRACTSUBTRACT 2 FROM 5GREATER THANLESS THANSHORT/SHORTER/SHORTESTTALL/TALLER/TALLESTARRANGE THE NUMBER FROM THE GREATEST TO THE SMALLESTARRANGE THE STRINGS FROM THE SHORTEST TO THE LONGERSTREAD THE QUESTIONS CAREFULLY

KNOW YOUR KEY WORDSLABEL THE FOLLOWINGEVALUATEHEAVY/HEAVIER/HEAVIESTNUMBER SEQUENCEHOW MUCH MONEY I LEFT?1 MORE THAN 103 LESS THAN 10HOW MANY MARBLE HAD SHE LEFT?HOW MUCH MORE MONEY JOHN HAVE THAN MARY?PRODUCT

KNOW YOUR KEY WORDSFACTORSMULTIPLES OF 2, 3NUMBER LINESPOSITIVE NUMBERNEGATIVE NUMBERINTEGERS3 TO THE POWER OF 2PRIME NUMBERVENN DIAGRAMINEQUALITIESMULTIPLY

KNOW YOUR KEY WORDSDIVIDEADD TWO NUMBER UP TO THREE DIGITSFACTIONMIXED NUMBERIMPROPER FRACTIONCONVERT THE FOLLOWING FRACTION TO DECIMALSEQUILATERALISOSCELESRIGHT ANGLE TRIANGLENUMBERATORDENOMINATOR

FACTORS AND MULTIPLESWe can write a whole number greater than 1 as a product of two whole numbers.E.g.18=1x1818=2x918=3x6Tip : Note that 18 is divisible by each of its factors.Therefore, 1, 2, 3, 6, 9 and 18 are called factors of 18.The common factors of two numbers are the factors that the numbers have in common. Factors of a number are whole numbers which multiply to give that number.E.g.Factors of 12: 1, 2, 3, 4, 6, 12Factors of 21: 1, 3, 7, 21The common factors of 12 and 21 are 1 and 3.

FACTORS AND MULTIPLESWhen we multiply a number by a non-zero whole number, we get a multiple of the number.E.g.1 x 3 = 31 x 5 = 5 2 x 3 = 62 x 5 = 10 3 x 3 = 9 Multiples 3 x 5 = 15 Multiples 4 x 3 = 12 of 3 4 x 5 = 20 of 5 5 x 3 = 155 x 5 = 25Therefore, the multiples of 3 are 3, 6, 9, 12, 15, andthe multiples of 5 are 5, 10, 15, 20, 25, The first three common multiples of 4 and 6 are 12, 24 and 36.The common multiple of two numbers is a number that is a multiple of both numbers.E.g.Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, ...Multiples of 6 are 6, 12, 18, 24, 30, 36, ...

PRIME NUMBERS,PRIME FACTORISATIONA prime number is a whole number greater than 1 that has exactly two different factors, 1 and itself. E.g. 5 = 1 x 5

Since 5 has no other whole number factors other than 1 and itself, it is a prime number.A composite number is a whole number greater than 1 that has more than 2 different factors. The numbers 2, 3, 5, 7, 11, 13, 17, are prime numbers.E.g.6 = 1 x 66 = 2 x 3

Therefore, 6 is a composite number.4 Factors

PRIME NUMBERS,PRIME FACTORISATIONThe numbers 4, 6, 8, 9, 10, 12, 14, 15, 16, are composite numbers. In other words, all whole numbers greater than 1 that are not prime numbers are composite numbers.Tip: 0 and 1 are neither prime nor composite numbers.E.g. The factors of 18 are 1, 2, 3, 6, 9, and 18. The prime factors of 18 are 2 and 3.Prime factors are factors of a number that are also prime.We can use either the factor tree or repeated division to express a composite number as a product of its prime factors. The process of expressing a composite number as the product of prime factors is called prime factorisation.

PRIME NUMBERS, PRIME FACTORISATIONWORKED EXAMPLE 1:Express 180 as a product of prime factors. 180 2 x 90

2 x 2 x 45

2 x 2 x 3 x 15

2 x 2 x 3 x 3 x 5

Therefore, 180= 2 x 2 x 3 x 3 x 5= 22 x 32 x 5SOLUTION:Method I (Using the Factor Tree)Steps:Write the number to be factorised at the top of the tree.Express the number as a product of two numbers.Continue to factorise if any of the factors is not prime.Continue to factorise until the last row of the tree shows only prime factors.A quicker and more concise way to write the product is using index notation.

