Mathematics 3º ESO - IES Jovellanos

Mathematics 3º ESO

María Isabel Muñoz Molina Juan Francisco Antona Blázquez

Real Instituto de Jovellanos

1st TERM Numbers Polynomials Equations

Numbers

E xpressing N um bers and Operations in E nglish 1. Numbers 1 2 3 4 5 6 7 8 9 10

one two three four five six seven eight nine ten

11 12 13 14 15 16 17 18 19 20

Cardinal numbers from 1 through 1 000 000 eleven 21 twenty-one 31 twelve 22 twenty-two 40 thirteen 23 twenty-three 50 fourteen 24 twenty-four 60 fifteen 25 twenty-five 70 sixteen 26 twenty-six 80 seventeen 27 twenty-seven 90 eighteen 28 twenty-eight 100 nineteen 29 twenty-nine 1 000 twenty 30 thirty 1 000 000

thirty-one forty fifty sixty seventy eighty ninety one hundred one thousand one million

Examples: 305

three hundred and five

754 1 507

seven hundred and fifty-four one thousand, five hundred and seven

2 328 1 800

two thousand, three hundred and twenty-eight one thousand, eight hundred or eighteen hundred

1 436 209

one million, four hundred thirty-six thousand, two hundred and nine

2 700 400

two million, seven hundred thousand, four hundred

(NOTE: The British use and before the last two figures, but the Americans usually leave the and out).

The number 0: Nought (UK English) Zero (US English) Oh (like the letter O): used when reading out numbers figure by figure (e.g. telephone numbers, credit card numbers, etc).

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E xpressing N um bers and O perations - 3 o E SO

2 Decimal numbers:

In English we use a full stop (.) instead of a comma (,) to separate the main part of a number from the decimal part: 0.1

nought point one (UK) or

0.75 1.263

nought point seven five (UK) one point two six three

zero point one (US) or

zero point seven five (US)

You can use a comma (,) in large numbers to separate the hundreds, thousands, and millions. For example, 5,074 means five thousand and seventy four. In UK English, spaces are sometimes used instead of commas (5 074).

Place value: In our decimal number system, the value of each digit depends on its place in the number. Each place is 10 times the value of the next place to its right. Millions Hundred Thousands Ten Thousands Thousands Hundreds Tens Units Tenths Hundredths Thousandths Ten Thousandths

1 000 000 100 000 10 000 1 000 100 10 1 0.1 0.01 0.001 0.0001

Ordinal numbers:

1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th

first second third fourth fifth sixth seventh eighth ninth tenth

11th 12th 13th 14th 15th 16th 17th 18th 19th 20th

Ordinal numbers from 1 through 1 000 000 eleventh 21st twenty-first 31st twelfth 22nd twenty-second 40th thirteenth 23rd twenty-third 50th fourteenth 24th twenty-fourth 60th fifteenth 25th twenty-fifth 70th sixteenth 26th twenty-sixth 80th seventeenth 27th twenty-seventh 90th eighteenth 28th twenty-eighth 100th nineteenth 29th twenty-ninth 1 000th twentieth 30th thirtieth 1 000 000th

D pto. M atemáticas. IES Jovellanos. 2011

thirty-first fortieth fiftieth sixtieth seventieth eightieth ninetieth one hundredth one thousandth one millionth

E xpressing N um bers and O perations - 3 o E SO

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2. Operations Operation symbols +

plus

− × or ·

minus times

/ or :

divided by

= 6 =

equals or is equal to is not equal to

< >

is less than is greater than

≤ ≥

is less than or equal to is greater than or equal to

Addition (to add): 2+3 =5

Two plus three equals five

The answer is called the sum or total. Subtraction (to subtract): 7−3 =4

Seven minus three equals four

The answer is called the difference. Multiplication (to multiply): 5 · 3 = 15 Five times three equals fifteen or Five multiplied by three. . . The answer is called the product. Division (to divide): 8:2=4

9

2

1

4

Eight divided by two equals four 9: 2: 4: 1:

dividend divisor quotient remainder

✁✃✁✃✁✃✁✃✁✃✁✃✁✃


Rational N um bers 1. Natural Numbers The natural numbers are the counting numbers from one to infinity. We use the letter N to refer to the set of all natural numbers: N = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, . . .} The set of natural numbers is endless. As there is no largest counting number, we say that the set of natural numbers is infinite. The factors of a natural number are all the natural numbers which divide exactly into it, leaving no remainder. For example, the factors of 10 are: 1, 2, 5 and 10. 5 is a factor of 20 because 5 divides exactly into 20; we can also say that 20 is divisible by 5. The even numbers are the natural numbers which are divisible by 2. The sequence of even numbers is: 2, 4, 6, 8, 10, 12, 14 . . . and so on. The odd numbers are the natural numbers which are not divisible by 2. The sequence of odd numbers is: 1, 3, 5, 7, 9, 11, 13 . . . and so on. A prime number is a natural number which has exactly two distinct factors, itself and 1. For example, 17 is a prime number since it has only two factors, 1 and 17. A composite number is a natural number which has more than two factors. For example, 26 is a composite number since it has more than two factors: 1, 2, 13 and 26. Notice that these definitions indicate that one (1) is neither prime nor composite. Every composite number can be factorised as a product of prime factors in one and only one way (apart from order). For example,

72 = 2 · 2 · 2 · 3 · 3

or

72 = 23 · 32

(in exponent form).

A multiple of any natural number is obtained by multiplying it by another natural number. For example, the multiples of 3 are: 3, 6, 9, 12, 15, 18,. . . and these are obtained by multiplying 3 by each of the natural numbers in turn: 3 · 1 = 3,

3 · 2 = 6,

3 · 3 = 9,

1

3 · 4 = 12 etc.

R ational N um bers - 3 o E SO

2

Example 1 a) Find the largest multiple of 9 less than 500. b) Find the smallest multiple of 11 greater than 1000. a) We begin dividing 500 by 9: 500 5

9 55

We have 5 remainder, so 9 ·55 is smaller than 500. Hence, as 9 · 55 = 495, the largest multiple of 9 less than 500 is 495.

b) We begin dividing 1000 by 11: 1000

11

10

90

We have 10 remainder, so 11 · 90 is smaller than 1000: 11 · 90 = 990. Hence, the smallest multiple of 11 greater than 1000 is 11 · 91 =1001.

The HCF (highest common factor) of two or more natural numbers is the largest factor which is common to both of them. For example: The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24. The factors of 40 are: 1, 2, 4, 5, 8, 10, 20, 40. Hence, the HCF of 24 and 40 is 8. In general, to find the HCF of two or more natural numbers, we write them as a product of prime factors in exponent form, and we multiply the common factors raised to the smallest power. For example: 24 = 23 · 3 and 40 = 23 · 5, so HCF is 23 = 8.

The LCM (lowest common multiple) of two or more natural numbers is the smallest multiple which is common to both of them. For example: The multiples of 6 are: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, . . . The multiples of 8 are: 8, 16, 24, 32, 40, 48, 56,. . . Hence, the LCM of 6 and 8 is 24. In general, to find the LCM of two or more natural numbers, write them as a product of prime factors in exponent form, and multiply common and non common factors raised to the highest power. For example: 6 = 2 · 3 and 8 = 23 , so LCM is 23 · 3 = 24.



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Exercises - Set A 1. Find the largest multiple of 7 which is less than 1000. 2. Find the smallest multiple of 13 which is greater than 1000. 3. Find the largest multiple of 17 which is less than 2000. 4. Find the smallest multiple of 15 which is greater than 10 000. 5. Find the HCF of 24, 72, 120. 6. Find the LCM of 12, 18, 27. 7. Three bells chime at intervals of 4, 5 and 6 seconds respectively. If they all chime at the same instant, how long before they all chime together again?

2. Integers and order of operations The negative whole numbers, zero and the positive whole numbers (not fractions or decimals) form the set of all integers . We use the letter Z to refer to the set of integers: Z = {· · · − 5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5 . . .} We can show these numbers on a number line . Zero is neither positive nor negative. b

−5

b

−4

b

−3

b

b

−2

−1

b

b

b

b

b

b

0

1

2

3

4

5

Remember the rules for handling addition and subtraction of integers: + − + −

(positive) = (positive) (positive) = (negative) (negative) = (negative) (negative) = (positive)

The following rules apply to multiplication and division with integers: (positive) · (positive) = (positive) (positive) · (negative) = (negative) (negative) · (positive) = (negative) (negative) · (negative) = (positive)

(positive) : (positive) = (positive) (positive) : (negative) = (negative) (negative) : (positive) = (negative) (negative) : (negative) = (positive)

Order of operations with integers: • Perform the operations within brackets first. • Calculate any part involving exponents (powers, roots). • Starting from the left, perform all divisions and multiplications as you come to them. • Finally, restart from the left and perform all additions and subtractions as you come to them. The word BEDMAS may help you remember this order: Brackets Exponents Division Multiplication Addition Subtraction D pto. M atemáticas. IES Jovellanos. 2011


4 Rules for brackets:

• If an expression contains one set of grouping symbols, i.e., brackets, work that part first. • If an expression contains two or more sets of grouping symbols one inside the other, work the innermost first. • The division line of fractions also behaves as a grouping symbol. This means that the numerator and the denominator must be found separately before doing the division.

Exercises - Set B 1. Simplify: a) −5 + 4 · (−2 + 1)3 − (−9 + 6)2 b) 12 − 2 · [25 : (−4 − 1) + (−2) − (−6 − 10)] c) −7 − (−3) + (−8) · (−1) − (−12) : (−4) d) −5 − 4 · [−8 : 2 − 2 · (−3)] e) 10 − 10 · [−6 + 5 · (−4 + 7 − 3)] f) 6 − 5 · [−4 − 1 + (−2)2 − 32 ] 2. Simplify: a)

12 + (5 − 7) 18 : (6 + 3)

b)

57 7 − (2 · 3)

c)

3·8+6 6

d)

(3 + 8) − 5 3 + (8 − 5)

3. Insert grouping symbols if necessary, to make the following true: a) 120 : 4 · 2 = 15 d) 5 · 7 − 3 − 1 = 19

b) 120 : 4 · 2 = 60 e) 5 · 7 − 3 − 1 = 33

c) 5 · 7 − 3 − 1 = 15 f) 3 + 2 · 8 − 4 = 36

g) 3 + 2 · 8 − 4 = 11

h) 3 + 2 · 8 − 4 = 15

i) 8 − 6 · 3 = 6

Using the calculator: Most modern calculators have the order of operations rules built into them. They also have grouping symbols keys (left hand bracket and right hand bracket). 12 For example, if we consider and key in 12 : 4 + 2 = the calculator gives an answer 4+2 of 5, which is incorrect. However, if we type 12 : (4 + 2) = we obtain the answer 2, which is correct. Remember to use the sign change key before a negative number.

Exercises - Set C 1. Evaluate each of the following using your calculator: a) 17 + 23 · 15

b) (17 + 23) · 15

d) 128 : (8 + 8)

e)

89 + (−5) −7 · 3


c) 128 : 8 + 8 f)

−15 − 5 6−8:4


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3. Rational numbers (fractions) A rational number is a number which can be written as a fraction, that is, in the form and b are both integers, and b 6= 0. 4 −1 2 0 Examples: , , , , 7 10 3 1

a , where a b

−2 ... 1

A fraction indicates a part of a unit or a part of a quantity; it consists of two whole numbers, a numerator and a denominator , separated by a bar symbol: • the denominator, b, shows how many equal parts the whole has been split into. • the numerator, a, tells us how many of those equal parts are being described. The set of rational numbers is represented by the letter Q: o na , a, b ∈ Z, b = 6 0 Q= b Remember that in general:

−a a a = =− b −b b

However:

−a a = −b b

How to say a fraction: Here are some examples: 1 a half or one half 2 2 two fifths or two over five 5 8 eight over thirty-five 35

1 a third or one third 3 4 four sevenths or four over seven 7 54 fifty-four over seventy-two 72

1 4

a quarter

Proper and improper fractions: A fraction which has numerator less than its denominator is called a proper fraction . 3 For example, is a proper fraction. 4 A fraction which has numerator greater than or equal to its denominator is called an improper fraction . 7 For example, is an improper fraction. 4 When an improper fraction is written as a whole number and a fraction, it is called a mixed number . 7 3 For example, can be written as the mixed number 1 43 , as it is really 1 + . 4 4 Example 2 Write the improper fraction

29 as a mixed number. 3

First, we divide 29 by 3: 29 2

3 9

Hence:

29 2 = 9 + = 9 32 3 3 D pto. M atemáticas. IES Jovellanos. 2011


6 Simplifying fractions:

We can simplify a fraction by cancelling common factors in the numerator and denominator. When a fraction is written as a rational number with the smallest possible denominator, we say it is in lowest terms . We can simplify a fraction to its lowest terms dividing both numerator and denominator by their HCF.

Example 3 Simplify:

16 24

As HCF of 16 and 24 is 8, dividing both numerator and denominator by 8 we get: 16 2 = , 24 3

which is the fraction in its lowest terms

Two fractions are equal or equivalent if they can be written in the same lowest terms. We can convert a fraction to an equivalent fraction by multiplying or dividing both the numerator 3 6 9 12 and denominator by the same non-zero number: = = = = ... 4 8 12 16 We can also test if two fractions are equivalent by cross-multiplying their numerators and denomi12 24 nators. For example, and are equivalent because 12 · 40 = 20 · 24 = 480. 20 40 Comparing fractions: To compare two or more fractions: • Find the common denominator , which is the LCM of the original denominators. • Work out the equivalent fractions. • Write the fractions in ascending or descending order. Remember that if two fractions have the same denominator, the greater one is the fraction with the highest numerator.



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Example 4 Which fraction is bigger,

4 7 or ? 5 9

The common denominator of 4 36 = 5 45 7 35 = 9 45

4 7 and is 45, the LCM of 5 and 9. 5 9

(multiply numerator and denominator by 9). (multiply numerator and denominator by 5).

The order is

36 35 4 7 > , so > . 45 45 5 9

Exercises - Set D 1. Express with denominator 12: a)

2 3

b)

3 4

5 6

c)

d)

6 18

e)

15 45

d)

24 28

e)

18 42

e)

33 77

2. Express with numerator 12: a)

3 7

b)

6 5

4 9

c)

3. Express in lowest terms: a)

6 10

6 18

b)

c)

25 10

d)

14 35

f)

48 72

g)

78 117

h)

125 1000

4. Simplify: a)

8−2·5 4·3

b)

−1 + 8 : 2 8−5

c)

32 − 2 · 5 8 − 32

d)

−5 · (−6) −11 − 4

e)

6−3:3 2 + 10 : 2

f)

12 + 8 : 2 12 − 8 · 2

g)

6 + 22 6 − 23

h)

11 − 3 16 : 4

5. Place these fractions in ascending order: a)

2 3 3 17 , , , 5 8 4 40

b)

4 7 5 3 6 , , ,− ,− 3 5 7 4 11

c)

1 2 3 1 3 ,− , ,− ,− 8 3 11 6 4

6. Place these fractions in descending order: a)

5 11 7 5 , , , 6 24 12 8

b)

2 3 4 5 6 , , , , 5 7 9 13 10

5 1 4 7 6 c) − , − , − , − , − 8 2 7 11 13

7. Write as a mixed number: a)

45 4

b)

11 2

a)

37 5



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4. Operations with fractions Addition and subtraction: To add or subtract fractions: • If necessary, convert the fractions so they have the lowest common denominator. • Add or subtract the new numerators. The denominator stays the same. For example:

3 2 1 9 8 6 9−8+6 7 − + = − + = = 4 3 2 12 12 12 12 12

When adding or subtracting mixed numbers, you can first convert them to improper fractions and then perform the operation. However you can also add the whole numbers and fractions separately, then combine the result.

