Haryana Board Class 7 Maths Solutions For Chapter 1 Integers

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Haryana Board Class 7 Maths Solutions For Chapter 1 Integers

  • Properties of integers under addition:
    1. Closure property: The sum of any two integers is also an integer. It is called “closure property” under addition in integers.
      Example: 2 + (-5) = -3,2 + 3 = 5 etc.
    2. Associative property: In general for any three integers a, b and c we have a + (b + c) = (a + b) + c
    3. Commutative property: for any two integers a, b we have a + b = b + a.
    4. Additive identity: There exists a number ‘0’ in integers such that for any integer a, we have a + 0 = a = 0 + a. ‘0’ is called the additive identity.
    5. Additive inverse: To each integer a, there is an integer -a such that a + (-a).= 0
      = -a + a.
      ‘-a’ is called the additive inverse of ‘a’.
  • Properties of integers under multiplication:
    1. Closure property:
       If a,b are two integers then a.b is also an integer.
    2. Associative property:
      If a, b and c are any three integers, then a(b. c)=(a .b). c
    3. Commutative property:
       If a, b are any two integers, we have a.b = b.a.
    4. Distributive property:
       If a, b,c are any three integers then we have
      a.(b + c)= a.b +a.c (Left distributive law)
      (a + b).c = a.c + b.c(Right distributive law)
    5. Multiplicative identity: There exists an element 1 in integers such that for any integer a, we have a.1 =1.a = a ‘1’ is called the multiplicative identity.
    6. Multiplication by zero: The product of any integer and zero is zero.
    7. Multiplicative inverse: To each integer ‘a’, there is an integer 1/a such that a x 1/a = 1/a x a = 1. ‘1/a’ is called multiplicative inverse of ‘a’.
  • Closure property under subtraction: The difference of any two integers is also an integer. i.e., a, b are integers ⇒ a -b is also an integer
  • The subtraction of integers is not commutative.
    i.e., a -b ≠ b – a where a, b are integers
  • Division of integers is not closed because 4 and 8 are integers but 4/8 = 1/2 is not an integer.
  • Division of integers is neither commutative nor associative.
  • Division with zero is not defined.
  1. We now study the properties satisfied by addition and subtraction.
    1. Integers are closed for addition and subtraction both. That is, a + b and a -b are again integers, where a and b are any integers.
    2. Addition is commutative for integers, i.e., a + b = b + a for all integers a and b.
    3. Addition is associative for integers, i.e., (a + b) + c = a + (b + c) for all integers a, b and c.
    4. Integer 0 is the identity under addition. That is, a + 0 = 0 + a = a for every integer a.
  2. We studied,how integers could be multiplied, and found that product of a positive and a negative integer is’a negative integer, whereas the product of two negative integers is a positive integer.
    For example, – 2 x 7 =- 14 and – 3 x – 8 = 24.
  3. Product of even number of negative integers is positive, whereas the product of odd number of negative integers is negative.
  4. Integers show some properties under multiplication.
    1. Integers are closed under multiplication. That is, a x b is an integer for any two integers a and b.
    2. Multiplication is commutative for integers. That is,a x b= b x a for any integers a and b.
    3. The integer 1 is the identity under multiplication, i.e.,l x a = a x l=a for any integer a.
    4. Multiplication is associative for integers, i.e., (a x b) x c = a x (b x c) for any three integers a, b and c.
  5. Under addition and multiplication, integers show a property called distributive property.
    That is, a x (b + c) = (a x b) + (a x c) for any three integers a, b and c.
  6. The properties of commutativity, associativity under addition and multiplication, and the distributive property help us to make our calculations easier.
  7. We also learnt how to divide integers. We found that
    1. When a positive integer is divided by a negative integer, the quotient obtained is negative and vice-versa.
    2. Division of a negative integer by another negative integer gives positive as quotient.
  8. For any integer a, we have
    1. a + 0 is not defined
    2. a +1= a