PRIME NUMBERS, PRIME FACTORISATIONWORKED EXAMPLE 1:Express 180 as a product of prime factors.

2 1802 903 453 155 5 1

Therefore, 180= 2 x 2 x 3 x 3 x 5= 22 x 32 x 5SOLUTION:Method II (Using Repeated Division)Steps:Start by dividing the number by the smallest prime number. Here, we begin with 2.Continue to divide using the same or other prime numbers until you get a quotient of 1.The product of the divisors gives the prime factorisation of 180.

INDEX NOTATIONIf the factors appear more than once, we can use the index notation to represent the product.E.g. 3 x 3 x 3 x 3 x 3 = 35In index notation, 3 is called the base and the number at the top, 5 is called the index.35 is read as 3 to the power of 5E.g. 2 x 2 x 2 x 5 x 5 = 23 x 5235indexbaseThe answer is read as 2 to the power of 3 times 5 to the power of 2.

HIGHEST COMMON FACTOR (HCF)The largest common factor among the common factors of two or more numbers is called the highest common factor (HCF) of the given numbers.E.g.Factors of 12 are 1, 2, 3, 4, 6, and 12.Factors of 18 are 1, 2, 3, 6, 9, and 18.Another method to find the HCF of two or more numbers is by using prime factorisation which is the more efficient way. The common factors of 12 and 18 are 1, 2, 3 and 6. The highest common factor (HCF) of 12 and 18 is 6.We can also repeatedly divide the numbers by prime factors to find the HCF.

HIGHEST COMMON FACTOR (HCF)WORKED EXAMPLE 1:Find the highest common factor of 225 and 270.SOLUTION:Therefore, the HCF of 225 and 270 is 45.225 =32x 52270 = 2 x33 x 5HCF =32 x 5 = 45Find the prime factorisation of each number first.To get the HCF, multiple the lowest power of each common prime factor of the given numbers.

LOWEST COMMON MULTIPLE (LCM)The smallest common multiple among the common multiples of two or more numbers is called the lowest common multiple (LCM) of the given numbers.E.g.Multiples of 8: 8, 16, 24, 32, 40, 48, ...Multiples of 12: 12, 24, 36, 48, 60, ...Another method to find the LCM of two or more numbers is by using prime factorisation which is the more efficient way. The common multiples of 8 and 12 are 24, 48, ... The lowest common multiple (LCM) of 8 and 12 is 24.We can also repeatedly divide the numbers by prime factors to find the LCM.

LOWEST COMMON MULTIPLE (LCM)WORKED EXAMPLE 1:Find the lowest common multiple of 24 and 90.SOLUTION:Therefore, the LCM of 24 and 90 is 360.24= 23x 390= 2x 32 x 5LCM= 23x 32 x 5 = 360To get the LCM, multiple the highest power of each set of common prime factors. Also include any uncommon factors

SQUARES AND SQUARE ROOTSWhen a number is multiplied by itself, the product is called the square of the numberThe numbers whose square roots are whole numbers are called perfect squares.5 is the positive square root of 25.Tip :22 = 4and 4 = 232 = 9and 9 = 342 = 16and 16 = 4

E.g.5 x 5 = 25or52 = 25E.g. 25 = 5E.g.1, 4, 9, 16, 25, ... are perfect squares.

SQUARES AND SQUARE ROOTSWORKED EXAMPLE 1:Using prime factorisation, find the square root of 5184.

5184= 26 x 342 51845184= 26 x 34 2 2592= 23 x 322 1296= 8 x 92 648= 722 3242 1623 813 273 93 3 1SOLUTION:

CUBES AND CUBE ROOTSWhen a number is multiplied by itself thrice, the product is called the cube of the numberThe numbers whose cube roots are whole numbers are called perfect cubes.125 is the cube of 5 and 5 is the cube root of 125.Tip :23 = 8and 8 = 233 = 27and 27 = 343 = 64and 64 = 4

E.g.5 x 5 x 5 = 125or53 = 125E.g. 125 = 5E.g.1, 8, 27, 64, 125, ... are perfect cubes.

CUBES AND CUBE ROOTSWORKED EXAMPLE 1:Using prime factorisation, find the cube root of 1728.

