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# CH 6  –  Integers

#### Integers

Exercise 6.1

1. Write opposites of the following:

(a)    Increase in weight (b) 30 km north (c) 326 BC (d) Loss of Rs 700 (e) 100 m above sea level

Solution

1. a)Decrease in weight
2. b)30 km south
3. c)326 A.D.
4. d)Gain of Rs. 700
5. e)100 m below sea level
1. Represent the following numbers as integers with appropriate signs.
2. a)An aeroplane is flying at a height two thousand metre above the ground.
3. b)A submarine is moving at a depth, eight hundred metre below the sea level.
4. c)A deposit of rupees two hundred.
5. d)Withdrawal of rupees seven hundred.

Solution

1. a)+2000 m
2. b)–800  m
3. c)+ 200
4. d)–700
1. Represent the following numbers on a number line :

(a)    + 5 (b) – 10 (c) + 8 (d) – 1 (e) – 6

Solution

1. a)

1. b)

1. c)
2. d)

1. e)

1. Given figure is a vertical number line, representing integers. Observe it and locate the following points:

(a)    If point D is + 8, then which point is – 8?

(b)   Is point G a negative integer or a positive integer?

(c)    Write integers for points B and E.

(d)   Which point marked on this number line has the least value?

(e)    Arrange all the points in decreasing order of value.

Solution

Let us complete the number line.

1. a)F
2. b)G is a negative integer
3. c)B is 4 and E is -10
4. d)E has the least value of -10
5. e)D, C, B, A, O, H, G, F, E
1. Following is the list of temperatures of five places in India on a particular day of the year.

1. a)Write the temperatures of these places in the form of integers in the blank column.
2. b)Following is the number line representing the temperature in degree Celsius.

Plot the name of the city against its temperature.

1. c)Which is the coolest place?
2. d)Write the names of the places where temperatures are above 10°C.

Solution

1. a)

The temperature below 0°C is negative and the temperatures above 0°C are positive.

 Place Temperature Siachin 10°C below 0°C –10°C Shimla 2°C below 0°C –2°C Ahmedabad 30°C above 0°C +30°C Delhi 20°C above 0°C +20°C Srinagar 5°C below 0°C –5°C
1. b)

1. c)The coolest place has the lowest temperature. From the number line above, we see that the lowest temperature is -10°C, Siachin.

So, Siachin is the coolest place.

1. d)Ahmedabad and Delhi has temperatures more than 10°C.

1. In each of the following pairs, which number is to the right of the other on the number line?

(a)    2, 9 (b) – 3, – 8 (c) 0, – 1 (d) – 11, 10 (e) – 6, 6 (f) 1, – 100

Solution

The larger number is always to the right of the other on a number line.

1. a)9
2. b)-3
3. c)0
4. d)10
5. e)6
6. f)1
1. Write all the integers between the given pairs (write them in the increasing order.)

(a)    0 and – 7 (b) – 4 and 4 (c) – 8 and – 15 (d) – 30 and – 23

Solution

1. a)-6, -5, -4, -3, -2, -1
1. b)-3, -2, -1, 0, 1, 2, 3
1. c)-14, -13, -12, -11, -10, -9
1. d)-29, -28, -27, -26, -25, -24
1. a) Write four negative integers greater than – 20.

(b) Write four negative integers less than – 10.

Solution

1. a)–19, –18, –17, –16
1. b)–11, –12, –13, –14
1. For the following statements, write True (T) or False (F). If the statement is false, correct the statement.

(a) – 8 is to the right of – 10 on a number line.

(b) – 100 is to the right of – 50 on a number line.

(c) Smallest negative integer is – 1.

(d) – 26 is greater than – 25.

Solution

1. a)True (-8 is greater than -10)
1. b)False (-50 is greater than -100 and hence false)

Correct Statement: -50 is to the right of -100 on a number line

1. c)False

Correct Statement: -1 is the greatest negative integer.

1. d)False

Correct Statement: -25 is greater than -26.

1. Draw a number line and answer the following:
2. a)Which number will we reach if we move 4 numbers to the right of – 2.
3. b)Which number will we reach if we move 5 numbers to the left of 1.
4. c)If we are at – 8 on the number line, in which direction should we move to reach –13?
5. d)If we are at – 6 on the number line, in which direction should we move to reach – 1?

Solution

1. a)
1. b)

1. c)We have to move to the left to reach -13.
2. d)We have to move to the right to reach -1.

Exercise 6.2

1. Using the number line write the integer which is:
2. 3 more than 5
3. 5 more than –5
4. 6 less than 2
5. 3 less than –2

Solution

1. a)To find the integer that is 3 more than 5, we start with 5 and proceed 3 steps to the right as shown below:

3 more than 5 is 8.

1. b)To find the integer that is 5 more than -5, we start with -5 and move 5 steps to the right as shown below.