Example 5 Find: 2 13 − 3 12 + 1 41 First, we convert the mixed numbers into improper fractions: 1 6 1 7 1 6 1 7 1 4 1 5 2 13 = 2 + = + = ; 3 12 = 3 + = + = ; 1 14 = 1 + = + = 3 3 3 3 2 2 2 2 4 4 4 4 Hence: 2 31 − 3 21 + 1 41 =

7 7 5 28 42 15 1 − + = − + = 3 2 4 12 12 12 12

Multiplication: To multiply two fractions, we multiply the two numerators to get the new numerator, and multiply the two denominators to get the new denominator: a c a·c · = b d b·d To help make multiplication easier, we can cancel any common factors in the numerator and denominator before we multiply. For example:

4 3 4·3 4·1 4 · = = = 9 5 9·5 3·5 15

Two numbers are reciprocals of each other if their product is one. For any fraction

a a b a b , we notice that · = 1. So, the reciprocal of is . b b a b a

Division: To divide two fractions, multiply the first by the reciprocal of the second: a c a d a·d : = · = b d b c b·c For example: 1 31 : 3 21 =

4 7 4·2 8 : = = 3 2 3·7 21



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Exercises - Set E 1. Find: a) 1 32 − 2

b) 3 43 − 1 21

c)

3 − 2 21 4

d) 1 23 + 3 14

e) 4 31 + 2 61

f) 2 32 − 5 56

g) −2 41 + 3 18

h) 4 15 − 2 16

2. Find: a)

2 1 3 · · 3 4 5

3 5 4 − · 5 2 3 2 3 2 − · 1 32 g) 3 4 d)

b)

3 4 2 · − · − 8 3 5

c)

2 3 2 + · 3 4 3

e)

3 1 2 1 · + · 5 3 3 4

f)

4 1 1 2 · − · 3 2 6 3

d)

4 :3 5

h) 4 · 1 13 − 5 ·

2 7

3. Find: 2 1 a) : 3 6 e)

1 − 2 3 g) 1 21 : − 4

5 1 b) : 7 3

1 : 1 23 4

f) 2 34 :

3 c) : 4

2 3

h) 3 15 : 1 31

4. Find: 2 3 4 a) − : 3 2 5

5 1 4 b) : + 3 2 3

1 2 3 6 c) · − : 2 5 4 5

2 d) : 5

1 3 2 − + · 2 4 5

5. Find and give your answers in their simplest form:

a)

3 7 7 + − 4 6 8

:

25 12

3 1 (−3) · − 5 3 d) 4 6 (−2) · − 3 5

b)

7 9 13 13 − · + − 15 25 22 33

1 3 2 3− · − 4 5 15 e) 4 1 3 6+ · − 25 2 4

c)

1 − 2

3 −1 4

3 +1 4 2 5 3 5 − · − 3 9 4 6 f) 7 5 4 − · +1 12 6 3

5. Problem solving To find a fraction of a quantity we multiply the quantity by the numerator, and divide the result by the denominator. 3 3 3 · 85 For example: of 85 e are: · 85 = = 51 e. 5 5 5



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Example 6 If

3 of a shipping container holds 2100 identical cartons, how many cartons will fit into: 8 a)

5 of the container 8

b) the whole container?

3 1 of the container holds 2100 cartons; hence of the container holds 8 8 2100 : 3 = 700 cartons. 5 holds 700 · 5 = 3500 cartons 8 8 b) The whole is which is 700 · 8 = 5600 cartons. 8 a) So,

Example 7 Rob eats remains?

1 3 of a watermelon one day and of it the next day. What fraction of the watermelon 3 8

We let the whole watermelon be represented by 1. The fraction remaining will be: 24 8 9 7 1 3 1− − = − − = 3 8 24 24 24 24

Exercises - Set F 1. Millie calculated that her bicycle cost $, what did her bicycle cost?

1 of the cost of her father’s car. If the car cost 38 014 83

1 1 1 , and of the brickwork of his new garage. What 3 5 4 fraction must he complete on the fourth and final day? 2. Over three successive days Colin builds

3. 200 kg of sugar must be poured into packets so there is packets will be filled?

2 kg of sugar per packet. How many 5

4. 2400 kg of icecream is put into plastic containers which hold containers are needed?

3 kg each. How many plastic 4

5. John says that his income is now 3 21 times what it was 20 years ago. If his current annual income is 63 000 e , what was his income 20 years ago? 1 1 1 1 of its weekly budget on rent, on food, on clothes, on entertainment 3 4 8 12 and the remainder is banked. How much is banked if the weekly income is 864.72 e?

6. A family spends



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Exercises - Set G 1. Renee used pipe is left? 2.

1 2 of a length of pipe and later used of what remained. What fraction of the 3 3

2 1 of an amount of money is 519 $. Find: a) of the money 3 3

b) the whole amount.

3 of a field was searched for truffles and 39 were found. How many truffles would we expect 13 to find in the remainder of the field?

3.

4. The seven tenths of a quantity are 210. What is the quantity? 5. Alfredo sent

2 2 of his potato crop to market last week. This week he sent of the remainder. 5 3

a) What fraction of his crop has now gone to market? b) If he has 860 kg remaining, what was the original weight of the crop? 2 9 of the weight of a loaf of bread comes from the flour used in making the bread. If of 10 9 the weight of the flour is protein, what fraction of the weight of a loaf of bread is protein?

6.

7. A tree is losing its leaves. Two thirds fall off in the first week, two thirds of those remaining fall off in the second week and two thirds of those remaining fall off in the third week. Now there are 37 leaves. How many leaves did the tree have originally? 8. Out of a group of 40 people in a shop, 32 were aged thirty or over. What fraction of the people were under thirty? 4 1 of the cars on the road are saloons. Of these saloons are red. What fraction of the cars 5 8 on the road are red saloons? 9.

13 of the distance between college and home. She wants to split 28 the remaining distance into 5 equal parts. What fraction of the whole journey is each part?

10. Sadie has already driven

1 of his time when awake playing on his 8 computer. What fraction of his day does Marc spend playing on the computer? 11. Marc spends a third of the day asleep. He spends

12. A 9 tonne load of top soil is divided into can be filled?

3 tonne plant containers. How many containers 25

2 4 of a bar of chocolate. Linda eats of what remains. What fraction of the bar 5 9 of chocolate have they eaten between them? 13. John eats

14. In his will a man leaves half his estate to his wife and the rest is shared equally among his five sons. What fraction of his estate does each son inherit? 15. Out of a deposit of oil you empty one half. Out of what remains, you empty one half again, 11 and then you empty of what remains. Finally, there are 36 litres left. How many litres were 15 there at the beginning?

✁✃✁✃✁✃✁✃✁✃✁✃✁✃ D pto. M atemáticas. IES Jovellanos. 2011

D ecim al N um bers 1. What are decimal numbers? 6 3 + , which can also be written as an improper The number 4.63 is a quick way of writing 4 + 10 100 463 63 fraction or as a mixed number 4 100 . 100 Numbers such as 4.63 are commonly called decimal numbers . 4 is the whole part and 63 is the decimal part . You say: four point six three, or four units six tenths and three hundredths, or four units and sixtythree hundredths. 2 6 + (expanded fractional form). You 100 1000 say: fourteen point oh six two, or fourteen units six hundredths two thousandths, or fourteen units sixty-two thousandths. Likewise, 14.062 is the quick way of writing 14 +

Exercises - Set A 1. Write the following in expanded fractional form: a) 2.5

b) 2.05

c) 2.0501

d) 4.0052

e) 0.0106

2. Write the following in decimal form: a) 3 + d)

2 10

7 9 + 100 1000

b)

7 8 + 10 100

e) 4 +

1 10 000

c)

6 3 + 10 1000

f) 5 +

3 2 + 100 10 000

2. Types of decimal numbers There are three different types of decimal numbers: terminating, recurring and irrational numbers. A terminating or exact decimal is one which does not go on forever, so you can write down all its digits. For example: 0.125. A recurring decimal is a decimal number which does go on forever, but where some of the digits are repeated over and over again. For example: 0.12525252525 . . . is a recurring decimal, where 25 1

D ecim al N um bers - 3 o E SO

2

is repeated forever. Sometimes recurring decimals are written with a bar over the digits which are repeated, or with dots over the first and last digits that are repeated. ˙ 4˙ For example: 3.2014014014 · · · = 3.2014 = 3.201 Irrational numbers are those which go on forever and don’t have digits which repeat. For example: √ 2 = 1.4142135 . . . , π = 3.14159265 . . . All rational numbers (fractions) can be expressed as either terminating decimals or recurring decimals, dividing the numerator by the denominator. For example:

7 = 0, 175; 40

8 = 0.72727272 . . . ; 11

5 = 0.4166666 . . . 12

Conversely, all terminating and recurring decimals can be expressed as fractions. However, irrational numbers cannot be expressed as fractions, they are not rational. All rational and irrational numbers form the set of real numbers , which is represented by the letter R. How to convert a terminating decimal into a fraction: • Write the decimal as a fraction with denominator 10, 100, 1000,. . . , according to the number of decimal places. 45 For example: 0.45 = 100 • Simplify the fraction to its lowest terms:

0.45 =

45 9 = 100 20

How to convert a recurring decimal into a fraction: Look at these examples: 5.454545. . . 1. Let x = 5.454545 . . .

(A)

2. Multiply by 100 (because there are two recurring figures; if there were three recurring figures, you would multiply by 1000): (B)

100x = 545.454545 . . . 3. Subtract B−A:

100x = x = 99x = 4. Divide by 99 and simplify: x =

545.454545 . . . 5.454545 . . . 540

540 60 = 99 11

2.5636363. . . 1. Let x = 2.5636363 . . . 2. Multiply by 10 (because there is one figure between the whole part and the recurring figures; if there were two figures between the whole part and the recurring figures, you would multiply by 100): 10x = 25.636363 . . .

(A)



3

3. Multiply (A) by 100 (because there are two recurring figures): (B)

1000x = 2563.636363 . . . 4. Subtract B−A:

1000x = 10x =

2563.636363 . . . 25.636363 . . .

990x = 5. Divide by 990 and simplify: x =

2538

2538 141 = 990 55

Exercises - Set B 1. Write as decimals, and state in each case whether they are terminating or recurring decimals: a)

7 50

b)

2 125

c)

5 9

d)

f)

87 60

g)

11 3

h)

33 22

i)

4 11 3 20

e)

44 7

j)

3 5

2. Write as a fraction in its lowest terms: a) 0.15

b) 0.046

c) 0.7˙

d) 0.5˙ 4˙

e) 0.12˙

f) 3.407˙

˙ 5˙ g) 0.30

h) 0.4˙ 8˙

i) 4.2˙

j) 1.2˙ 7˙

3. Calculate. Firstly you must express decimals as fractions: 2 3 1 3 a) 0.4˙ 6˙ − + 3.6 b) · 2.4˙ − c) 5 − 2 · − 0.1˙ 4˙ 5 3 5 5 1 to a decimal, and I got the answer 0.07692308. 13 There is no repeating pattern, so the decimal does not recur.” Explain why Shula is wrong. 4. Shula says, “I used my calculator to change

3. Approximation and rounding Rounding a number is a way of writing it approximately. Sometimes we don’t need to write all the figures in a number, as an approximate one will do. For example: for a population of 27 653 the number is large and will change daily. It is better to round up and say 28 000. When rounding numbers to a given degree of accuracy, look at the next digit: • If it is 5 or more then we have to round up , that is, increase the previous digit by one. • Otherwise we round down , that is, leave the previous digit unchanged. For example, to round 7.365 to 2 decimal places, look at the thousandths digit: 7.365. The thousandths digit is 5, so we round up to 7.37: 7.365≈7.37 (to 2 decimal places). Numbers can be rounded: D pto. M atemáticas. IES Jovellanos. 2011


4 • to decimal places (d.p.) :

4.16 = 4.2 to 1 decimal place

• to the nearest unit, ten, hundred, thousand,. . . 32 559 = 33 000 to the nearest thousand • to significant figures (s.f.) : the first non-zero digit in a number is the 1st significant figure; it has the highest value in the number. When rounding to significant figures, count from the first non-zero digit. For example: 54.76 ≈ 55 (to 2 s.f.) 0.00405 ≈ 0.0041 (to 2 s.f.) 6.339 ≈ 6.34 (to 3 s.f.) Exercises - Set C 1. Round these decimal numbers to the nearest whole number: a) 5.8

b) 21.67

c) 39.175

2. Round these numbers to the nearest thousand: a) 2239

b) 12 563

c) 155 669

3. Round these numbers to two decimal places: a) 0.317

b) 15.304

c) 16.445

4. Use a calculator to work these out. Write your answers correct to 3 s.f. where necessary: a) (16.8 + 12.4) · 17.1 d)

27.4 18.6 − 3.2 16.1

b) 37.4 − 16.1 : (4.2 − 2.7) e)

27.9 − 17.3 + 4.7 8.6

c)

16.84 7.9 + 11.2

f)

0.0768 + 7.1 18.69 − 3.824

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Percentage and Ratio 1. Percentage Percentages are comparisons with the whole amount (which we call 100 %). % reads percent which is short for per centum, which loosely translated from Latin, means out of every hundred. 1 25 1 10 Thus 10 % means 10 out of every 100 or = . Likewise, 25 % means = . 100 10 100 4 x In general, x % = (we can simplify the fraction). 100 We often need to calculate a percentage of a quantity. To do this, we multiply the quantity by the percentage expressed either as a fraction or as its decimal equivalent. For example: 9 % of 24 =

9 · 24 216 9 · 24 = = = 2.16 100 100 100

37 % of 58 = 0.37 · 58 = 21.46 (0.37 is the decimal equivalent of 37 %). To convert any fraction to a percentage, we write it as a decimal and multiply by 100. For example: 2 3 = 0.75 ; 0.75 · 100 % = 75 % = 0.4 ; 0.4 · 100 % = 40 % 4 5 Exercises - Set A 1. Copy and complete these tables (simplify any fractions). Percentage 12 %

Fraction

Decimal

Percentage 75 %

Fraction

0.36

Decimal 0.8

3/10

4/25

25 %

85 % 0.4

0.15

1

3/5

2. In a football squad of 24 players, 5 of the players are goalkeepers. What percentage of the football squad are not goalkeepers? 3. Leon scores 68 % in his French exam and gets he do the best?

37 in his German exam. In which subject did 54

4. A train journey is 395 km long. Lola is traveling on a train that has completed 23 % of the journey. How many kilometres has Lola’s train traveled?

1

Percentage and R atio - 3 o E SO

2

2. Ratio A ratio is an ordered comparison of quantities of the same kind. For example: the ratio of cordial to water in a bottle is 1 : 5. This means that we have 1 part of cordial and 5 parts of water (6 parts in total).

Example 1 Write as a ratio, without simplifying your answer: a) Jack has 5 e and Jill has 50 cents. b) Mix 200 ml of cordial with 1 l of water. a) Jack : Jill = 5 e : 50 cents = 500 cents : 50 cents = 500 : 50 b) cordial : water = 200 ml : 1 l = 200 ml : 1000 ml = 200 : 1000

If we have a ratio and multiply or divide both parts by the same non-zero number, we obtain an equal ratio . For example: 45 : 15 = 3 : 1 (dividing both parts by 15) 0.4 : 1.4 = 4 : 14 = 2 : 7 (multiplying both parts by 10 and dividing by 2 later). Ratios are equal if they can be expressed in the same simplest form. For example: 3 : 5 = 6 : 10

15 : 20 = 3 : 4 = 12 : 16

Example 2 The ratio of walkers to guides on a demanding bushwalk is to be 9 : 2. How many guides are required for 27 walkers? walkers : guides = 9 : 2 = 27 : ? If 9 parts is 27, then 1 part is 27 : 9 = 3, so 2 parts is 3 · 2 = 6. Hence, 9 : 2 = 27 : 6, and 6 guides are needed.

Quantities can be divided in a particular ratio by considering the number of parts the whole is to be divided into.


Percentage and R atio - 3 o E SO

3

Example 3 An inheritance of 60 000 $ is to be divided between Donny and Marie in the ratio 2 : 3. How much does each receive? There are 2 + 3 = 5 parts. 2 2 Donny gets of 60 000 $ = · 60 000 = 24 000 $ 5 5 Marie gets

3 3 of 60 000 $ = · 60 000 = 36 000 $ 5 5

Exercises - Set B 1. Write as a ratio, simplifying your answer if possible: a) 10 $ is to 7 $

b) 2 l is to 5 l

c) 80 kg is to 50 kg

d) 2 $ is to 50 cents

e) 500 ml is to 2 l

f) 800 m is to 1.5 km

2. Express the following ratios in simplest form: a)

3 1 : 4 4

b) 0.5 : 0.2

c) 18 : 24

d) 2 12 : 1 21

e) 1.5 : 0.3

3. A hospital employs nurses and doctors in the ratio 7 : 2. If 84 nurses are employed, how many doctors are employed? 4. A farmer has pigs and chickens in the ratio 3 : 8. If she has 360 pigs, how many chickens does she have? 5. The price of a TV is reduced from 500 e to 400 e. A DVD player costing 1250 e is reduced in the same ratio as the TV. What does the DVD player sell for? 6. Divide: a) 50 $ in the ratio 1 : 4

b) 35 $ in the ratio 3 : 4

7. A fortune of 400 000 $ is to be divided in the ratio 5 : 3. What is the larger share? 8. The ratio of girls to boys in a school is 5 : 4. If there are 918 students at the school, how many are girls? 9. A glass contains alcohol and water in the ratio 1 : 4. A second glass has the same quantity of liquid but this time the ratio of alcohol to water is 2 : 3. each glass is emptied into a third glass. What is the ratio of alcohol to water for the final mixture? 10. One full glass contains vinegar and water in the ratio of 1 : 3. Another glass of twice the capacity of the first has vinegar and water in the ratio 1 : 4. If the contents of both glasses were mixed together what is the ratio of vinegar to water? 11. A glass contains alcohol and water in the ratio 1 : 3. Another glass of the same capacity contains alcohol and water in the ratio 3 : 5. What is the ratio of alcohol to water if we mix both glasses? 12. The same as before, but the ratios are 1 : 4 and 2 : 5.

