Unit 1 – Number System

Chapter 1 – Knowing Our Numbers

Chapter 2 – Fractions

Chapter 3 – Decimals

Chapter 4 – Rational Activity

Chapter 5 – Powers and Exponents

Chapter 1 – Knowing Our Numbers

You will Learn

1. Integers

2. Multiplication of Integers

3. Division of Integers

4. Properties of Operations on Integers

Integers

In our day–to–day life, we often come across with many situations where we are

involved with the use of integers.

We use integers to express temperature, height of a place or person from earth’s

surface or specific place, profit, loss, monitory matters, etc.

All the numbers which are Less than ZERO on the number line are known as

NEGATIVE numbers and that of Greater than ZERO are POSITIVE.

The set of negative numbers along with the set of whole numbers are known as

INTEGERS.

Once a direction is chosen as Positive, the opposite direction must be considered

as Negative.

If North is taken as Positive, then South would be Negative

In the same way, East – West

Right – Left

Profit – Loss

Height – Depth

And If Above Zero is Positive then Below Zero would be Negative.

For every Positive Integers there is a Negative Integer on Number Line at the same

distance from Zero.

These reflections are known as Opposite or Additive Inverse of each other.

For example : –2 is additive Inverse or Opposite of 2.

The sum of Additive inverse or opposite integers are always ZERO.

The whole number ‘0’ is neither a positive integer nor a negative integer. As the

negative of 0 is also 0.

On number line of Integers , every number on the Right is greater than that of

Left side and vise versa.

Remember

Addition and Subtraction of IntegersRules for Addition and Subtraction of Integers

When adding of two or more integers with same sign, JUST add them and put therespective sign before the outcome.+3 +7 = +10–3 –7 = –10

When adding two integers of different signs, Subtract them and put the sign oflarger integer before the outcome.+3 –7 = –4

During Addition and Subtraction signs are placed side by side without any numberin between, the two opposite signs give negative sign.– 3 + (– 7) = – 3 – 7 = – 10

When two Negative signs are placed side by side with no numeral in between, thetwo like signs give a Positive sign.– 3 – (– 7) = – 3 + 7 = +4

Generally : Positive numbers are written without ‘+’ sign before them.

Addition and Subtraction of Integers

Solved Examplesa).

734 + 69 + 131 – 234= 734 – 234 + 69 + 131= 500 + 200= 700

b).937 + (–37) + 100 – (–200) + 300

= 900 + 100 + 200 + 300= 1500

c).Subtract 64 from –80

= –80 – (+65) = –80 –65 = –145d).

Subtract –30 from –70= –70 – (–30) = –70 + 30 = – 40

e).–11 – (– 14) + 8= – 11 + 14 + 8= – 11 + 22 = 11

f).Subtract –80 from 24= 24 – (–80) = 24 + 80= 104

g).Subtract –265 from –75

= –75 – (– 265) = –75 +265= 190

h).Subtract –23 from –133

= –133 – (–23) = –133 + 23= – 110

Multiplication of Integers

Multiplication is a simpler form of repeated addition of integers.

Multiplication : 4 x 2 = 4 + 4 = 8 OR 3 x 6 = 3 x 3 x 3 x 3 x 3 x 3 = 18

Rules for Multiplication of Integers1) + x + = +2) – x – = +3) – x + = –4) + x – = –

Multiplication of ‘0’ with any integer, even with ‘0’ gives the result as ZERO(0)0 x 0 = 0 , 0 x (–5) = 0 , 4 x 0 = 0

a) 6 x 8 = 48b) (–6) x (–2) = +12c) (–25) x 5 = –125d) 4 x (–8) = –32e) (0) x (– 6) = 0f) (–3) x (15) = – 45

Learn the Tables Minimum 2 to 15

2 3 4 5 6 7 8 9 10 11 12 13 14 15

4 6 8 10 12 14 16 18 20 22 24 26 28 30

6 9 12 15 18 21 24 27 30 33 36 39 42 45

8 12 16 20 24 28 32 36 40 44 48 52 56 60

10 15 20 25 30 35 40 45 50 55 60 65 70 75

12 18 24 30 36 42 48 54 60 66 72 78 84 90

14 21 28 35 42 49 56 63 70 77 84 91 98 105

16 24 32 40 48 56 64 72 80 88 96 104 112 120

18 27 36 45 54 63 72 81 90 99 108 117 126 135

20 30 40 50 60 70 80 90 100 110 120 130 140 150

Division of IntegersDivision is reverse process of Multiplication.