Solutions To Try These

1. Write a pair of integers whose sum gives

  1. a negative integer.
    Solution: -24 and -15
  2. zero
    Solution: -10 and 10
  3. an integer smaller than both the integers.
    Solution: -8 and -5
  4. an integer smaller than only one of the integers.
    Solution: -5 and 9
  5. an integer greater than both the integers.
    Solution: 6 and 10

HBSE Class 7 Integers Solutions

2. Write a pair of integers whose difference gives

  1. a negative integer.
    Solution: 5 and 7
  2. zero.
    Solution: 8 and 8
  3. an integer smaller than both the integers.
    Solution: -3 and1
  4. an integer greater than only one of the integers.
    Solution: 11 and 3
  5. an integer greater than both the integers.
    Solution: 9 and -5

Exercise 11

1. Write down a pair of integers whose:

  1. sum is -7
  2. difference is -10
  3. sum is 0

Solution:

  1. -15 and 8
  2. 8 and 18
  3.  -9 and 9

2. 1) Write a pair of negative integers whose difference gives 8.

Solution: -18 and -26

2) Write a negative integer and a positive integer whose sum is -5.

Solution: -8 and 3

3) Write a negative integer and a positive integer whose difference is -3.

Solution: -5 and 2

Haryana Board Class 7 Maths Integers solutions

3. In a quiz, team A scored – 40, 10, 0 and team B scored 10, 0, – 40 in three successive rounds. Which team scored more? Can we say that we can add integers in any order?

Solution:

Total score of team A = (- 40) + 10 + 0

=(- 40) + 10 = -30

Total score of team B = 10 + 0 + (- 40)

10 + (- 40)= -30

Both teams A and B scored equally.

Yes, we can add integers in any order.

Key Questions in Integers for Class 7 HBSE

4. Fill in the blanks to make the following statements true:

Solution:

  1. (-5) + (-8)= (-8) + (-5)
  2. -53 + 0 = -53
  3. 17 + (-17) = 0
  4. [13 (-12)] + (-7) = 13 + [(-12) + (-7)]
  5. (-4) + [15 + (-3)] = [-4 + 15] + (-3)

Solutions To Try These

1) Find 4 x (-8) using number line.

Solution:

4 x (-8) using number line

∴ 4 x (-8) = -32

2) Find 8 x (-2) using number line

Solution:

8 x (-2) using number line

∴ 8x (-2) = -16

3) Find 3 x (-7) usingnumber line.

Solution:

3 x (-7) usingnumber line

∴ 3X (-7) = -21

Practice Problems Integers Class 7 Haryana Board

4) Find 10 x (-1) using number line

Solution:

10 x (-1) using number line

∴ 10 x (-1) = -10

Solutions To Try These

Find:

1) 6 x (-19)

Solution: 6x (-19) = -(6×19) = -114

2) 12 x (-32)

Solution: 12 x (-32) =- (12 x 32) = -384

3) 7 x (-22)

Solution: 7 x (-22) = – (7 x 22) = -154

Solutions To Try These

1. Find:

1) 15 x (-16)

Solution.

15 x (-16) = – (15 x 16) = -240

2) 21 x (-32)

Solution: 21 x (-32) = – (21 x 32) = -672

3) (- 42) x 12

Solution. (- 42) x 12 = – (42 x 12) = -504

4) -55×15

Solution. (-55) x 15 = – (55 x 15) = -825

2. Check if

1) 25 x (-21) = (-25) x 21

Solution:

25 x (-21) = -(25 x 21) = -525

(-25) x 21 = -(25 x 21) = -525

∴ 25 x (-21) = (-25) x 21

2) (-23) x 20 = 23 x (-20)

Solution:

(-23) x 20 = -(23 x 20) = -460

23 x (-20) = -(23 x 20) = -460

∴  (-23) x 20 = 23 x (-20)

Other examples:

1) 10 x (-43) =43 x (-10)

2) (-29) x 25 = 29 x (-25)

3) (-63) x 37 = 63 x (-37)

4) 40 x (-28) = (-40) x 28

5) 30 x (-19) = (-30)x 19 .

Solutions To Try These

1) Starting from (-5) x 4, find (-5) x (-6)

Solution:

(-5) x 4 = -20 .