The integer that is 5 more than -5 is 0.

1. c)To find 6 less than 2, we start from 2 and move 6 steps to the right.

6 less than 2 is -4.

1. d)To find 3 less than -2, we start from -2 and move left by 3 steps.

The integer that is 3 less than -2 is -5.

1. Use number line and add the following integers :
2. 9 + (– 6)

On the number line, we first move 9 steps to the right from 0. Then we move 6 steps to the left from 9 and reach 3.

Hence 9 + (-6) = 3

1. 5 + (– 11)

On the number line, we first move 5 steps to the right from 0. Then we move 11 steps to the left from 5 and reach 6.

5 + (– 11) = -6

1. (– 1) + (– 7)

On the number line, we first move 1 step to the left from 0(as the integer is negative). Then we move 7 steps to the left from -1 and reach -8.

(– 1) + (– 7) = (-8)

1. (– 5) + 10

On the number line, we first move 5 steps to the left from 0. Then we move 10 steps to the left from -5 and reach 5.

(– 5) + 10 = 5

1. (– 1) + (– 2) + (– 3)

On the number line, we first move 1 step to the left from 0. Then we move 2 steps to the left from -1 and reach -3. Then we move 3 steps to the left from -3 and reach -6.

(– 1) + (– 2) + (– 3) = -6

1. (– 2) + 8 + (– 4)

On the number line, we first move 2 steps to the left from 0. Then we move 8 steps to the right from -2 and reach 6. Then we move 4 steps to the left from 6 and reach 2.

(– 2) + 8 + (– 4) = 2

1. Add without using number line :
2. 11 + (– 7)
3. (– 13) + (+ 18)
4. (– 10) + (+ 19)
5. (– 250) + (+ 150)
6. (– 380) + (– 270)
7. (– 217) + (– 100)

Solution

1. 11 + (– 7)

When one positive and one negative integers are added we subtract them and put the sign of the bigger integer. The bigger integer is decided by ignoring the signs of the integers [e.g. (+4) + (–3) = + 1 and (–4) + ( + 3) = – 1].

Subtracting the integers we get 11 – 7 = 4.

Now put the sign of the bigger integer. The bigger integer is 11 with a positive sign.

So, 11 + (– 7) = 4

1. (– 13) + (+ 18)

When one positive and one negative integers are added we subtract them and put the sign of the bigger integer. The bigger integer is decided by ignoring the signs of the integers [e.g. (+4) + (–3) = + 1 and (–4) + ( + 3) = – 1].

Subtracting the integers we get 18 – 13 = 5.

Now put the sign of the bigger integer. The bigger integer is 18 with a positive sign.

So, (– 13) + (+ 18) = 5

1.  (– 10) + (+ 19)

When one positive and one negative integers are added we subtract them and put the sign of the bigger integer. The bigger integer is decided by ignoring the signs of the integers [e.g. (+4) + (–3) = + 1 and (–4) + ( + 3) = – 1].

Subtracting the integers we get 19 – 10 = 9.

Now put the sign of the bigger integer. The bigger integer is 19 with a positive sign.

So, (– 10) + (+ 19) = 9

1. (– 250) + (+ 150)

When one positive and one negative integers are added we subtract them and put the sign of the bigger integer. The bigger integer is decided by ignoring the signs of the integers [e.g. (+4) + (–3) = + 1 and (–4) + ( + 3) = – 1].

Subtracting the integers we get 250 – 150 = 100.

Now put the sign of the bigger integer. The bigger integer is 250 with a negative sign.

So, (-250) + (+150) = (-100)

1. (– 380) + (– 270)

When we have the same sign, add and put the same sign. When two negative integers are added, we get a negative integer [e.g. (–2) + ( – 1) = – 3].

Adding the numbers we get 380 + 270 = 650

As both are negative integers, the sum is also a negative integer.

So, (– 380) + (– 270) = (– 650)

1. (– 217) + (– 100)

When we have the same sign, add and put the same sign. When two negative integers are added, we get a negative integer [e.g. (–2) + ( – 1) = – 3].

Adding the numbers we get 217 + 100 = 317

As both are negative integers, the sum is also a negative integer.

So, (– 217) + (– 100) = (– 317)

1. Find the sum of :

(a)    137 and – 354

(b)   – 52 and 52

(c)    – 312, 39 and 192

(d)   – 50, – 200 and 300

Solution

1. a)137 and – 354

When one positive and one negative integers are added we subtract them and put the sign of the bigger integer.

137 + (– 354) = –217

[Subtract the numbers: 354 – 137 = 217. Put the sign of the bigger integer.]

1. b)– 52 and 52

When one positive and one negative integers are added we subtract them and put the sign of the bigger integer.

-52 + 52 = 0

[Subtract the numbers: 52 – 52 = 0. ]

1. c)– 312, 39 and 192

When one positive and one negative integers are added we subtract them and put the sign of the bigger integer.