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E rrors and accuracy of m easurem ent 1. Measurement The measurement of length, area, volume and capacity is of great importance. Builders, architects, engineers and manufacturers need to measure the sizes of objects to considerably accuracy. The most common system of measurement is the Systeme International (SI) . Important units that you should be familiar with include: Measurement of Length Mass Capacity Time Temperature Speed

Standard unit metre gram litre hours, minutes, seconds degrees Celsius metres per second (ms−1 )

What it means How long or how far How heavy an object is How much liquid or gas is contained How long it takes How hot or cold How fast it is travelling

The SI uses prefixes to indicate an increase or decrease in the size of a unit: Prefix tera giga mega kilo hecto

Symbol T G M k h

Meaning 1 000 000 000 000 1 000 000 000 1 000 000 1 000 100

Prefix centi milli micro nano pico

Symbol c m µ n p

Meaning 0.01 0.001 0.000 001 0.000 000 001 0.000 000 000 001

2. Errors Whenever we take a measurement, there is always the possibility of error. Errors are caused by inaccuracies in the measuring device we use, and in rounding off the measurement we take. They can also be caused by human error, so we need to be careful when we take measurements. There are two types of error: • The absolute error due to rounding or approximation is the difference between the actual or true value and the measured value, regardless the sign: εa = |true value − rounded value| • The percentage or relative error is the absolute error compared with the true value, expressed as a percentage: εr =

εa · 100 true value 1

E rrors - 3 o E SO

2

Example 1 The crowd at a tennis tournament was 14 869, but in the newspaper it was reported as 15 000. Find the absolute and percentage errors in this approximation.

εa = |15 000 − 14 869| = 131

εr =

131 · 100 ≈ 0.881 % 14 869

3. Accuracy When a measurement is written, it is always written to a given degree of accuracy. The real measurement can be anywhere within ± half a unit. For example: A man walks 23 km (to the nearest km). Because the real measurement has been rounded, it can lie anywhere between 22.5 km (minimum) and 23.5 (maximum). When we take measurements, we are usually reading from some sort of scale. The scale of a ruler may have millimetres marked on it, but when we measure an object’s length it is likely to lie between two marks. So, when we round or estimate to the nearest millimetre, our answer may be inaccurate by up to half a millimetre. We say the ruler is accurate to the nearest half a millimetre. In general:

A measurement is accurate to ±

1 of the smallest division on the scale. 2

Example 2 Rod’s height was measured using a tape measure with centimetre graduations. It was recorded as 188 cm. For this measurement state the absolute and the percentage error. The tape measure is accurate to ±

1 cm. 2

The absolute error is 0.5 cm. 0.5 The percentage error is: · 100 ≈ 0.266 %. 188


E rrors - 3 o E SO

3

Exercises - Set A 1. Find the absolute error and percentage error in saying that: a) there were 300 people at the conference when there were actually 291. b) 2.95 can be approximated by 3. c) 31 823 $ can be rounded to 32 000 $. 2. State the accuracy possible when using: a) a ruler marked in mm. b) a set of scales marked in kg. c) a tape measure marked in cm. d) a jug marked with 100 ml increments. 3. Su-Lin’s height was measured using a tape measure with centimetre markings. Her height was recorded as 154 cm. a) State the range of possible heights in which her true height lies. b) Find the absolute error in the measurement. c) Find the percentage error. 4. When we measured the height of a building we got 14,48 m, but the true height actually is 14,39. When measuring the height of a tree, we got 7,85 m, whereas the true height is 7,92 m. a) Find the absolute and percentage errors in each measurement. b) Which measurement is more accurate? Why?

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N um ber sets and interval notation 1. Number sets A set is a collection of objects or things. For example:

V = {vowels} = {a,e,i,o,u}

E = {even numbers} = {2, 4, 6, 8, 10 . . . } V is a finite set as it has a finite number of elements. E is an infinite set as it has infinitely many elements.

are both sets.

We use the symbol ∈ to mean “is a member of ” or “is in”. So, a ∈ V and 28 ∈ E, but h 6∈ V or 119 6∈ E. We use: • N to represent the set of all natural numbers {0, 1, 2, 3, 4, 5, 6 . . . } • Z to represent the set of all integers {0, ±1, ±2, ±3, ±4, ±5 . . .} p • Q to represent the set of all rational numbers , p, q ∈ Z, q 6= 0 q • R to represent the set of all real numbers; these are the numbers which can be placed on the number line, that is, all decimal numbers (exact, recurring or irrational).

Exercises - Set A 1. True or false? a) −3 ∈ N

b) 6 ∈ Z

e) −

f) 2 31 ∈ Z

1 6∈ Q 4

2. Which of these are rational? 3 a) 8 b) −7 c) 4 e) π f) 2 15 g) 9, 176

3 ∈Q 4 g) 0, 3684 ∈ R c)

√ 3 √ h) 400 d)

1

d) h)

√ 2 6∈ Q

1 ∈Z 0, 1

N um ber sets and interval notation - 3 o E SO

2

2. Interval notation Interval or set notation allows us to quickly describe sets of numbers using mathematical symbols only. For example: {x | − 3 < x ≤ 2, x ∈ R} including 2”.

reads “the set of all real x such that x lies between minus 3 and 2,

Unless stated otherwise, we assume we are dealing with real numbers, thus the set can also be written as {x | − 3 < x ≤ 2}. not included

included b

We can represent the set on a number line as

|

|

−3

2

Sometimes we want to restrict a set to include only integers or rationals. For example: {x | − 5 < x < 5, x ∈ Z} minus 5 and 5”.

reads “the set of all integers x such that x lies between

We can represent the set on a number line as

b

|

b

b

b

b

−5

b

b

0

1. Write verbal statements for the meaning of: b) {x | x ≤ 5}

d) {x | 1 ≤ x ≤ 4}

c) {y | 0 < y < 8}

e) {t | 2 < t < 7}

f) {n | n ≤ 3 or n > 6}

2. Sketch the following number sets: a) {x | 4 ≤ x < 8, x ∈ N}

b) {x | − 5 < x ≤ 4, x ∈ Z}

c) {x | − 3 < x ≤ 5, x ∈ R}

d) {x | x > −5, x ∈ Z}

e) {x | x ≤ 6}

f) {x | − 5 ≤ x ≤ 0}

3. Write in set notation: a)

c)

e)

|b

0

b

|

|

0

3

b

b

|

|

−1

2

b

b

3

b

b

b)

d) b

|

|

2

5

|b

b

b

b

b

0 |b

6

f)

b

|

5 |

0

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b

|

5

Exercises - Set B

a) {x | x > 4}

b

Index Law s 1. Index notation Rather than write 2 · 2 · 2 · 2 · 2, we write such a product as 25 .

25 reads “two to the power of five”, “two raised to five” or “two with index five”. If n is a positive integer, then an is the product of n factors of a i.e., an = |a · a · a{z· a . . . a} n factors

a is the base. n is the power, index or exponent. Other examples:

52 reads “five to the power of two”, “five raised to two” or “five squared”. 73 reads “seven to the power of three”, “seven raised to three” or “seven cubed”.

Exercises - Set A 1. What is the last digit of 3100 ? (Hint: Consider 31 , 32 , 33 , 34 , 35 , 36 . . . and look for a pattern.) 2. What is the last digit of 7200 ?

2. Negative bases So far we have only considered positive bases raised to a power. We will now look at negative bases. Consider the statements below: (−1)1 = −1 2

(−1) = (−1) · (−1) = 1 (−1)3 = (−1) · (−1) · (−1) = −1

(−1)4 = (−1) · (−1) · (−1) · (−1) = 1

(−2)1 = −2

(−2)2 = (−2) · (−2) = 4 (−2)3 = (−2) · (−2) · (−2) = −8

(−2)4 = (−2) · (−2) · (−2) · (−2) = 16

In he pattern above it can be seen that: A negative base raised to an odd power is negative; whereas a negative base raised to an even power is positive. 1

Index law s - 3 o E SO

2 Be careful: to a power.

(−2)4 = 16, but −24 = −16, so always use brackets when raising a negative base

Exercises - Set B 1. Simplify:

a) (−1)17

b) −54

c) (−2)5

a) 2, 86

2. Find using your calculator:

b) (−5)5

d) −(−3)2 c) −94

d) (−1, 14)23

3. Find using your calculator, correct to 3 decimal places: a) (2, 6 + 3, 7)4 3 3, 2 + 1, 92 d) 1, 47

b) 8, 63 − 4, 23 4 0, 52 e) 0, 09·, 14

c) 12, 4 · 10.74 a)

648 3, 624

3. Index laws Recall the following index laws where the indices m and n are positive integers: am · an = am+n

To multiply numbers with the same base, keep the base and add the indices.

am = am−n an

To divide numbers with the same base, keep the base and subtract the indices.

(am )n = am·n

When raising a power to a power, keep the base and multiply the indices.

(a · b)n = an · bn a an = n b b

The power of a product is the product of the powers. The power of a quotient is the quotient of the powers.

Exercises - Set C 1. Simplify using the index laws (leave your answer in index form): a) 22 · 24

b) 114 · 11

c) 1315 · 136

d) 172 · 175

e)

f)

g) 149 : 144

h) 98 : 92

25 22

56 5

i) (22 )4

j) (104 )2

k) (93 )7

l) (74 )5

m) 63 · 53

n) 84 : 24

o) 57 · 37

p) (63 )2 · 96

q) (83 )3 · 82

r) (102 )5 : 210

s) (62 )3 · 6

t) (73 )4 : (72 )4

4. Zero and negative indices Look at this example:

23 8 = =1 23 8



3 23 = 23−3 = 20 . 23

But, applying the second index law: 20 = 1.

So, we deduce that:

a0 = 1, for all a 6= 0.

In general:

72 7·7 1 1 = = = 3 75 7·7·7·7·7 7·7·7 7 72 But, applying the second index law: = 72−5 = 7−3 . 75 1 So, we deduce that: 7−3 = 3 . 7

Now, look at this example:

In general, if a is any non-zero number, and n is an integer, then: • a−n =

1 an

(i.e., an and a−n are reciprocals of one another).

• In particular, a−1 =

1 a

and

a −n b

=

n b . a

Example 1 Simplify, giving answers in simplest rational form: −2 3 −2 a) 5 b) c) 80 − 8−1 5

a) 5−2 =

1 1 = 52 25

b)

−2 2 3 5 52 25 = = 2 = 5 3 3 9

c) 80 − 8−1 = 1 −

1 7 = 8 8

Exercises - Set D 1. Simplify, giving answers in simplest rational form: a) 5 − 70

b) 60 − 20

c) (6 − 2)0

d) 4−1

e) 3−2 −1 1 i) 3

f) 3−3 −1 2 j) 5

h) 2−4

m) 20 + 21 + 2−1

n) 1 21

g) 10−5 −2 3 k) 4 −2 1 o) + 2−1 3

−3

l) 50 − 5−1 p)

−2 4 4 1 − 3 2

2. Write as powers of 2, 3 or 5: a) 8 f)

1 125

b)

1 8

g) 32

c) 9 h)

1 32

d)

1 9

i) 81

e) 125 j)

1 81



4

Exercises - Set E 1. Simplify: a) 73 · 72

b) 54 · 53

c) a7 · a2

d) a4 · a

e) b8 · b5

f) a3 · an

g) b7 · bm

h) m4 · m2 · m3

k) 77 : 74

l)

59 52 b10 m) 7 b

1113 119 p5 n) m p

i)

j)

o)

a6 a2

ya y5

p) b2x : b

q) (32 )4

r) (53 )5

s) (24 )7

t) (a5 )2

u) (p4 )5

v) (b5 )n

w) (xy )3

x) (a2x )5

2. Express in simplest form with a prime number base: a) 8

b) 25

c) 27

d) 43

e) 92

f) 3a · 9

g) 5t : 5

h) 3n · 9n

k) (54 )x−1

l) 2x · 22−x

3x+1 3x−1 4y n) x 8

16 2x 2y m) x 4 i)

j)

o)

3x+1 31−x

p)

2t · 4t 8t−1

3. Remove the brackets of: a) (a · b)3

b) (a · c)4

c) (b · c)5

d) (a · b · c)3

e) (2a)4

f) (5b)2

g) (3n)4

i) (4abc)3

j)

h) (2bc)3 5 2c l) d

a 3

k)

b

m 4 n

4. Express the following in simplest form, without brackets: a) (2b4 )3 3 4 m e) 2n2

b) (5a4 b)2 4 2 4a f) b2

c) (−6b2 )2 3 −2a2 g) b2

d) (−2a)3 2 −3p2 h) q3

5. Rational indices Look at this example: 1

1

1

1

Since 3 2 · 3 2 = 3 2 + 2 = 31 = 3, and √ 1 3 2 = 3.

√ √ 3 · 3 = 3 also, then, by direct comparison:

√ √ √ 1 1 1 Likewise, 2 3 · 2 3 · 2 3 = 21 = 2, compared with 3 2 · 3 2 · 3 2 = 2, suggests: √ 1 2 3 = 3 2. √ √ 1 In general: an = n a ( n a reads “the nth root of a”) m

1

Also: a n = (am ) n =

√ n am



5

Example 2 Write as a single power of 2: √ 1 a) 3 2 b) √ 2

a)

√ 3

c)

√ 5 4

1 1 1 b) √ = 1 = 2− 2 2 22

1

2 = 23

c)

√ √ 2 5 5 4 = 22 = 2 5

Exercises - Set F 1. Write as a single power of 2: √ √ 1 a) 5 2 b) √ c) 2 2 5 2 √ √ 4 f) 2 · 3 2 g) √ h) ( 2)3 2 2. Write as a single power of 3: √ √ 1 a) 3 3 b) √ c) 4 3 3 3

√ d) 4 2 1 i) √ 3 16 √ d) 3 3

1 e) √ 3 2 1 j) √ 8 1 e) √ 9 3

3. Write the following in the form ax where a is a prime number and x is rational: √ √ √ √ √ a) 3 7 b) 4 27 c) 5 16 d) 3 32 e) 7 49 1 f) √ 3 7

1 g) √ 4 27

1 h) √ 5 16

1 i) √ 3 32

1 j) √ 7 49

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Standard form (Scientific notation) 10 000 1 000 100 10 1

Observe the pattern. As we divide by 10, the exponent (power) of 10 decreases by one.

1 10 1 100 1 1000

= = = = = =

104 103 102 101 100 10−1

= 10−2 = 10−3

etc.

We can use this pattern to simplify the writing of very large and very small numbers. For example, 5 000 000 = 5 · 1 000 000 = 5 · 106 , and 1 3 =3· = 3 · 10−6 0, 000 003 = 1 000 000 1 000 000 Standard form (or scientific notation) involves writing any given number as a number between 1 and 10, multiplied by an integer power of 10, i.e., a · 10n

where a lies between 1 and 10.