(-5) x 3 = -15 = [-20 + 5]

(-5) x 2 = -10 = [-15 + 5]

(-5) x 1 = -5 = [—10 + 5]

(-5) x 0 = 0 = [-5 + 5]

(-5) x (-1) = 5 = [0 + 5] ,

(-5) x (-2) = 10 = [5 + 5]

(-5) x (-3) = 15 = [10 + 5]

(-5) x (-4) =20 = [15 + 5] ‘

(-5) x (-5) =25 = [20 + 5]

(-5) x (-6) = 30 = [25 + 5]

Important Concepts Integers Class 7 HBSE

2) Starting from (- 6) x 3, find (- 6) x (-7)

Solution:

(- 6) x 3 = -18

(- 6) x 2 = -12 = [-18 + 6]

(- 6) x 1 = -6 = [-12 + 6]

(- 6) x 0 = 0 = [-6 + 6]

(- 6) x (-1) = 6 = [0 + 6]

(- 6) x (-2) = 12 = [6 + 6]

(- 6) x (-3) = 18 = [12 + 6]

(-6) x (-4) = 24 = [18 + 6]

(- 6) x (-5) = 30 = [24 + 6]

(- 6) x (-6) = 36 = [30 + 6]

(- 6) x (-7) = 42 = [36 + 6]

The product of two negative integers is a positive integer. We multiply the two negative integers as whole numbers and put the positive sign before the product.

(Negative integer) x (Negative integer) = Positive integer

Solutions To Try These

Find: (-31) x (-100), (-25) x(-72), (-83) x (-28).

1) (-31) x (-100)

Solution: (-31) x (-100) = 31 x 100 = 3100

2) (-25) x (-72)

Solution: (-25) x (-72) = 25 x 72 = 1800

3) (-83) x (-28)

Solution: (-83) x (-28) = 83 x 28 = 2324

Addition and subtraction of integers Class 7 HBSE

Solutions To Try These

1) Is 10 x [6 + (-2)] = 10 x 6 + 10 x (-2) ?

Solution:

10 x [6 + (-2)] =10 x 4 = 40

10 x 6 + 10 x (-2) = 60 -20 =40

∴ 10 x [6 + (-2)] =10 x 6 + 10 x (-2)

2) Is (-15) x [(-7) + (-1)] = (-15) x (-7) + (-15) x (-1) ?

Solution:

(-15) x [(-7) + (-1)] = (-15) x (-8) = 120

(-15) x (-7) + (-15)x (-1) = 105 + 15 = 120

∴ (-15) x [(-7) +,(-1)] = (-15) x (-7) + (-15) x (-1)

Solutions To Try These

1) Is 10 x [6-(-2)] = 10 x 6 – 10 x (-2) ?

Solution: 10 x [6 -(-2)] =10 x 8 = 80

10 x 6 -10 x (-2)= 60-(-20) = 80

∴ 10 x [6-(-2)] = 10 x 6-10 x (-2)

HBSE Class 7 Maths Chapter 1 Guide

2) Is (-15) x [(-7- (-1)] = (-15) x (-7) – (-15) x (-1) ?

Solution: (-15) x [(-7) – (-1)] = (-15) x (- 6) = 90

(-15) x (-7) – (-15) x (-1) =105-15 = 90

∴ (-15) x [(-7)- (-1)] = (-15)x (-7)-(-15) x (-1)

Exercise – 1.2:

1. Find each of the following products:

1) 3 x (-1)

Solution: 3 x (-1) = -(3X1)= -3

2) (-1) x 225

Solution: (-1) x 225 = – (1 x 225) = -225

3) (-21) x (-30)

Solution: (-21) x (-30) = 21 x 30 = 630

4) (-316) x (-1)

Solution: (-316) x (-1) = 316 x 1 = 316

5) (-15) x 0 x (-18)

Solution: (-15) x 0 x (-18)- [(-15) x 0] x (-18)= 0 x (-18) = 0

6) (-12) x (-11) x (10)

Solution: (-12)x(-11) x 10 = [(-12) x (-11)] x 10 = 132 x10 = 1320

7) 9 x(-3)x(-6)