1728= 26 x 332 17281728= 26 x 33 2 864= 22 x 32 432= 4 x 32 216= 122 1082 543 273 93 3 1 SOLUTION:

REAL NUMBERSNumbers with the negative sign ( - ) are called negative numbers.E.g.-1, -2, -3, -4, -5, ...Positive integers are whole numbers that are greater than zero.Zero is an integer that is neither positive nor negative.Integers refer to whole numbers and negative numbers.E.g...., -3, -2, -1, 0, 1, 2, 3, 4, ... are integers.E.g.1, 2, 3, 4, 5, ... Negative integers are whole numbers that are smaller than zero.E.g.-1, -2, -3, -4, -5, ...

REAL NUMBERSA number line showing integers is shown below:Every number on the number line is greater than any number to its left.The arrows on both ends of the number line show that the line can be extended on both ends.E.g.2 is greater than -3 and is denoted by 2 > -3We can also write -3 is smaller than 2 and is denoted by -3< 2 -5 -4 -3 -2 -1 0 1 2 3 4 5Negative Integers Positive Integers-5 -4 -3 -2 -1 0 1 2 3 4 5

REAL NUMBERS>, and < are called inequality signs.1, 2, 3, 4, 5, 6, 7, ... are called natural numbers. The natural numbers are also called positive integers.> means is greater than< means is smaller than> means is greater than or equal to< means is smaller than or equal toE.g.|2| = 2, |0| = 0, |-2| = 2The numerical or absolute value of a number x, denoted by |x|, is its distance from zero on the number line.Since distance can never be negative, the numerical or absolute value of a number is always positive.

ADDITION OF INTEGERSRules for adding two integers:

Sign of numbersMethodBoth numbers have the same signs

(+a) + (+b) = +(a + b)(-a) + (-b) = -(a + b)Add the numbers while ignoring their signs.Write the sum using their common sign.E.g. (+3) + (+8) = +11 = 11E.g. (-3) + (-8) = -(3 + 8) = -11Both numbers have different signs

(+a) + (-b) = +(a b) if a>b(+a) + (-b) = - (b a) if b>a(-a) + (+b) = - (a b) if a>b(-a) + (+b) = +(b a) if b>aSubtract the numbers while ignoring their signs.The answer has the same sign as the number having the larger numerical value.E.g. 12 + (-4) = 12 4 = 8E.g. 5 + (-11) = -(11 5) = -6E.g. -8 + 3 = -(8 3) = -5E.g. -9 + 15 = 15 9 = 6

SUBTRACTION OF INTEGERSTo subtract integers, change the sign of the integer being subtracted and add using the addition rules for integers.a b = a + (-b)E.g. 8 15 = 8 + (-15) = -(15 8) = -7 -11 7 = -11 + (-7) = -(11 + 7) = -18-6 (-10) = -6 + 10 = 10 6 = 4 3 (-13) = 3 + 13 = 16

MULTIPLICATION OF INTEGERSRules for multiplying integers:

MultiplicationExamples(+a) x (+b)= +(a x b)(- a) x (- b)= +(a x b)(+a) x (- b)= - (a x b)(- a) x (+b)= - (a x b)3 x 4 = 12(-5) x (-6) = +(5 x 6) = 308 x (-3) = -(8 x 3) = -24(-12) x 4 = -(12 x 4) = -48

Rules for signs:( + ) x ( + ) = ( + )The product of two positive integers is a positive integer( - ) x ( - ) = ( + )The product of two negative integers is a positive integer( + ) x ( - ) = ( - )( - ) x ( + ) = ( - )The product of a positive and a negative integer is a negative integer.

DIVISION OF INTEGERSRules for dividing two integers:

DivisionExamples(+a) (+b)= +(a b)(- a) (- b)= +(a b)(+a) (- b)= - (a b)(- a) (+b)= - (a b)16 2 = 8(-20) (-5) = +(20 5) = 436 (-4) = -(36 4) = -9(-24) 8 = -(24 8) = -3

Rules for signs:( + ) ( + ) = ( + )The quotient of two positive integers is a positive integer( - ) ( - ) = ( + )The quotient of two negative integers is a positive integer( + ) ( - ) = ( - )( - ) ( + ) = ( - )The quotient of a positive and a negative integer is a negative integer.