If the original number is > 10, the power of 10 is positive. If the original number is < 1, the power of 10 is negative. If the original number is between 1 and 10, leave it as it is. More examples: 37 600 = 3, 76 · 104

3, 2 · 102 = 320

0, 00086 = 8, 6 · 10−4

5, 76 · 10−5 = 0, 0000576

Note about the use of calculators: If you perform on your calculator 2 300 000·400 000 your calculator will display 9.211 or 9.2E11 , depending on the model, which actually represents 9, 2 · 1011 . If you get something like 2.4−10 or 2.4E − 10 , this actually represents 2, 4 · 10−10 . Numbers which are already represented in standard form can be entered into the calculator using the EXP key. 1

Scientific notation - 3 o E SO

2

Exercises - Set A 1. Express the following in scientific notation: a) 259

b) 259 000

c) 25, 9

d) 0, 259

e) 0, 000 259

f) 40, 7

g) 4070

h) 0, 0407

i) 407 000 000

j) 0, 000 040 7

2. Express the following in scientific notation: a) The distance from the Earth to the Sun is 149 500 000 000 m. b) Bacteria are single cell organisms, some of which have a diameter of 0, 0003 mm. c) A speck of dust is smaller than 0, 001 mm. d) The probability that your six numbers will be selected for Lotto on Friday night is 0, 000 000 141 62. e) The central temperature of the Sun is 15 million degrees Celsius. f) A single red blood cell lives for about four months and during this time it will circulate around the body 300 000 times. 3. Write as an ordinary decimal number: a) 4 · 103

b) 2, 1 · 10−3

c) 4, 33 · 107

d) 8, 6 · 10−1

e) 3, 8 · 105

f) 2, 9 · 10−5

g) 5, 86 · 109

h) 5, 86 · 10−9

i) 5 · 10−2

j) 6, 3 · 108

4. Write as an ordinary decimal number: a) The wave length of light is 9 · 10−7 m. b) The world population in the year 2000 was 6, 130 · 109 . c) The diameter of our galaxy, the Milky Way, is 1 · 105 light years. d) The smallest visuses are 1 · 10−5 mm in size. e) The mass of a bee’s wing is 10−7 kg. 5. Use your calculator to find correct to 2 decimal places, and giving your answer in scientific notation: a) 0, 06 · 0, 002 : 4000 d) 320 · 600 · 51 400

b) 426 · 760 · 42 000 e) 0, 004 28 : 120 000

c) 627 000 · 74 000 f) 0, 026 · 0, 0042 · 0, 08

6. If a missile travels at 5400 km/h how far will it travel in: a) 1 day b) 1 week c) 2 years? (Give your answers in standard form with decimal part correct to 2 places and assume that 1 year=365,25 days.) 7. Light travels at a speed of 3 · 108 metres per second. How far will light travel in: a) 1 minute b) 1 day c) 1 year? (Give your answers in standard form with decimal part correct to 2 places and assume that 1 year=365,25 days.) 8. Find, with decimal part correct to 2 places: a) (3, 42 · 105 ) · (4, 8 · 104 )

b) (6, 42 · 10−2 )2

d) (9, 8 · 10−4 ) : (7, 2 · 10−6 )

e)


1 3, 8 · 105

c)

3, 16 · 10−10 6 · 107

f) (1, 2 · 103 )3

Scientific notation - 3 o E SO

3

Example 1 Simplify the following, giving answers in standard form: a) (5 · 104 ) · (4 · 105 ) b) (8 · 105 ) : (2 · 103 ) a) (5 · 104 ) · (4 · 105 ) = (5 · 4) · (104 · 105 ) = 20 · 109 = 2 · 1010 b) (8 · 105 ) : (2 · 103 ) =

8 105 · = 4 · 102 2 103

Exercises - Set B 1. Simplify the following, giving your answer in standard form: a) (3 · 103 ) · (2 · 102 )

b) (5 · 102 ) · (7 · 105 )

c) (5 · 104 ) · (6 · 103 ) e) (5 · 104 )2

d) (3 · 103 )2 f) (8 · 10−2 )2

g) (8 · 104 ) : (4 · 102 )

h) (8 · 105 ) : (2 · 103 )

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Polynomials

Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona

Algebraic notation and Polynomials

Algebraic Notation

The ability to convert worded sentences and problems into algebraic symbols and to understand algebraic notation is essential in the problem solving process. Notice that:

• • •

2 + 3 2 + 3 = 8 2 + 3 > 28

Is an algebraic expression, whereas Is an equation, and Is an inequality or inequation.

When we simplify repeated sums, we use product notation: For example:

+

= 2 ‘lots’ of =2 × =2

and ++ = 3 ‘lots’ of =3 × =3

When we simplify repeated products, we use index notation: For example:

× =

and

× × =

EXAMPLE 1 Write, in words, the meaning of:

a) − 5

b) +

c) 3 + 7

a) Is “5 less than x” b) Is “the sum of a and b” or “b more than a” c) Is “7 more than three times the square of x”

EXAMPLE 2 Write the following as algebraic expressions:

a) The sum of p and the b) The square of the sum of p square of q a) + b) ( + ) c) 2 −

and q

c) b less than double a

1



2

EXAMPLE 3 Write, in sentence form, the meaning of:

a) =

b)

a) D is equal to the product of c and t

=

b) A is equal to a half of the sum of b and c, or, A is the average of b and c.

EXAMPLE 4 Write “S is the sum of a and the product of g and t” as an equation. The product of g and t is

The sum of a and is + So, the equation is

= +

TO PRACTICE EXERCISE 1 Write in words, the meaning of: a. 2

b.

e. − 3

f.

b+c

g. 2 +

h. (2 )

j.

−

k. +

l.

i.

2

c. √

d.

(a + b)

EXERCISE 2 Write the following as algebraic expressions: a. The sum of " and #

i.

The difference between p and q if p>q

b. The sum of p, q and r

j.

a less than the square of b

c. The product of a and b

k. Half the sum of a and b

d. The sum of r and the square of s

l.

e. The square of the sum of r and s

m. A quarter of the sum of a and b

f.

n. The square root of the sum of m and n

The sum of the squares of r and s

g. The sum of twice a and b h. The sum of x and its reciprocal

The sum of “a” and a quarter of “b”

o. The square root of the sum of the squares of x and y



3

EXERCISE 3 Write in sentence form: a. $ = +

b. % =

d. ) =

e. * =

f.

+ =

h. = √ +

i.

=

-

g. % = , .

c. ' = 3(

&

&

EXERCISE 4 Write the following as algebraic equations: a. S is the sum of p and r b. D is the difference between a and b where b>a

The difference between two numbers is the larger one minus the smaller one

c. A is the average of k and m d. M is the sum of a and its reciprocal e. K is the sum of t and the square of s f.

N is the product of g and h

g. Y is the sum of x and the square of d and e

Algebraic Substitution

To evaluate an algebraic expression, we substitute numerical values for the unknown, then calculate the result. Consider the number crunching machine alongside:

Input x 5x - 7 calculator Output

If we place any number / into the machine, it calculates 0/ − 1. So, / is multiplied by 5, and then 7 is subtracted: For example: if = 2

5 − 7

=5×2−7 = 10−7 =3

and

if = −2 ;

5 − 7

= 5 × (−2) − 7 = −10 − 7 =−17



4

Notice that when we substitute a negative number such as −2, we place it in brackets. This helps us to get the sign of each term correct. TO PRACTICE EXERCISE 5 If 4 = 3, = 1 and 5 = −2, evaluate: 7

b.

78

c.

7:98

f.

978

g.

a.

8

e.

(7:8)

9

7:9

9:7 8

9:87

78: 9

d.

;8: 9

h.

7

7:9

9

−4

EXERCISE 6 If = −3, = −4 and = −1, evaluate: a.

b.

e. +

f.

( + )

c. +

d. ( + )

g. 2

h. (2 )

EXERCISE 7 If = 4, = −1 and = = 2, evaluate: a. > +

b. > +

e. >= −

f.

> +

c. >= −

d. > −

g. > + = + 2

h. >2 − 5=



DEFINITIONS: product, factors, sum, terms A product is an expression where the last operation is multiplication. In a product, the things being multiplied are called the factors. A sum is an expression where the last operation is addition. In a sum, the things being added are called the terms.

As an example, consider the expression

then here is the order that computations would be done: 1. Add b and c . 2. Multiply this sum by a .

( + )

Notice that the last operation done is multiplication. Thus, the expression ( + ) is a product. The factors are and + . As a second example, consider the expression

here is the order that computations would be done: 1. Multiply a and b . 2. Add this result to c .

+ .

Notice that the last operation done is addition. Thus, + is a sum. The terms are and . EXAMPLES The expression 35 is a product. The factors are @, /, A The expression −4( + 2) is a product. The factors are −B, /, / + C The expression 5 − 5 + 1 is a sum. The terms are 0/, −A, D The expression 2 + 253 − 7 is a sum. The terms are /2, CA3, −1 EXERCISE 8 In the following expressions, how many terms are there? And each term has how many factors? a) 2 + 4 + 5 ( + ) b) E + 5FG + 2

5



6

Algebraic expressions An algebraic expression is an expression that contains one or more numbers, one or more variables, and one or more arithmetic operations. It doesn't include an equal sign. Algebraic expressions can be or many forms, for example:

3x 2 − 2 x − 1

3x 2 + 2 x +

1 x

− 6x 2 y − 2

3x 2 + 2 x x2 − 3

A term consist of products of numbers and letters, so 3x2, -2x, -1, -6x2y, etc. are terms The number multiplying the letters is the coefficient of the term. 3x2 (x2 term. Coefficient is 3) -2x (x term. Coefficient is -2) -1 (constant term is -1)

Polynomials Polynomials are algebraic expresions. A polynomial in is a sum of terms, each of the form k , where:

is a real number, H is a nonnegative integer. That is, H = {0 , 1 ,2, 3 , . . . } .

DEFINITION: standard form of polynomials; degree; leading coefficient The standard form of a polynomial is:

"n /n + "n-1 /n-1+. . . + "1/ + "0

Here, n denotes the highest power to which is raised; this highest power is called the degree of the polynomial. Thus, in standard form, the highest power term is listed first, and subsequent powers are listed in decreasing order.



Notice that in the notation i i (read as " " KLM N ONPQK / OR OSQ N "), the number i denotes the coefficient of the i term. The number n , which is the coefficient of the highest power term, is called the leading coefficient of the polynomial.

Note that a constant (like 5 ) can be written as 5x0 . This is why the power is allowed to equal zero in the definition of polynomial—to allow for constant terms. EXAMPLE 5 The expression 5x4-x3-3x2+7x-5 coefficient and degree.

is a polynomial. Find its terms, coefficients, leading

The terms are: 5x4, -x3, -3x2, 7x, and -5 . Comparing each term with the required form ax k , we have:

term

writing in the form ax k

a

k

5x4

5x4

a=5

k=4

-x3

(-1)x3

a=-1

k=3

-3x2

-3x2

a=-3

k=2

7x

7x

a=7

k=1

-5

-5x0

a=-5

k=0

Notice that every value of a is a real number, and every value of k is a nonnegative integer. The standard form of this polynomial is: 54−3−3 2+7 − 5 Here, the highest power term is written first, and subsequent terms decrease in power. The degree is 4 , since this is the highest power. The leading coefficient is 5 , since this is the coefficient of the highest power term. Notice that the leading coefficient actually leads (comes at the beginning of) the polynomial, WHEN the polynomial is written in standard form.

EXAMPLE 6 The following expressions are NOT polynomials. Why? a)

1x + x- 1

b) x - x1/2

c) 7x2 - 7x + 7x1/ 2

7



8

a) 1x + x- 1 is not a polynomial; no negative powers are allowed. b) x - x1/2 is not a polynomial; the number 1/2 is not an allowable power. c) 7x2 - 7x + 7x1/ 2 is not a polynomial; the number 1/2 is not an allowable power.

DEFINITION: monomial, binomial, trinomial A polynomial with exactly one term is called a monomial. A polynomial with exactly two terms is called a binomial. A polynomial with exactly three terms is called a trinomial.

DEFINITION: quadratic, cubic, quartic A polynomial of degree A polynomial of degree A polynomial of degree A polynomial of degree

• •

• •

1 2 3 4

is called linear is called quadratic. is called a cubic. is called a quartic.

Polynomials have beautiful smooth graphs—no breaks and no kinks. The higher the degree of a polynomial, the more it is allowed to "turn" (change direction). Indeed, it can be shown easily (using calculus) that a polynomial of degree n can have at most n-1 turning points. The graph below is the polynomial T() = 3 − . Notice that this polynomial has degree 3 and has 2 turning points.



EXERCISE 9 1) What is the degree of these polynomials? a) 2x16+80x8 b) 7769x-97x7-56x9+31x19 c) 21x4-12x11+6x2-1710x10+4171x14 d) 60x3+65x5-3425x15 e) 8+27x11+234x18 2) What name is given to a polynomial with exactly one term? 3) What is the leading coefficient of these polynomials? a) b) c) d)

85-54x20 100x19-57x20+10029x6+2x4 61x4-6x3-7726x12+1425x11+45x19 9643x3-45x16-97x19

4) A polynomial is a sum of terms, each of a particular form. What is this form? 5) Is −28 an allowable term in a polynomial? 6) Is 2954 an allowable term in a polynomial? 7) Is −7x70 an allowable term in a polynomial? 8) Is the term x−9 an allowable term in a polynomial? 9) What is a quartic function? 10) What name is given to a polynomial with exactly two terms? 11) What is a trinomial? 12) What name is given to a polynomial with exactly one term? 13) Suppose that a polynomial has degree 8 . What (if anything) can be said about the number of turning points for this polynomial? 14) Suppose that a polynomial has 5 turning points. What (if anything) can be said about the degree of this polynomial? 15) Write the following polynomial in standard form: a) b) c) d) e)

-3x14+8x27+6x5+4x6 4x2-7x4+5x6 -6x26+3x9-7x6 5x-2x25-2x2 2x3+8x6+8x7+4x-3x2

16) Can the graph of a polynomial have a break in it? 17) Can the graph of a polynomial have a kink in it?

9

Departamento de Matemáticas. Real Instituto de Jovellanos. J. F. Antona

Adding and subtracting polynomials

1

Adding and subtracting polynomials Only like terms (those with identical letters and powers) can be added or subtracted.

Collecting like terms In algebra, Like terms are terms which contain the same variables (or letters) to the same indices.

For example: • and −2 are like terms • 2 and 3 are unlike terms because the power of x are not the same Algebraic expressions can often be simplified by adding or subtracting like terms. We call this collecting like terms. Consider

2 + 3 = + + + + = 5

2 lots of a

3 lots of a

5 lots of a

EXAMPLE 1 Where possible, simplify by collecting like terms: a) 4 + 3

b) 5 − 2

c) 2 − 1 +

d) − 2

e) 2−4

a) 4 + 3 = 7 b) 5 − 2 = 3 c) 2 − 1 + = 3 − 1

(since 2 and are like terms)

d) − 2 = −

(since and −2 are like terms)

2

e) −4

cannot be simplified since 2 and −4 are unlike terms.

EXAMPLE 2 Simplify: a) 3 2 − 7 2

b) 3 2−7

c) 3 3−7 3−2 + 2 6−3 6−4 2

a) 3 2 − 72 = −42 b) 3 2−7 cannot be expressed as a single term. c) 3 3−7 3−2 + 2 6−3 6−4 2= −4 3−2 − 6−4 2

Departamento de Matemáticas. Real Instituto de Jovellanos. J. F. Antona

Adding and subtracting polynomials

2

EXERCISE 1 Simplify, where possible, by collecting like terms: a)

5++4

b)

e)

+−3

i) + 2 m)

2 + 3 − 5

c)

−2+5

d)

+1+

f) 5 +

g)

5 −

h)

− 5

j) + +

k)

5 + 5

l) − 5 + 5

o)

4 +

p)

n)

6+3+

2 + 3 −

3 −

EXERCISE 2 Simplify, where possible: a)

7 − 7

b)

7 −

c)

7 − 7

d)

+ 2

e)

− 2

f)

4 −

g)

+ 3 + 2 + 4

h)

2 + + 3 − 4

i)

2 − + 3 + 3

j)

3 + 2 − −

k)

+ 4 − 3 + 2

l)

+ 2 − − 5

n)

+ + + 4

o)

2 − 3 − − 7

m) + 5 + 2 − 3

EXERCISE 3 Simplify, where possible: a)

4 + 6 − − 2

b)

2 + − 2

c)

3 − 2 +

d)

+ 2 + 2 − 5

e)

− 6 + 2 − 1

f)

3 + 7 − 2 − 10

g)

−3 + 2 − −

h)

+ 2 −

i)

2 − − + 3

j)

4 − −

k)

+ +

l)

4 − 2 − −


Multiplication of polynomials

1

Multiplication and expansion of polynomials

A. PRODUCT NOTATION In algebra we agree: To leave out the “×” signs between any multiplied quantities provided that at least one of them is an unknown (letter) • To write numerals (numbers) first in any product • Where products contain two or more letters, we write them in alphabetical order For example: • 3 is used rather than 3 × or 3 • 2 is used rather than 2 •

ALGEBRAIC PRODUCTS The product of two or more factors is the result obtained by multiplying them together Consider the factors −3 22 . Their product −3 × 22 can be simplified by following the steps below: • • •

Step 1: Find the product of the signs. Step 2: Find the product of the numerals or numbers. Step 3: Find the product of the variables or letters.

