Bahan Ajar VII Mathematics

AL. Kristiyanto,S.Pd

[Grade 7]SMP 21 Semarang

Mathematics for

Junior High School

2 AL. Kristiyanto,S.Pd | Mathematics for Junior High School – Year 7

CHAPTER 1 INTEGERS

Standard Competence 1. Understanding the properties of mathematical operation of numbers and their application in problem solving Basic Competence 1. 1 Doing mathematical operation of integers and apply their properties in problem solving Indicators 1. Students are able to mention the member of integers 2. Students are able to mention quantities in daily activities used a negative number 3. Students are able to locate an integer in a number line 4. Students are able to do mathematical operation of integers (addition, subtraction, multiplication, division,

powering, root and its combination) 5. Students are able to find the properties of mathematical operation of integers (addition, subtraction,

multiplication, division, powering, root and its combination) 6. Students are able to apply the properties of mathematical operation of integers in problem solving 7. Students are able to find the prediction value of the product of multiplication and division 8. Students are able to find the prediction value of the product of multiplication and division Hello, Math Lovers! Do you like to watch Doraemon during your holiday? Have you ever imagined when Doraemon played snow with Nobita? What do you think about the weather? Off course, it’s too cold. The temperature is under zero level. Then, we have a problem. How to state a number under zero level? Yes, you are right. We need negative numbers. A. The Definition of Integers and Its Symbol

Integers are a set of numbers that consists of negative numbers, zero and positive numbers. Integers = { },...4,3,2,1,0,1,2,3,4..., −−−− . Three dots (…) mean the members are infinite. Negative Numbers = …, – 4, – 3, – 2, – 1 ⇒ the sign is minus ( - ) Zero = 0 (neutral numbers) Positive Numbers = 1, 2, 3, 4, … ⇒ the sign is plus ( + ), but it’s not need to be written It can be shown on a number line

Negative direction = left← → right = positive direction

Numbers are getting smaller numbers are getting greater If a is located to the right of b then a > b → a is greater (more) than b If a is located to the left of b then a < b → a is smaller (less) than b Examples: – 5 < 4 because – 5 is located to the left of 4 – 7 > – 10 because – 7 is located to the ………. of ……………

0 1 2 3 4 5 -1 -2 -3 -4 -5


Beside Integers, there are the other set of numbers such as Whole Numbers = { },...4,3,2,1,0 , Natural numbers (N) = { },...4,3,2,1 Rational Numbers (Q) = a set of numbers that consist of all numbers that can be

statedba , a, b are the element of Integers, and b ≠ 0. For examples = 9.0,7,

25− , 5 etc

Irrational numbers = a set of all numbers that are not rational numbers, for examples = 3 41.2 − , π, etc

Real Numbers = a set that consist of rational numbers and irrational numbers EXERCISE 1A 1. Compare these two integers by giving sign “ < ”, “ > “ between them. Give your

reasons! a. – 3 …. 4 c. – 2 …. – 5 e. – 451 …. – 541 b. 4 …. 6 d. 6 …. – 6 f. 0 …. – 3

2. Compare these three integers by giving sign “ < ”, “ > “ between them. Give your reasons! a. 3 …. 4 ..…. 5 c. – 21 …. – 25 …. – 30 b. – 3 …. 6 …. 7 d. – 17 …. 0 …. 8

3. Determine the possible value of x. x is the element of integers! a. 14 ≤≤− x c. 94 ≤< x e. 72 −≥>− x b. 51 <<− x d. 2521 −>>− x

B. Mathematical Operation of Integers

1. Addition of Integers and the Properties Example: 1. – 3 + 4 = …. (Start from 0, move 3 steps to the left until – 3, move 4 steps to the

right, the end point is 1. So, the result is 1.

–3 + 4 = 1

2. 4 + (- 5) = ….. (Start from 0, move …. steps to the ….....….. until ….., move ……. steps to the ………, the end point is ………. So, the result is ………...

4 + (-5) = ……

The properties of addition of integers are a. Study these cases!

2 + 3 = 5 ⇒ 2 and 3 are the element of integer. The result is …, is also the element of integer

– 1 + 1 = …. ⇒ – 1 and 1 are the element of integer. The result is …, is also the element of integer For any a and b are the element of integer then the sum of them is also the element of integer. It’s called enclosed property.

0 1 2 3 4 5 -1 -2 -3 -4 -5

–3 4

0 1 2 3 4 5 -1 -2 -3 -4 -5


b. Study these cases!

........73.....37.....73

+=+⎭⎬⎫

=+=+

........92.....)2(9.....92

+=+−⎭⎬⎫

=−+=+−

For any a and b are the element of integer then a + b = … + …. It’s called commutative property

c. Study these cases! [ ]

[ ] [ ] [ ].....(.....)....3)5(2.......3)5(2.......3)5(2

++=+−+⎭⎬⎫

=+−+=+−+

....)(........1)72(.......)17(2.......1)72(

++=++−⎭⎬⎫

=++−=++−

For any a, b and c are element of integer then (a + b) + c = ….+ (…. + ….) It’s called associative property

d. Study these cases!

.....5005.....50......05

=+=+⎭⎬⎫

=+=+

.....)2(002.....)2(0

......02=−+=+−

⎭⎬⎫

=−+=+−

For any a is the element of integer then a + 0 = 0 + a = …. Zero (0) is called identity element of addition

e. Study these cases!

.....66....)2(2

=+−=−+

For any p is the element of integer then there’s q so that p + q = q + p = 0 q is called the addition inverse of p. For examples: the inverse of 4 is -4, the inverse of -7 is 7.

EXERCISE 1B 1) By using a number line, find the sum of:

a. -3 + 7 d. 2 + (-6) b. -5 + (-3) e. 0 + (-3) c. 3 + 8 f. (-5) + 5

2) Find the sum of: a. 12 + (-21) d. 123 + 501 b. -25 + (-34) e. (-48) + 67 c. 0 + (-23) f. 223 + (-249)

3) Given a = 2, b = -7 and c = -5. Find the value of: a. a + b c. a + c b. b + c d. a + b + c


2. Subtraction of Integers and the Properties Subtraction of two numbers means addition with the inverse of the second number. For examples:

1. 3 – 5 = 3 + (-5) = -2 (the addition inverse of 5 is -5) 2. 4 – (-6) = 4 + ….= ….. (the addition inverse of -6 is ….) 3. -7 – 2 = -7 + ….= ….. (the addition inverse of 2 is ….) 4. -3 – (-9) = -3 + ….= ….. (the addition inverse of -9 is ….)

EXERCISE 1C

1. Find the result of subtraction: a. 9 – 4 e. -123 – (127) b. 24 – 35 f. -205 – 205 c. -21 – (-45) g. 400 – (-645) d. -20 – 43 h. (-27) – (-27)

2. Given a = 4, b = -2 and c = -5. Find the value of: a. a – b e. a + b – c b. b – c f. b – c – a c. a – b – c g. b – a + c d. c – b – a h. c – b + a

3. Find the difference of 9 and -4 4. Determine the difference of -6 and -23 5. Find the difference of -12 and 5

3. Multiplication of Integers and the Properties Study this process!

.....33......32......31

.....30.....31

63332933333

12333334

=×−=×−=×−

=×=×

=+=×=++=×

=+++=×

WORK IN A GROUP! Make a group consists of 4 students! You have to discuss about the properties of subtraction. Answer these questions! Ask your teacher if you get trouble! 1. Does subtraction of integer have enclosed property? Explain! 2. Does subtraction of integer have commutative property? Explain! 3. Does subtraction of integer have associative property? Explain! 4. Does subtraction of integer have identity element? Explain!

The results of those multiplications are always subtracted by 3.


.....)3(3......)3(2......)3(1

0)3(03)3(1

6)3(3)3(2

=−×−=−×−=−×−

=−×−=−×

−=−+−=−×

Conclusion: Positive number x Positive number = Positive number Negative number x Negative number = Positive number Negative number x Positive number = Negative number Positive number x Negative number = Negative number The properties of multiplication: 1. Multiplication of integers is enclosed. (Prove it by your self!) 2. Multiplication of integers is commutative. (Prove it by your self!) 3. Multiplication of integers is associative. (Prove it by your self!) 4. The identity element of multiplication of integers is 1. (Prove it by your self!) 5. Multiplication of integer has distributive property

a. Distributive toward addition For any a, b and c are the element of integer then )()()( cabacba ×+×=+×

b. Distributive toward subtraction For any a, b and c are the element of integer then )()()( cabacba ×−×=−×

EXERCISE 1D

1. Find the product of: a. 2 x (-5) d. -36 x 11 b. (-30) x -21 e. 5 x (-56) c. (-23) x (-5) f. 0 x (-53)

2. Given a = -5, b = 4 and c = -3. Determine the value of: a. a x b f. 4 x b x c b. a x c g. -5 x c x a c. b x c h. – 24 x b x a d. -2 x a i. a x b x c e. -6 x a x b j. – 4 x c x b x a

3. Find the product of these operations! (Use associative properties to make it easier) a. 2 x 9 x 5 d. 16 x 25 x 26 b. 4 x 11 x 5 e. 80 x 13 x 125 c. 15 x 12 x 4 f. 32 x 34 x 25

4. Find the product of these operations! (Use distributive properties to make it easier) a. 13 x 9 d. 8 x 126 b. 25 x 101 e. 6 x 599 c. 14 x 99 f. 24 x 999

}The results of those multiplications are always added by 3.


4. Division of Integers and the Properties Study these examples: 1. 6 : 2 = 3 because 2 x 3 = 6 2. 12 : 4 = …. because 4 x …… = 12 3. 18 : 3 = …. because 3 x …… = 18 Division is the inverse operation of multiplication. These below properties are also satisfied in division. Positive number : Positive number = Positive number Negative number : Negative number = Positive number Negative number : Positive number = Negative number Positive number : Negative number = Negative number Division involving ZERO

0 : 2 = ….. because 2 x … = 0. The result of that division is 0 because 2 x 0 = 0 2 : 0 = ….. because 0 x … = 2. The result of that division is undefined because

there’s no number multiplied by 0 so that the result is 2. 0 : 0 = ….. because 0 x … = 0. The result is infinite because all number multiplied

by 0, the result is 0. EXERCISE 1E 1. Find the result of:

a. 120 : (-5) d. -316 : 4 b. (-36) : -2 e. -438 : (-6) c. (-225) : (-15) f. 0 : (-3)

2. Given a = -120, b = 4 and c = -3. Determine the value of: a. a : b d. 24 : b b. a : c e. -295 : c c. a : b : c f. – 204 : c

5. Powering of Integers and the Properties

Study these examples: 1. 23 = 2 x 2 x 2 = 8 2. 35 = 3 x 3 x …. x …. x ….= ….. 3. a4 = a x a x a x a 4. 44 344 21

factorsnfor

n aaaaa ××××= .....

5. (-2)4 = (-2) x …..x…..x……=……. 6. -24 = – (2 x 2 x…..x…..) = – …..

WORK IN A GROUP! Make a group consists of 4 students! You have to discuss about the properties of division. Answer these questions! Ask your teacher if you get trouble! 1. Does division of integer have enclosed property? Explain! 2. Does division of integer have commutative property? Explain! 3. Does division of integer have associative property? Explain! 4. Does division of integer have identity element? Explain!


The properties of powering: Study these cases!

1. 22 x 23 = (2 x 2) x ( … x … x ….) = …. x…..x…..x…..x…. = ................ 22 +=

33 x 34 = (….x…..x…..) x (….x…..x….x…..) = ….x…..x…..x…..x……x……x….. = .................. 33 +=

Generally, for any a, m, n are integers then am x an = a….+……

2. 23 : 22 = 22

222×××

= ............

= ................ 22 −=

35 x 33 =..................

...........................××

××××

= ….x….. = .................. 33 −=

Generally, for any a, m, n are integers then am : an = a…. –……

3. (22)3 = (22) x (22) x (22) = (2 x 2) x (…..x…...) x (….x…..) = ….x….x….x….x….x…. = ................ 22 ×=

(32)4 = (32) x (32) x (32) x (32) = (3 x 3) x (…..x…...) x (….x…..) x (…..x…..) = ….x….x….x….x….x….x…..x….. = ................ 33 ×=

Generally, for any a, m, n are integers then (am)n = a….x……

4. (2 x 3)3= (2 x 3) x (2 x 3) x (2 x 3) = 2 x 3 x 2 x 3 x 2 x 3 = 2 x 2 x 2 x …x….x…. = 2….. x 3…….

(2 x 5)4= (2 x 5) x (2 x 5) x (2 x 5) x (2 x 5) = 2 x 5 x 2 x 5 x 2 x 5 x 2 x 5 = 2 x 2 x 2 x 2 x 2 x …x….x….x….x….. = 2….. x 5…….

Generally, for any a, b, n are integers then (a x b)n = a…..x b……

5. 23 : 22 = .........2 − = 2…… = ……

34 : 33 = .........3 − = 3…… = …… Generally, for any a is the element of integer then a1 = ….


6. 22 : 22 = ..........22

×× = 1, beside that we also can obtain

22 : 22= .........2 − = 2…… , so we can conclude 2…… = ….

42 : 42 = ......................

×× = 1, beside that we also can obtain

42 : 42= .........4 − = 4…… , so we can conclude 4…… = …. Generally, for any a is the element of integer then a0 = …..