Multiplication : 4 x 2 = 8 Division : 8 ÷ 2 = 4

Rules for Division of Integers1) + ÷ + = +2) – ÷ – = +3) – ÷ + = –4) + ÷ – = –

• For integers with like signs (+ and + or – and –), the quotient is POSITIVE.

• For integers with unlike signs (+ and – or – and +), the quotient is NEGATIVE.

• When ‘0’ is multiplied or divided by any positive or negative integer, it gives the result as

ZERO(0).

a) 56 ÷ 8 = 7b) (–6) ÷ (– 2) = +3c) (–25) ÷ 5 = –5d) 64 ÷ (–8) = –8e) (0) ÷ (– 6) = 0f) (–300) ÷ (15) = – 20

Exercise 1.1

a). (+8) by (+2)= (+8) ÷ (+2) = +4

b). (–24) by (–8)= (–24) ÷ (–8)= +3

d). (–36) by (+6)= (–36) ÷ (6) = –6

e). 0 by (+29)= 0 ÷ (+29) = 0

l). (+126) by (–21)= (+126) ÷ (–21)= –21

1. Multiply the following

a). (–3) with (+5)= –3 x 5 = –15

b). (–7) with (–3)= (–7) x (–3)= 21

e). (–1861) with 0= (–1861) x (0) = 0

h). 0 with 0= 0 x 0 = 0

i). (–1000) with (+7)= (–1000) x (+7)= –7000

2. Divide the following Integers

B O D M A S

B Bracket { } or [ ]O Of ( )D Division ÷M Multiplication xA Addition +

S Subtraction –

Priority to solve the equation, with multiple operations on Integers

c). (– 21) x [(+16) + (– 13)] = ______= (– 21) x [16 – 13]= (– 21) x (+3)= – 63

d). (+103) x [(– 3) – (– 5)] = ______= (+103) x [– 3 +5]= (+103) x (+2)= +206

g). If (–254) x (+146) = (–37084), then (–37084) ÷ (+146) = ________

As, (–254) x (+146) = (–37084)

Therefore

3. Fill in the Blanks to complete

a). (– 28) x (+33) = ________= (– 28) x (+33) = – 924

b). (+496) x (– 213) = ________= (+496) x (– 213) = – 105648

28

x 33

8 4

8 4 x

9 2 4

(–) x (+) = –

496x 213

1 4 8 84 9 6 x

9 9 2 x x1 0 5 6 4 8

–37084+146

= (–254)

4. State TRUE or FALSE for the below mentioned statements

a). The difference of two integers can never be zero. (FALSE as it can be zero)

b). The sum of a Positive Integer and a Negative Integer is always a Negative Integer. (FALSE,

It can be Positive or Negative, it depends on the sign of larger number.

c). The absolute value of an integer is always greater than or equal to the value of the

integer. (TRUE)

d). The value of the sum of two integers is always greater than the value of their difference.

(FALSE, it can be Positive, Negative or Zero)

e). The value of the product of two integers is always greater than the value of their sum.

(FALSE, it can be Positive, Negative or Zero)