−2 × 3 = −6

So.

× =

− × + = − 3 × 2 = 6

EXAMPLE 1 Simplify the following products:

a) −3 × 4

b) 2 × −

c) −4 × −2

a) −3 × 4

b) 2 × −

c) −4 × −2

= −12

= −2

= 8

For −2, the sign is −, the numeral is 2, and the variable is



2

EXERCISE 1 Write the following algebraic products in simplest form: a) ×

b) × 2 ×

c) ×

d) × 2

e) 2 × 3

f) 4 × 5

g) −2 × 7

h) 3 × −2

i) 2 ×

j) 3 × 2

k) −2 ×

l) −3 × 4

m) −2 × (−)

n) −3 ×

o) − × (−2)

p) 3 × (−2 )

q) (−)

r) (−2)

s) 2 ×

t) × (−3)

EXERCISE 2 Simplify the following: a)

2 × 5 + 3 × 4

b)

5 × 3 − 2 ×

c)

3 × + 2 × 4

d)

× 2 + × 3

e)

4 × − 3 ×

f)

3 × − 2 × 2

g)

3 × + 2 × 2

h)

4 × − 3 × 2

i)

3 × − 2 ×

•

Multiplication is often written using the multiplication sign or cross symbol "×" between the factors: ×

•

In algebra multiplication is sometimes denoted by a middle dot ∙

•

In algebra, multiplication involving variables is often written as a juxtaposition

•

This notation can also be used for quantities that are surrounded by parentheses ()()

EXERCISE 3 Simplify the following: (−5 ) ∙ (4 ) ∙ ( )

a)

(2) ∙ (3)

b)

d)

(12) ∙ (6 ) ∙ ( 23)

e) (−3 3) ∙ (−26 5)

c)

(3 ) ∙ (−5 )

f)

(−4 736) ∙ (332 5)



3

B. THE DISTRIBUTIVE LAW Consider the expression 2( + 3). We say that 2 is the coefficient of the expression in the brackets. We can expand the brackets using the distributive law:

( + ) = + The distributive law says that we must multiply the coefficient by each term within the brackets, and add the results. EXAMPLE 2 Expand the following:

a) 3(4 + 1) a)

b) 2(5 − 2)

c) −2( − 3)

b)

c)

3(4 + 1)

2(5 − 2)

= 3 × 4 + 3 × 1

= 2 × 5 + 2 × (−2)

= 12 + 3

= 10 − 4

2

−2( − 3) = (−2) × + (−2) × (−3) 2

= −2 + 6

With practice, we do not need to write all of these steps. EXAMPLE 3 Expand and simplify:

a) 2(3 − 1) + 3(5 − ) a)

b) (2 − 1) − 2(5 − ) b)

2(3 − 1) + 3(5 − )

(2 − 1) − 2(5 − )

= 6 − 2 + 15 − 3

= 2 − − 10 + 2

= 3 + 13

= 4 − 11

Notice in b) that the minus sign in front of 2 affects both terms inside the following brackect

EXERCISE 4 Expand and simplify: a) 3( + 1)

b) 2(5 − )

c) – ( + 2)

d) ( + 3)

e) −3( + 2)

f) −2( − )

g) (2 − 1)

h) 2( − − 2)

i) 1 + 2( + 2)

j) 13 − (4( + 3)

k) 3( − 6) + 2

l) ( − 1) +

m) 2(3 − ) +

n) 4 − 5(2 − 3)

o) 7 − 4( + 2)

p) 3( − ) + 5



4

EXERCISE 5 Expand and simplify: a) 3( − 5) + 2(5 + )

b) 2 + (5 − 7)

c) 2 − (3 − 6)

d) 3( + 2) + 5(4 − )

e) 6( − 2) − 4(3 + 5)

f) 4 − 3(2 − 3)

g) 2 − (3 − 2)

h) 2 (4 − 3) − 3(5 + 3) i) −3 ( − 4)

j) (5 − )

k) 3 (2 − 4 )

C. THE PRODUCT

(! + ")(# + $)

l) −4 (2 − 5) − 2(3 − )

( + )( + ) = + + +

To remember

FOIL rule

= is the product of First terms of each bracket = is the product of Outer terms of each bracket = is the product of Inner terms of each bracket = is the product of Last terms of each bracket

EXAMPLE 4 Expand and simplify:

a) ( + 2)( + 3) a)

b) (2 + 3)(3 − 5) b)

( + 2)( + 3)

(2 + 3)(3 − 5)

= + 3 + 2 + 6

= 6 − 10 + 9 − 15

= + 5 + 6

= 6 − − 15



5

EXERCISE 6 Expand and simplify: a) ( − 6)( + 7)

b) ( + 3)( − 2)

c) ( + 5)( − 5)

d) (2 + 2)(4 − 6)

e) ( − 2)(1 − 3)

f) (6 − 1)(6 − 1)

g) (4 − 6)(3 − 2)

h) (4 − 3)(5 + 3)

i) ( − 3)( − 4)

j) (2 − 5)(3 + 4)

k) (3 + 4 )(2 − 5)

l) (5 − 4)(7 + 5)


a) ( + 5)( − 5) a)

( + 5)( − 5)

What do you notice about the two middle terms?

b) (4 − 3)(4 + 3) b)

(4 − 3)(4 + 3)

= − 5 + 5 − 25

= 16 + 12 − 12 − 9

= − 25

= 16 − 9

EXERCISE 7 Expand and simplify: a) ( − 6)( + 6)

b) ( + 3)( − 3)

c) (2 + 5)(2 − 5)

d) (4 + 6)(4 − 6)

e) (3 − 2)(2 + 3)

f) (6 + 1)(6 − 1)


a) (2 + 5) a)

(2 + 5)

b) (4 − 3)

What do you notice about the two middle terms?

b) (4 − 3)

= (2 + 5)(2 + 5)

= (4 − 3)(4 − 3)

= 4 + 10 + 10 + 25

= 16 − 12 − 12 + 9

= + 20 + 25

= 16 − 24 + 9

EXERCISE 8 Expand and simplify doing the product: a) ( + 6)

b) ( − 3)

c) (2 − 5)

d) (4 − 3)

e) (3 − 2)

f) (6 + )


D.


6

DIFFERENCE OF TWO SQUARES

If and are perfect squares then − is called the difference of two squares Notice that

( + )( − ) = − + − = − The middle two terms add to zero

Thus,

( + )( − ) = −


a) (3 + 5)(3 − 5) a)

b) (4 − 3)(4 + 3)

(3 + 5)(3 − 5)

b)

(4 − 3)(4 + 3)

= (3) − (5)

= (4) − (3)

= 9 − 25

= 16 − 9

EXERCISE 9 Expand and simplify using the rule (! + ")(! − ") = !& − "& : a) ( − 6)( + 6)

b) ( + 3)( − 3)

c) (2 + 5)(2 − 5)

d) (4 + 6)(4 − 6)

e) (3 − 2)(2 + 3)

f) (6 + 1)(6 − 1)

g) (2 − 1)(2 + 1)

h) (2 + 3)(2 − 3)

i) (2 + 5)(2 − 5)

j) (4 + 6)(4 − 6)

k) (3 − 2)(2 + 3)

l) (6 + )(6 − )

m) (4 + 5)(4 − 5)

n) (3 + )(3 − )

o) (2 + 4)(2 − 4)

E. PERFECT SQUARES EXPANSION ( + ) and ( − ) are called perfect squares. Notice that

( + ) = ( + )( + ) = + + + using FOIL = + 2 +

Notice that the middle two terms are identical



7

Thus, we can state the perfect square expansion rule:

( + ) = + 2 + We can remember the rule as follows: Step 1: Square the first term. Step 2: Add twice the products of the first and last terms. Step 3: Add on the square of the last term.

( − ) = ( − )( − )

Notice that

= − − + using FOIL = − 2 + So, we can state:

( − ) = − 2 + EXAMPLE 8 Expand and simplify:

a) (5 + 3) a)

b) (6 − 4)

(5) + 2 ∙ 5 ∙ 3 + 3

= 25 + 30 + 9

b) 6 − 2 ∙ 6 ∙ 4 + (4) = 36 − 48 + 16

EXERCISE 10 Expand and simplify using the perfect square expansion rule: a) ( + 6)

b) ( − 3)

c) (2 − 5)

d) (4 − 3)

e) (3 − 2)

f) (6 + )

g) (3 + 4)

h) (2 − )

i) (7 + 2)

j) (12 − 3)

k) (3 − 4)

l) (25 − 2)


a) (3 + 5)

a) (3 + 5)

b) 6 − (4 + 3)

b) 6 − (4 + 3)

= (3 ) + 2 ∙ 3 ∙ 5 + 5

= 6 − (4 + 2 ∙ 4 ∙ 3 + (3))

= 9 + 30 + 25

= 6 − (16 + 24 + 9 ) = 6 − 16 − 24 − 9 = −10 − 24 − 9

Notice the use of square brackets in the second line. These remind us to change the signs inside them when they are removed



8

EXERCISE 11 Expand and simplify: a) ( + 4)

b) (2 − )

c) (7 + 2 )

d) (5 − 3 )

e) (2 − 3)

f) (2 − 5 )

g) ( − 2)

h) (2 − 4 )

i) (2 − 6)

EXERCISE 12 Expand and simplify: a) 3 + 1 − ( + 4)

b) 5 − 3 + ( − 2)

c) ( + 1)( − 1) − ( + 4)

d) ( + 3)( − 3) − ( − 2)

e) (2 − 3) + ( + 5)( − 5)

f) (3 + 2) − ( + 4)( − 4)

g) (3 + 2)(3 − 2) − (5 − )

h) (2 + 9)(2 − 9) − (4 − 6)

i) ( − 7) + ( + 8)

j) (4 + 2) − ( − 9)

F. FURTHER EXPANSION In this section we will expand more complicated expressions by repeated use of the expansion laws. ( + )( + + ') Consider the expansion of

( + )( + + ') = + + ' + + + '

Notice that there are 6 terms in this expansion and that each term within the first bracket

is multiplied by each term in the second

2 terms in the first bracket × 3 terms in the second bracket = 6 terms in the expansion



9


(2 + 4)( + 3 + 5)

(2 + 4)( + 3 + 5) = 2 + 6 + 10 + 4 + 12 + 20 = 2 + 10 + 22 + 20

{Collecting like terms}


( + 2)

( + 2) = ( + 2)( + 2) = ( + 2)( + 4 + 4) = ( + 2)( + 4 + 4)

{each term within the first bracket is multiplied by each term in the second bracket}

= + 4 + 4 + 2 + 8 + 8 = + 6 + 12 + 8



a)

( + 1)(2 − 3 + 4)

a) ( + 1)(2 − 3 + 4)

b) ( + 2)( − 3)( + 3)

b) ( + 2)( − 3)( + 3) = ( + 2)( − 3)( + 3) = ( + 2)( − 9) {difference of two squares}

= ( + )(2 − 3 + 4) = − 9 + 2 − 18 = 2 − 3 + 4 + 2 − 3 + 4 = 2 − + + 4


Always look for ways to make your expansions simpler. In b) we can use the difference of two squares

= + 2 − 9 − 18



10

EXERCISE 13 Expand and simplify: a) ( + 3)( + + 5)

b) (2 − 4)(3 − 5 + 2)

c) (3 − 1)(5 + 4 − 6)

d) (4 + 3)(6 − 2 − 8)

e) ( − 2)

f) (2 + 3)

g) (2 + 1)

h) (3 − 2)

i) ( + 3)( + 3)

j) ( − 3)( + 1)

k) 2( − 4)( + 4)

l) ( + 2)( − 2)(2 + 3)

m) −3(2 − 1)( + 2)

n) (1 − 3)(2 + 5)

o) 2 ( − 1)

p) (3 − 2)( − 2)( + 1)

q) ( − 3)(1 − )(2 − 2)

r) ( − 1)( − 2)( + 2)

HARDER EXTENSION EXERCISE 14 Expand and simplify: a) ( + 3 )(4 ) + 2 + 5 )

b) (3 − 2 )( − 3 − 5 − 1)

c) (2 + + 3 − 2)( − 3 + 1)

d) ( − 7 + − 1)( + 4 − 6 + 2)

G.

THE BINOMIAL EXPANSION

Consider

( + )* .

• • •

We note that:

+ is called a binomial as it contains two terms Any expression of the form ( + )* is called a power of a binomial The binomial expansion of ( + )* is obtained by writing the expression without brackets.

Now (! + ")+ = (! + ")(! + ")& = (! + "),!& + &!" + "& = !+ + &!& " + !"& + "!& + &!"& + "+ = !+ + +!& " + +!"& + "+



11

So, the binomial expansion of

(! + ")+ = !+ + +!& " + +!"& + "+ Doing the same process with (! − ")+ = (! − ")(! − ")& = (! − "),!& − &!" + "& - = !+ − &!& " + !"& − "!& + &!"& − "+ = !+ − +!& " + +!"& − "+ So, the binomial expansion of

(! − ")+ = !+ − +!& " + +!"& − "+ EXAMPLE 12 Expand and simplify using the binomial expansion:

a)

( + 2)

b) (2 − 4)

a) {Using ( − ) = − 3 + 3 − }

b)

We substitute = and = 2

{Using ( − ) = − 3 + 3 − } We substitute = (2) and = 4

∴ ( + 2) = + 3 2 + 32 + 2

∴ (2 − 4) = (2) − 3(2) 4 + 3(2)4 − 4

= + 6 + 12 + 8

= 8 − 48 + 96 − 64

We use brackets to assist our substitution

b)

{Using ( + ) = + 3 + 3 + } We substitute = (2) and = (−4)

∴ (2 − 4) = (2) + 3(2) (−4) + 3(2)(−4) + (−4) = 8 − 48 + 96 − 64

EXERCISE 15 Use the binomial expansion for (! + ")+ or (! − ")+ to expand and simplify: a) ( + 1)

b) (3 − 2)

c) ( − 5)

d) (2 + 3)

e) (4 − 6)

f) (2 + 1)

g) ( − 2)

h) ( − 4 )

i) (3 + )



EXERCISE 15 (Extension) Find, in terms of x, the surface area of the following figures: a)

c)

b)

d)

EXERCISE 16 (Extension) Find, in terms of x, the surface area of the orange zone of the following figures: a)

EXERCISE 17 (Extension) Find, in terms of x, the volume of the following figure: a)

b)

12



EXERCISE 18 (Extension) a) Find, in terms of /, the surface area and the volume, of the following cylindrical water tank trailer, with hemispherical endings. b) Find the surface and volume if / = + m.

10 m

EXERCISE 19 (Review) a) Expand and simplify: 1.

4 ∙ 8

2.

6 ∙ 2

3.

2 ∙ 5 ∙ (−3 ) ∙

4.

(−4)(−7)

5.

5 2 ∙ (−32 3)

6.

(5 − 6)

7.

4 − 2(2 + 3)

8.

2( − 6) + 3(2 − )

9.

5(2 − 3) − 8(4 − 7)

10.

3 (4 ) − 7 )

11.

−6 (3 − 4) − 3(5 − 2)

12.

(3 + 2)( − 4)

13.

( − 5)

14.

(3 + 4)

15.

(2 − 6)

16.

(2 − 6)(2 + 6)

17.

( − 5)( + 5)

18.

( − 3)( + 3)

13


19.

– 5( + 2)( − 2)

20.

−( + 3)

21.

(2 − 5)( − 1)( + 1)

22.

( − 5)( − 1)( − 1)

23.

( − 1)

24.

5 + 2 − ( + 2)

25.

(3 − 2)( − 2 + 7)

26.

( − 1)( + 2)( − 3)

27.

( + 2)

28.

( + 1)( + 1)( − 1)

29.

( + 5)( − 5)(2 + 3)

30.

(2 − )(3 − + 5 − 2)

31.

(3 − 4)


b) Find, in terms of x, the surface area of the orange zone of the following figures: 1.

2.