EXERCISE 1F 1. Find the value of:

a. 112 c. (-36)2 b. (-14)2 d. -352

2. Find the value of: a. 63 c. -54 b. 26 d. (-3)5

3. Find the value of: a. (16 – 28)2 c. (-18 – (-14))2 b. (-7 + 14)2 d. (-17 + 16)5

6. Root of Integers a. Square root

abba =⇔= 22 , a is positive integer. 2 a is usually symbolized a Examples: 1. 13169 = because 132 = 169 6. ......2704 = because ….2 = 2704 2. ....729 = because ….2 = 729 7. ......10000 = because ….2 = 10000 3. ......1849 = because ….2 = 1849 8. ......576 = because ….2 = 576 4. ......961 = because ….2 = 961 9. ......676 = because ….2 = 676 5. ......625 = because ….2 = 625 10. ......1369 = because ….2 = 1369

b. Cube root abba =⇔= 33

Generally, abba nn =⇔= Examples: 1. 4643 = because 43 = 64 6. ......1253 =− because ….3 = (-125) 2. ....83 =− because ….3 = (-8) 7. ...10000006 = because ….6 = 1000000 3. 5325 = because 25 = 32 8. ......125.03 = because ….3 = 0.125 4. ......273 = because ….3 = 27 9. ......5123 =− because ….3 = (-512) 5. ......814 = because ….4 = 81 10. ......2435 =− because ….5 = (-243)

c. How to estimate the value of square root Quadratic numbers = 1, 4, 9, 16, 25, 36, 49, … n is a number located between a and b where a and b are quadratic numbers than


bnanan

−−

+≈

Example: Find the prediction value of 86 Answer: n = 86, is located between 81 and 100. It means a = 81 and b = 100.

bnanan

−−

+≈

263.9263.09

1959

81100818681

≈+≈

+≈

−−

+≈

EXERCISE 1G 1. Find the root of:

a. 324 c. 3 729− b . 3 6859− d. 6 64

2. Find the prediction value of 125 3. Find the prediction value of 35 4. Find the prediction value of 222 5. Find the prediction value of 876 6. Find the prediction value of 1005

7. The combination of mathematical operation Combination of mathematical operation means mathematical operation involving addition, subtraction, multiplication, division, powering and root. Multiplication and Division have to be done first before addition and multiplication. Examples: 1. =×+− 356 -6 + 15 = 9 (you have to multiply 5 and 3 first, then do the addition) 2. 4 + 6 : (-3) + 3 x (-4) = 4 + ……+ …..= ….. EXERCISE 1H 1. 67 – 3 x (-5) + 39 : 13 6. 3 – 5 x (-4) + 27 : (-9) 2. 45 + 27 : 9 x 5 – 40 7. 125 : 25 – 10 x 2 + (-15) 3. 105 – 25 + 15 x 5 : 25 8. (-4) x 6 + 45 : (-3) – 12 4. 5 x (45 – 21) : (4 + 8) 9. 21 – 4 x 6 : 12 + 3 5. 240 : 12 x (8 + 6) 10. -5 – 7 x (-2) – 12 : (-4)

C. The Application of Integers in Daily Life Study these sentences! 1. The temperature in Japan when it was autumn is 60C under 00C.

60C under 00C means -60C


2. A fish swims at 200 m under sea water level. 200 m under sea water level means -200 m

Keyword: Decrease = drop = left direction = negative = minus increase = up = right direction = positive

EXERCISE 1I Some numbers are as the example then you continue to finish it by your self.

1. During the afternoon the temperature in a desert is 180C. During the night the temperature decreases by 220C. What is the temperature of the desert at the night? Known : temperature in afternoon = 180C

Decreases 220C in the night Question : temperature in the night Answer : Decrease → minus, then Temperature in the night = temperature in afternoon – decreased temperature

= 180C – 220C = – 40C So temperature in the desert in the night is – 40C. 2. A diver is 10 m under sea water level. He dives 5 m further and swims up 12 m. Find

his present position! 3. An airplane is flying at altitude 2 500 m over sea water level, then moves to altitude

825 m. How many meters lower is it? 4. A bus is running with 45 passengers. At a bus stop, 20 of them gets off and 23 more

gets on. How many passengers are there in the bus now? 5. Dita is charged Rp. 75 000 for telephone bill and Rp. 60 000 for electricity bill. She

provides notes of Rp. 100 000 and Rp. 50 000. How much change will she get? 6. A bird is at 12 m above sea water level while a fish is directly below it at 9 m below

sea water level. Determine the distance between the bird and the fish! 7. Tina buys 4 books. The price is Rp. 1500 for a book. She provides a note of Rp. 10

000. Determine the total price and the change that she will get! 8. The length of a ribbon is 124 cm. It is cut into 4 pieces. How long is each piece of

ribbon? 9. A printer prints 6 copies per minute. How many copies are printed in an hour? 10. The temperature on top Mount Jaya Wijaya is -20C. When the weather changes, It rises

30C. What is the new temperature then? 11. The temperature in a desert increases 40C for every hour in the day. The temperature at

8.00 am was 300C. What was the temperature at 11.00 am? 12. The temperature of ice is 00C. When some salt is added, the temperature drops by 80C.

Then the ice is heated and the temperature rises by 170C. What is the temperature now? 13. Sita is 13 years old. Her father is 4 times older than her. How old will her father be in

five years later? 14. Only 605 people were present during a competition, although 753 people had

registered. How many people were absent? 15. Rio’s money is Rp. 12 000. Then, his mother gives him Rp. 8.000 and his father gives

him Rp. 10.000. He buys a pencil box at price Rp. 15.000. Then he gives the rest to his 3 brothers. Everyone gets equal part. How many Rp does each of his brothers get?


D. Prediction Value Place Values and Digit Values

Given the number 1 234 567 Number Place Value Digit Value

1 Millions 1 000 000 2 Hundred thousands 200 000 3 Ten thousands 30 000 4 Thousands 4 000 5 Hundreds 500 6 Tens 60 7 Ones 7

The rule for rounding off For any integers a, If 5<a then delete a and the number in the left of a is not changed. If 5≥a then delete a and add 1 into the number in the left of a.

For examples: a. Round off until 2 decimal places!

1. 2.334 ≈ 2. 33 ( 54 < , then delete 4 and the number in the left of 4, i.e. 3 is not changed)

2. 123.056 ≈ 123.06 ( 56 ≥ , then delete 6 and add 1 to the number in the left of 6, i.e. 5, so 5 is changed into 5 + 1, it is 6)

b. Round off until the nearest tens. 1. 235 175 ≈ 23518 ( 55 ≥ , then delete 5 and add 1 to the number in the left of 5, i.e.

7, so 7 is changed into 7 + 1, it is 8) 2. 2 501 152 ≈ ….. ...................................................................................................... ……………………………………………)

c. Round off the result into the nearest tens. 1. 235 x 1247 ≈ 240 x 1250 ≈ 3000 2. 44 x 1324 ≈ 40 x ……. ≈ …………

EXERCISE 1J 1. Round off each of the following numbers to the nearest hundreds.

a. 2 335 601 b. 399 099 2. Estimate the prediction value of these following multiplication and division to the

nearest tens! a. 152 x 179 c. 238 : 84 b. 346 x 71 d. 175 x 59

3. Estimate the prediction value of these following multiplication and division to the nearest hundreds! a. 1262 x 179 c. 2318 : 854 b. 3460 x 871 d. 1758 x 459


TEST FOR CHAPTER 1 INTEGERS A. CONCEPTS UNDERSTANDING

1. By using a number line, find the result of these addition: (a) – 6 + 9 (b) 4 + (– 7)

2. Arrange these numbers in ascending order: (a) – 7, 0, – 12, 13, 5 (b) 1, – 1, 4, – 4, – 6

3. Find the result of: (a) 25 + (– 37) (b) – 201 – 81 (c) 30 – 42 + 13

4. Find the result of : (a) – 5 x (– 14) x 4 (b) 1240 : 8 : (– 4)

B. LOGIC AND COMMUNICATION 1. If a = 2, b = – 4, c = – 1 then determine the value of:

(a) 2a – b + c (b) a + 3b – 2c 2. Determine the result of:

(a) 323 832 −− (b) 223 5)4(27 −−+− 3. Determine the result of

a. [ ] )123()2(85 +−−−−+ c. 18 : (– 6) + 2 x 3 + (– 10) b. 2 x (– 3 + 5) – 3 x (– 3)

C. PROBLEM SOLVING 1. The temperature in a desert increases 30C for every hour at the day light.

The temperature at 10 a. m was 350C. a. Determine the temperature at 12 a. m b. Determine the temperature at 9 a. m

2. Rio has 36 marbles. He lost 6 marbles. Then he gives the remains to his 5 friends. If every one gets same parts, how many marbles are received by each of them?

3. A fish is at 15 m underwater. It swims up 7 m and finally dives 2 m. Find the new position of the fish!

4. There are 20 questions in a test. The score for true answer is 2; score for false answer is -1 and 0 for unanswered. Ina can answer 14 questions correctly and 2 questions are unanswered. Determine her total score!


CHAPTER 2

FRACTIONS Standard Competence 1. Understanding the properties of mathematical operation of numbers and their application in problem solving Basic Competence 1. 2 Doing Mathematical operation of fractions and apply their properties in problem solving Indicators 1. Students are able to understand the meaning of fraction 2. Students are able to mention the forms of fractions 3. Students are able to convert a fraction to the other forms 4. Students are able to arrange fractions and locate them in a number line 5. Students are able to do mathematical operation of fractions (addition, subtraction, multiplication, division

and powering) and apply them in problem solving 6. Students are able to round off a decimal 7. Students are able to write a scientific notation of a fraction or a decimal 8. Students are able to predict the product of mathematical operation of fraction A. The Meaning of fraction and Its symbol

Hello, Math Lovers! Do you like pizza? If you want to eat it, off course you have to divide it into some parts like the picture below!

The pizza is divided into 8 parts. If you take a part of it, then what is the suitable number to represent that part? Yes, you are right. We need a fraction.

For example81 , it means 1 part of all (8 parts).

Fraction is a number to show a part of all. It is usually written 0, ≠bba . a is called

numerator and b is called denominator. The position of fraction in a number line is between two integers. 21 is located in the middle of 0 and 1. You divide the distance of 0 and 1 into 2 equal parts Negative fraction is fraction that is located to the left of zero.

ba

ba

ba

ba

−−

≠−

=−

=−

0 1 2 3 4 5 -1 -2 -3 -4 -5 21

0 1 2 -1 -2 21−3

11−321− 3

11 321


Remember: The Least Common Multiple (LCM) and The Greatest Common Factor (GCF) You have studied these matters in elementary school. a) Find the LCM and GCF of 18 and 120

First, you have to find its prime factors: 18 = 2 x 32 24 = 23 x 3 x 5 LCM = you write all the factors, choose the highest power = 23 x 32 x 5 = 8 x 9 x 5 = 360 GCF = you write the same factors, choose the lowest power = 2 x 3 = 6

b) Find the LCM of 16 and 36 16 = ………. 36 = ………. LCM = ……………. = ……………. = ……….. GCF = ……… = ……….

The equivalence of fractions An equivalent fraction can be formed by multiplying or dividing the numerator and the denominator by the same number. Examples:

42

2221

21

=××

=

248

124

93

62

31

====

41

16641616

6416

=÷÷

= (We divide by 16 because 16 is the GCF of 16 and 64)

Compare two fractions You have to makes the denominator of those fractions equal by finding the least common multiple (LCM) Examples: Give the sign “<”, “>” or “=” between these fractions!

1) 41 ……

43 . The denominators are equal and 1 < 3. So,

41 <

43

2) 21 ……

65 . The denominators are different. We must change the fraction into

equivalence fraction whose denominators are equal. It’s 6 (the LCM of 2 and 6)

63 ……

65 . The denominators are equal and 3 < 5. So,

21 <

65


3) Put two fractions between 133 and

136 !

The denominators are equal i. e 13. Numbers located between 3 and 6 are 4 and 5. So,

two fractions between 133 and

136 are

134 and

135 .

4) Put two fractions between 31 and

21 !

The denominators are not equal, so we make them equal by finding the LCM. The LCM of 2 and 3 is 6.

63......

62

21.......

31

⇒ , there’s no natural number located between 2 and 3. Then,

126......

124

21.......

31

⇒ , there’s only 1 natural number located between 4 and 6. Then,

189......

186

21.......

31

⇒ . Numbers located between 6 and 9 are 7 and 8. So, those 2

fractions are 187 and

94

188= .

EXERCISE 2A 1) Our basket of fruit contains 4 bananas, 3 apples, and 3 oranges. What fraction of the

fruit in the basket is bananas? 2) What fraction of the days in a week is Friday? 3) Dani has 30 pieces of marbles. He lost 6. What is the fraction remains? 4) Out of 60 students in the class, 12 are boys. What fraction are girls? 5) What fraction is it to show the number of months having 31 days as part of the year? 6) Write down the equivalents of the following fractions with denominators less than 20.

a) 32 b)

43 c)

52 d)

65

7) Fill in the blanks to form the equivalent of the following fractions!

a) .......12

20......

.....3

41

=== c) 147......

......16

35......

72

===

b) 50......

.....27

.....6

53

=== d) ......33

......27

32......

83

===

8) Simplify the following fractions!

a) 3616 b)

13545 c)

23418 d)

72981

9) Put two fractions located between the following pairs:

a) 112 and

116 b)

31 and

43 c)

41 and

43 d)

53 and

43

10) Insert sign >, = or < between these two fractions!

a) 31 ……

52 b)

72........

83 c)

54.......

43 d)

107.......