RULES FOR OPERATING ON INTEGERSAddition and multiplication of integers obey the Commutative Law.Addition and multiplication of integers obey the Associative Law.E.g. 1 2 + (-10) = (-10) + 2 = -8E.g. 2 2 x (-10) = (-10) x 2 = -20Commutative Law of Addition of Integers:a + b = b + aCommutative Law of Multiplication of Integers:a x b = b x aAssociative Law of Addition of Integers:(a + b) + c = a + (b + c)Associative Law of Multiplication of Integers:(a x b) x c = a x (b x c)E.g. 1 [3 + (-5)] + 8 = 3 + [(-5) + 8] = 6E.g.2 [3 x (-5)] x 8 = 3 x [(-5) x 8] = -120

RULES FOR OPERATING ON INTEGERSMultiplication of integers is distributive overa) additionb) subtractionThe order of operation on integers is the same as those for whole numbersE.g. 1 -2 x (-3 + 5) = -2 x (-3) = (-2) x 5 = -4E.g. 2 -2 x (-8 + 6) = -2 x (-8) = (-2) x 6 = 28Distributive Law of Multiplication over Addition of integers:a x (b + c) = (a x b) + (a x c)Distributive Law of Multiplication over Subtraction of Integers:a x (b c) = (a x b) (a x c)Order of operationsSimplify expressions within the brackets first.Working from left to right, perform multiplication or division before addition or subtraction.

RULES FOR OPERATING ON INTEGERSWORKED EXAMPLE 1:Evaluate each of the following.25 36 (-4) + (-11)(-10) (-6) + (-9) 3{-15 [15 + (-9)]2} (-3)(3 5)3 x 4 + [(-18) + (-2)] (-3)2SOLUTION:

25 36 (-4) + (-11)= 25 (-9) + (-11)= 25 + 9 11= 23

RULES FOR OPERATING ON INTEGERSSOLUTION:

(-10) (-6) + (-9) 3= (-10) (-6) + (-3)= -10 + 6 3= -7(3 5)3 x 4 + [(-18) + (-2)] (-3)2= (-2)3 x 4 + (-18 2) (-2)2= (-2)3 x 4 + (-20) (-2)2= (-8) x 4 + (-20) 4= -32 + (-5)= -32 5= -37{-15 [15 + (-9)]2} (-3)= [-15 (15 9)2] (-3)= (-15 62) (-3)= (-15 36) (-3)= (-51) (-3)= 17

INTRODUCTION TO ALGEBRAUsing Letters to Represent NumbersA variable is a letter that is used to represent some unknown numbers/quantity .E.g. There are n apples in a bag.If there are 5 bags, then the total number of apples is 5 x n.5 x n can be any whole number value.It can be 5, 10, 15, depending on the value of n. i.e. n = 1, 2, 3, Here, n is called the variable and 5 x n is called the algebraic expression. E.g. x, y, z, a, b, P, Q, are variablesIn algebra, we use letters (e.g. x, y, z, a, b, P, Q, ) to represent numbers.

INTRODUCTION TO ALGEBRAUsing Letters to Represent NumbersE.g. 3x + y, a2 ab, 2x2 + 3x 4.An algebraic expression is a collection of terms connected by the signs +, -, x, .Tip:An algebraic expression does not have an equal sign (=).An algebraic expression is different from an algebraic equation. An equation is a mathematical statement that says that two expressions are equal to each otherE.g. A = lb is an equation. A and lb are algebraic expressions.

ALGEBRAIC NOTATIONSWe use the signs +, -, x, and = in Algebra the same way as Arithmetic. The examples below show how we rewrite mathematical statements as algebraic expressions.

WordsAlgebraic ExpressionAdd a to bSum = a + b = b + a Subtract c from dDifference = d cNote: d c c d Multiple e by fProduct = e x f = f x e = efDivide g by h(h 0)Quotient = g h = g/h

ALGEBRAIC NOTATIONSMore examples below show how we rewrite mathematical statements as algebraic expressions.

WordsAlgebraic ExpressionAdd 3 to the product of p and qpq + 3 The total cost of x books and y magazines if each book cost $4 and each magazine costs $5.Cost of x books= $4 x x= $4xCost of y magazines= $5 x y= $5yTotal Cost= $(4x + 5y)

ALGEBRAIC NOTATIONSIn Algebra, we use the same index notation as in Arithmetic.Index NotationRecall: 5 x 5 x 5 = 53

53 is read as 5 to the power of 3

In Algebra,a x a = a2 (read as a squared)a x a x a = a3 (read as a cubed)a x a x a x a x a = a5 (read as a to the power of 5) baseindex

ALGEBRAIC NOTATIONSWORKED EXAMPLE 1:3x x 4y 6z2a x 3b x a5p 10q + 7s x 2SOLUTION:3x x 4y 6z= 3 x x x 4 x y 6z= 12xy 6z= 12xy/6z= 2xy/z