14



KEY POINTS Useful products:

(In Spanish: “productos notables”)

( + ) = + ( + )( + ) = + + +

( + )( − ) = − ( + ) = + 2 + ( − ) = − 2 + ( + ) = + 3 + 3 + ( − ) = − 3 + 3 − Expansion and factorisation relation: Expansion

( + ) = + 2 + Factorisation Note: !& + "& cannot be factorised. !& − "& is the difference of squares

MORE EXERCISES TO PRACTICE AT HOME: EXERCISE 20 1)

2)

15


3)

4)

5)

6)

7)

8)

9)


16

Equations

Linear equations Remember: • x2 + 3x • x2 + 3x = 8

• x2 + 3x > 28

is an algebraic expression, whereas is an equation, and is an inequality or inequation.

Many mathematical situations can be described using an equation. This is a formal statement that one expression is equal to another. The expressions are connected by the equals sign =. An algebraic equation is an equation which involves at least one variable. For example, 3x + 5 = 8 is an algebraic equation. We read the symbol = as “equals” or “is equal to”. By following a formal procedure, we can solve the equation to find the value of the variable which makes it true. In this case we see that x must be 1, since 3 + 5 = 8.

1. Linear equations A linear equation contains at least a variable which can only be raised to the power 1. For example, 3x + 5 = 8, whereas

1 2x

x2 − 3x + 2 = 0,

−3=2

and

1 √ = x, and x

1+x =x 3 x3 = 8

are linear equations

are not linear equations.

The left hand side (LHS) of an equation is on the left of the = sign. The right hand side (RHS) of an equation is on the right of the = sign. For example,

3x + 5 = |{z} 8 | {z } RHS LHS

The solution of a linear equation is the value of the variable which makes the equation true. In other words, it makes the left hand side (LHS) equal to the right hand side (RHS). In the example true is x = 1.

3x + 5 = 8

above, the only value of the variable x which makes the equation

1

Linear equations - 3 o E SO

2

2. Maintaining balance For any equation the LHS must always equal the RHS. We can therefore think of an equation as a set of scales that must always be in balance. The balance of an equation is maintained provided we perform the same operation on both sides of the equal sign.

Any mathematical operation we perform on one side of an equation we must also perform on the other side.

Example 1 What equation results when: 3x − 2 a) both sides of = −1 are multiplied by 4 4 b) both sides of 5x = −15 are divided by 5? 3x − 2 3x − 2 = −1 ⇒ 4 · = 4 · (−1) ⇒ 3x − 2 = −4 4 4 5x −15 b) 5x = −15 ⇒ = ⇒ x = −3 5 5 a)

Exercises - Set A 1. What equation results from adding: a) 6 to both sides of x − 6 = 4

b) 7 to both sides of x − 7 = 6

c) 11 to both sides of 2x − 11 = 3

d) 13 to both sides of 3x − 13 = −2?

2. What equations results from subtracting: a) 2 from both sides of x + 2 = 3 c) 5 from both sides of 2x + 5 = 03

b) 4 from both sides of x + 4 = −2

d) 9 from both sides of 3x + 9 = −1?

3. What equation results from multiplying both sides of: a) x = −3 by 3 d)

x = −1 by 4 4

b) 3x = 1 e)

by 5

x = −2 by −3 −3

4. What equation results from dividing both sides of:

x = 4 by 2 2 x f) = 4 by −10? −10 c)

a) 3x = 12 by 3

b) −5x = 30 by −5

c) 2x + 8 = 0 by 2

d) 3x − 9 = 24 by 3

e) 4(x + 2) = −12 by 4

f) −6(x − 1) = −24 by −6?



3

3. Formal solution of linear equations When we use the = sign between two algebraic expressions we have an equation which is in balance. Whatever we do to one side of the equation, we must do the same to the other side to maintain the balance.

Technique for solving equations: Do the same to

both sides

Since both sides of an equation are equal, any operation performed on both sides will not affect the equality. For example, if one side is multiplied by 2, so must the other side, to maintain equality. Compare the balance of weights:

As we perform the inverse operations necessary to isolate the unknown, we must perform the same operations on the other side to maintain the balance. Consider how the expression has been built up and then isolate the unknown by using inverse operations in reverse order.

Example 2 Solve for x:

3x + 7 = 22

3x + 7 − 7 = 22 − 7 3x = 15

3x 15 = 3 3 x= 5 Check:

(subtract 7 from both sides) (simplify) (divide both sides by 3) (simplify)

LHS= 3 · 5 + 7 = 22 ⇒ LHS=RHS


11 − 5x = 26

11 − 5x − 11 = 26 − 11 −5x = 15

−5x 15 = −5 −5 x = −3 Check:

(subtract 11 from both sides) (simplify) (divide both sides by −5) (simplify)

LHS= 11 − 5 · (−3) = 26 ⇒ LHS=RHS D pto. M atemáticas. IES Jovellanos. 2012


4


x +2−2 3 x 3 x ·3 3 x

x + 2 = −2 3

= −2 − 2

(subtract 2 from both sides)

= −4

(simplify)

= −4 · 3

(multiply both sides by 3)

= −12

(simplify)


5·

4x + 3 = −2 5

4x + 3 = −2 · 5 5 4x + 3 = −10

(multiply both sides by 5) (simplify)

4x + 3 − 3 = −10 − 3 4x = −13

4x 13 = − 4 4 x = −3 41

(subtract 3 from both sides) (simplify) (divide both sides by 4) (simplify)

Exercises - Set B 1. Solve for x: a) 6 − x = −5

b) −4x = 15

d) 5 − 4x = −7 2. Solve for x: x a) = 5 2 x d) − 2 = 1 3 3. Solve for x: 2x + 7 a) =0 5 3x − 1 d) =4 2

c) 3 − 2x = 7

e) 3 − 7x = −2 2x = −4 3 x−1 e) =1 2 b)

1 (1 − 3x) = −1 2 1 e) (4 − 3x) = −1 5 b)


f) 17 − 2x = −1 x + 1 = −3 4 x+5 f) = −1 3 c)

1 + 4x =3 7 1 f) (1 + 2x) = −2 4 c)


5

4. Equations with a repeated unknown Equations in which the unknown appears more than once need to be solved systematically. We use the following procedure: Step 1: If necessary expand any brackets and collect like terms. Step 2: If necessary, remove the unknown from one side. Aim to be left with a positive unknown on one side. Step 3: Use inverse operations to isolate the unknown and maintain balance. Step 4: Check that your solution satisfies the equation, i.e. LHS=RHS.


5(x + 1) − 2x = −7

5x + 5 − 2x = −7 3x + 5 = −7

3x + 5 − 5 = −7 − 5 3x = −12

3x 12 =− 3 3 x = −4

(expand brackets) (collect like terms) (subtract 5 from both sides) (simplify) (divide both sides by 3) (simplify)


15 − 2x = 11 + x

15 − 2x + 2x = 11 + x + 2x 15 = 11 + 3x

15 − 11 = 11 + 3x − 11 4 = 3x

4 3x = 3 3 x = 1 31

(add 2x to both sides) (simplify) (subtract 11 from both sides) (simplify) (divide both sides by 3) (simplify)



6 Exercises - Set C 1. Solve for x: a) 4(x − 1) − 2x = 2

c) 4(x − 2) − (x + 1) = 6

b) 2(x − 3) + 3x = 9

d) 5(3 + 2x) + 2(x + 1) = −7

2. Solve for x: a) 3x + 2 = x + 10 d) 3x − 1 = 6x − 4

b) 7x + 2 = 4x − 8

c) 8x + 3 = 4x + 4

e) 4x − 2 = 8x + 10

f) 5 + 2x = 18 + 4x

b) 6 + 2x = 15 − x

c) 6 + x = 13 − 2x

3. Solve for x: a) 3x + 5 = 1 − x

d) 9 − 2x = 12 − 7x

e) 4 − 3x = 8 − x

f) 9 − 3x = x + 2

4. Solve for x: a) 3(x + 1) − x = 15

b) 5(x + 2) − 2x = 11

e) 4(3x + 1) + 18 = x

f) 7x − 2(x + 1) = 13

c) 6(1 − 2x) = −4 − 7x g) 14x − 5(2x + 5) = 7

i) 2x − 4(4 − 3x) = x + 10

k) 2(x − 6) + 7x = 3(3x − 4)

d) 2(x + 4) + 8(x − 2) = 12 h) 19 − (3 − x) = 5x

j) 2(x + 4) = 7 − (3 + x)

l) 5(2x − 4) = 4x − 2(1 − 3x)

5. Fractional equations More complicated fractional equations can be solved by: • writing all fractions with the same lowest common denominator (LCD), and then • equating numerators.


2 x = 3 5 x·5 2·3 = 3·5 5·3 5x = 6 x =

1 51

x 2 = 3 5

has LCD of 15 (to achieve a common denominator) (equating numerators) (divide both sides by 5)

In this case, as we have only one fraction on the LHS and only one fraction on the RHS, we get the same result cross-multiplying the fractions: x 2 = ⇒ 5x = 2 · 3 ⇒ 5x = 6 ⇒ x = 1 51 3 5



7


2x + 3 x−2 = 4 3 (cross-multiply)

3(2x + 3) = 4(x − 2)

(expand brackets)

6x + 9 = 4x − 8

(use inverse operations)

6x − 4x = −9 − 8

(simplify)

2x = −17 17 x= − 2

(divide both sides by 2)


2x + 1 x − 2 − =5 3 2

2x + 1 x − 2 − = 5 3 2 2 · (2x + 1) 3 · (x − 2) 6·5 − = 2·3 3·2 6

has LCD of 6 (to achieve a common denominator)

2(2x + 1) − 3(x − 2) = 30

(equating numerators)

x + 8 = 30

(collecting like terms)

4x + 2 − 3x + 6 = 30

(expanding brackets)

x = 22 Note the use of brackets in the original fractions.

Exercises - Set D 1. Solve for x: 6x 3 a) = 4 2

b)

2 x = 3 5

c)

4x 1 = 5 2

d)

1 5x = 5 3

e)

2x + 3 2x = 5 3

f)

x+1 2x + 5 = 3 4

g)

4−x x+1 = 2 3

h)

3x + 8 =x−5 3

i)

x+2 5−x = 3 4

2. Solve for x: x x a) + = 2 2 5 d)

3x 2x 1 + =− 4 3 2

b)

x 2x 5 − = 2 3 6

c)

3x x − = 11 2 8

e)

5x x 3 − = 2 6 8

f)

x 3x − =4 7 2



8

Exercises - Set E 1. Solve for x: x 2x + 1 =0 a) + 3 6

b)

2x + 3 x − 1 3 + = 5 2 4

c)

2x − 3 x + 4 + = −2 5 2

d)

x − 2 2x + 1 + =2 3 12

e)

4x − 3 x + 2 1 + =− 5 3 4

f)

x − 4 2x − 3 5 + = 3 8 6

2. Solve for x: x + 1 2x − =4 a) 2 5

b)

3x − 2 x − 1 1 − = 3 2 2

c)

4x x − 2 − = −3 5 2

d)

2x − 1 x + 2 1 − =− 4 3 3

e)

4x + 1 2x − 3 − =1 2 6

f)

x−5 x+1 − = −2 6 4

Exercises - Set F 1. Solve for x: 30 − x 5 a) = 20 + x 4 c) 3(2x − 5) + 8x − 6 =

x − (5x + 3) 2

b)

x x x − = − 11 5 9 3

d)

x−3 x+5 x−1 = − 4 6 9 3x + 1 = −5x − 3 2

e)

2x 3x − 5 x − = −3 15 20 5

f) −4(2x − 1) +

g)

8−x 4(3 − x) 4x + 10 − =2−x− 5 2 10

h)

20 − x 3x − 5 x + 1 1 − 2x − = + 3 4 12 4

j)

3x − 1 1 − 4x 1 − x 14 − x − = − 4 5 4 6

i)

2x − 1 5x + 2 2x − 3 − = +1 3 12 4

k)

x−4 (−4x + 2) 5x + 6 − 4(−2x + 1) − = 2(x − 3) + 5 10 2

✁✃✁✃✁✃✁✃✁✃✁✃✁✃


W ord problem s 1. Problem solving Many problems can be translated into algebraic equations. When problems are solved using algebra, we follow these steps: Step 1: Read the problem. Step 2: Decide on the unknown quantity and allocate a variable. Step 3: Decide which operations are involved. Step 4: Translate the problem into an equation. Step 5: Solve the equation by isolating the variable. Step 6: Check that your solution does satisfy the original problem. Step 7: Write your answer in sentence form. Remember, there is usually no variable in the original problem. Example 1 When a number is trebled and subtracted from 7, the result is −11. Find the number. Let x be the number, so 3x is the number trebled. 7 − 3x is this number subtracted from 7. So, 7 − 3x = −11 ⇒ −3x = −18 ⇒ x = 6 So, the number is 6.

Check:

7 − 3 · 6 = 7 − 18 = −11

X

Example 2 Sarah’s age is one third her father’s age. In 13 years’ time her age will be a half of her father’s age. How old is Sarah now? Let Sarah’s present age be x years, so her father’s present age is 3x years. So, 3x + 13 = 2(x + 13)

Table of ages: Sarah Father

Now x 3x

3x + 13 = 2x + 26 3x − 2x = 26 − 13

13 years time x + 13 3x + 13

x = 13 Sarah’s present age is 13 years.

1

W ord problem s - 3 o E SO

2

Exercises - Set A 1. When three times a certain number is subtracted from 15 the result is −6. Find the number. 2. Five times a certain number, minus 5, is equal to 7 more than three times the number. What is the number? 3. Three times the result of subtracting a certain number from 7 gives the same answer as adding eleven to the number. Find the number. 4. I think of a number. If I divide the sum of 6 and the number by 3, the result is 4 more than one quarter of the number. Find the number. 5. The sum of two numbers is 15. When one of these numbers is added to three times the other, the result is 27. What are the numbers? 6. What number must be added to both the numerator and denominator of the fraction get the fraction 87 ?

2 5

to

7. What number must be subtracted from both the numerator and denominator of the fraction 3 1 4 to get the fraction 3 ? 8. Eli is now one quarter of his father’s age. In 5 years’ time his age will be one third of his father’s age. How old is Eli now? 9. When Maria was born, her mother was 24 years old. At present, Maria’s age is 20 % of her mother’s age. How old is Maria now? 10. Five years ago, Jacob was one sixth of the age of his brother. In three years’ time his age doubled will match his brother’s age. How old is Jacob now?

2. Money and investment problems Problems involving money can be made easier to understand by constructing a table and placing the given information into it.

Example 3 Britney has only 2-cent and 5-cent stamps. Their total value is $1,78, and there are two more 5-cent stamps than there are 2-cent stamps. How many 2-cent stamps are there? If there are x 2-cent stamps then there are (x + 2) 5-cent stamps. 2x + 5(x + 2) = 178 Type 2-cent 5-cent

Number x x+2

Value 2x cents 5(x + 2) cents

(equating values in cents)

2x + 5x + 10 = 178 7x = 168 x = 24 So, there are 24 2-cents stamps.



3

Exercises - Set B 1. Michaela has 5-cent and 10-cent stamps with a total value of 5,75 e. If she has 5 more 10-cent stamps than 5-cent stamps, how many of each stamp does she have? 2. The school tuck-shop has milk in 600 ml and 1 litre cartons. If there are 54 cartons and 40 ml of milk in total, how many 600 ml cartons are there? 3. Aaron has a collection of American coins. He has three times as many 10 cent coins as 25 cent coins, and he has some 5 cent coins as well. If he has 88 coins with a total value $11,40, how many of each type does he have? 4. Tickets to a football match cost 8 e, 15 e or 20 e each. The number of 15 e tickets sold was double the number of 8 e tickets sold. 6000 more 20 e tickets were sold than 15 e tickets. If the gate receipts totalled 783 000 e, how many of each type of ticket were sold? 5. Kelly blends coffee. She mixes brand A costing $6 per kilogram with brand B costing $8 per kilogram. How many kilograms of each brand does she have to mix to make 50 kg of coffee costing her $7,20 per kg? 6. Su Li has 13 kg of almonds costing $5 per kilogram. How many kilograms of cashews costing $12 per kg should be added to get a mixture of the two nut types costing $7,45 per kg?

3. Motion problems Motion problems are problems concerned with speed, distance travelled, and time taken. These variables are related by the formulae: speed =

distance time

distance = speed × time

time =

distance speed

Speed is usually measured in either kilometres per hour (denoted km/h or km h−1 ) or metres per second (denoted m/s or m s−1 ). Example 4 A car travels for 2 hours at a certain speed and then 3 hours more at a speed 10 km h−1 faster than this. If the entire distance travelled is 455 km, find the car’s speed in the first two hours of travel. Let the speed in the first 2 hours be s km h−1 . First section Second section

Speed (km h−1 ) s (s + 10)

Time (h) 2 3 Total

Distance (km) 2s 3(s + 10) 455

So, 2s + 3(s + 10) = 455 2s + 3s + 30 = 455 5s = 425 s = 85 The car’s speed in the first two hours was 85 km h−1 .