43


B. The Form of Fraction There are some forms of fraction. 1. Proper fraction

Proper fraction is a fraction stated in baba

<, . For examples 1312,

71,

52− etc

2. Improper fraction

Improper fraction is a fraction stated in baba

>, . For examples 1322,

79,

57− etc

3. Mixed Numbers It is a fraction consists of integer and proper fraction. Examples: 2 3

1 , 5211− etc

4. Decimal It is a fraction that the denominator is 10 or a power of 10. Examples: 0.1, 23. 457, -12.5 etc. The place and digit values of a decimal Given a number 1. 234

Number Place Value Digit Value 1 ones 1 2 Tenth 0.2 3 Hundredth 0.03 4 Thousandth 0.004

5. Percent (%)

It is a fraction that the denominator is 100. Examples: 2%, 3.5%, 100%, %3 3

1 etc 6. Per mil

It is a fraction that the denominator is 1000 Examples: 3‰, 2.5‰ etc

C. How to Convert A fraction to the Other Form 1) Improper fraction and Mixed Number

How to convert an improper fraction to be mixed number Example:

Convert 5

13 into mixed numbers!

513 means 12 : 5 = 2 residue 3. So,

532

513

=

How to convert mixed number to be fraction Example:

Convert 532 into improper fraction!

The numerator = 2 x 5 + 3 = 13 The denominator = 5

So, 532 =

513


2) Decimal and Fraction How to convert decimal into fraction The Rule: One decimal place means the denominator is 101 = 10 Two decimal places means the denominator is 102 = 100 Three decimal places means the denominator is 103 = 1000 n decimal places means the denominator is 10n Example a) Convert 0.7 into a fraction!

0.7 = 107 (One decimal place, so the denominator is 10)

b) Convert 0.25 into a fraction!

0.25 = 10025 (Two decimal places, so the denominator is 100)

= 41 (Both of numerator and denominator are divided by 25)

How to convert fraction into decimal The rule: change the denominator into 10 or a power of 10. If the denominator cannot be changed into 10 or a power of 10 then divide the numerator by the denominator manually. Example:

a. 4.0104

2522

52

==××

=

b. 75.010075

254253

43

==××

= (the denominator changes into 102 = 100)

c. 2857.07:272

==

d. 3.0.....33333.03:131

=== It’s called repeated fraction.

The method to round off a decimal is equal to the method to round off integers.

3) Percent and Fraction How to convert percent into fraction Percent means a fraction that the denominator is 100. Example:

a) 2% = 501

1002

=

b) 12.5% = 81

1000125

10100105.12

1005.12

==××

=

How to convert fraction into percent The rule: change the denominator into 100 or multiply that fraction by 100% Example:

a) %3510035

52057

207

==××

=

b) %3133%

3100%100

31

31

==×=


4) Per mil and Fraction How to convert per mil into fraction Per mil means a fraction that the denominator is 1000. Example:

a) 200‰ = 51

1000200

=

b) 2.5‰ = 4001

1000025

101000105.2

10005.2

==××

=

How to convert fraction into per mil The rule: change the denominator into 1000 or multiply that fraction by 1000‰ Example:

a) 151000

155200

532003

==××

= ‰

b) ‰3266‰

3200‰

302000‰1000

302

302

===×=

EXERCISE 2B 1) Convert these improper fractions into mixed numbers!

a) 38 b)

419 c)

1517

− d) 621

− e) 12126

2) Convert these mixed numbers into improper fractions!

a)837 b)

5112 c)

319− d)

6511− e)

12714

3) Convert these fractions into decimals, round off to 2 decimal places!

a) 54 b)

87 c)

412 d)

92 e)

5635

4) Convert these decimals into fractions! a) 0.12 b) 0.05 c) – 0.625 d) 0.0125 e) 2.02

5) Convert these fractions into percents!

a) 153 b)

87 c)

412 d)

61 e) 0.025

6) Convert these percents into fractions! a) 20% b) 3

216 % c) %3 41 d) 37.5% e) 2.2%

7) Convert these per mil into fractions! a) 350‰ b) 3

216 ‰ c) ‰2 41 d) 0.5‰ e) 112.5‰

8) Convert these fractions into per mil!

a) 259 b)

4015 c)

435 d)

31 e) 0.25

D. Mathematical Operation of Fraction 1. Addition

The rule: you have to make the denominator equal by finding the LCM.


Examples:

a) 65

62

63

31

21

=+=+

b) 12114

121113

121313

129

12413

43

31)211(

432

3111 =+=+=++=+++=+

c) 12.231 + 592.07 = 604.07 12.231 592.07 604.301 EXERCISE 2C Simplify these fractions

1) 157

152+ 7)

83

873

437 ++

2) 247

83+ 8)

1254

416

925 ++

3) 218

125+ 9) 5.52 + 286.6

4) 651

625 + 10) 45.3 + 28.75 + 1.274

5) 52

107

43

++ 11) 0.25 + 51

6) 83

65

41

++ 12) 1.5 + 20% + 241

2. Subtraction

The rule: you have to make the denominator equal by finding the LCM. Examples:

a 61

63

64

21

32

=−=− (The LCM of 2 and 3 is 6)

b121111

1211

121212

121113

121313

129

12413

43

31)211(

432

3111 =−=−=−=−+=−+−=−

(The LCM of 3 and 4 is 12) c. 231.04 -125.9 = 105.14

231.04 125.90 105.14

EXERCISE 2D Simplify these fractions!

1) 83

87− 8)

411

325

617 −−

2) 81

851 − 9) 30% - 2.5%

3) 125

85− 10)

973

212

659 +−

+

–


4) 83

1274 − 11) 28.15 – 9.09

5) 12112

976 − 12) 200.75 – 99.89

6) 312

611

858 −− 13) 2.85 – 200‰ – 75%

7) 21

43

65

−− 14) %31338.0

411 +−

3. Multiplication

The rule:

dbca

dc

ba

××

=×

You multiply numerator by numerator, and multiply denominator by denominator. Examples:

a) 161

483

12413

121

43

==××

=×

b) .....................

1552

15

525

52

=−=××

−=×−=×−

c) 1.5 x 1.1 = 1.65

d) 20% x 5000 = ..............................................

.......................

15000

10020

===×

EXERCISE 2E Determine the product of these mathematical operations!

1) ⎟⎠⎞

⎜⎝⎛−×

83

54 6) 250‰ x 600

2) 612

53× 7) 24000%

3216 ×

3) 94

411

32

×× 8) 2. 35 x 0.6

4) 85

41

83

21

×+× 9) 5.12521 ×

5) 143

31

76

32

×−× 10) 41%20

72

−×

4. Division

The rule:

cbda

cd

ba

dc

ba

××

=×=÷

Division means multiplication with the inverse of multiplication of the second number Examples:

a) ............

.............

.........................

12

83

21:

83

==××

=×=


b) .......................

............

............34

23

43:

211 =−=

××

−=×−=−

c) 2.25 : 0.5 = 225 : 50 = 4.5 EXERCISE 2F Simplify these fractions!

1) 72:

74 8) 24, 12 : 3

2) 149:

185 10) 120 : 0.4

3) 4 : 81 11) 5.0:

212

4) 24 : 323 12) %120:

432

5) 312:

313 13)

212:

411:

4312

6) 311:

515 13)

87

65:

31

+

7) 4:21 14)

32:

651

512 +

5. Powering The Rule:

nnn

n

nn

bababa

ba

×=×

=⎟⎠⎞

⎜⎝⎛

)(

Examples:

a) 278

32

32

3

33

==⎟⎠⎞

⎜⎝⎛

b) 0.24 = 0016.010000

16102

102

4

44

===⎟⎠⎞

⎜⎝⎛

EXERCISE 2G

1) 3

56⎟⎠⎞

⎜⎝⎛ 6) (2.3)2

2) 4

31⎟⎠⎞

⎜⎝⎛ 7) ( )35.1−

3) 5

21⎟⎠⎞

⎜⎝⎛− 8) 4)3.0(

4) 2

523 ⎟⎠⎞

⎜⎝⎛ 9) 4)2.0(−


5) 2

524 ⎟⎠⎞

⎜⎝⎛− 10) (5 x 0.01)3

6. Root

The Rule:

n

nn

nnn

ba

ba

baba

=

×=×

Examples:

a) 32

94

94

==

b) 81

81

81

81

81

3

333 −=

−=

−=

−=−

c) 2.0102

1000016

10000160016.0

4

444 ====

EXERCISE 2H

1) 3

271 6) 21.1

2) 3

1258

− 7) 3 008.0

3) 5

321

− 8) 0196.0

4) 729121 9) 3 000125.0−

5) 1691 10) 4 00000081.0

E. Application of Fraction in Daily Life

1. The Application of Addition and Subtraction We often face some problems around us relating to addition and subtraction of fraction. EXERCISE 2I Solve these questions carefully!

1) The weight of Maria is 4941 kg while her sister Tania is 35.4 kg. Determine the

sum of their weight!

2) Mrs. Mirna buys a cloth measuring 20 m in length. She uses 121115 m of it to make

curtain and another 312 m to make pillowcases. How much cloth is left?

3) The weight of Albert is 52.5 kg and Amy is 49.25 kg. Determine the difference of their weight!


4) A box contains of 4 types of cakes. 31 of the cakes are chocolate cakes,

41 of the

cakes are durian cakes, 61 of the cakes are orange cakes and the rest are strawberry

cakes. Determine the part of strawberry cakes!

5) Tony did 21 of his task in the morning then he continued

31 of his task in the

afternoon. He finished the rest in the night. How many parts of his task did at the night?

2. The Application of Multiplication

EXERCISE 2J Solve these questions carefully! 1) Ria’s weight is 40. 5 kg. Her father is 2.2 times heavier than her. What is her father

weight?

2) Anton takes 43 hours to walk to his school. But he takes only

53 of that time if he

goes to school by a bicycle. How many minutes does Anton take to go to school by a bicycle?

3) Given the price for 1 kg of rice is equal to a half of the price of 1 kg of sugar. How ever, the price for 1 kg of sugar is 0.75 of the price of 1 kg of eggs. If the price for 1 kg of eggs is Rp. 8 000, then determine the price of 1 kg of rice!

4) 52 of actors are musician.

41 of musician are not smoking. How many musician

are smoking?

3. The Application of Division EXERCISE 2K Solve these questions carefully!

1) Devi has 1832 kg of salt. She packs every 1

61 kg of salt into a packet. How many

packets of salt does Devi have? 2) 44.81 kg of nuts A is mixed with 56.39 kg of nuts B. Then it packed into 8

packets. What is the weight of each packets nuts?

3) The length of a rope is 5921 m. It cuts into some pieces each measuring 2

65 m.

How many pieces of rope are there?

4) Mr. George has $ 3750. 61 of his money is divided equally between his sons. If

each of them receives $ 7881 , how many sons does Mr. George have?

5) Mr. Smith wants to plant some palms in the edge of a street measuring 28 m long. If the distance between 2 trees is 3.5 m, then determine the number of palms does Mr. Smith need?


4. Contextual Questions for fraction EXERCISE 2L Solve these questions carefully!

1) There are 2500 students in a university. 51 of them are Australians,

43 of them are

Indonesians, and the rest is Japanese. a. What is the fraction for Japanese students? b. How many Japanese students are there?

2) There are 300 boys in a field. 31 of them wear blue shoes,

53 of them wear black

shoes and the rest wear white shoes. a. What is the fraction for boys wearing white shoes? b. How many students wear white shoes?

3) Rp. 36000 is divided for 3 boys. 2 boys get 61 and

95 of the money.

a. Determine the part of third boy! b. How much does the third boy get?

E. Scientific Notation Hello, Math Lover? Have you ever imagined what the total number of Indonesian citizen is? May be the number of Indonesian Citizen is more than 200 000 000 people. Have you ever imagined how small a virus is? Perhaps it is 0. 000 000 000 000 001. We need too much zero to represent that number. In Mathematics there’s a simple way to state a very large number or a very small number. It is called scientific notation. The general form of scientific notation is

101,10 <≤× aa n , n is the element of integer. Remember: 103 = 1 000 10-1 = 0.1 102 = …. 10-2 = 0.01 101 = …. 10-3 = ……….. 100 = …. 10-4 = ……….. Examples Write the scientific notation, round off till one decimal place. 1. 200 000 000 = 2 x 100 000 000 = 2 x 108 2. 450 000 000 000 = 4.5 x 100 000 000 000 = 4.5 x 10……. 3. 123 000 000 = 1.23 x ……………… = 1.23 x 10……… = 1.2 x …….. 4. 0. 000 000 3 = 3 x 0. 000 000 1 = 3 x 10-7 5. 0. 012 = 1.2 x 0.01 = 1.2 x 10….. 6. 0. 000567 = 5.67 x ………. = 5.67 x 10…….= 5.7 x ……….


EXERCISE 2M Write the scientific notation, round off till one decimal place. 1. The distance between Earth and Mars is 149 600 000 km. 2. The distance between Mars and Sun is 227 900 000 km. 3. The weight of the smaller spider is 0. 0001235 kg. 4. The weight of an electron is 0. 000 000 000 000 000 000 000 1625 kg 5. Avogadro number is equal to 0. 000 000 000 000 000 000 000 06023 6. State these numbers in scientific notation, round off till 2 decimal places

a. 4 985 000 c. 0. 000 000 000 1016

b. 3165 trillion d. 161

7. State the numbers in decimal forms a. 1.733 x 10-3 c. 6.104 x 1012 b. 7. 8901 x 10-5 d. 9.7850 x 107

TEST FOR CHAPTER 2 FRACTION A. Concepts Understanding

Answer these questions carefully!

1. Arrange these fractions 2419,

127,

43 in ascending order!

2. Put two fractions located between43

32 and !

3. Simplify 12048 into the simplest form!

4. Convert %3183 into the simplest fraction!

5. Convert 35‰ into decimals!

6. Convert 74 into decimal, round off to 2 places decimals!

B. Logic and Communication

Find the result of these mathematical operations!

1. 32

61

43

++ 4. 3524:

522−

2. 325

417 − 5.

256

321

107

×+

3. 5315.0 ×


C. Problem Solving Answer these questions completely!

1. There’re 181122 liters of petrol in the fuel tank of a car. In a journey to Kudus,

9422 liters of

petrol are used. Then, another 2115 liter of petrol is added into the fuel tank. How many

liters of petrol does the fuel tank contain right now?