2a x 3b x a= 6a2b

5p 10q + 7s x 2= 5p/10q + 14s= p/2q + 14s

ALGEBRAIC NOTATIONSWORKED EXAMPLE 2:Subtract 3 from the sum of 5a and 4b.Add the product of c and d to the cube of e.Multiple 2 to the quotient of f divided by g.SOLUTION:Sum of 5a and 4b= 5a + 4bRequired expression= 5a + 4b 3 (ans)

Product of c and d= c x d = cdCube of e = e x e x e = e3Required expression= cd + e3 (ans)Quotient of f divided by g = f/gRequired expression = 2 x f/g = 2f/g (ans)

EVALUATION OF ALGEBRAIC EXPRESSIONS AND FORMULATo evaluate an algebraic express, we substitute a number for the variable and carry out the computation.WORKED EXAMPLE 1:3a + 2b 4c,a(2b c) 3b2,a/b (a+b)/ac,given that a = 4, b = 2, c = -3.SOLUTION:3a + 2b 4c= 3(4) + 2(2) 4(-3)= 12 + 4 + 12= 28

EVALUATION OF ALGEBRAIC EXPRESSIONS AND FORMULA

a/b (a+b)/ac = 4/2 (4+2)/4(-3)= 2 (6/-12)= 2 + = 2 SOLUTION:a(2b c) 3b2= 4[2(2) (-3)] 3(2)2= 4(4+3) 3(4)= 4(7) 12= 28 12= 16

ALGEBRAIC EXPRESSIONSa)Find the total cost of m cups and n plates if each cup cost $3 and each plate costs $4. 1 cup = $3 1 plate= $4m cups= m x $3n plates= n x $4= $3m= $4nTotal Cost = $3m + $4n = $(3m + 4n) (ans)b)Find the total cost of 7 bars of wafers at p cents each and q packets of sweets at $1 each. 1 bar = p cents 1 packet= 100 cents 7 bars= 7 x p centsq packets= q x 100 cents= 7p cents= 100q centsTotal Cost = 7p cents + 100q cents = (7p + 100q) cents

ALGEBRAIC EXPRESSIONSc)John has $100, He bought n comic books at $9 each. How much money had he left? 1 book= $9 n books= n x $9= $9nAmt left = $100 - $9n = $(100 9n) (ans)d)The cost of 3 caps is $x. Find the cost of 5 caps. Each cap costs the same. 3 caps= $x 1 cap= $x 3 = $x/3 5 caps= $x/3 x 5 = $5x/3 (ans)

RATIONAL NUMBERSA rational number is any number that can be written as a ratio of two integers. In other words, a number is a rational number if it can be written as a fraction where both the numerator and denominator are integers. E.g.3 and -6 are rational nos. since 3 = 3/1 and -6 = -6/1 E.g.0.5 and 3.2 are rational nos. since 0.5 = 5/10 and 3.2 = 32/10 E.g. -3/5, , 5/3, 12/3, are rational numbers.A rational number can be written in the form a/b where a and b are integers and b 0All integers are rational numbers since each integer, n can be written as n/1. Most decimals can be expressed as rational numbers too.

RATIONAL NUMBERSRecall:

5/10 = 5/10 and are equivalent fractions.

is said to be in its simplest form or in its lowest terms. Here, the numerator and denominator have no common factors. 31/5 is called a mixed number. It represents the sum of whole number and a proper fraction. E.g. , 3/7, and 5/9 are proper fractions.32/10 = 16/5 = 31/5

16/5 is called an improper fraction. The numerator is greater than or equal to the denominator in an improper fraction.A proper fraction has its numerator smaller than its denominator.

ADDITION AND SUBTRACTION OF RATIONAL NUMBERSTo add or subtract rational numbers, express the rational numbers as equivalent fractions in the same denominators first WORKED EXAMPLE 1:61/6 23/4,b) (-51/4) + (-12/5) + (-) SOLUTION:61/6 23/4 = (5 2) + (11/6 )= 3 + (7/6 )= 3 + (14/12 9/12)= 3 + 5/12= 35/12SOLUTION:(-51/4) + (-12/5) + (-)= - (51/4 + 12/5 + )= - (55/20 + 48/20 + 10/20)= - (563/20)= - (5 + 33/20)= -83/20