Solution:

9 x (-3) x (-6) = 9 x [(-3)x(-6)]

= 9 x 18 = 162

8) (-18) x (-5) x (-4)

Solution: (-18) x (-5)x (-4) = [(-18)x (-5)]x (-4)= 90 x (-4) = -360

9) (-1) x (-2) x (-3) x 4

Solution: (-1) x (-2) x (-3) x 4

= [(-1) x (- 2)] x [(-3) x 4]

= 2 x (-12) = -24

Integer Operations Class 7 Haryana Board

10) (-3) x (-6) x (-2) x (-1)

Solution:

(-3) x (-6) x (-2) x (-1) =

[(-3) x (-6)]x[(-2)x(-l)]

=18 x 2 =36

2. Verify the following:

1) 18 x [7 + (-3)] = [18 x 7] + [18 x(-3)]

Solution:

18 x [7 + (-3)] = 18 [7-3] = 18x 4 = 72

[18 x 7] +[18 x (-3)] = 126 + (-54) = 72

∴ 18 x [7 +(-3)] = [18 x 7] + [18x (-3)]

2) (-21) x [(-4) + (-6)] = [(-21) x(-4)] + [(-21) x (-6)]

Solution:

(-21) x [(-4) + (-6)]= (- 21)x(-10) = 210

[(-21) x (-4)] +[(-21) x (-6)]

= 84 + 126 = 210

∴ (-21) X [(-4) + (-6)] = [(-21) x (-4)] + [(-21) x (-6)]

3. 1) For any integer a, what is (-1) x a equal to?

Solution: (-1) x a =-a

Multiplication and division of integers Class 7

2) Determine the integer whose product with (-1) is

  1. -22
  2. 37
  3. 0

Solution:

  1. (-1) x 22 = -22;
  2. (-1) x (-37)=37;
  3. (-1)x 0 = 0

4. Starting from (-1) x 5, write various products showing some pattern to show (-1) x (-1) = 1.

Solution:

(-1) x 5 = -5

(-1) x 4 = -4 = [(-5) + 1]

(-1) x 3 = -3 = [(-4) +l ]

(-1) x 2 = -2 = [(-3) +1 ]

(-1) x 1 = -1 = [(-2) + 1]

(-1) x 0 = 0 = [(-1) + 1]

(-1) x (-1)=1 = [0 + 1]

Very Short Answer Questions

1. Write four negative integers greater than -20.

Solution: -19, -18, -17 and -16.

2. Write any four negative integers less than -10.

Solution: -11, -12, -13 and -14.

3. Find: 50 -(-40) -(-2)

Solution: 50 – (-40) – (-2) =50 + 40 + 2 = 90 + 2 = 92

4. Write the properties of integers under addition.

Solution

  1. Closure property
  2. Commutative property
  3. Associative property
  4. Additive identity

5. Write down a pair of integers whose sum is 0.

Solution: (-25) + 25 = 0

6. Give an example for ‘Subtraction is not commutative for integers’.

Solution: Consider the integers 9 and (-5)

9- (-5) = 9 + 5 = 14 and

(-5) -9 = -5-9 = -14

∴ 9-(-5)=(-5)-9

7. Find (-45) x (-100).

Solution: (-45) x (-100) =4500

8. Find (-72) + 8.

Solution: (-72) + 8 = -9

9. Write the additive identity and multiplicative identity for integers.

Solution:

Additive Identity for integers is ‘0’.

MultiplicativeIdentity for integers is ‘1’.

HBSE 7th Class Integer Rules and Properties

10. The product of two integers is -165. If one number is -15, Find the other integer.

Solution:

The product of two integers = – 165

First number = -15

Second number = “x”

(-15) × “x” = -165

“x” = \(\frac{-165}{-15}=\frac{165}{15}\)

“x” =11

∴ Second number (integer) = 11

11. Because of COVID-19 a company locked down for 6 months and got loss of 1,32,000 in the year 2020. Find the average loss of each month.