2. There are 20 dancers on a stage. 41 of them wear red skirt;

51 of them wear yellow skirt

and the rest wear green skirt. a. What is the fraction for the dancer wearing green skirt? b. How many dancers are there that wear green skirt?

3. Toni’s father weight is 2172 kg while Toni’s weight is

43 of his father weight. Determine

Toni’s weight!

4. The length of a rope is 3.6 m. Andy takes 95 of it. Then, he cuts it into 8 equal pieces. How

many centimeters is the length of each pieces?


MID SEMESTER TEST Subject : Mathematics Grade : VII Date : Time Allocation : 07.00 – 09.00 (120 minutes)

GENERAL INSTRUCTIONS 1. Write your name, your class and your test number in the given column on the answer

sheet! 2. Read the instructions carefully before you start the test! 3. Answer all questions on the answer sheet! 4. Do not use pencil to write the answers 5. Do not activate your HP and calculators 6. Do not see your friend’s answer and don’t open your note books 7. Check your answers before you submit your answer sheet! SPECIFIC INSTRUCTIONS A. Concepts Understanding

Choose the correct answer by giving (X) on your answer sheet! 1. Study these mathematical sentences!

i. – 3 > 2 iii. 4 < – 5 ii. – 2 > – 10 iv. – 7 < – 6 The right mathematical sentences are …. a. i and iii b. ii and iv c. i, ii, and iii d. i, ii, and iv

2. – 5 + 3 – 6 = …. a. – 5 b. – 6 c. – 7 d. – 8

3. – 2 x (– 21) x 5 = …. a. 210 b. 110 c. 105 d. 42

4. Study these properties of mathematical operations! i. Enclosed iii. Associative ii. Commutative iv. The identity element is 0 The properties of addition of integers are …. a. ii, iv and iv b. i, iii and iv c. i, ii, and iv d. i, ii, iii, and iv

5. The result of 235 )22( ÷ is …. a. 23 b. 24 c. 25 d. 26

6. Which one of these statements is true?

a. 25% < 0.75 <85 b.

85 < 25% < 0.75 c. 25% <

85 < 0.75 d.

85 < 0.75 <

25%

7. The result of 2413 ÷ is ….

a. 851 b.

216 c.

813 d.

211


8. ....2524

3635

=⎟⎠⎞

⎜⎝⎛−×

a. 157

− b. 1514

− c. 151

− d. 21

−

9. ....000256.0 = a. 16 b. 1.6 c. 0.16 d. 0.016

10. 40% of Rp. 50 000 is …. a. Rp. 40 000 b. Rp. 30 000 c. Rp. 20 000 d. Rp. 10 000

B. Logic and Communication Fill in the blanks with the correct answer! 11. The difference between – 3 and 4 is …. 12. The result of 4 + 3 x (- 2) is …. 13. The result of 15 x 11+ 15 x 9 is …. 14. The result of 54 ÷ (- 9) + 8 is …. 15. The result of 25 + 3 8− + 5 is ….

16. A fraction located between 43 and

54 is ….

17. %3216 of 1200 is ….

18. 25‰ of 80 is ….

19. ....8.1212 =×

20. ....434

616 =−

C. Problem Solving

Solve these questions carefully! 21. Andy has 16 marbles. His father gives him 8 marbles and his mother gives him 6

marbles. Then, Andy gives his 20 of his marbles to his brother. How many marbles does Andy have now?

22. A bird is 10 m above sea water level while a fish is directly below it, at 6 m below sea water level. What is the distance between the bird and the fish?

23. ....32

98

651 =÷+

24. 200‰ – 0.25 + ....53=

25. There are 60 persons in a room. 52 of them are teachers,

31 of them are scientists and

the rest are students. a. What is the fraction for students? b. How many students are there?


CHAPTER 3 ALGEBRA FORM

Standard Competence 2. Understanding algebra form, linear equation in one variable and linear in equation in one variable Basic Competence 2. 1 Recognizing algebra form and its parts 2. 2 Doing mathematical operation in algebra form Indicators 1. Students are able know the definition of algebra form and its parts (term, coefficient, variables, constant,

like term and unlike term)Students are able to mention the forms of fractions 2. Students are able to do mathematical operation of like terms and unlike terms (addition, subtraction,

multiplication, division and powering) 3. Students are able to apply the properties of algebra form multiplication for solving a question 4. Students are able to do mathematical operation of algebra form in fraction that the denominator is one term

(addition, subtraction, multiplication, division and powering) 5. Students are able to simplify the result of mathematical operation of fraction of algebra form A. The definition and the Part of Algebra Form

Hello, Math Lovers! Can you see the contents of your pencil box? There are 3 pens, 2 pencils and an eraser. If we suppose pens a, pencils b and eraser c, we will obtain that your pencil box contains 3a + 2b + c. This form is called Algebra Form. 3a means 3 x a = 3 . a 4n means 4 x n = 4 . n y means 1 x y = 1 . y = 1y = y - y means -1 x y = -1 . y = -1y = - y Algebra form is a mathematical sentence involving variable, coefficient and constant. Variable is something that the value is unknown. It is usually stated in an alphabet such as a, b, x, y, z, p, q, etc. Coefficient is a number sticking on a variable. Constant is an exact number. The parts of algebra form are terms, variable, coefficient and constant. Examples: 1) 3a + 2b + c

There are 3 terms. They are 3a, 2b and c There are 3 variables. They are a, b and c


The coefficient of a is 3. The coefficient of b is 2. The coefficient of c is 1. (1 x c = 1c = c) There’s no constant.

2) 2x – 5y + 6 There are 3 terms. They are 2x, – 5y, and 6 There are 2 variables. They are x and y The coefficient of x is 2. The coefficient of y is – 5. The constant is 6.

3) 4p – q + r + s21

There are 4 terms. They are ……., ………., …….. and …….. There are …… variables. They are ……………………….. The coefficient of p is ……. The coefficient of q is ……. The coefficient of r is ……. The coefficient of s is ……. There’s no constant.

4) 12 −− xx There are ……. terms. They are …………….. There are 1 variable. It is x. The coefficient of x2 is …... The coefficient of x is ………... The constant is ……..

Like Terms and Unlike Terms

Like terms are terms that have similar variable and the power of the variable is equal. Unlike terms are terms that don’t have similar variable and the power of the variable is

not equal. Study these examples. Identify whether it is like term or unlike term. 1. 2x and – 3x (They are like terms, because the variable are equal, and also its power) 2. 4x and 4y (They are unlike terms, because the variable is different) 3. – 2x2 and 11x (They are unlike terms, because the power of its variable is different) 4. 4ab2 and 6a2b. (………………………………………………………………………..) 5. – x2y and xy (………………………………………………………………………..) 6. 4ab3c and -4ab3c (……………………………………………………………………) EXERCISE 3A 1) Identify the parts (terms, variables, coefficients and constant) of these algebra forms!

a. 3x – y d. 151

21

+− qp

b. a – 2b – 3 e. 6a2 + 3ab – b2 +5 c. 4p – 5q +2 f. 2xxy −

2) Identify whether it is like term or unlike term) a. 4xy and -4xy c. 2a and 2a2 b. 5x2y3 and 5xy d. -3a3 and -3a


B. Mathematical Operation of Algebra Form 1. Multiplication and Powering

Remember: ..........+=+ aaa nm (a x b)n = a…..x b…… a x b = b x a (………………… properties) (a x b)x c = a x (b x c) (………………….properties) Examples: a) 3 x a x b = 3ab b) 4 x 5 x a = 20a c) 3a x 5a3 = 3 x 5 x a x a2 = 15a1+3 = 15a3 d) – 4x3y x 15x2y6 = - 4 x 15 x x3 x x2 x y x y6 = - 60x3+2y1+6 = - 60x5y7 e) (2a)4 = 24 x a4 = 16a4 f) – (3a)2 = 222 9)3( aa −=×− EXERCISE 3B Simplify these algebra forms! 1) 4 x (-2) x a x b2 6) (12k)2 2) )2(3 mnm −× 7) 4)( ab− 3) )8()5(2 2 prxqrp −−× 8) 32 )5( y− 4) )2(46 322 abcbaab −××− 9) 32 )2( pqqp × 5) )12(4 24223 cbacba −× 10) 32 )2(3 baab×

2. Addition Remember about distributive toward addition:

a x (b + c) = a x b + a x c acabcba +=+⇔ )(

)()(

)(

baxbxaxbaxxbxa

cbaacab

+=+⇔+=+

+=+⇔By, commutative property we obtain,

Rule of Addition: two terms or more can be added when they are like terms. Examples: a) aaaaa 99)72(72 =⋅=+=+ b) 22222 44.)51(5 aaaaa ==+−=+− c) 22 5454 xxxx +=+ (We cannot make it simple, because they are unlike terms) d) 2(4x + 3) = 2. 4x + 2 x 3 = 8x + 6 e) -3 (-2a + 5) = -3. (-2a) + (-3) . 5 = 6a + (-15) = 6a – 15

EXERCISE 3C Simplify the form of these algebra forms! 1) aa 48 + 7) xxxx 512 22 +++ 2) )2(7312 baba −+++ 8) Find the sum of 2a + 1 and -2a + 5 3) )53(5 +x 9) Find the sum of -3a + 6 and 12a - 3 4) )72(2 +− x 10) 2 (x + 2y) + 4x + 6y 5) )57(4 +− x 11) (2x)2 + 3 (x2 – 2)


6) Given A = x + 2 and B = 3x + 4. Determine: a) A + B c) 3A + 2B b) 2A + B d) A + 4B

3. Subtraction

Remember about distributive toward subtraction: a x (b - c) = a x b - a x c

acabcba −=−⇔ )(

)()(

)(

baxbxaxbaxxbxa

cbaacab

−=−⇔−=−

−=−⇔By, commutative property we obtain,

Rule of Addition: two terms or more can be subtracted when they are like terms. Examples: a) aaaaa 77)512(512 =⋅=−=− b) 22222 4)4.()31(3 ababababab −=−=−−=−− c) 22 8282 xxxx −=− (We cannot make it simple, because they are unlike terms) d) 2 (x – 4) = 2 . x – 2 . 4 = 2x – 8 e) – 3 (3a – 2) = - 3 . 3a – (-3) . 2 = - 9a – (-6) = -9a + 6 EXERCISE 3D Simplify these algebra forms! 1) 7a + 5b – 2a 2) 5x + 7y – 3x – 2y 3) 7x2 – 4x2 + 3y – 6y 4) 15 + 3 (x – 2) 5) 4 (3p + 2) – 6p 6) – 3 (x – 4y) + 4 (5x – 3y) 7) 5 (x – 2y) – 3 (2x – y) 8) 4 (3x2 + 2x) – (4x)2 – 6x 9) Subtract

a) 2a – 5 from 8a + 4 c. 4x2 + 2x – 7 from x2 + 3x + 2 b) – 3a2 – 9a from a2 + 5a d. 5x2 – xy – 2y2 from 8x2 + 4xy + y2

10) Given A = 2x – 5 and B = 3x + 2. Determine: a) A – B c) 2A – B e) 3A – 2B b) B – A d) – A – B f) A – 3B

4. Division

Remember: nmnm

n

m

aaaaa −=÷=

a1 = a a0 = 1 Examples:

a) 2353

5

3

5

662

122

12 aaaa

aa

=⋅=⋅= −

b) aabababb

aa

abba 41444

416

416 011112

22

=⋅===⋅⋅= −−

c) 4ab : 2b = (4 : 2) . (ab : b) = 2 . a = 2a


EXERCISE 3E Simplify these algebra forms!

1) abc : ac 6) )4(29

18 yxxxy

−+−

2) 12a2 : 6a 7) 48p5q8r6 : (-6p2q3r3) : 4pq2r2

3) 6x8y5 : 3x2y3 8) xxxx 1012

2

+−

4) 223

242

48

zyxzyx

− 9) – 5 (x + 3y) + (14xy : 7x) – (16xy : -2y)

5) 2332456 9327 bcacbacba ÷÷ 10) – 4 (2x – 3y) + (24xy : 8x) – 2 (36xy : -12y)

5. Substituting a number into a variable in algebra form Substituting a number into a variable in algebra form means you change the value of the variable. Examples: 1) if x = 2, then determine the value of 2x – 4

2x – 4 = 2 . 2 – 4 = 4 – 4 = 0

2) If x = 3, then determine the value of 2x2 – x – 1 2x2 – x – 1 = 2. 32 – 3 – 1 = 2. 9 – 3 – 1 = 18 – 3 – 1 = 15 – 1 = 14

EXERCISE 3F 1) Determine the value of these algebra form for k = 4

a) 3k c) 2k – 9 e) k3 + 2k2 b) – 5k d) - 3k2 – k f) 3k2 + 5k – 8

2) If p = 3 and q = 2, then determine the value of these algebra forms a) p2 d) p3 – q3 g) 2p3 + pq b) p2 + q2 e) (p – q)3 h) 3p2q – 2pq2 c) (p + q)2 f) 5p2 – q

6. The Least Common Multiple and The Greatest Common Factor of Algebra Form Remember the rule of LCM and GCF LCM: write all factors, choose the highest power GCF: ………………………………………………… Examples: Find the LCM and GCF of 12a3bc2 and 20ab5 12a3bc2 = 22 x 3 x a3 x b x c2 20ab5 = 22 x 5 x a x b5 LCM = write all factors, choose the highest power = 22 x 3 x 5 x a3 x b5 x c2 = 4 x 3 x 5 x a3 x b5 x c2 = 60a3b5c2 GCF = write the same factors, choose the lowest power = 22 x a x b = 4 x a x b = 4ab


EXERCISE 3G Find the LCM and GCF of these algebra forms! a) 18ab and 9b2c b) 15ab2 and 20a2b3c c) 12p2q, 15q2r and 24pqr2 d) 15pq2r, 25qr2s and 75prs e) 24pqr2, 30pq3r and 12prs3

C. Fraction in Algebra Form

It is a fraction which numerator, denominator or both of them are algebra form.

aaa

21

221

≠=

1. Addition and subtraction

The rule: you make the denominator equal by finding the LCM) Example:

aaaaaaaaaa65

65

623

62

63

3231

21

==+

=+=+=+

EXERCISE 3H Simplify these algebra forms!