5. Simplifya). (–6) x (–6) = +36c). (–8) x (2) = –16d). 0 x (–9) = 0

6. Simplifya). (–4) x (–4) x (–4)

= (+16) x (–4)= –64

c). (–1) x (–3) x (+6)= (+3) x (+6)= +18

7. Simplifya). 18 ÷ 3 = 6b). (–18) ÷ 3 = –6c). 15 ÷ (–3) = –5d). (–12) ÷ (–4) = +3

–48–8

–300–10

–7209

8. Simplify

= +6

= +30

= –80

9. Multiply the Followinga). (+9) x (–3) = –27c). (–8) x (–4) = +32

10. Divide the Followinga). (–81) ÷ (9) = –9c). (–72) ÷ (–8) = +9d). (–49) ÷ (–7) = +7

11. Solve the Followinga). (–20) x 25 x (–10)

= (–500) x (–10)= +5000

b). (–70) x (–35) x 0 x (–63)= 0

c). 1673 x 99 – (–1673)= 1673 x 99 + 1673= 165627 + 1673= 167300

d). 40 x (–23) + 40 x (–17)= –920 + (–680)= –920 – 680= –1600

Properties of Integers

Properties of Addition

1. Closure Property :– If a and b are two integer then their sum will always be an integer suppose cOr a + b = c. Therefore, integers are closed under addition.(+5) + (+3) = +8 , (–5) + (+3) = –2 , (+5) + (–3) = +2 , (–5) + (–3) = –8

2. Commutative Property:– If a and b are two integers then a + b = b + a (while adding the integers, order ofintegers do not matter)(–6) + (+8) = (+8) + (–6) , (–4) + (–3) = (–3) + (–4)

3. Associative Property:– If a, b and c are three integers then (a + b) + c = a + (b + c). Means, while adding,when grouping of the integers changed, the result does not change.[7 + 9] + 6 = 7 + [9 + 6]16 + 6 = 7 + 1522 = 22, which is true

4. Additive Identity:– By adding 0 to any integer, it does not change the value of the integer. Means 0 is theadditive identity of the integer.Therefore, a + 0 = a or 0 + a = a

5. Additive Inverse:– For every integer there is an equal and opposite integer on number line. If a is an integerthen (–a) will be its additive inverse. While addition of integer and its inverse, the result will always be Zero.Means, (+a) + (–a) = 0. For example, (–7) + (+7) = 0 , (+2) + (–2) = 0

6. Property of One:– Addition of One to any integer gives its successor. Means, (a+1) is successor of a.For example, (+3) + 1 = 4 (here 4 is successor of 3) , (–6) + 1 = –5 (here –5 is successor of –6).

Properties of Integers

Properties of Subtraction

1. Closure Property :– If a and b are two integer then during subtraction, the result always will be an integersuppose cOr a – b = c. Therefore, integers are closed under subtraction as well as addition.(+5) + (+3) = +8 , (–5) + (+3) = –2 , (+5) + (–3) = +2 , (–5) + (–3) = –8

2. Commutative Property:– If a and b are two integers then a – b ≠ b – a (while subtracting the integers, orderof integers does matter). Hence, subtraction in integers is not Commutative.(–6) – (+8) ≠ (+8) – (–6) , (–4) – (–3) ≠ (–3) – (–4)

3. Associative Property:– If a, b and c are three integers then (a – b) – c ≠ a – (b – c). Means, while subtracting,when grouping of the integers changed, the result does change.[7 – 9] – 6 ≠ 7 – [9 – 6]–2 – 6 ≠ 7 – 3–8 ≠ 4, which is true

4. Property of Zero:– By subtracting 0 from any integer, it does not change the value of the integer. Means 0 isthe subtractive identity of the integer.Therefore, a – 0 = a or 0 – a = –a

5. Property of One:– When One is subtracted from any integer, it gives its predecessor. Means, (a–1) ispredecessor of a.For example, (+3) – 1 = 2 (here 2 is predecessor of 3) , (–6) – 1 = –7 (here –7 is predecessor of –6).

Exercise 1.2

a). (–18 + 4) + 6 = –18 + (6 + 4)–14 + 6 = –18 + 10–8 = –8 (which is true)

b). 19 + (–3+9) = [19 + (–3)] + 919 + 6 = 16 + 925 = 25 (which is true)