Solution: A company locked down for 6 months because of covid-19

Loss of company in the year 2020 is 1,32,000

The average loss of each month = 1,32,000 ÷ 6 = 22,000

12. Write the additive inverses of 5, -8,1 and 0.

Solution:

The additive inverse of 5 is = -(+5) =-5

The additive inverse of -8 is = -(-8) = 8

The additive inverse of 1 is = -(+1)=-1

The additive inverse of 0 is = -(0) = 0

Integer number line Class 7 Haryana Board

13. Compute the following.

1) -36 ÷ (-4)

2) (-201) ÷ (-3)

3) (-325) ÷ (-13)

Solution:

1) -36 ÷ (-4) = 9

2) (-201) ÷ (-3) =67

3) (-325) ÷ (-13) =25

14. Write the following integers in ascending order (smallest to biggest).

1) -5, 2,1, -8
2) -4, -3, -5, 2
3) -10, -15, -7

Solution:

1) -8, -5,1, 2 (or) -8 < -5 <1 < 2
2) -5, -4, -3, 2 (or) -5 < -4 < -3 <2
3) -15, -10, -7 (or) -15 < -10 < -7

15. Write the following integers in descending order (biggest to smallest)

1) -2,-3,-5
2) -8,-2,-1
3) 5, 8, -2

Solution:

1)-2, -3, -5 (or) -2> -3> -5

2) -l,-2-8 (or) -1>-2>-8

3) 8,5,-2 (or) 8>5>-2

16. Multiply the following.

1) 5×7

Solution: 5×7=35

2) (-9) x (6)

Solution: (-9) x (6) =-(9×6) =-54

3) (9) x (-4)

Solution: (9) x (-4) = -(9 x 4) =-36

4) (8) x (-7)

Solution: (8) x (-7) =-(8×7) =-56

5) (-124) x (-1)

Solution: (-124) x(-1) =124

6) (-12) x (-7)

Solution: (-12) x (-7) =84

7) (-63) x 7

Solution: (-63) x 7= -(63 x 7) = -441

8) 7 x (-15)

Solution: 7 x (-15) = -(7 x 15) = -105

Sample Problems Integers Haryana Board Class 7

17. Write the pair of integers whose product will be

1) A negative integer

Solution: (-4) x 6 =-(4 x 6) = -24

3 x (-7) = -( 3 x 7)=- 21

2) A positive integer

Solution: 5 x 6 = 30 (-4) x (-5) = (4 x 5) = 20

3) Zero

Solution: 0 x 5 = 0 ; 3  x 0= 0

18. During the summer, the level of water in a pond decreases by 5 inches every week due to evaporation. What is the change in the level of the water over a period of 6 weeks?

Solution:

The level of water that decreases ever week = 5 inches

The level of water that decreases 6 weeks = 6 x 5 inches

= 30 inches

The change in the level of the water over a period of 6 weeks

= The water in a pond decreases by 30 inches

= -30 inches

19. A green grocer earns a profit of 7 per kg on tomato and got loss of per kg on brinjal by selling. On Monday he got neither profit nor loss, if he sold 68 kgs of tomato. How many kgs of brinjal did he sell?

Solution:

Profit on tomato per1 kg that a green grocer earns = 7

Loss on brinjal per kg = 4

Number of kgs of tomato he sold on Monday = 68 kg

Let number of kgs of brinjal he sold on Monday = x kg

On Monday he gets neither profit nor loss

68 x 7 = x × 4

x = \( \frac{68 \times 7}{4} \)

x = 119

Number of kgs of brinjal he sold on Monday = 119 kgs

20. In a test, + 3 marks are given for every correct answer and-1 mark is given for every incorrect answer. Sona attempted all the questions and scored + 20 marks though she got 10 correct answers.

1) How many incorrect answers she attempted?

2) How many questions were given in the test?

Solution:

In a test,

Marks given for every correct answer = + 3

Marks given for each incorrect answer =-1

Marks scored by Sona =20

Number of correct answers she got =10

Marks scored for 10 correct answers

= 10 x (+3) = + 30

The difference of marks

= 20 -30 = -10

1) The number of incorrect answers she attempted = -10/-1 =10

Thus Sona attempted 10 incorrect answers.