1) 7

47

aa+ 6)

yxx −

2) 4

244

3 qqq−+ 7)

83

41 ++

− aa

3) 3

24

kk+ 8)

xx

xx

32

61 −−

+

4) 2

45mm

+ 9) 3

46

14

3 ++

−−

+ bbb

5) 6

54

3 pp− 10)

yx

yxy 33

2

+−

2. Multiplication, Division and Powering

Remember:

bdac

dbca

dc

ba

=××

=×

bcad

cbda

cd

ba

dc

ba

=××

=×=÷

n

nn

ba

ba

=⎟⎠⎞

⎜⎝⎛

Examples:

a) bab

aab

aab

a 248

488

4==

××

=×

b) qpq

pppqppq

8324

3212

23:12

==×=


c) ( )( ) 6

3

3.2

33

32

33

2

8222yx

yx

yx

yx

===⎟⎟⎠

⎞⎜⎜⎝

⎛

EXERCISE 3I Simplify these algebra forms!

a) 3

2 ba× f)

pqqp

524

8

2

× k) 33⎟⎠⎞

⎜⎝⎛

bca

b) 152

27 ab× g) 2

2

36

49

mkn

kmn

× l) 42

2 ⎟⎟⎠

⎞⎜⎜⎝

⎛ yx

c) p

p29

154

× h) 146

2 xyy

x÷ m)

3

23

5⎟⎠⎞

⎜⎝⎛−

ba

d) 154

52 qq

÷ i) 2158

6 zxy

zxy

÷ n) 53

⎟⎟⎠

⎞⎜⎜⎝

⎛−

cba

e) 214

514 q

q÷ j) 2

2

1512 yxz

zyx÷ o) 2

2 23abab

÷⎟⎠⎞

⎜⎝⎛

D. The application of algebra form

EXERCISE 3J Solve these questions carefully! 1. The length and width of a rectangle respectively are (x + 3) cm and (x – 2) cm.

Determine: a. the perimeter of the rectangle stated in x b. the area of the rectangle stated in x c. the perimeter and the area of the rectangle if x = 9

2. There’s a square ABCD. AB = (2a + 5) cm. Determine: a. the perimeter of the square stated in a b. the area of the square stated in a c. the perimeter and the area of the square if a = 9

3. Look at this right triangle ABC. Determine: AB = (x + 1) m, BC = 2x cm, AC = (x + 3) cm.

a. the perimeter of the triangle b. the area of the rectangle c. the perimeter and the area of the triangle when x = 5

4. Rio wants to make a garden in his land. The length and the width of the land is 30 m and 20 m. Determine: a. the length and width of the garden. b. the perimeter and area of the garden c. the perimeter and area of the garden if x = 2

A B

C

GARDEN

x

x

x x


5. Diagonals of a rhombus are (5x + 8) cm and (3x – 2) cm. a. state the area of a rhombus in algebra form b. determine the area when x = 4

TEST FOR CHAPTER 3 ALGEBRA FORM A. Concepts Understanding

Choose the correct answer! 1. Given algebra form 5x2 – 3x + 2y + 5. The coefficient of x is ….

a. 5 b. – 3 c. 2 d. 5 2. The variables of algebra form in questions number 1 are ….

a. 5, - 3 and 2 b. x2 c. x and y d. x2 and y 3. These are the examples of like terms, except ….

a. 5xy and –5xy b. 2x2y and 2xy c. 4ab2 and 3a2b d. –a2b and a2b 4. The simple form of 2x + 4 – 4x + 5 is ….

a. – 2x + 9 b. 2x + 9 c. – 6x + 9 d. 6x + 9 5. The simple form of 5a – b + 3a – 4b is ….

a. 2a – 5b b. 8a – 5b c. 2a + 3b d. 8a – 3b 6. The result of 2x2 x 0.5xy is ….

a. x3y b. 2.5x3y c. x2y d. 10xy 7. =÷− 34 6156 xx …..

a. 26x b. 26x2 c. – 26x d. – 26x2 8. The value of –3x2 + x – 4 for x = 2 is ….

a. 6 b. – 6 c. – 8 d. – 14 9. – 2(3x – 4) = …..

a. – 6x – 8 b. – 6x + 8 c. 6x + 8 d. 6x – 8 10. (-4a2b)2 = …..

a. –16a2b2 b. –16a4b2 c. 16a4b2 d. 16a2b2 B. Logic and Communication

Answer these questions carefully! 1. Given A = - 3x + 2 and B = x – 5. Determine:

a. A + B c. A – 2B b. 2A + 3B d. – A – 3B

2. Simplify these algebra forms!

a. 4132

21

+−+ xx c. 3

132

8 ++

− aa

b. 3

2:9

123

26 22 aabaab

−× d. 9

34

52 +−

− xx

C. Problem Solving

Answer these questions completely! 1. There’s a square PQRS. PQ = (4a – 3 ) cm. Determine:

a. the perimeter of the square stated in a b. the area of the square stated in a c. the perimeter and the area of the square if a = 9


2. Study these figures!

AB = 3x – 7 BC = 2x + 1 CD = 4x – 11 AD = 7x – 5 Determine the perimeter of that figure!

3. Simplify this algebra form! (x + 2)2 – 3(2x – 1)

A

B C

D


CHAPTER 4

LINEAR EQUATION IN ONE VARIABLE (LEOV) AND LINEAR INEQUATION IN ONE

VARIABLE (LiEOV) A. LINEAR EQUATION IN ONE VARIABLE (LEOV)

Standard Competence 2. Understanding algebra form, linear equation in one variable and linear in equation in one variable Basic Competence 2. 3 Solving linear equation in one variable 2. 4 Solving linear in equation in one variable Indicators 1. Students are able to recognize linear equation in one variable in some forms and variables 2. Students are able to find the equivalent form of LEOV by doing addition, subtraction, multiplication

and division for both of 2 sides 3. Students are able to find the root of LEOV

I. The Definition of Linear Equation in One Variable (LEOV) A statement is a sentence that the value is known whether it is true or false. The examples of statements are: 1. Zero is the member of whole numbers. (True statement) 2. Prime numbers are always odd numbers. (…………………..) An open sentence is a statement that contains variables so that the value is unknown. The examples of open sentences are: 1. x is one of the factor of 6. 2. a + 3 = 7 3. x < 17 Linear Equation in One Variable (LEOV) is an open sentence that use equal sign (“=”), it has one variable and the power of its variable is one (1). Study these examples whether it is LEOV or not! 1. 512 =+x (LEOV, it has one variable i. e x, its power is 1, the sign is =) 2. 432 =+x ( it is not LEOV because the power of the variable is 2) 3. 5=+ yx (……………………………………………………………………………..) 4. – a – 2 = 0 (……………………………………………………………………………..) 5. 2a + 3 > 7 (……………………………………………………………………………..)


6. 2 + 3 = 5 (It’s not LEOV, it is not an open sentence but it is true sentence, it is called equality) Equality is a true statement that the sign of the relation is equal (=)

The equivalence of LEOV The symbol of equivalence is ⇔ Two or more equations are equivalent when both of two sides are added, subtracted, multiplied or divided by same numbers. Examples:

622

122

421242

=+⇔

=+

⇔

=+

x

xx

(Both of two sides are divided by 2)

1132112142

1242

=+⇔−=−+⇔

=+

xx

x(Both of two sides are subtracted by 1)

1352112142

1242

=+⇔+=++⇔

=+

xx

x(Both of two sides are added by 1)

36126123)42(3

1242

=+⇔×=+⇔

=+

xx

x(Both of two sides are multiplied by 3)

II. Finding the Root of Linear Equation in One Variable (LEOV)

Root of LEOV is the value of the variable so that the value of the equation is true. Root has the same meaning with solution. The graphic of root of LEOV is a point in a number line. There are two methods to find the root of LEOV. 1. Finding the root of LEOV by substitution method

Substitution method means finding the root by substituting the possible value for x. The value of x that makes true equation is called the root of equation. Example

Find the root of 3x – 1 = 8, x is the member of natural numbers. Answer: Natural numbers = 1, 2, 3, 4, … For x = 1, then 3 . 1 – 1 = 8 ⇔ 3 – 1 = 8 ⇔ 2 = 8 (False statement) For x = 2, then 3 . 2 – 1 = 8 ⇔ 6 – 1 = 8 ⇔ 5 = 8 (False statement) For x = 3, then 3 . 3 – 1 = 8 ⇔ 9 – 1 = 8 ⇔ 8 = 8 (True statement) For x = 4, then 3 . 4 – 1 = 8 ⇔ 12 – 1 = 8 ⇔ 11 = 8 (False statement)

So, the root of the equation is x = 3.


The graphic is

EXERCISE 4A Determine the root of these equations and make the graphic!

1) 2x – 3= 3, x is the member of whole numbers. 2) – 2a + 4 = 8, a is the member of Negative numbers less than – 5.

3) 3221

=+x , x is the member of whole numbers.

4) 21

21

31

−=+− x , x is the member of natural numbers.

5) – a – 7= – 11 , a is the member of natural numbers. 2. Finding the root of LEOV by adding, subtracting, multiplying and dividing

with same number. Examples: a. 64 −=−x

2

204644

−=⇔−=+⇔

+−=+−⇔

xxx

(Both of two sides are added by 4)

The root is x = -2.

b. 722 +=+ aa 27222 −+=−+⇔ aa (Both of two sides are subtracted by 2) 52 +=⇔ aa 52 +−=−⇔ aaaa (Both of two sides are subtracted by a) 5=⇔ a So, the root is a = 5.

c. 2x + 3 = 7

37332 −=−+⇔ x (Both of two sides are subtracted by 3) 42 =⇔ x 2:42:2 =⇔ x (Both of two sides are divided by 2) 2=⇔ x So, the root is x = 2.

0 1 2 3 4 5 -1 -2 -3 -4 -5

0 1 2 3 4 5 -1 -2 -3 -4 -5

0 1 2 3 4 5 -1 -2 -3 -4 -5


d. 3421

−=−x

....3....421

+−=+−⇔ x (Both of two sides are added by ……..)

.....21

=⇔ x

........2212 ×=×⇔ x (Both of two sides are multiplied by 2)

......=⇔ x

e. 125=+

xx

125)5

(5 ⋅=+⋅⇔xx (Both of 2 sides are multiplied by 5)

605 =+⇔ xx 606 =⇔ x ....=⇔ x (Both of 2 sides are divided by 6) So, the root is x = …..

f. 7

523

1 +=

− xx

75221

3121 +

⋅=−

⋅⇔xx (Both of two sides are multiplied by the LCM of 3 and

7 i. e 21) )52(3)1(7 +=−⇔ xx

⇔ 7x – 7 = 6x + 15 ⇔ 7x = 6x + 15 + 7 (Both of two sides are added by 7, the left side – 7 + 7 = 0)

22....7 =−⇔ x ................. =⇔

So, the root is x = ….. EXERCISE 4B Find the root of these equations!

1. 3a – 4 = a 11. 75

73

53

=−x

2. 231

21

=− aa 12. 4532

21

+=− aa

3. aa213

43

=− 13. 01432

=+x

4. 2x + 15 = 27 – 4x 14. 314

32

−=+ xx

5. 3(2a + 3) = 4a + 3 15. 06

54

32=

−−

+ xx

6. a + 4 = 7 – a 16. 212512

+=+xx

7. 2(q + 3) + (3q – 4) = 9 17. 5110

511

53

541 +=− xx


8. 4x + 3(x – 2) – (5 – 4x) = 0 18. 5(2y + 4) – (10y – 16) = 6 – 4y

9. 125=+

xx 19. (2x + 5) (x – 6) = x(2x + 3)

10. xx 234

25

+= 20. (4y – 3) (y + 6) = (2y + 3)2

III. The Application of LEOV

EXERCISE 4C Solve these questions below carefully. Ask you teacher if you get some problems! 1. The sum of three ordered number is 36.

a. make the linear equation in mathematical sentence b. determine those three numbers

2. The length of a rectangle is twice of the width. The perimeter is 54 cm. a. make the linear equation of that problem b. determine the length and width of that rectangle

3. State 0.333333…….. as a fraction! 4. State 0.454545…….. as a fraction! 5. Michelle is 30 years younger than her father. Five years later, the sum of their ages is

46 years old. a. How old is Michelle right now? b. How old is her father right now?

6. Twice of a number then subtracted by 15 is 117. a. Suppose that number is x, then make the linear equation in x b. Determine that number

7. The sum of three ordered odd numbers is 117. Determine those numbers! 8. The sum of the ages of Fandi and Ahmad is 38. Seven years ago, Fandi was three

times as old as Ahmad. Find their present ages!