4

Exercises - Set C 1. Joe can run twice as fast as Pete. They start at the same point and run in opposite directions for 40 minutes. The distance between them is now 16 km. How fast does Joe run? 2. A car leaves a country town at 60 km per hour. Two hours later, a second car leaves the town; it catches the first car after 5 more hours. Find the speed of the second car. 3. A boy cycles from his house to a friend’s house at 20 km h−1 and home again at 25 km h−1 . 9 If his round trip takes 10 of an hour, how far it is to his friend’s house? 4. A motor cyclist makes a trip of 500 km. If he had increased his speed by 10 km h−1 ,he could have covered 600 km in the same time. What was his original speed? 5. Normally I drive to work at 60 km h−1 . If I drive at 72 km h−1 I cut 8 minutes off my time for the trip. What distance do I travel?

Revision problems (1) 1. One-half of Heather’s age two years from now plus one-third of her age three years ago is twenty years. How old is she now? 2. A piece of 16-gauge copper wire 42 cm long is bent into the shape of a rectangle whose width is twice its length. Find the dimensions of the rectangle. 3. A collection of 33 coins, consisting of nickels, dimes, and quarters, has a value of $3.30. If there are three times as many nickels as quarters, and one-half as many dimes as nickels, how many coins of each kind are there? 4. A wallet contains the same number of pennies, nickels, and dimes. The coins total $1.44. How many of each type of coin does the wallet contain? 5. An aircraft carrier made a trip to Guam and back. The trip there took three hours and the trip back took four hours. It averaged 6 km/h on the return trip. Find the average speed of the trip there. 6. A passenger plane made a trip to Las Vegas and back. On the trip there it flew 432 mph and on the return trip it went 480 mph. How long did the trip there take if the return trip took nine hours? 7. A cattle train left Miami and traveled toward New York. 14 hours later a diesel train left traveling at 45 km/h in a effort to catch up to the cattle train. After traveling for four hours the diesel train finally caught up. What was the cattle train’s average speed? 8. Jose left the White House and drove toward the recycling plant at an average speed of 40 km/h. Rob left some time later driving in the same direction at an average speed of 48 km/h. After driving five hours Rob caught up with Jose. How long did Jose drive before Rob caught up? 9. A passenger train leaves the train depot 2 hours after a freight train left the same depot. The freight train is traveling 20 mph slower than the passenger train. Find the rate of each train, if the passenger train overtakes the freight train in three hours. 10. Two cyclists start at the same time from opposite ends of a course that is 45 miles long. One cyclist is riding at 14 mph and the second cyclist is riding at 16 mph. How long after they begin will they meet?



5

Revision problems (2) 1. A boat travels for three hours with a current of 3 mph and then returns the same distance against the current in four hours. What is the boat’s speed in calm water? How far did the boat travel one way? 2. A spike is hammered into a train rail. You are standing at the other end of the rail. You hear the sound of the hammer strike both through the air and through the rail itself. These sounds arrive at your point six seconds apart. You know that sounds travels through air at 1100 feet per second and through steel at 16500 feet per second. How far away is that spike? 3. The sum of two consecutive integers is 15. Find the numbers. 4. A golf shop pays its wholesaler $40 for a certain club, and then sells it to a golfer for $75. What is the markup rate? 5. A shoe store uses a 40 % markup on cost. Find the cost of a pair of shoes that sells for $63. 6. An item is marked down 15 %; the sale price is $127,46. What was the original price? 7. 2 m3 of soil containing 35 % sand was mixed into 6 m3 of soil containing 15 % sand. What is the sand content of the mixture? 8. 5 fl. oz. of a 2 % alcohol solution was mixed with 11 fl. oz. of a 66 % alcohol solution. Find the concentration of the new mixture. 9. 9 lbs. of mixed nuts containing 55 % peanuts were mixed with 6 lbs. of another kind of mixed nuts that contain 40 % peanuts. What percent of the new mixture is peanuts? 10. Working alone, Ryan can dig a 10 ft by 10 ft hole in five hours. Castel can dig the same hole in six hours. How long would it take them if they worked together? 11. It takes Trevon ten hours to clean an attic. Cody can clean the same attic in seven hours. Find how long it would take them if they worked together. 12. Working together, Paul and Daniel can pick forty bushels of apples in 4,95 hours. Had he done it alone it would have taken Daniel 9 hours. Find how long it would take Paul to do it alone.

✁✃✁✃✁✃✁✃✁✃✁✃✁✃


More exercises to practice with answers:

J.F. Antona. Maths Dep.

3º ESO

Word Problems

(involving linear equations)

"Age" Word Problems 1) One-half of Heather's age two years from now plus one-third of her age three years ago is twenty years. How old is she now?

Geometry Word Problems 1) Suppose a water tank in the shape of a right circular cylinder is thirty feet long and eight feet in diameter. How much sheet metal was used in its construction? 2) A piece of 16-gauge copper wire 42 cm long is bent into the shape of a rectangle whose width is twice its length. Find the dimensions of the rectangle.

"Coin" Word Problems 1) A collection of 33 coins, consisting of nickels, dimes, and quarters, has a value of $3.30. If there are three times as many nickels as quarters, and one-half as many dimes as nickels, how many coins of each kind are there? 2) A wallet contains the same number of pennies, nickels, and dimes. The coins total$1.44. How many of each type of coin does the wallet contain?

"Distance" Word Problems

5) A passenger train leaves the train depot 2 hours after a freight train left the same depot. The freight train is traveling 20 mph slower than the passenger train. Find the rate of each train, if the passenger train overtakes the freight train in three hours.

6) Two cyclists start at the same time from opposite ends of a course that is 45 miles long. One cyclist is riding at 14 mph and the second cyclist is riding at 16 mph. How long after they begin will they meet?

= Problems to try for homework


3º ESO

7) A boat travels for three hours with a current of 3 mph and then returns the same distance against the current in four hours. What is the boat's speed in calm water? How far did the boat travel one way?

8) With the wind, an airplane travels 1120 miles in seven hours. Against the wind, it takes eight hours. Find the rate of the plane in still air and the velocity of the wind.

9) A spike is hammered into a train rail. You are standing at the other end of the rail. You hear the sound of the hammer strike both through the air and through the rail itself. These sounds arrive at your point six seconds apart. You know that sound travels through air at 1100 feet per second and through steel at 16,500 feet per second. How far away is that spike?

"Investment" Word Problems (simple interest) 1) You put $1000 into an investment yielding 6% annual interest; you left the money in for two years. How much interest do you get at the end of those two years? 2) You invested $500 and received $650 after three years. What had been the interest rate? 3) You have $50,000 to invest, and two funds that you'd like to invest in. The You-Risk-It Fund (Fund Y) yields 14% interest. The Extra-Dull Fund (Fund X) yields 6% interest. Because of college financial-aid implications, you don't think you can afford to earn more than $4,500 in interest income this year. How much should you put in each fund?" 4) An investment of $3,000 is made at an annual simple interest rate of 5%. How much additional money must be invested at an annual simple interest rate of 9% so that the total annual interest earned is 7.5% of the total investment? 5) A total of $6,000 is invested into two simple interest accounts. The annual simple interest rate on one account is 9%; on the second account, the annual simple interest rate is 6%. How much should be invested in each account so that both accounts earn the same amount of annual interest? 6) An investor deposited an amount of money into a high-yield mutual fund that returns a

9% annual simple interest rate. A second deposit, $2,500 more than the first, was placed in a certificate of deposit that returns a 5% annual simple interest rate. The total interest earned on both investments for one year was $475. How much money was deposited in the mutual fund? 7) The manager of a mutual fund placed 30% of the fund's available cash in a 6% simple interest account, 25% in 8% corporate bonds, and the remainder in a money market fund that earns 7.5% annual simple interest. The total annual interest from the investments was $35,875. What was the total amount invested?

"Number" Word Problems 1) The sum of two consecutive integers is 15. Find the numbers. 2) The product of two consecutive negative even integers is 24. Find the numbers.



3º ESO

3) Twice the larger of two numbers is three more than five times the smaller, and the sum of four times the larger and three times the smaller is 71. What are the numbers?

"Percent of" Word Problems 1) A golf shop pays its wholesaler $40 for a certain club, and then sells it to a golfer for$75. What is the markup rate? 2) A shoe store uses a 40% markup on cost. Find the cost of a pair of shoes that sells for$63. 3) An item originally priced at $55 is marked 25% off. What is the sale price? 4) An item that regularly sells for $425 is marked down to $318.75. What is the discount rate? 5) An item is marked down 15%; the sale price is $127.46. What was the original price?

Mixture Word Problems

Work Word Problems


Solutions to Word Problems (involving linear equations) _____________________________________________________________________________________

"Age" Word Problems ____________________________________________________________________________________ 1) One-half of Heather's age two years from now plus one-third of her age three years ago is twenty years. How old is she now? This problem refers to Heather's age two years in the future and three years in the past. So I'll pick a variable and label everything clearly: age now: H age two years from now: H + 2 age three years ago: H – 3 Now I need certain fractions of these ages: one-half of age two years from now: ( 1/2 )(H + 2) = H/2 + 1 one-third of age three years ago: ( 1/3 )(H – 3) = H/3 – 1 The sum of these two numbers is twenty, so I'll add them and set this equal to 20: H

/2 + 1 + H/3 – 1 = 20 /2 + H/3 = 20 3H + 2H = 120 5H = 120 H = 24 H

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Geometry Word Problems ___________________________________________________________________________________ 1) Suppose a water tank in the shape of a right circular cylinder is thirty feet long and eight feet in diameter. How much sheet metal was used in its construction? What they are asking for here is the surface area of the water tank. The total surface area of the tank will be the sum of the surface areas of the side (the cylindrical part) and of the ends. If the diameter is eight feet, then the radius is four feet. The surface area of each end is given by the area formula for a circle with radius r: A = (pi)r2. (There are two end pieces, so I will be multiplying this area by 2 when I find my total-surface-area formula.) The surface area of the cylinder is the circumference of the circle, multiplied by the height: A = 2(pi)rh.

Side view of the cylindrical tank, showing the radius "r".

An "exploded" view of the tank, showing the three separate surfaces whose areas I need to find.

Then the total surface area of this tank is given by: 2 ×( (pi)r2 ) + 2(pi)rh (the two ends, plus the cylinder) = 2( (pi) (42) ) + 2(pi) (4)(30) = 2( (pi) × 16 ) + 240(pi) = 32(pi) + 240(pi) = 272(pi) Since the original dimensions were given in terms of feet, then my area must be in terms of square feet: the surface area is 272(pi) square feet. ___________________________________________________________________________________ 2) A piece of 16-gauge copper wire 42 cm long is bent into the shape of a rectangle whose width is twice its length. Find the dimensions of the rectangle. Do I care that the wire is made of copper, or that the wire is a length of sixteen-gauge? No; all I care is that the length is forty-two units, that the units are centimeters, that the rectangle is twice as long in one direction as the other, and that I'm supposed to find the values of each of these directions. I can ignore the other information. Since the wire is 42 centimeters long, then the perimeter of the rectangle is 42 centimeters. That is: 2L + 2W = 42 I also know that the width is twice the length, so: W = 2L Then: Copyright © Elizabeth Stapel 2000-2008 All Rights Reserved 2L + 2(2L) = 42 (by substitution for W from the above equation) 2L + 4L = 42 6L = 42 L=7 Since the width is related to the length by W = 2L, then W = 14, and the rectangle is 7centimeters long and 14 centimeters wide.

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"Coin" Word Problems ____________________________________________________________________________________ 1) A collection of 33 coins, consisting of nickels, dimes, and quarters, has a value of $3.30. If there are three times as many nickels as quarters, and one-half as many dimes as nickels, how many coins of each kind are there? I'll start by picking and defining a variable, and then I'll use translation to convert this exercise into mathematical expressions. Nickels are defined in terms of quarters, and dimes are defined in terms of nickels, so I'll pick a variable to stand for the number of quarters, and then work from there: number of quarters: q number of nickels: 3q number of dimes: (½)(3q) = (3/2)q There is a total of 33 coins, so: q + 3q + (3/2)q = 33 4q + (3/2)q = 33 8q + 3q = 66 11q = 66 q=6 Then there are six quarters, and I can work backwards to figure out that there are 9 dimes and 18 nickels. _________________________________________________________________________________ 2) A wallet contains the same number of pennies, nickels, and dimes. The coins total$1.44. How many of each type of coin does the wallet contain? Since there is the same number of each type of coin, I can use one variable to stand for each: number of pennies: p number of nickels: p number of dimes: p The value of the coins is the number of cents for each coin times the number of that type of coin, so: value of pennies: 1p value of nickels: 5p value of dimes: 10p The total value is $1.44, so I'll add the above, set equal to 144 cents, and solve: 1p + 5p + 10p = 144 16p = 144 p=9 There are nine of each type of coin in the wallet.

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"Distance" Word Problems "Distance" word problems, often also called "uniform rate" problems, involve something travelling at some fixed and steady ("uniform") pace ("rate" or "speed"), or else moving at some average speed. Whenever you read a problem that involves "how fast", "how far", or "for how long", you should think of the distance equation, d = rt, where d stands for distance, r stands for the (constant or average) rate of speed, and t stands for time. Warning: Make sure that the units for time and distance agree with the units for the rate. For instance, if they give you a rate of feet per second, then your time must be in seconds and your distance must be in feet. Sometimes they try to trick you by using the wrong units, and you have to catch this and convert to the correct units. ____________________________________________________________________________________

_____________________________________________________________________________________ 5) A passenger train leaves the train depot 2 hours after a freight train left the same depot. The freight train is traveling 20 mph slower than the passenger train. Find the rate of each train, if the passenger train overtakes the freight train in three hours.

passenger train freight train total

d d d ---

r r r – 20 ---

t 3 3+2=5 ---

(As it turns out, I won't need the "total" row this time.) Why is the distance just "d" for both trains? Partly, that's because the problem doesn't say how far the trains actually went. But mostly it's because they went the same distance as far as I'm concerned, because I'm only counting from the depot to wherever they met. After that meet, I don't care what happens. And how did I get those times? I know that the passenger train drove for three hours to catch up to the freight train; that's how I got the "3". But note that the freight train had a two-hour head start. That means that the freight train was going for five hours.

passenger train freight train total

d d = 3r d = 5(r – 20) ---

r r r – 20 ---

t 3 3+2=5 ---

Now that I have this information, I can try to find my equation. Using the fact that d = rt, the first row gives me d = 3r (note the revised table above). The second row gives me: d = 5(r – 20) Since the distances are equal, I will set the equations equal: 3r = 5(r – 20) Solve for r; interpret the value within the context of the exercise, and state the final answer. _____________________________________________________________________________________ 6) Two cyclists start at the same time from opposite ends of a course that is 45 miles long. One cyclist is riding at 14 mph and the second cyclist is riding at 16 mph. How long after they begin will they meet?

slow guy fast guy total

d d 45 – d 45

r 14 16 ---

t t t ---

Why is t the same for both cyclists? Because I am measuring from the time they both started to the time they meet somewhere in the middle. And how did I get "d" and "45 – d" for the distances? Because once I'd assigned the slow guy as having covered d miles, that left 45 – dmiles for the fast guy to cover: the two guys together covered the whole 45 miles. Using "d = rt", I get d = 14t from the first row, and 45 – d = 16t from the second row. Since these distances add up to 45, I will add the distance expressions and set equal to the given total: 45 = 14t + 16t Solve for t. ___________________________________________________________________________________ 7) A boat travels for three hours with a current of 3 mph and then returns the same distance against the current in four hours. What is the boat's speed in calm water? How far did the boat travel one way?

downstream upstream total

d d d 2d

r b+3 b–3 ---

t 3 4 7

(It may turn out that I won't need the "total" row.) I have used "b" to indicate the boat's speed. Why are the rates "b + 3" and "b – 3"? Because I actually have two speeds combined into one on each trip. The boat has a certain speed (the