2) Number of questions were given in the test

= Number of correct answered questions + Number ofincorrect answered questions.

= 10 + 10 = 20

Word problems on integers for Class 7 HBSE

21. Verify -3 x [ (-4)- 2] = [(-3) x (-4)]– [(-3)x 2].Is multiplication distributive over subtraction of integers? Write your observations.

Solution: LHS = (-3) x [(-4) -2]

= (-3) x (-6)

= 18

RHS = [(-3)x(-4)]-[(-3)x2]

= (3×4) -[-(3×2)]

= 12 -(-6)

= 12 + 6 = 18

Observation: LHS = RHS

Yes, multiplication is distributive over subtraction of integers.

22. Identify the laws in the following statements:

1) -3 + 5 = 5 + (-3)

Solution: Commutative property for addition.

2) -2 x1 =1 x (-2) = -2

Solution: Multiplicative Identity.

3) [(-5) x 2)] x 3 = (-5) x [(2 x 3)]

Solution: Associative property for multiplication.

4) 18 x [7+ (-3)] = [18 x 7] + [18 x (-3)]

Solution: Distributive over addition and multiplication.

5) -5×6 = -30

Solution: Closure property for multiplication

6) -3 + 0 = 0 + (-3) = -3

Solution: Additive Identity

23. Simplify the following using suitable laws.

1) -11 x (-25) x (-4)

Solution:

= -11 x[+(25×4)]      ( Closureproperty)

= -11×100

= -(11×100)

= -1100

2) 3 x (-18) +3 x (-32)

Solution:

= 3x[-18(-32)]        ( Distributive property)

= 3x[-18-32]

= 3x[-50]

= -150

24. Sankar, a fruit vendor sells 100kg of oranges and 75 kg of pomegranates. If he makes a profit of 11 per one kg of pomegranates and loss of 8 per one kg oranges, what will be his overall profit or loss?

Solution: Profiton pomegranates per one kg= 11

Loss on oranges per one kg = 8

Number of kgs of pomegranates sold = 75

Number of kgs of oranges sold = 100

The amount of profiton pomegranates = 75×11 =825

The amount of loss on oranges = 100 x 8 = 800

As the profit is more than loss he got profit.

Profit = 825- 800

= 25.

25. Solve the following.

1) 17 -(-14)

2) 13 -(-8)

3) 19 -(-5)

4) 15-28

5) 25-33

6) 80 -(-50)

7) 150-25

8) 32 -(-18)

Solution:

1) 17 -(-14) =17 + 14 = 31

2) 13 – (-8) = 13 + 8- 21

3) 19- (-5) = 19 + 5 = 24

4) 15 -28 = -13

5) 25 – 33 = -8

6) 80 -(-50) = 80 + 50 = 130

7) 150 – 75 = 75

8) 32 -(-18) =32 + 18 = 50

Properties of integers Class 7 HBSE Maths

26. A merchant on selling rice earns a profit of 10 per bag of basmati rice sold and a loss of 5 per bag of non-basmati rice.

1) He sells 3, 000 bags of basmati rice and 5,000 bags of non-basmati rice in a month. What is his profit or loss in a month?

2) What is the number of basmati rice bags he must sell to have neither profit nor loss,if the number of bags of non-basmati rice sold is 6,400 ?

Solution:

Profit that the merchant gets per bag of Basmati rice = 10.

Loss that the merchant gets per bag ofnon basmati rice = 5.

1) No. of Basmati rice bags he sold = 3,000

No. of Non-Basmati rice bags he sold = 5,000

∴ The total result = 3,000 x 10- 5,000 x 5

= 30,000- 25,000

= 5,000

Since the result is positive, the merchant get a profit of 5,000

2) No. of non-basmati rice bags sold = 6400

Let the No. of basmati rice bags sold = x

The no.of basmati rice bags he must sell to have neither profit nor loss,

= x × 10 = 6400 x 5

x = 6400×5/10

x = 3200

∴ Merchant has to sell 3200 basmati rice bags to have neither profit nor loss.