B. LINEAR INEQUATION IN ONE VARIABLE (LiEOV)

Standard Competence 2. Understanding algebra form, linear equation in one variable and linear in equation in one variable Basic Competence 2. 4 Solving linear in equation in one variable Indicators 1. Students are able to use notations ≥≤>< ,,, 2. Students are able to recognize linear in equation in one variable in some forms and variables 3. Students are able to find the equivalent form of LIEOV by doing addition, subtraction, multiplication and

division for both of 2 sides 4. Students are able to find the root of LIEOV

I. The Definition of Linear inequation in one variable (LiEOV)

Linear inequation in one variable (LiEOV) is an open sentence that use relation <, >, ,≤ or ≥ , it has one variable and the power of its variable is 1 (one) Study these examples whether it is LiEOV or not! 1. 2−≤x (It is LiEOV, it has one variable i. e x, its power is 1, and the sign is ≤ ) 2. 40 <≤ a (……………………………………………………………………………...) 3. 722 ≤−x (……………………………………………………………………………..) 4. 4≥+ yx (……………………………………………………………………………...) 5. 2x + 5 = - 2 (……………………………………………………………………………) 6. 2 – 1 < 3 (It is not LiEOV, because it’s a true statement. It’s called inequality) Inequality is a true statement that use one of the sign <, >, ,≤ or ≥ . The Equivalence of LiEOV Two or more inequations are equivalent (⇔ ) when: a) both of two sides are added or subtracted by same number

2 < 3 (true), if we add both of two sides by same numbers then we obtain: 2 + 1 < 3 + 1 ⇔ 3 < 4 (True)

b) both of two sides are multiplied or divided by positive numbers 2 < 3 (true), if we multiply both of two sides by positive numbers then we obtain: 2 x 4 < 3 x 4 ⇔ 8 < 12 (True)

c) both of two sides are multiplied or divided by negative numbers but the sign should be changed, means: > is changed into <, < is changed into >, ≥ is changed into ≤ , ≤ is changed into ≥ . 2 < 3 (true), if we multiply both of two sides by negative numbers then we obtain:

2 x (-3) < 3 x (-3) ⇔ -6 < -9 (False), the true statement is -6 > -9. (< is changed into >)


II. Finding the root of LiEOV Root of LiEOV means all values of the variables so that the value of the inequation is true. The graphic of the root of LiEOV is a point or points in a number line. The method to find the root is by adding, subtracting, multiplying or dividing with same number) Examples: a) x + 2 < 4, x is the member of whole numbers

2422 −<−+⇔ x (Both of two sides are subtracted by 2) 2<⇔ x

x = 0, 1 The graphic is

b) 3x – 4 ≤ 11, x is the member of natural numbers 4113 +≤⇔ x (Both of two sides are added by 4, the left side -4 + 4 = 0)

153 ≤⇔ x

315

≤⇔ x (Both of two sides are divided by 3, the left side 3:3 = 1)

5≤⇔ x x = 1, 2, 3, 4, 5, the graphic is

c) 732 ≥+x , x is the member of real numbers 372 −≥⇔ x (Both of two sides are subtracted by 3, the left side 3 – 3 = 0)

42 ≥⇔ x 2≥⇔ x (Both of two sides are divided by 2)

So, the root is 2≥x for x is the members of real numbers. The graphic is

d) 1024 ≤−< x , x is the member of real numbers

210

22

24

−≥

−−

>−

⇔x (For every side is divided by – 2, so the sign is changed)

52 −≥>−⇔ x So, the root is 52 −≥>− x for x is the member of real numbers. The graphic is

e) 31

23

31

>+− x , x is the member of real numbers

631)

23

31(6 ⋅>+−⇔ x (Both of two sides are multiplied by the LCM of 2 and 3, i. e 6)

233)1(2 >⋅+−⋅⇔ x 292 >+−⇔ x 922 −>−⇔ x

0 1 2 3 4 5 -1 -2 -3 -4 -5

0 1 2 3 4 5 -1 -2 -3 -4 -5

0 1 2 3 4 5 -1 -2 -3 -4 -5

0 1 2 3 4 5 -1 -2 -3 -4 -5


72 −>−⇔ x

27

−−

<⇔ x (Both of two sides are divided by negative number, i. e – 2)

27

<⇔ x , The graphic is

EXERCISE 4D 1. Find the root of these inequations, show the solution in a graphic!

a) 3x + 4 < 22, x is the member of natural numbers b) 2x + 3 ≤ 7, x is the member of whole numbers c) – 2x + 5 ≥ -3, x is the member of real numbers

d) 1332

≥+− x , x is the member of whole numbers

2. Find the root of these inequations!

a) yy 453 ≥− h) 9321

>+x

b) 1)1(2 >+p i) 5)13(21

<−p

c) )32(3)12(2 +<− nn j) )12(31)15(

41

+>− yy

d) )5(4)34(2 −≤− nn k) 23

21

512

+−

<+ mm

e) 117)4(5 −>− pp l) 123

2>−

tt

f) 12 < 2x + 2 < 24 m) xx 2

534>−

g) – 5 ≤ - 3x + 1 7≤ n) 2

32)2(31 xx +>+

III. The Application of LiEOV

EXERCISE 4E Solve these questions below carefully! 1. The width and the length of a rectangle are (2n – 1) cm and (3n + 2) cm respectively.

The perimeter is not more than 82 cm. Write down the LiEOV of the perimeter and solves it!

2. Sides of a parallelogram are (3n – 4) cm and (13 + n) cm. The perimeter is not more than 98 cm. Write down the LiEOV of the perimeter and solves it!

3. Twice of Anton’s height adds with 160 cm is not more than 4 m subtracts by three times Anton’s height. If Anton’s height is t, then write down the LiEOV of the perimeter and solves it!

4. A cuboid PQRS. TUVW has length, width and height are (n + 5) cm, (n + 2) cm and n cm respectively. The sum of the edges is not more than 100 cm. Write down the LiEOV of the perimeter and solves it!

0 1 2 3 4 5 -1 -2 -3 -4 -5 27


TEST FOR CHAPTER 4 LINEAR EQUATION IN ONE VARIABLE (LEOV) AND

LINEAR INEQUATION IN ONE VARIABLE (LiEOV)

A. Concepts Understanding Choose the correct answer! 1. The solution of 3x – 4 = 17 is ….

a. 5 b. 6 c. 7 d. 8 2. The solution of 3 (2 – x) = - 12 is ….

a. 5 b. 6 c. 7 d. 8 3. Study these equations:

i. 2p +4 = 10 iii. p + 2 = 5 ii. p + 5 = 8 iv. 3p + 5 = 11 The equivalent equations are …. a. i, ii, iii b. i, ii, iv c. ii, iii, iv d. i, ii, iii, iv

4. The solution of 2x – 4 = x – 16 is …. a. – 4 b. – 12 c. – 16 d. – 18

5. The solution set of 31532

=−p is ….

a. {-27} b. {-18} c. {18} d. {27} 6. The solution set of 14 – 2x > 6, for x is natural numbers is ….

a. {4, 3, 2, 1} b. {3, 2, 1} c. {3, 2, 1, 0} d. {3, 2, 1, 0, … } 7. The solution of 4x + 7 > 3x – 14 is ….

a. x > - 21 b. x > 7 c. – 3 < x d. – 1 < x 8. Study these inequations:

i. 2 < 5 < - 10 iii. 30 gr > 5 ons > 1 kg ii. 25 cm < 5 dm < 10 m iv. – 2 > - 3 > - 7 The true inequations are …. a. i and ii b. i and iii c. ii and iii d. ii and iv

9. The solution set of 45 – 3x 276 −≤ x , for x is whole numbers is …. a. {8, 9, 10, 11, … } b. {9, 10, 11, … } c. {0, 1, 2, … , 8} d. {0, 1, 2, … , 7}

10. The solution set of 8 – (3x – 7) < 2x is …. a. x < - 3 b. x < 3 c. x > 3 d. x > - 3

B. Logic and Communication Answer the question carefully!

1. Determine the solution of )2(21)1(

31

−≤− xx , x is whole numbers. Draw the solution

on the graphic! 2. Determine the solution of 3 (2x + 1) – 4 (2x – 3) = 5. Draw the solution on the

graphic!

3. The solution of 13

142

42=

−−

+ mm is ….

4. The solution of 2x + 10 > 15 + 3x, x is whole numbers are ….


C. Problem Solving Answer these questions completely!

1. When a number is added to two thirds of it self, the result is 45. Find the number! 2. Find 3 consecutive odd numbers whose sum is 57. 3. Joe and Ahmad have 80 marbles altogether. Ahmad has 4 times as many marbles as

Joe. How many marbles does each boy have? 4. Twice of Andy’s height then added by 160 cm is not more than 4 cm subtracted by 3

times Andy’s height. If Andy’s height is t cm then determine: a. inequation based on that problems b. Andy’s maximum height.


CHAPTER 5 SOCIAL ARITHMATICS

Standard Competence 3. Applying algebra form, linear equation and linear in equation in one variable, and comparison in problem solving Basic Competence 3. 1 Making and solving mathematical models of problems related to equation in one variable 3. 2 Making and solving mathematical models of problems related to in equation in one variable 3. 3 Applying the concepts of algebra in a simple social arithmetic Indicators 1. Students are able to make mathematical models of daily problems related to equation in one variable 2. Students are able to solve mathematical models of daily problems related to equation in one variable 3. Students are able to make mathematical models of daily problems related to linear in equation in one

variable 4. Students are able to solve mathematical models of daily problems related to linear in equation in one

variable 5. Students are able to do simulation of social arithmetic in economical activities in daily life 6. Students are able to calculate total value, a half and the value for each unit 7. Students are able to calculate the value and the percentages of profits, losses, selling price, purchase price,

discount, net, bank interest in economical activities A. Purchase Price, Selling Price, Profits and Loss

Purchase Price = cost = buying price

It is the price paid by a seller to a retailer (the capital) Selling Price

It is the price determined by a seller or the money received by a seller on a given transaction

Profits A seller gets profits when the selling price is greater than the purchase price. Profits = selling price – purchase price

Loss A seller suffers loss when the selling price is less than the purchase price. Loss = purchase price – selling price

EXAMPLE A farmer buys 10 chicken for Rp. 138 000. He sells the chicks for Rp 13 500 each. Does he make loss or profit? How much is it?

Known: Purchase price for 10 chicken = Rp. 138 000 Selling price = Rp 13 500/ chick Question: loss/ profit? How much is it? Answer: Selling price for 10 chicken = 10 x 13 500 = 135 000


In fact, selling price < purchase price. So, the farmer suffers loss. Loss = purchase price – selling price = 138 000 – 135 000 = 3 000 So, the farmer suffers loss Rp. 3 000

EXERCISE 5A 1. Evaluate the loss or profit if the purchase price is Rp. 6 700 000 and the selling price is

Rp. 4 500 000. 2. Evaluate the loss or profit if the purchase price is Rp. 4 000 and the selling price is Rp.

4 800 000. 3. Evaluate the loss or profit if the purchase price is Rp. 7 650 000 and the selling price is

Rp. 7 650 000. 4. A dozen forks costs Rp. 30 000. It is sold for Rp. 3 000 each. What is the profit it the

forks are sold out? 5. A milkman buys 200 bottles of milk at dairy farm. Totally, he pays the farm Rp. 1 000

000. The transportation from the dairy farm to his place cost Rp. 7 000. He sells the milk for Rp. 6 000 per bottle. Evaluate the profit the milkman makes!

6. A baker needs Rp. 1 500 000 as capital to make 100 chocolate cakes. He sells the cakes for Rp. 25 000 each. Evaluate the profit he makes

7. A school cooperative shop purchases 4 dozen books for Rp. 425 per book. It sells the book for Rp. 450 each. Evaluate the possible loss or profit

8. A toy seller purchases 75 plastic toy cars valued Rp. 300 000. Each toy car is sold for Rp. 4 250. Five toys are damaged and unsold. Does he make a loss or profit?

9. A grocer buys 200 kg of eggs for Rp. 6 500 per kg and has to spend Rp. 25 000 for transportation fee. He sells the eggs for Rp. 6 800 per kg. If 5 kg are rotten and unsold, calculate the loss or profit!

10. Arif has bought 5 cartons of instant noodles valued Rp. 48 000. Each carton contains 24 packages. He plans to sell the noodle for Rp. 500 per package. Calculate the profit he will make if he can sell all packages!

Identifying Purchase price and selling price b. Profit Condition

Profits = selling price – purchase price Purchase price = selling price – ……… Selling price = …………………. + …………………

c. Loss Condition Loss = purchase price – selling price Purchase price = selling price + ………. Selling price = …………… – ………………

EXAMPLE The purchase price of a product is Rp. 15 000. When it is sold, the seller makes a profit valued Rp. 3 000. Find out the selling price! Known: purchase price = Rp. 15 000 Profit = Rp. 3 000 Question Selling price?


Answer: Selling price = purchase price + profit = 15 000 + 3 000 = 18 000 So, the selling price is Rp. 18 000. EXERCISE 5B 1. Complete the table below!

No Purchase Price Selling Price Profit Loss 1 Rp. 7 000 ……………….. Rp. 250 - 2 Rp. 16 750 Rp. 18 300 3 …………………… Rp. 5 700 - Rp. 450 4 Rp. 97 600 ………………….. Rp. 6 500 -

2. To make 60 cookies, Dina needs the following: b. 8 eggs valued Rp. 3 000 c. 1 kg of flour valued Rp. 5 500 d. 0.5 kg 0f sugar valued Rp. 1 200

After the cookies are sold out, she finds out that she makes a loss Rp. 700 Evaluate: a. the selling price for 60 cookies b. the selling price for each cookie

B. Percentage of Profit or Loss

Percentage of Profit = %100)(Pr×

pricepurchase

Rpofit

Profit (Rp) = Percentage of profit x Purchase price

Percentage of Loss = %100)(×

pricepurchase

RpLoss

Loss (Rp) = Percentage of loss x Purchase Price Example Nancy buys a television at price Rp. 800 000. Then she sells it at price Rp. 900 000. Evaluate the percentage of loss or profit! Known: purchase price = Rp. 800 000 Selling price = Rp. 900 000 Question: percentage of profit or loss Answer: She gets profits because selling price > purchase price. Profit = selling price – purchase price

= 900 000 – 800 000 = 100 000

Percentage of profit = %100)(Pr×

pricepurchase

Rpofit

= %5.12%100800000100000

=×

So, the percentage of her profit is 12.5%


EXERCISE 5C 1. Evaluate profit or loss percentage!

a. purchase price is Rp. 16 000, selling price is Rp. 17 500 b. purchase price is Rp. 80 000, selling price is Rp. 90 000 c. 12 dozen hairpins are bought for Rp. 57 600, they are sold for Rp. 1 750 per 5

hairpins d. The purchase price of 200 mugs is Rp. 150 000. The selling price is Rp. 3 000 per

5 mugs e. The purchase price of a mobile phone is Rp. 1 400 000. It is sold for Rp. 1 540 000

2. Mr. Raharjo purchases 50 liters of gasoline at Rp. 1 200 per liter. He sells it for Rp. 1 300 per liter. What is the percentage of the profit?