-312 + 39 + 192 = -81

[Subtract the numbers: 312 – 39 = 273. Put the sign of the bigger integer, (-312). So, -273 + 192 = -81]

1. d)– 50, – 200 and 300

(a) When we have the same sign, add and put the same sign.

(b) When one positive and one negative integers are added we subtract    them and put the sign of the bigger integer.

(– 50) + (– 200) + (300) = 50

[The integers 50 and 200 have the same sign. So we add and put the same sign as -250. Now (-250) and 300 are added. So, we subtract them and put the sign of the bigger integer as +50.]

1. Find the sum :

(a)    (– 7) + (– 9) + 4 + 16

(b)   (37) + (– 2) + (– 65) + (– 8)

Solution

1. (– 7) + (– 9) + 4 + 16 = (-16) + (20) = 4
1. (37) + (– 2) + (– 65) + (– 8) = (37) + (-75) = (-38)

Exercise 6.3

1. Find

(a) 35 – (20)

(b) 72 – (90)

(c) (– 15) – (– 18)

(d) (–20) – (13)

(e) 23 – (– 12)

(f) (–32) – (– 40)

Solution

(a)    35 – (20) = 15 [Subtract 35 and 20 and put the sign of the bigger number 35, which is positive]

(b)   72 – (90) = -18 [Subtract 72 and 90 and put the sign of the bigger number 90, which is negative.]

(c)    (– 15) – (– 18)

To subtract two integers, we add the additive inverse of the integer that is being subtracted.

Additive inverse of (-18) is +18.

(– 15) – (– 18) = (– 15) + (18) = +3

(d)   (–20) – (13) = -33 [Both the integers have the same sign, So, we add the integers and put the same sign.]

(e)    23 – (– 12)

To subtract two integers, we add the additive inverse of the integer that is being subtracted.

Additive inverse of (-12) is +12.

23 – (– 12) = 23 + 12 = 35

(f)    (–32) – (– 40)

To subtract two integers, we add the additive inverse of the integer that is being subtracted.

Additive inverse of (-40) is +40

(–32) – (– 40) = (–32) + 40 = 8 [Subtract 40 and 32 and put the sign of the bigger integer 40, which is positive.]

1. Fill in the blanks with >, < or = sign.

(a) (– 3) + (– 6) ______ (– 3) – (– 6)

(b) (– 21) – (– 10) _____ (– 31) + (– 11)

(c) 45 – (– 11) ______ 57 + (– 4)

(d) (– 25) – (– 42) _____ (– 42) – (– 25)

Solution

(a)    (– 3) + (– 6) ______ (– 3) – (– 6)

L.H.S

(– 3) + (– 6) = (-9) [Both the integers have the same sign. So, we add the integers and put the same sign.]

R.H.S

(– 3) – (– 6) = (-3) + 6 = 3 [To subtract two integers, we add the additive inverse of the integer that is being subtracted.]

(-9) < 3

(– 3) + (– 6)  <  (– 3) – (– 6)

(b)   (– 21) – (– 10) _____ (– 31) + (– 11)

L.H.S

(– 21) – (– 10) = (-21) + 10 = -11 [To subtract two integers, we add the additive inverse of the integer that is being subtracted.]

R.H.S

(– 31) + (– 11) = (-41) [Both the integers have the same sign. So, we add the integers and put the same sign.]

(-11) > (-41)

(– 21) – (– 10) > (– 31) + (– 11)

(c)    45 – (– 11) ______ 57 + (– 4)

L.H.S

45 + 11 = 56

R.H.S

57 + (-4) = 53

56 > 53

45 – (– 11) > 57 + (– 4)

(d)   (– 25) – (– 42) _____ (– 42) – (– 25)

L.H.S

(-25) – (-42) = (-25) + 42 = 17

R.H.S

(-42) – (-25) = (-42) + 25 = (-17)

17 > (-17)

(– 25) – (– 42) > (– 42) – (– 25)

1. Fill in the blanks.

(a) (– 8) + _____ = 0

(b) 13 + _____ = 0

(c) 12 + (– 12) = ____

(d) (– 4) + ____ = – 12

(e) ____ – 15 = – 10

Solution

1. Find

(a) (– 7) – 8 – (– 25)

(b) (– 13) + 32 – 8 – 1

(c) (– 7) + (– 8) + (– 90)

(d) 50 – (– 40) – (– 2)

Solution

(a)    (– 7) – 8 – (– 25) = (-15) – (-25) = (-15) + 25 = 10

(b)   (– 13) + 32 – 8 – 1 = 19 – 8 – 1 = 11 – 1 = 10

(c) (– 7) + (– 8) + (– 90) = (-15) + (-90) = (-105)

(d) 50 – (– 40) – (– 2) = 50 + 40 – (-2) = 90 – (-2) = 90 + 2 = 92

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