MULTIPLICATION AND DIVISION OF RATIONAL NUMBERSTo multiply two rational numbers:a)Convert all mixed numbers to improper fractions first.b)Simplify the fractions first by crossing out the common factors of the numerators and denominators.c)Multiply the numerators, then the denominators. d)Reduce answer to its simplest form.a/b x c/d = a x c/b x d , where a, b, c, d are integers and b 0, d 0To divide a rational number by another number :a)Convert all mixed numbers to improper fractions first .b)Invert the second fraction by interchanging its numerator and denominator .c)Multiply the numerators, then the denominators . d)Reduce answer to its simplest form.a/b c/d = a/b x d/c = a x d/b x c , where a, b, c, d are integers andb 0, c 0, d 0

MULTIPLICATION AND DIVISION OF RATIONAL NUMBERSWORKED EXAMPLE 1:(-21/2) x 31/5,c) (1/5 1/3) (-1/4 x 2/9)b) (-2/11) (-10/33) SOLUTION:(-21/2) x 31/5 = - 5/2 x 16/5= -8

(-2/11) (-10/33)= - 2/11 x (-33/10)= 3/5SOLUTION:(3/15 5/15) (-1/4 x 2/9)= - 2/15 (- 1/18)= - 2/15 x (- 18/1)= 36/15= 22/5

TECHNIQUE & STRATEGIES IN SOLVING MATHEMATICS WORD PROBLEM SUM1/2

SHARING HOW TO GO ABOUT TEACHING

SYNOPSISIn solving math problem sum at secondary school level, it is widely acknowledged that heuristics strategies play a major role. By using suitable heuristics, it could greatly enhance pupils problem solving performance. Heuristics, referred to the method or strategies of achieving a solution to a given problem sumModel Drawing is just one of the methods that can be used. Of course there are also various strategies that can be used. Through-out this seminar, I will share with you the various types of strategies used to solve word problems sum.The reason why Model drawing is used is because it is one of the most common heuristics used to solve word problems in Mathematics. It is recognized internationally as an effective way for young children to solve word problems and to be exposed early to algebraic concepts.

TECHNIQUES AND OTHER HEURISTICS STRATEGIES:Guess & Check methodMaking a TableMake a List (Listing method)Draw a PictureFind a PatternWorking BackwardsModel Drawing

COMMON DIFFICULTIES IN MATHEMATICAL PROBLEM SOLVINGInability to read the problemLack of comprehension of the problem posedLack of strategy knowledgeInappropriate strategy usedInability to translate the problem into a mathematical formComputational error

4 - Step in solving problem sum:I dentify the problem(what is the questions exactly asking for?) D evise a plan(model method)E xecute the plan(work it out)A nswer check(number sense)

4 - Step in solving problem sum:I dentify the problemAfter reading the problem sum, what is the questions exactly asking for?D evise a planBy drawing models, pupils can represent the mathematical relationships in a problem pictorially. This helps them understand the problem and plan the steps for the solution

4 - Step in solving problem sum:E xecuting the planIn your plan, you might required to use one or more of the strategies (heuristics) listed below to help you solve the words problem sum.

A nswer checkAnswer must be check to be able to satisfy the condition of the question.

Strategy 5 : Find a PatternQ1.A few children had to share a plate of chicken wings. If each of them took 5 chicken wings, there would be 4 left. In the end, they decided to take 6 chicken wings each, leaving 1 chicken wing on the plate. How many children shared the chicken wings? What was the original number of chicken wings on the plate?STEP 1 : Draw a Table and determine the information that needs to be found.STEP 2 : I find the patterns present in the data to complete the table.From the table, I can see that number 19 satisfies both conditions of the question.Ans: There were 3 children sharing the chicken wings. There were 19 chicken wings.12345Checking:( 3 x 5 ) + 4 = 19( 3 x 6 ) + 1 = 19 ------- Correct914192429713192531

No. of ChildrenMultiple of 5 and add 4

Multiple of 6 and add 1

Strategy 5 : Find a PatternQ2.Mr. Tom wanted to distribute his stamps equally among a few of his students. If he were to give each of them 5 stamps, he would have 4 left. If he gave each of them 6 stamps, he would have 1 left. How many students and stamps did he have?STEP 1 : Draw a Table and determine the information that needs to be found.STEP 2 : I find the patterns present in the data to complete the table.From the table, I can see that number 19 satisfies both conditions of the question.Ans:There were 3 students sharing the stamps. There were 19 stamps.12345Checking:( 3 x 5 ) + 4 = 19( 3 x 6 ) + 1 = 19 ------- Correct914192429713192531