3. A grocer buys 200 kg of rice A for Rp. 2 250. He mixes it with 300 kg of another kind of rice (rice B) which cost Rp. 1 500 per kg. Then he sells is Rp. 2 000 per kg. Find out the loss or profit percentage!

Determining the purchase price and selling price when percentage is given Consider that the percentage of purchase price is 100%

The seller gets profit p% Purchase price = 100% Selling price = 100% + p% = (100 + p)%

Selling price (Rp) = ×+

100100 p purchase price (Rp)

Purchase price = ×+ p100

100 selling price (Rp)

The seller gets loss l%

Purchase price = 100% Selling price = 100% – l% = ( 100 – I)%

Selling price (Rp) = ×−

100100 l purchase price (Rp)

Purchase price = ×− l100


EXAMPLES 1. Nanny sold her shoes for Rp. 120 000. She suffered loss 20%. How much is the

purchase price? Known: selling price = Rp. 120 000 Loss = 20 % Question purchase price … ? Answer:

Purchase price = ×− l100


= ×− 20100

100 120 000

= 15000012000080

100=×

So, the purchase price for her shoes is Rp. 150 000


2. Mrs. Nancy sold her sewing machine Rp. 2 000 000. She gets profit 25%. How much did she buy that sewing machine? Known: selling price = Rp. 2 000 000 Profit = 25% Question purchase price? Answer:

Purchase price = ×+ p100


= ×+ 25100

100 2 000 000

= ×125100 2 000 000

= 1 600 000 So, she bought that sewing machine for Rp. 1 600 000.

EXERCISE 5D 1. The purchase is Rp. 4 000. The profit is 5%. What is the selling price? 2. A farmer sells 35 chicken for Rp 1 610 000. He makes 15% profit. Find out the

purchase price for a chick! 3. A dozen pens are bought for Rp. 18 000. The profit is 20%. How much should each

pen sold? 4. The purchase price of a China mug is Rp. 5 000. If there’s 2.5% loss, how much is it

sold? 5. Dani buys 20 bicycles for Rp. 120 000 each. His profit is 18%. How much does he sell

each bicycle? 6. The selling price of a product is Rp. 108 000. The seller gets profit 20%. Determine

the purchase price of that product! 7. A seller gets loss 12.5%, if she sold a bag for Rp. 12 250. Determine the purchase

price of that bag! 8. A greengrocer buys 100 kg of tomatoes for Rp. 112 000. Determine the selling price

for 1 kg of tomatoes when he makes 35% profit! 9. The purchase price of a T Shirt is Rp 300 000. It is found that the loss is 5%. What is

the selling price? 10. The school cooperative purchases 50 packages of books containing 20 books for every

package. Each book cost Rp. 600. If the transportation fee is Rp. 15 000 and the cooperative makes 25% profit, how much is each book sold?

C. Discount, Gross, Tare and Nett

Discount Discount is reduction in the cost of goods we are buying. It is usually written in percentage form. Discount (Rp) = percentage of discount x the original price

Discount (%) = %100)(×

priceoriginal

Rpdiscount


EXAMPLE A night gown costs Rp. 180 000. Because there’s a big sale, it is given 35% discount. What is the price of the gown now? Known: the original price = Rp. 180 000 Discount = 35% Question: the new price Answer: Discount = percentage of discount x original price = 35% x Rp. 180 000

= 5700018000010035

=×

The new price = 180 000 – 57 000 = 123 000 So, the price of the gown is Rp. 123 000 now.

EXERCISE 5E 1. Complete the following table

No The original price discount Nett price 1 Rp. 80 000 15% ……………….. 2 Rp. 750 000 …………….. Rp. 700 000 3 …………… 12.5% Rp. 31 500 4 Rp. 15 000 ……………... Rp. 14 625 5 ………………. 25% Rp. 33 000

2. The original price of a hand phone is Rp. 900,000.00. It is given 10% discount. What is the new price of that hand phone?

3. The original price of a tape recorder is Rp. 500,000.00. Rio pays only Rp. 450,000.00 because he gets discount. What is the percentage of the discount?

4. Tia pays Rp. 80,000.00 for a new bag because she gets discount 20%. What is the original price of that bag?

Gross, Tare and Nett Gross is the total weight of something, including its wrapping Tare is the weight of wrapping material in which good are packed Nett is the weight of something without its container or its wrapping Gross = Tare + Nett Tare = Gross – Nett Nett = Gross – Tare Tare is usually written in percentage form. Tare = percentage of tare x gross

Percentage of tare = %100×grosstare

EXAMPLE In a sack of rice, it is written GROSS = 200 kg and tare = 2 %. Determine the nett! Known: Gross = 200 kg Tare = 2% Question: Nett……….?


Answer: Tare = percentage of tare x gross = 2% x 200

= 4200100

2=×

Nett = gross – tare = 200 – 4 = 196 So, the nett of that sack is 196 kg.

EXERCISE 5F 1. Complete the following table!

No Gross Tare Nett 1 300 kg 5% ……………….. 2 80 kg …………….. 76 kg 3 …………… 12.5% 175 kg 4 71.5 kg ……………... 70 kg 5 1.5 ton ……………… 1400 kg

2. A grocer purchases a sack of flour with the gross weight 92.2 kg. The sack weighs 0.7 kg. How much will he pay if 1 kg of flour costs Rp. 3 800?

3. A grocer purchases 5 sacks of rice with the gross and nett weight of every sack are 72 kg and 1% respectively. How much will he have to pay if every kg of rice costs Rp. 2 000?

4. Mr. Darto buys a box of oranges which the weight is 25 kg and tare is 8%. The cost for a kilogram of oranges is Rp. 4,000.00. How much does Mr. Darto pay for those oranges?

5. Mrs. Arum buys a sack of sugar for Rp. 200,000.00. The gross is 20 kg and tare is 5%. Then she sells that sugar for Rp. 10,000.00 per kg. Evaluate she gets profit or loss? How much is it?

D. Bank Interest

Extra money from a bank after a customer save some money for certain time. Extra money given to the bank after a customer borrow some money from the bank

Bank interest is stated in percentage and usually the time is per year. Bank interest (Rp) = percentage of interest x capital x time New capital = capital + interest (Rp)

Bank interest (%) = %100)(int×

capitalRperest

EXAMPLE Farida has a saving account at a national bank amounting to Rp. 100 000. The bank gives interest of 18% annually. How much is Farida’s saving after 1 year? Known: capital = Rp. 100 000 Interest per year = 18% Question: Farida’s saving after 1 year


Answer: Interest in 1 year = percentage of interest x capital x time

= 18 % x 100 000 x 1

= 1800010000010018

=×

Farida’s saving after 1 year = interest + capital = 18 000 + 100 000 = 118 000

So, Farida’s saving after 1 year is Rp. 118 000.

EXERCISE 5G 1. Fany saves her money in a bank amounting to Rp. 1 500 000 with interest 15% per

year. What is the amount of her saving after a full year? 2. Dika deposits Rp. 500 000 into a bank at a simple interest 2% per year. Determine:

a. Dika’s money after 1 year b. Dika’s money after 2 years c. Dika’s money after 9 months

3. Tyas saves her money in a bank with interest of 15% per year. After 3 months, her money increases Rp. 9 000. Determine her first deposit!

4. Mr. Tarjo borrows Rp. 20 000 000 from a bank with interest 11% per year. He will pay it step by step for 12 months. How much does he pay to the bank for each month?

5. On April 21, Budi opens an account at Vira Bank with interest 15 % paid daily. His first saving is Rp. 150 000. What is his saving on August 21 in the same year?

E. Tax

Money that is paid to government Tax is usually stated in percentage

There are some kinds of tax. 1. Income tax (PPh)

Income tax (Rp) = percentage of tax x original salary Nett salary = original salary – income tax (Rp)

2. Sales tax (PPn) Sales tax (Rp) = percentage of tax x original price New price = Original price + Sales Tax (Rp)

EXAMPLE Tariff for an SMS is Rp. 2 000 and PPn 10%. How much does Ina pay for an SMS? Known: original price = Rp 2 000 PPn = 10% Question: payment for an SMS Answer: PPn = percentage of tax x original price = 10% x 2 000

= 200200010010

=×


Payment for an SMS = PPn + original price = 200 + 2 000 = 2 200 So, Ina pays Rp 2 200 for an SMS EXERCISE 5H 1. Mr. Budi’s original salary is Rp. 900 000 per month. He gets PPh 10%. How much is

the salary that he receives every month? 2. The cost for a meal is Rp. 24 000 and PPn 10%. How much does a customer pay for 2

packets of meal? 3. Mr. Andy buys a new TV for Rp. 1 300 000 and the sales tax is 10%. But he gets

discount 5% because he pays it cash. How much does he pay for that TV?

TEST FOR CHAPTER 5 SOCIAL ARITHMATICS

A. Concepts Understanding Choose the correct answer! 1. A bag costing Rp. 28 000 is sold for Rp. 35 000. The percentage profit is ….

a. 10% b. 15% c. 20% d. 25% 2. A vase costing Rp. 60 000 is sold for Rp. 50 000. The percentage loss is ….

a. %3212 b. 12% c. %

3216 d. 16%

3. A book seller gains 30% by selling a book for Rp. 65 000. Find the cost of the book! a. Rp. 50 000 b. Rp. 40 000 c. Rp. 30 000 d. Rp. 20 000

4. A watch priced at Rp. 160 000 is sold for Rp. 140 000. The percentage discount is …. a. 12.5% b. 2.5% c. 12% d. 5%

5. A man borrows Rp. 1 000 000 for 3 years at a rate of 6% per annum. What is the simple interest he has to pay? a. Rp. 150 000 b. Rp. 180 000 c. Rp. 200 000 d. Rp. 210 000

6. In a sack of flour, it’s written gross = 200 kg, tare = 2.5%. The weight of flour is …kg a. 195 b. 5 c. 190 d. 10

7. John bought an AC priced at Rp. 800 000 but was given a discount of 12.5%. Calculate the price he paid! a. Rp. 500 000 b. Rp. 600 000 c. Rp. 700 000 d. Rp. 800 000

8. Andy buys a bicycle at price Rp. 810 000 and transportation fee Rp. 15 000. He sold it again at price Rp. 820 000. The true statement is …. a. profit, Rp. 10 000 c. loss, Rp. 5 000 b. profit, Rp. 25 000 d. loss, Rp. 10 000

9. A seller sells 10 goats at price Rp. 200 000/ each. He gets profit Rp. 300 000. The buying price is … a. Rp. 1 700 000 b. Rp. 2 300 000 c. Rp. 2 800 000 d. Rp. 3 200 000

10. Mrs. Louise sold her sewing machine at price Rp. 1 500 000 and got loss 25%. The buying price is …. a. Rp. 1 100 000 b. Rp. 1 125 000 c. Rp. 1 875 000 d. Rp. 2 000 000


B. Logic and Communication Answer these questions carefully! 1. Susan sells her car at a loss of 6%. What is her selling price when she paid

Rp. 18 400 000 for it? 2. A shopkeeper bought a radio from a wholesaler for Rp. 250 000. In addition, he paid a

value added tax of 15% on the cost price. Then he sold the radio for Rp. 315 000. Calculate the cash profit made by the shopkeeper!

3. A shopkeeper buys 300 identical vases at a total cost Rp. 1 500 000. He fixes the selling price of each vase at 20% above the cost price and sells 260 vases at this price. As for the remaining vases, he sells them at 50% of the selling price. Calculate the shopkeeper’s total profit!

C. Problem Solving

Answer these questions completely!

1. To make a profit of %3133 , a bicycle must be sold for Rp. 240 000. What is the

purchase price for the bicycle? 2. By selling a book for Rp. 16 500, a book seller suffers loss 12%. What is the purchase

price of the book? 3. A florist bought 360 roses at Rp 10 000 per dozen. If he sold them at Rp. 1 100 each,

what is his percentage profit? 4. During a sale a shopkeeper reduced the price of all goods by 15%. Calculate the

original selling price of a calculator which was sold for Rp. 238 000 during the sale! 5. A shopkeeper buys some eggs at Rp. 150 each. Six of them are broken while the rest

are sold at Rp. 200 each. If he makes a profit of Rp. 48 000, find how many eggs he bought!