"speed in calm water" that I'm looking for; this is the speed that registers on the speedometer), and the water has a certain speed (this is the "current"). When the boat is going with the current, the water's speed is added to the boat's speed. This makes sense, if you think about it: even if you cut the engine, the boat would still be moving, because the water would be carrying it downstream. When the boat is going against the current, the water's speed is subtracted from the boat's speed. This makes sense, too: if the water is going fast enough, the boat will still be going downstream (a "negative" speed, because the boat would be going backwards at this point), because the water is more powerful than the boat. (Think of a boat in a cartoon heading toward a waterfall. The guy paddles like crazy, but he still goes over the edge.) d d = 3(b + 3) d = 4(b – 3) 2d

downstream upstream total

r b+3 b–3 ---

t 3 4 7

Using "d = rt", the first row (of the revised table above) gives me: d = 3(b + 3) The second row gives me: d = 4(b – 3) Since these distances are the same, I will set them equal: 3(b + 3) = 4(b – 3) Solve for b. Then back-solve for d. In this case, I didn't need the "total" row. ____________________________________________________________________________________ 8) With the wind, an airplane travels 1120 miles in seven hours. Against the wind, it takes eight hours. Find the rate of the plane in still air and the velocity of the wind.

tailwind headwind total

d 1120 1120 2240

r p+w p–w ---

t 7 8 15

(I probably won't need the "total" row.) Just as with the last problem, I am really dealing with two rates together: the plane's speedometer reading, and the wind speed. When the plane turns around, the wind is no longer pushing the plane to go faster, but is instead pushing against the plane to slow it down. Copyright © Elizabeth Stapel 1999-2009 All Rights Reserved The first row gives me: 1120 = 7(p + w) The second row gives me: 1120 = 8(p – w) The temptation is to just set these equal, like I did with the last problem, but that just gives me:

7(p + w) = 8(p – w) ...which doesn't help much. I need to get rid of one of the variables. I'll take that first equation: 1120 = 7(p + w) ...and divide through by 7: 160 = p + w Then, subtracting w from either side, I get that p = 160 – w. I'll substitute "160 – w" for "p" in the second equation: 1120 = 8([160 – w] – w) 1120 = 8(160 – 2w) ...and solve for w. Then I'll back-solve to find p. _____________________________________________________________________________________ 9) A spike is hammered into a train rail. You are standing at the other end of the rail. You hear the sound of the hammer strike both through the air and through the rail itself. These sounds arrive at your point six seconds apart. You know that sound travels through air at 1100 feet per second and through steel at 16,500 feet per second. How far away is that spike?

air steel total

d d = 1100t d = 16,500(t – 6) ---

r 1100 16,500 ---

t t t–6 6

However long the sound took to travel through the air, it took six seconds less to propagate through the steel. (Since the speed through the steel is faster, then that travel-time must be shorter.) I multiply the rate by the time to get the values for the distance column. (Once again, I didn't need the "total" row.) Since the distances are the same, I set the distance expressions equal to get: 1100t = 16,500(t – 6) Solve for the time t, and then back-solve for the distance d by plugging t into either expression for the distance d. _____________________________________________________________________________________

"Investment" Word Problems Investment problems usually involve simple annual interest (as opposed to compounded interest), using the interest formula I = Prt, where I stands for the interest on the original investment, P stands for the amount of the original investment (called the "principal"), r is the interest rate (expressed in decimal form), and t is the time.

For annual interest, the time t must be in years. If they give you a time of, say, nine months, you must first convert this to 9/12 = 3/4 = 0.75 years. Otherwise, you'll get the wrong answer. The time units must match the interest-rate units. If you got a loan from your friendly neighborhood loan shark, where the interest rate is monthly, rather than yearly, then your time must be measured in terms of months. Investment word problems are not generally terribly realistic; in "real life", interest is pretty much always compounded somehow, and investments are not generally all for whole numbers of years. But you'll get to more "practical" stuff later; this is just warm-up, to prepare you for later. In all cases of these problems, you will want to substitute all known information into the "I = Prt" equation, and then solve for whatever is left. _____________________________________________________________________________________ 1) You put $1000 into an investment yielding 6% annual interest; you left the money in for two years. How much interest do you get at the end of those two years? In this case, P = $1000, r = 0.06 (because I have to convert the percent to decimal form), and the time is t = 2. Substituting, I get: I = (1000)(0.06)(2) = 120 I will get $120 in interest. Another example would be: Copyright © Elizabeth Stape6-2008 All Rights Reserved _____________________________________________________________________________________ 2) You invested $500 and received $650 after three years. What had been the interest rate? For this exercise, I first need to find the amount of the interest. Since interest is added to the principal, and since P = $500, then I = $650 – 500 = $150. The time is t = 3. Substituting all of these values into the simple-interest formula, I get: 150 = (500)(r)(3) 150 = 1500r 150 /1500 = r = 0.10 Of course, I need to remember to convert this decimal to a percentage. I was getting 10% interest.

The hard part comes when the exercises involve multiple investments. But there is a trick to these that makes them fairly easy to handle. _____________________________________________________________________________________ 3) You have $50,000 to invest, and two funds that you'd like to invest in. The You-Risk-It Fund (Fund Y) yields 14% interest. The Extra-Dull Fund (Fund X) yields 6% interest. Because of college financial-aid implications, you don't think you can afford to earn more than $4,500 in interest income this year. How much should you put in each fund?" The problem here comes from the fact that I'm splitting that $50,000 in principal into two smaller amounts. Here's how to handle this:

Fund X Fund Y total

I ? ? 4,500

P ? ? 50,000

r 0.06 0.14 ---

t 1 1 ---

How do I fill in for those question marks? I'll start with the principal P. Let's say that I put "x" dollars into Fund X, and "y" dollars into Fund Y. Then x + y = 50,000. This doesn't help much, since I only know how to solve equations in one variable. But then I notice that I can solve x + y= 50,000 to get y = $50,000 – x. THIS TECHNIQUE IS IMPORTANT! The amount in Fund Y is (the total) less (what we've already accounted for in Fund X), or 50,000 – x. You will need this technique, this "how much is left" construction, in the future, so make sure you understand it now.

Fund X Fund Y total

I ? ? 4,500

P x 50,000 – x 50,000

r 0.06 0.14 ---

t 1 1 ---

Now I will show you why I set up the table like this. By organizing the columns according to the interest formula, I can now multiply across (right to left) and fill in the "interest" column.

Fund X Fund Y total

I 0.06x 0.14(50,000 – x) 4,500

P x 50,000 – x 50,000

r 0.06 0.14 ---

t 1 1 ---

Since the interest from Fund X and the interest from Fund Y will add up to $4,500, I can add down the "interest" column, and set this sum equal to the given total interest: 0.06x + 0.14(50,000 – x) = 4,500 0.06x + 7,000 – 0.14x = 4,500 7,000 – 0.08x = 4,500 –0.08x = –2,500 x = 31,250 Then y = 50,000 – 31,250 = 18,750. I should put $31,250 into Fund X, and $18,750 into Fund Y. Note that the answer did not involve "neat" values like "$10,000" or "$35,000". You should understand that this means that you cannot always expect to be able to use "guess-n-check" to find your answers. You really do need to know how to do these exercises. f you set up your investment word problems so everything is labeled and well-organized, they should all work out fairly easily. Just take your time and do things in an orderly fashion. I've done the set-up (but not the complete solutions) for a few more examples: _____________________________________________________________________________________ 4) An investment of $3,000 is made at an annual simple interest rate of 5%. How much additional money must be invested at an annual simple interest rate of 9% so that the total annual interest earned is 7.5% of the total investment?

first additional total

I (3,000)(0.05) = 150 0.09 x (3,000 + x)(0.075)

P 3,000 x 3,000 + x

r 0.05 0.09 0.075

t 1 1 1

First I fill in the P, r, and t columns with the given values. Then I multiply across the rows (from the right to the left) in order to fill in the I column. Then add down the I column to get the equation 150 + 0.09 x = (3,000 + x)(0.075). To find the solution, I would solve for the value of x. _____________________________________________________________________________________ 5) A total of $6,000 is invested into two simple interest accounts. The annual simple interest rate on one account is 9%; on the second account, the annual simple interest rate is 6%. How much should be invested in each account so that both accounts earn the same amount of annual interest?

9% account 6% account total

I 0.09x (6,000 – x)(0.06) ---

P x 6,000 – x 6,000

r 0.09 0.06 ---

t 1 1 ---

In this problem, I don't actually need the "total" row at all. First I'll fill in the P, r, and t columns, and multiply to the left to fill in the I column. From the interest column, I then get the equation 0.09x = ($6,000 – x)(0.06), because the yields are required to be equal. Then I'd solve for the value of x, and back-solve to find the value invested in the 6% account. (This exercise's set-up used that "how much is left" construction, mentioned earlier.) _____________________________________________________________________________________ 6) An investor deposited an amount of money into a high-yield mutual fund that returns a9% annual simple interest rate. A second deposit, $2,500 more than the first, was placed in a certificate of deposit that returns a 5% annual simple interest rate. The total interest earned on both investments for one year was $475. How much money was deposited in the mutual fund? The amount invested in the CD is defined in terms of the amount invested in the mutual fund,so I will let "x" be the amount invested in the mutual fund.

mutual fund cert. of deposit total

I 0.09x (x + 2,500)(0.05) 475

P x x + $2,500 2x + $2,500

r 0.09 0.05 ---

In this problem, I don't actually need the "total" for the "rate" or "time" columns. First I'll fill in the P, r, and t columns, multiplying to the left to fill in the I column. Then I'll add down the I column to get the equation 0.09x + (x + 2,500)(0.05) = 475.

t 1 1 ---

Then I'd solve for the value of x. Copyright © Elizabeth Stape999-2010 All Rights Reserved ____________________________________________________________________________________ 7) The manager of a mutual fund placed 30% of the fund's available cash in a 6% simple interest account, 25% in 8% corporate bonds, and the remainder in a money market fund that earns 7.5% annual simple interest. The total annual interest from the investments was $35,875. What was the total amount invested? For this problem, I'll let "x" stand for the total amount invested.

6% account 8% bonds 7.5% fund total

I (0.30x)(0.06) = 0.018x (0.25x)(0.08) = 0.02x (0.45x)(0.075) = 0.03375x $35,875

P 0.30x 0.25x 0.45x x

r 0.06 0.08 0.075 ---

t 1 1 1 ---

Once 30% and 25% was accounted for in the 6% and 8% accounts, then there is 100% – 30% – 25% = 45% left for the third account. I can use this information to fill in the "principal" column. Then I'll fill out the "rate" and "time" columns, and multiply to the left to fill in the "interest" column. From the interest column, I get the equation 0.018x + 0.02x + 0.03375x = 35,875. Then I'd solve for the value of x. _____________________________________________________________________________________

"Number" Word Problems _____________________________________________________________________________________ 1) The sum of two consecutive integers is 15. Find the numbers. "The numbers are 7 and 8."

_____________________________________________________________________________________ 2) The product of two consecutive negative even integers is 24. Find the numbers. They have told me quite a bit about these two numbers: the numbers are even and they are negative. (The fact that they are negative may help if I come up with two solutions — a positive and a negative — so I'll know which one to pick.) Since even numbers are two apart (for example,–4 and –2 or 10 and 12), then I also know that the second number is two greater than the first. I also know that, when I multiply the two numbers, I will get 24. In other words, letting the first number be "n" and the second number be "n + 2", I have: (n)(n + 2) = 24 n2 + 2n = 24 n2 + 2n – 24 = 0 (n + 6)(n – 4) = 0 Then the solutions are n = –6 and n = 4. Since the numbers I am looking for are negative, I can ignore the "4" and take n = –6. Then the next number is n + 2 = –4, and the answer is The numbers are –6 and –4.

In the exercise above, one of the answers was one of the solutions to the equation; the other answer was the negative of the other solution to the equation. Warning: Do not assume that you can use both solutions if you just change the signs to be whatever you feel like. While this often "works", it does notalways work, and it's sure to annoy your teacher. Throw out invalid results, and solve properly for valid ones. _____________________________________________________________________________________ 3) Twice the larger of two numbers is three more than five times the smaller, and the sum of four times the larger and three times the smaller is 71. What are the numbers? The point of exercises like this is to give you practice in unwrapping and unwinding these words, and turning the words into algebraic equations. The point is in the solving, not in the relative "reality" of the problem. That said, how do you solve this? The best first step is to start labelling: the larger number: x the smaller number: y twice the larger: 2x three more than five times the smaller: 5y + 3 relationship between ("is"): 2x = 5y + 3 four times the larger: 4x three times the smaller: 3y relationship between ("sum of"): 4x + 3y = 71 Now I have two equations in two variables: 2x = 5y + 3 4x + 3y = 71 I will solve, say, the first equation for x: x = (5/2)y + (3/2) Then I'll plug the right-hand side of this into the second equation in place of the "x": 4[ (5/2)y + (3/2) ] + 3y = 71 10y + 6 + 3y = 71 13y + 6 = 71 13y = 65 y = 65/13 = 5 Now that I have the value for y, I can solve for x: x = (5/2)y + (3/2) x = (5/2)(5) + (3/2) x = (25/2) + (3/2) x = 28/2 = 14 As always, I need to remember to answer the question that was actually asked. The solution here is not "x = 14", but is the following sentence: The larger number is 14, and the smaller number is 5. The trick to doing this type of problem is to label everything very explicitly. Until you become used to doing these, do not attempt to keep track of things in your head. Do as I did in this last example: clearly label every single step. When you do this, these problems generally work out rather easily.

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"Percent of" Word Problems _____________________________________________________________________________________ 1) A golf shop pays its wholesaler $40 for a certain club, and then sells it to a golfer for$75. What is the markup rate? First, I'll calculate the markup in absolute terms: 75 – 40 = 35 Then I'll find the relative markup over the original price, or the markup rate: ($35) is (some percent) of ($40), or: Copyright © Elizabeth Stapel 1999-2009 All Rights Reserved 35 = (x)(40) ...so the relative markup over the original price is: 35 ÷ 40 = x = 0.875 Since x stands for a percentage, I need to remember to convert this decimal value to the corresponding percentage. The markup rate is 87.5%. _____________________________________________________________________________________ 2) A shoe store uses a 40% markup on cost. Find the cost of a pair of shoes that sells for$63. This problem is somewhat backwards. They gave me the selling price, which is cost plus markup, and they gave me the markup rate, but they didn't tell me the actual cost or markup. So I have to be clever to solve this. I will let "x" be the cost. Then the markup, being 40% of the cost, is 0.40x. And the selling price of $63 is the sum of the cost and markup, so: 63 = x + 0.40x 63 = 1x + 0.40x 63 = 1.40x 63 ÷ 1.40 = x= 45 The shoes cost the store $45. _____________________________________________________________________________________ 3) An item originally priced at $55 is marked 25% off. What is the sale price? First, I'll find the markdown. The markdown is 25% of the original price of $55, so: x = (0.25)(55) = 13.75 By subtracting this markdown from the original price, I can find the sale price: 55 – 13.75 = 41.25

The sale price is $41.25. _____________________________________________________________________________________ 4) An item that regularly sells for $425 is marked down to $318.75. What is the discount rate? First, I'll find the amount of the markdown: 425 – 318.75 = 106.25 Then I'll calculate "the markdown over the original price", or the markdown rate: ($106.25) is (some percent) of ($425), so: 106.25 = (x)(425) ...and the relative markdown over the original price is: x = 106.25 ÷ 425 = 0.25 Since the "x" stands for a percentage, I need to remember to convert this decimal to percentage form. The markdown rate is 25%. _____________________________________________________________________________________ 5) An item is marked down 15%; the sale price is $127.46. What was the original price? This problem is backwards. They gave me the sale price ($127.46) and the markdown rate(15%), but neither the markdown amount nor the original price. I will let "x" stand for the original price. Then the markdown, being 15% of this price, was 0.15x. And the sale price is the original price, less the markdown, so I get: x – 0.15x = 127.46 1x – 0.15x = 127.46 0.85x = 127.46 x = 127.46 ÷ 0.85 = 149.952941176... This problem didn't state how to round the final answer, but dollars-and-cents is always written with two decimal places, so: The original price was $149.95. Note in this last problem that I ended up, in the third line of calculations, with an equation that said "eighty-five percent of the original price is $127.46". You can save yourself some time if you think of discounts in this way: if the price is 15% off, then you're only actually paying 85%. Similarly, if the price is 25% off, then you're paying 75%; if the price is 30% off, then you're paying 70%; and so on.

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Mixture Word Problems _____________________________________________________________________________________

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Work Word Problems ____________________________________________________________________________________