1. Use associative property to simplify

b). 89 + 36 + 64 + 11= (89 + 11) + (36 + 64)= 100 + 100 = 200

d). 64 + 26 + 74 + 36= (64 + 36) + (26 + 74)= 100 + 100 = 200

3. Calculate the value on both sides to check equality

2. Fill in the Blank

a). 5 + 3 = ____ + 5= 5 + 3 = 3 + 5

c). (–4) + (–5) = ____ + (–4)= (–4) + (–5) = (–5) + (–4)

d). [(–7) + (–a)] = ____ + (–7)= [(–7) + (–a)] = (–a) + (–7)

a). (18 – 6) + 3 = 18 – (6 + 3)12 + 3 = 18 – 915 ≠ 9 (Hence two integers are not equal)

b). (76 – 6) – 20 = 76 – (6 – 20)70 – 20 = 76 – (–14)50 = 76 + 1450 ≠ 90 (Hence two integers are not equal)

4. Prove the statements are false and give reason

Properties of Integers

Properties of Multiplication

1. Closure Property :– If a and b are two integer then their product will always be an integer, suppose it is c.Therefore, a x b = c. Hence, integers are closed under Multiplication.(+5) x (+3) = +15 , (–8) x (+2) = –16 , (+5) x (–3) = –15 , (–5) x (–3) = +15

2. Commutative Property:– If a and b are two integers then a x b = b x a (while multiplying the integers, order ofintegers do not matter). So, integers possess commutative property of multiplication.(–6) x (+8) = (+8) x (–6) , (–4) x (–3) = (–3) x (–4)

3. Associative Property:– If a, b and c are three integers then (a x b) x c = a x (b x c). Means, while multiplying,when grouping of the integers changed, the result does not change.(7 x 9) x 6 = 7 x (9 x 6)63 x 6 = 7 x 54378 = 378, which is true

4. Distributive Property:– If a, b and c are three integers then a (b + c) = ab + ac. Means, multiplication distributesover addition.–6 x (7 + 10) = (–6 x 7) + (–6 x 10)–42 – 60 = –102

5. Multiplicative Identity:– By Multiplying 1 to any integer, it does not change the value of the integer. Means 1 isthe Multiplicative identity of the integer.Therefore, a x 1 = a or 1 x a = a

6. Property of Zero:– If a is an integer and 0 is multiplied to it, then (a x 0) or (0 x a) is always will be Zero (0).

Properties of Integers

Properties of Division

1. Closure Property :– If a and b are two integers then their division will not always give quotient as an integer,suppose the quotient is c. Therefore, a ÷ b = c. Hence, integers do not have closure property for Division.Quotients may be integers like 2, –4, 9, etc, Fractions (not integers) –5½ , 8/9 , –1¾

2. Commutative Property:– If a and b are two integers then a ÷ b ≠ b ÷ a (while dividing the integers, order ofintegers does matter). So, integers do not possess commutative property of division.(–16) ÷ (+8) ≠ (+8) ÷ (–16)

3. Associative Property:– If a, b and c are three integers then (a ÷ b) ÷ c ≠ a ÷ (b ÷ c). Means, associative propertydoes not apply to the division of integers.(64 ÷ 8) ÷ 2 = 64 ÷ (8 ÷ 2)8 ÷ 2 = 64 ÷ 44 ≠ 16, which is true

4. Property of One:– If a is any integer when divided by one, always give the same number as result.–6 ÷ (1) = –6 , 9 ÷ (1) = 9

5. Property of Zero:– When the integer zero (0) is divided by any other integer, the result is always Zero. If a is aninteger, then (0 ÷ a = 0).

6. Division by Zero (0):– Division of any integer by Zero is meaningless, Hence, division by zero is not allowed.

Also, the result is expressed as ∞ (infinity)

• Integers are closed under Multiplication but not under Division.

• Commutative property holds under Multiplication but not under Division.

Means, a x b = b x a but a ÷ b ≠ b ÷ a

• Associative property holds under Multiplication but not under Division.