CHAPTER 6 RATIO

Standard Competence 3. Applying algebra form, linear equation and linear in equation in one variable, and comparison in problem solving Basic Competence 3. 4 Applying comparison in problem solving Indicators 1. Students are able to know the meaning of scale and how to calculate it 2. Students are able to solve some problems related to equivalent ratio and reverse ratio (comparison) A. Scale

Scale is usually found in a map or an atlas, for example: 1 : 200 000 means 1 cm on the map represents 200 000 cm in the fact (real distance)

Scale =cedisreal

maptheoncedis

tan

tan

Distance on the map = scale x real distance Real distance = distance on the map : scale EXAMPLE The distance from Bogor to Cirebon on the map is 4 cm. The scale of the map is 1 : 5 600 000. Determine the real distance between them! Known : distance on the map = 4 cm Scale = 1 : 5 600 000 Question : the real distance Answer: Real distance = distance on the map : scale

= 4 : (1 : 5 600 000)

= 1

56000004×

= 22 400 000 cm = 224 km

So, the real distance from Bogor to Cirebon is 224 km. Exercise 6A 1. The distance from Semarang to Jepara on the map is 15 cm. If the scale is 1 : 800 000

then determine the real distance of Semarang and Jepara! 2. The distance of 2 towns in a map is 4 cm. The real distance is 224 km. What is the

scale of that map?


3. The scale of a map is 1 : 200 000. Determine the distance of 2 towns in that map, if their real distance is 20 km!

4. A model ship is 0.5 meter long, and the real ship is 30 meter long. What is the scale of the model?

5. Mr. Tomo has a land. The length and the width respectively are 72 m and 66 m. Then he takes the picture of his land. The length and the width in the picture are 12 cm and 11 cm. Determine: c. the scale of that picture d. the comparison of the real area of the land and the area of its picture!

B. Ratio The ratio of a and b, 0≠b is written a : b or

ba . The result of that comparison is the

simplest form of that ratio. It means simplifying the fraction form. EXAMPLES 1. Simplify the comparison 45 : 75

Answer:

45 : 75 = 5:31575:

1545

=

2. The ratio of Anton’s height and Ian’s height is 8 : 9. Determine Anton’s height if Ian’s height is 180 cm! Known: Anton’s height : Ian’s height = 8 : 9 (It means that Anton’s part is 8 and

Ian’s part is 9) Ian’s height = 180 cm

Answer:

Anton’s height = 16018098

=×

So, Anton’s height is 160 cm.

Exercise 6B 1. Simplify these ratios!

a. 48 : 144 b. 4 cm : 1.6 m 2. There are 40 students in a class. 18 students are boys. Determine the ratio of:

a. the number of girl students and the number of boys students b. the number of girl students and all students in that class

3. There are 30 balloons. 8 balloons are yellow, 12 balloons are red and the rest is green. Determine the ratio of yellow balloons, red balloons and green balloons!

4. The ratio of Andy’s money and Shinta’s money is 3 : 5. If Andy has Rp. 18 000 then determine Shinta’s money!

5. The ratio of Rio’s marbles, Roy’s marbles and Ronnie’s marbles is 5: 4 : 3. The sum of their marbles is 60. Determine the marbles of each boy!

6. Mira is 8 years older than Tika. The ratio of their ages is 5 : 3. How old are Tika and Mira?


C. Proportion A proportion is a statement expressing the equivalence of two ratios. There are two types of proportion, i.e. direct proportion and inverse proportion.

1. Direct Proportion

Study this table!

No. of book 1 2 3 4 5 a b

Price (Rp) 2 000 4 000 6 000 8 000 10 000 c d

When the number of the book increases, the price for that book also increases. When the number of the book decreases, the price for that book also decreases. It is called direct proportion. In direct proportion a : b = c : d, where 0, ≠db then (i) cbdadcba ×=×⇔= ::

(ii) cbdadc

ba

×=×⇔= . It is called cross multiplication

EXAMPLE The price for 3 books is Rp. 36 000. How much will Ronnie pay for 8 books? Known: 3 books → Rp. 36 000

8 books → x Question: x = price for 8 books Answer:

33600083600083

:360008:3

×=

×==

x

xx

= 96 000 So, Ronnie will pay Rp. 96 000 for 8 books. Exercise 6C 1. A secretary can type 380 words for 20 minutes. How many words can be typed for

212 hours?

2. The prize of 2 dozen books is Rp. 28 800. Determine the price for 5 books! 3. The cost to stay in a hotel is Rp. 875 000 for a week. Determine the cost to stay there

for 4 days!

4. 95 of a piece of metal weighs 7 kg. What is the weight of

72 of the metal?

5. A worker receives Rp. 17 000 after working for 4 hours. Determine his wage for 7 hours!

6. A car takes 2 hours to reach 120 km. How many hours does it take to reach 50 km?


2. Inverse Proportion Study this table! This table show the time for students to finish eating 1 kg of popcorn.

No. of students 1 2 4 8 a b

Time (hour) 4 2 1 0.5 c d

When the number of students increases, the time decreases or when the number of students decreases, the time increases. It is called Inverse proportion.

In inverse proportion a : b = dc1:1 , then:

(i) dbcacb

da

cb

da

dcba ×=×⇔=⇔×=×⇔=

111:1:

(ii) the product of multiplication of each column is equal. 1 x 4 = 2 x 2 = 4 x 1= 8 x 0.5 = a x c = b x d EXAMPLE 10 men can finish a piece of job in 8 hours. How long will it take when 16 men work together? Known: 10 men → 8 hours 16 men → x hours Question: time taken by 16 men. Answer:

581016

81610

8116110

1:8116:10

=×=

=

×=×

=

xx

x

x

x

So, 16 men will take 5 hours to finish the job.

Exercise 6D 1. Ten men can dig a trench in 4 hours. How long will 5 man take to dig the same

trench? 2. Four pipes can fill a tank in 70 minutes. How long will it take to fill the tank if 7 pipes

are used? 3. A plane flies at an average speed of 770 km/h takes 15 hours to complete a journey.

Find the time taken for the plane to complete the same journey if its average speed is 660 km/h.

4. A contractor estimates that he would need 56 workers to complete a job in 21 days. If he asked to complete the job in 14 days, find the additional number of workers he has to employ.

5. A farmer has enough feed to last his 40 cattle 35 days. If he buys 10 more cattle, how long can the same feed last?


6. A contractor estimates his project will finish in 40 days with 48 workers. After 10 days, the project is stopped for 6 days. If he asked to finish the project on time, find the number of additional workers that is needed!

7. A project can be finished by 25 workers in 18 days. After they worked for 6 days, then the project is stopped for 2 days. How many workers are added so that the project finishes on time?

TEST FOR CHAPTER 6 RATIO A. Concepts Understanding

Choose the correct answer! 1. Distance between 2 towns in a map is 16 cm. If the real distance is 20 km then the

scale of the map is …. a. 1 : 125 000 b. 1 : 80 000 c. 1 : 12 500 d. 1 : 8 000

2. Father’s salary is Rp. 1 400 000 and mother’s salary Rp. 800 000. The comparison of father’s salary and mother’s salary is …. a. 2 : 1 b. 1 : 2 c. 4 : 1 d. 7 : 4

3. Which one of these conditions that represents inverse proportion? a. the number of books and its price c. distance and time b. velocity and time d. people and food needed

4. The height of a building in a magazine is 6 cm. The real height is 8 m. If the height of the door in the magazine is 1.5 cm, then the real height of the door is …m. a. 2.00 b. 1.50 c. 1.25 d. 1.00

5. If the price of 6 kg of biscuits is Rp. 27 000 then the price for 13 kg of biscuits is …. a. Rp. 55 500 b. Rp. 56 500 c. Rp. 57 500 d. Rp. 58 500

6. The graphic of direct proportion is …. a. b. c. d.

7. The prize of 1 dozen t-shirt is Rp. 300 000. The price of 3 t-shirts is …. a. Rp. 100 000 b. Rp. 75 000 c. Rp. 50 000 d. Rp. 25 000

8. The ratio of Intan’s money and Dita’s money is 3 : 5. If Dita’s money is Rp. 12 500, then Intan’s money is … a. Rp. 5 000 b. Rp. 7 500 c. Rp. 10 000 d. Rp. 12 500

9. The difference of Dita’s money and Intan’s money is Rp. 4 000. How much is Intan’s money? a. Rp. 12 000 b. Rp. 10 000 c. Rp. 8 000 d. Rp. 6 000

10. Rini wants to enlarge her picture measure 5 cm and 8 cm with ratio 2 : 1. The new area of her picture is …. cm2 a. 40 b. 60 c. 80 d. 160

B. Logic and Communication

Answer these questions carefully! 1. Twelve men take 6 hours to finish a piece of work. After they have worked for 1 hour,

the contractor decides to call in 8 more men so that the work can be completed earlier. How many more hours would 20 men take to complete the remaining work?

2. At a scout’s camp, there’s sufficient food to last 72 scouts for 6 days. If 18 scouts do not turn up for the camp, how much longer can the food last for the other scout?


C. Problem Solving Solve these problems completely! 1. A sum of money is divided among 3 persons, X, Y and Z in the ratio 10 : 7 : 5. If Y

gets Rp. 14 000 more than Z, how much will X get and what is the total sum of money?

2. A container is filled with 56 liters of pineapple juice. 8 liters of pineapple juice are extracted and the container refilled with mango juice. The content of the container is thoroughly mixed and 8 liters of the mixture are then extracted and the container is again refilled with mango juice. What is the ratio of mango juice to pineapple juice in the final mixture?

FINAL EXAMINATION OF FIRST SEMESTER Subject : Mathematics Grade : VII (Immersion Program) Date : Time Allocation :

GENERAL INSTRUCTIONS

1. Write your name, your class and your test number in the given column on the answer sheet!

2. Read the instructions carefully before you start the test! 3. Answer all questions on the answer sheet! 4. Do not use pencil to write the answers 5. Do not activate your HP and calculators 6. Do not see your friend’s answer and don’t open your note books 7. Check your answers before you submit your answer sheet!

SPECIFIC INSTRUCTIONS A. Concepts Understanding

Choose the correct answer by giving (X) on your answer sheet! 1. Study these properties!

i. Enclosed property iv. The identity element is 0 ii. Commutative property v. The identity element is 1

iii. Associative property vi. Distributive toward subtraction The properties of multiplication of integer are …. 4. i, ii, iii, iv, vi b. i, ii, iii, v, vi c. i, ii, iii, v d. i, ii, iv, v

2. (300 + 11) (300 – 11) = …. a. 89979 b. 89879 c. 89889 d. 88989

3. ....21

75

321 =+−

a. 42191 b.

42191− c.

42231 d.

42231−

4. The result of subtraction (2x + 5) from (5x – 2) is …. a. 3x – 7 b. 7x + 3 c. 3x + 3 d. 7x – 7

5. Which one of these equation is not equivalent to – 2x + 4 = 3

a. 12 −=− x b. 32 =+− x c. 342 −=−x d. 232 =−x


6. The root of – 4 (x + 2) + 5x = – 2 is …. a. – 6 b. – 10 c. 4 d. 6

7. Novi buys 2 blouses for Rp. 60 000 and the transportation fee is Rp. 5 000. Then she sells it for Rp. 35 000 and Rp. 29 000. Which one of this statement is true based on that problem? a. She gets loss Rp. 4 000 c. She gets loss Rp. 1 000 b. She gets profits Rp. 4 000 d. She gets profits Rp. 1 000

8. Mr. George buys a table for Rp. 400 000. He sells it and gets profits 10%. The selling price is …. a. Rp. 410 000 b. Rp. 420 000 c. Rp. 430 000 d. Rp. 440 000

9. Distance between Serang and Jakarta on the map is 3 cm. If the scale is 1 : 5 000 000, then the real distance between them are …. a. 100 km b. 150 km c. 1000 km d. 1500 km

10. The ratio of Andy’s money and Nina’s money is 2 : 5. If Nina’s money is Rp. 12 500, then how much is Andy’s money? a. Rp. 2 500 b. Rp. 5 000 c. Rp. 7 500 d. Rp. 10 000

B. Logic and Communication Fill in the blanks with the correct answer! 11. ....)3(4)5(:1520 =−×−−+ 12. Temperature in Tokyo is – 20C and temperature in Bogor is 180C. The difference

between those temperatures is ….

13. ....%6539

511625.0 =−+

14. The Least Common Multiple (LCM) of 8a3bc and 12a4b3 is …. 15. 152 ≥+− x . If x are the element of whole numbers then the possible value for x are

…. 16. Change 0.22222… into a fraction!

17. Find the root of 24

23

=−p

18. Tono sold his hand phone for Rp. 600 000. He got loss 20%. Determine the purchase price of that hand phone!

19. Mr. Sugiyarto saves his money Rp. 2 000 000 in a bank. After 2 years his money becomes Rp. 2 100 000. Determine the percentage of bank interest per year!

20. The comparison of Dewi’s money and Novi’s money is 3 : 7. If the sum of their money is Rp. 50 000, then determine Novi’s money!

C. Problem Solving

Solve these questions carefully! 21. In a test, there are 30 questions. The score for true answer is +4, score for false answer

is (– 1) and score for unanswered is 0. Michael answers 18 questions correctly and 5 questions are unanswered. Determine Michael’s total score!

22. There are 400 students in a school. 60% of them is girl. 52 of the girl like apple and

the rest like oranges. How many girls do like oranges? 23. Mrs. Hani buys a sack of rice for Rp. 450 000. The gross is 100 kg and the tare is 1 %.

Then she sells it Rp. 5 000 per kg. a. evaluate loss or profit b. how much is the loss or profit? c. What is the percentage of the loss or profit?


24. The perimeter of a rectangle is 52 cm. The width is 3 cm shorter than the length. Suppose the length is x then determine: e. linear equation based on that problem f. the value of x

25. 10 cows finish eating the grass in 12 days. If 5 cows is added, then how many days they will finish eating the grass?

BIBLIOGRAPHY Jones. Kamus Saku Matematika Collins Gem. Penerbit Erlangga. Jakarta. 1995 Seng, The Keng. New Syllabus Mathematics 1. Shinglee Publisher PTE ITD. Singapore.2001 Kassim, Mukmien. New Syllabus Mathematics Form 1. Preston Times Sdn Bhd. Selangor.

2003

Bahan Ajar VII Mathematics

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