No. of StudentsMultiple of 5 and add 4

Multiple of 6 and add 1

Strategy 5 : Find a PatternQ3.A Christmas tree was decorated with flashing light bulbs. The red bulbs flashed every 2 minutes. The yellow bulbs flashed every 3 minutes and the blue bulbs flashed every 4 minutes. At 8pm, all the light flashed simultaneously. Figure out the next time when all the bulbs will flash together?STEP 1 : Draw a Table and determine the information that needs to be found.STEP 2 : I find the patterns present in the data to complete the table.Looking at the pattern above, the starting time is 8.00pm.Ans:The next time all the bulbs will flash together is at 8.12pm.8.008.028.048.068.088.108.12Next time the bulbs will flash together 8pm + 12mins8.008.038.068.098.128.158.188.008.048.088.128.168.208.24

Red (every 2 mins)Yellow (every 3 mins)Blue (every 4 mins)

Strategy 6 : Working BackwardsQ1.On my way to the shopping centre, I found that I did not bring enough money for my shopping. I then went to the bank to withdraw $100. Next , I bought a pair of shoes for $40. Later, I paid for a T-shirt with half of the money I had left. I was left with $65. How much money did I have before I visited the bank to withdraw the money?To find out how much money I had at first, I have to work backward by starting at the end and undoing each step in reverse order.You can draw a flow chart or an arrow to show what happened.Step 1Step 2Step 3Amount IWithdrew $100Spent $40 on aSpent half of theAmountstarted withfrom bankpair of shoes money on T-Shirtleft?+ $100- $40 2$65

Strategy 6 : Working BackwardsQ1.contAns:I had $70 before I went to the bank to withdraw the money.Step 1Step 2Step 3Amount IWithdrew $100Spent $40 on aSpent half of theAmountstarted withfrom bankpair of shoes money on T-Shirtleft?+ $100- $40 2$65Amount I started withAmount before Step 2Amount before Step 3Amount left

Strategy 6 : Working BackwardsQ2.Alice, Billy and John each bought some drink. John poured his drink in a jug. Alice then added 0.7 of drink into the jug. After that, Billy added enough drink to double the amount in the jug. All of them drank 1.2 of it, leaving 1.3 in the jug. How much drink did John bring ?Ans: John brought 0.55 of drink.1.32.51.250.55+ 1.2 2- 0.7Amount left in the jug

Strategy 6 : Working BackwardsQ3.One evening, Lily baked some chocolate cookies. She put 44 chocolate cookies in a bag and packed another 24 in a tin. Then she divided the remainder equally among herself and 2 of her friends. She kept her share of 14 chocolate cookies in a jar. How many chocolate cookies did she bake that evening?Ans: She baked 110 chocolate cookies.144266110x 3+ 24+ 44Amount left in the jar

UNDERSTANDING THE 8 DIFFERENTS MODEL. (MODEL DRAWING)The model drawing/diagram is a very important strategy in secondary school mathematics. Using it correctly, a child will be able to solve many types of challenging problems sum easily.The 8 different types of model drawing is very useful as it can be used a Diagnostic Tool.The trainers or the teachers can straight away identified what kinds or types of problem sum a child has instead of spending time figuring out where is the weakness of the child.Therefore with constant practice on the model drawing it not only reinforce the understanding of the questions it also develop skill and the process thinking skill in solving word problems.

The 8 Different Models can be used in the following types of problem sum :ADDITIONSUBTRACTIONCOMPARISIONMULTIPLICATIONDIVISION1 STEP PROBLEM SUM2 STEP PROBLEM SUM3 STEP PROBLEM SUMCHALLENGING PROBLEM SUM INVOLVING BEFORE AND AFTER MODEL CONCEPT

Model Drawing : 2Q1.Jim and his brother share a sum of $150.If Jim gets $50 more than his brother.How much money do Jim and his brother get?$1501 Unit$50Step 1 :Look out for KEY PERSON (REFERENCE POINT) Key person or reference point has only 1 UNIT ( )

Step 2 :In this case, the key person is his brother.

Step 3 :Draw the MODEL drawing of his brother first as 1 unit.

BrotherJim

Model Drawing : 2Q1.ContAns:Therefore, Jim gets $100 and his brother gets $50.Step 4 :After finding your 1 unit, look back at the MODEL DRAWING. The one that has 1 unit is Brother. Step 5 :As for Jim, look at the model drawing it consist of 1 unit + $50$1501 Unit$50BrotherJim$150 - $50 = $100$100 2 = $50 1 Unit ( ) / Brother $50 + $50 = $100Therefore, the brother get $50$50

Model Drawing : 2Q2.Peter and David share a total of 300 sweets.If David gets 60 more than Peter.How many sweets do Peter and David get?3001 Unit60Step 1 :Identified KEY PERSON that has only 1 UNIT ( )CAN YOU FIGURE IT OUT?