Means, (a x b) x c = a x (b x c) but (a ÷ b) ÷ c ≠ a ÷ (b ÷ c)

• Distributive property : a (b x c) = ab + ac also, a (b – c) = ab – ac

Remember

Exercise 1.3

a). (8 x 9) x 5 = 8 x (9 x 5)72 x 5 = 8 x 45360 = 360 (which is true)

b). [(–7) x (–9)] x (3) = (–7) x [(–9) x (3)]63 x 3 = (–7) x (–27)189 = 189 (which is true)

4. Fill in the Blank

a). 2 + ____ = 0= 2 + (–2) = 0

c). 4 + (–4) = ____= 4 + (–4) = 0

e). ____ + 3 = 0= (–3) + 3 = 0

2. Calculate the value on both sides to check equality

3. Fill in the Blank

a). 5 x 7 = ____ x 5= 5 x 7 = 7 x 5

e). (–20) x (–3) = ____ x (–20)= (–20) x (–3) = (–3) x (–20)

f). 1 x 9 = ____ x 1= 1 x 9 = 9 x 1

a). 25(8 + 2) = 25 x 8 + 25 x 225 (10) = 200 + 50250 = 250 (Hence two integers are equal)

c). 48 x 76 + 48 x 24 = 48(76 + 24)3648 + 1152 = 48 (100)4800 = 4800 (Hence two integers are equal)

d). (–6) x 30 + (–6)20 = (–6) (30 + 20)(–180) + (–120) = (–6) (50)–300 = –300 (Hence two integers are equal)

6. Prove that both sides are equal

Word Problems

Solution –Let the ground level of earth assumed to be Zero.

Highest position of Pebble = +15 metersLowest position of Pebble = – 10 meters

Hence, Total distance = Difference of Highest and Lowest positions

= + 15 – (– 10)= + 15 + 10= + 25 Meters

Example 2 – A boy flung a pebble 15 meters high inthe air which fell and settled at the bottom of a pond10 meters deep. By how much distance did thepebble fall.

Solution –Moving point for lift = 3rd FloorFinishing point for lift = 15th Floor

Hence, number of floors to move= 15 – 3 = 12 Floors

Now, height of each floor = 5 MeterTotal distance to be moved = 12 x 5 = 60 meter

Distance moved by lift in one Second = 2 meter

So, the time taken by the lift to move from 3rd floor to 15th floor = 60 ÷ 2 = 30 Seconds

Example 3 – Every floor of a 20 storey is five meterhigh. If a lift moves 2 meter every second, how longit will take to move from 3rd floor to 15th floor.

Exercise 1.4

Solution –Mr. Nair’s Bank balance on 01.01.2008 = Rs. 2,500

Amount deposited in January = Rs. 1,250Amount in bank in January = 2500 + 1250 = Rs. 3750

Amount Withdrew in February= Rs. 750Amount in bank in February = 3750 – 750 = Rs. 3000

Amount deposited in March = Rs. 500Amount Withdrew in March = Rs. 300Amount in bank in March = 3000 + 500 – 300

= Rs. 3500 – 300= Rs. 3200

So, amount in bank on 01.04.2008 = Rs. 3,200

Problem 1. Mr. Nair had Rs. 2,500 in his bank on01.01.2008. He deposited Rs. 1,250 in January andwithdrew Rs. 750 in February. What was Mr. Nair’sbank balance on 01.04.2008 if he deposited Rs. 500and withdrew Rs. 300 in March?

Solution –Total distance of relay race = 2 x 50 = 100 metersPriya ran for distance = 50 metersDistance left = 100 – 50 = 50 meters

Rinki supposed to run = 50 meters

Now, Rinki ran in opposite direction = 10 meters

So, Total distance to be run by Rinki = 10 meters to cover the distance run in opposite direction + 50 meters to finish line

= 10 + 50 = 60 meters

Problem 2. In a 2 x 50 meters relay race for tiny tots,Priya ran the first 50 meters and handed the baton toRinki who instead of running towards finishing line,ran back 10 meters towards the starting line beforeshe was stopped. How many meters does Rinki haveto run now to finish the race?.

Solution –Suppose the other number is = aThen, a x (–7) = 105

Hence, a = 105 ÷ (–7) = –15

Therefore, the other number = –15

Problem 3. The product of two numbers is 105. If onenumber is –7, what is the other?