27. Find the product, using suitable properties.

1) 26 x (-48) + (-48) x (-36)

2) 8 x 53 x (-125)

3) 15 x (-25) x (-4) x (-10)

4) (-41) x102

5) 625 x (-35) + (-625) x 65

6) 7 x (50- 2)

7) (-17) x (-29)

8) (-57) x (-19) +57

Solution:

1) 26 x (-48) + (-48) x (-36)

= [26 + (-36)] x (-48) [Multiplicative Distributive property]

= (-10) x (-48)

= 480 [Product of two negatives is positive]

∴ 26 x (-48) + (-48) x (-36) =480

2) 8 x 53 x (-125) = [8 x 53] x (-125) [Associative property]

= 424 x (-125)

= -53000

Hint: To multiply 424 x 125, simply write 424 x 1000/8 which is easier

2) 15 x (-25) x (-4) x (-10) = 15 x [(-25) x (-4)] x -10 [Associative property]

= 15 x 100 x -10

= [15 x 100] x -10 [Associative property]

= 1500 x -10

= -15000

4) (-41) x 102 = (-41) x [100 + 2]

= (-41) x 100 + (-41) x 2 [Distributive property]

= -4100 + (-82)

= -4182

5) 625 x (-35) + (-625) x 65 = (-625) x 35 + (-625) x 65

= (-625) x (35 + 65) [Distributive property]

= -625×100

= -62500

6) 7 x (50 -2) = 7 x [50 + (-2)](Distributive property)

= 7 x 50 + 7 x (-2)

= 350 – 14 = 336

7) (-17) x (-29) = (-17) [-30 + 1]

= (-17) (-30) + (-17) (1) (Distributive property)

= 510-17 = 493

8) (-57) x (-19) +57 = (-57) x (-19) +(-57) x (-1)

= (-57) x [(-19) + (-1)] (Distributive property)

= (-57) x -20

= 1140

28. In a class test containing 15 questions, 4 marks are given for every correct answer and (-2) marks are given for every incorrect answer.

(1) Bharathi attempts all questions but only 9 answers are correct. What is her total score?

(2) One of her friends Hema answers only 5 questions correctly. What will be her total score?

Solution:

No. of questions in the test = 5

Marks given for correct answer = 4

Marks given for wrong answer = (-2)

1) No. of correct answers answered by Bharathi = 9

No. of incorrect answers =15-9 = 6 [ She attempted all questions]

Thus, the total score of Bharathi = 9×4 + 6 (-2)

= 36- 12 = 24 marks.

∴ The total score of Bharathi is 24 marks.

2) No. of correct answers answered by Hema = 5

No. of incorrect answers answered by Hema = 0

∴ Total score of Hema = 5 x 4 = 20 marks.

 

Fill in the blanks:

77. -4 + 6…………. 7 -1 (Use < or >)

Answer: <

78. If die number of negative integers in a product is……….then the product is a positive integer.

Answer: even

79. The product ofa negative integer and zero is………

Answer: zero

80. For any integer a, we have a + 0 is……..

Answer: not defined

81. -87 +……. = 87

Answer: -1

82.(Positive integer) x (Negative integer) =……..

Answer: negative integer

83. …………is not commutative for integers.

Answer: Subtraction

84. 7, 3, -1,-5,?………

Answer: -9

85. 17+ …… = 0

Answer: -17

86………….in his book Ankitung Zur Algebra (1770), was one of the first mathematicians to attempt to prove (-1) x (-1) =1

Answer: Euler

Match the following:

87.

1. For any two integers a and b,a + b is an integer                                  (   )             A) Associative property under addition

2. For any integers a, b and c a + (b + c) = (a + b) + c                            (   )              B) Multiplicative identity

3. For any integers a and b; a xb = b x a                                                   (   )              C) Distributive property

4. For any integers a, b and c a x (b + c) = a xb + a xc                            (   )              D) Closure under addition

5. For any integer a;1 x a = a x 1= a. The integer1 is called                    (   )              E) Commutative property for multiplication

Answer: 1. D 2. A 3. E 4. C 5. B

 

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