Step 2 :DRAW THE MODEL

PeterDavid300 60 = 240240 2 = 1201 unit ( ) / Peter 120 + 60 = 180 DavidAns:Peter gets 120 sweets and David gets 180 sweets.120

Hand-On : Exercise 2Q1.$400 is to be shared between Susan and Mary.If Susan gets $20 more than Mary.How much money do Susan and Mary get?$4001 Unit$20Step 4 :After finding your 1 unit, look back at the MODEL DRAWING. The one that has 1 unit is Mary. MarySusan$400 $20 = $380$380 2 = $190 1 unit ( ) / Mary $190 + $20 = $210 Susan$190Therefore, Mary gets $190Step 5 : As for Susan, look at the model drawing it consist of 1 unit + $20Ans:Therefore, Mary gets $190 and Susan gets $210.

Hand-On : Exercise 2Q2.Amy and John have 240 stickers altogether. If Amy has 80 stickers more than John. How many stickers does John have?2401 Unit80Step 4 :After finding your 1 unit, look back at the MODEL DRAWING. The one that has 1 unit is John. JohnAmy240 80 = 160160 2 = 80 1 unit ( ) / John 80 + 80 = 160Amy80Therefore, John gets 80Step 5 : As for Amy, look at the model drawing it consist of 1 unit + 80Ans:Therefore, John gets 80 and Amy gets 160.

Model Drawing : 3Q1.Sharon and Janet share a total of 180 beads.If Janet gets 30 bead less than Sharon.How many beads do Sharon and Janet get?1801 Unit30Step 1 :Look out for KEY PERSON (REFERENCE POINT) Key person or reference point has only 1 UNIT ( )

Step 2 :In this case, the key person is Sharon.

Step 3 :Draw the MODEL drawing of Sharon as 1 unit.

SharonJanetAns:Therefore, Sharon gets 105 and Janet gets 75.180 + 30 = 210210 2 = 105 1 unit ( ) / Sharon 105 - 30 = 75Janet105

Hand-On : Exercise 3Q1.$150 is shared between Judy and Susan.If Susan gets $30 less than Judy.How much money do Judy and Susan get?$1501 Unit$30JudySusanAns:Therefore, Judy gets $90 and Susan gets $60.$150 + $30 = $180$180 2 = $90 1 unit ( ) / Judy $90 - $30 = $60 Susan$90

Hand-On : Exercise 3Q2.David and John share a total of 360 stamps.If John gets 40 stamps less than David.How many stamps do David and John get?3601 Unit40DavidJohnAns:Therefore, David gets 200 and John gets 160.360 + 40 = 400400 2 = 200 1 unit ( ) / David 200 - 40 = 160 John200

Model Drawing : 4Q1.$300 is to be shared between Jason and Kevin. If Kevin gets twice as much as Jason. How much money do Jason and Kevin get?$3001 UnitJasonKevin$300 3 = $100 1 unit ( ) / Jason$100 x 2 = $200 Kevin$100Ans:Jason and Kevin each get $100 & $200 respectively.

Model Drawing : 4Q2.Ken and Joseph share a sum of $250.If Ken gets 4 times as much as Joseph.How much money do Ken and Joseph get?$2501 UnitJosephKen$250 5 = $50 1 unit ( ) / Joseph$50 x 4 = $200 Ken$50Ans:Joseph and Ken each get $50 & $200 respectively.

Model Drawing : 4Q3.Aaron has 4 times as many stamps as Jimmy. If he has 24 stamps more than Jimmy. How many stamps does Aaron have?1 UnitJimmyAaron24 3 = 8 1 unit ( ) / Jimmy8 x 4 = 32 Aaron8Ans:Jimmy and Aaron each get 8 & 32 stamps respectively.Step 1 :Who is the KEY PERSON?

Step 2 :Draw the model drawing of that person first.

24

Model Drawing : 4Q4.Sarah has 50 more stickers than Jenny. If Sarah has thrice as many stickers as Jenny. How many stickers does Sarah have?1 UnitJennySarah50 2 = 25 1 unit ( ) / Jenny25 x 3 = 75 Sarah25Ans:Jenny and Sarah each get 25 & 75 stickers respectively.50

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