Solution –Cost of one book = Rs. 96Mahesh brought the books = 60So, total cost of 60 books = 96 x 60 = Rs. 5760

Now, difference in cost of one book = Rs. 5So, total difference in the bill = 5 x 60 = Rs. 300

Problem 4. A book costs Rs. 96 and Mahesh bought60 such books. By mistake, the accountant at thestore made the bill by taking the cost of each book tobe Rs. 5 less. What is the difference in this bill fromwhat it would have been in reality?.

Solution –Rising point for Balloon = 36th FloorFinishing point for Balloon = 96th Floor

Hence, number of floors to rise by balloon= 96 – 36 = 60 Floors

Now, height of each floor = 4 MetersTotal distance to rise = 60 x 4 = 240 meters

Distance moved by balloon in 1 Second = 3 meters

So, the time taken by the balloon to rise from 36th

floor to 96th floor = 240 ÷ 3 = 80 Seconds OR 1 minute 20 Seconds

Problem 5 – Every floor of a 104 storey skyscraper is4 meter high. If a balloon rises 3 meters everysecond, how long it will take to rise from 36th floor tothe 96th floor.

Solution –Lalita dived first day to depth = 5 m

As per statement, she dived 5 m more deep on daily basis till fifth day.

Therefore, on fifth day she will dive deep to = 5 x 5= 25 m

Problem 6. Lalita is an enthusiastic student in herdiving class. On the first day she managed to dive to adepth of 5 m. From second day onwards, shemanaged to dive five meters deeper than theprevious day, and so on. How far did she dive on thefifth day?

Solution –Total marbles Ranjan have = 60Marbles gained in 8 games = (-5) x 8 = -40 marblesMarbles gained in 4 games = (4) x 4 = 16 marbles

Hence, marbles Ranjan have at the end of games= 60 + (– 40) + 16= 60 + 16 – 40= 76 – 40= 36 Marbles

Problem 7 – Ranjan went to play with 60 marbles. In8 games he gained -5 marbles each and in 4 gameshe gained 4 marbles each. When the games ended,how many marbles did Ranjan have?

Solution –Speed of submarine = -20m per minute

Distance covered by submarine in 7 minutes = (-20) x 7= -140 meters (deep from water surface)

Problem 8. A submarine moved from the water surface at a speed of -20m per minute. How far is thesubmarine from the water surface at the end of 7 minutes?

Solution –Daytime temperature = 33o CTemperature at night = 26o C

Now rise in temperature = Night temperature –daytime temperature

= 26o C - 33o C = - 7o C

Problem 1. The daytime maximum temperature inDelhi on a particular day was 33o C. At the nighttemperature became 26o C. What is the rise intemperature from daytime to night-time?

Solution –Assume the home is the base point of travelDistance travelled from east to home = 4 KmDistance travelled from home to west = 3 Km

Hence, total distance travelled = 4 + 3 = 7 Km

Problem 2 – I was 4Km from east of my home. Now Iam 3Km west of my home. How many kilometers didI travel?

More Problems on Chapter - 1

Solution –In first session, Satish gained = +10 secondsIn second session, Satish gained = +20 secondsIn third session, Satish lost = -60 seconds (1 min =

60 sec)In fourth session, Satish gained = +15 secondsIn fifth session, Satish lost = -27 secondsIn sixth session, Satish gained = +41 seconds

Hence, final result of Satish= +10 + 20 + (-60) + 15 + (-27) + 41= 10 + 20 + 15 + 41 – 60 – 27= +86 – 87= -1 second

So, Satish lost the race by 1 second

Problem 3 – Rajesh and Satish had a cycle race. Therace was conducted in six sections. In the firstsection, Satish gained 10 seconds. After that hegained 20 seconds, then lost 1 minute, gained 15seconds, lost 27 seconds and finally gained 41seconds over his friend Rajesh. Who lost the race?

1. What are Integers?

2. Positive and Negative Integers.

3. Addition and Subtraction of Integers.

4. Rules for Addition and Subtraction of integers.

5. Multiplication and Division of integers.

6. Rules for Multiplication and Division of integers.

7. Absolute value.

8. BODMAS rule to solve different operations on integers.

9. Properties of Addition, Subtraction, Multiplication and Division.

10. Statement problems.

Learning from Chapter - 1