In this chapter, we provide RD Sharma Class 7 ex 1.1 Solutions Chapter 1 Integers for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 7 ex 1.1 Solutions Chapter 1 Integers Maths pdf, Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 7 |

Subject | Maths |

Chapter | Chapter 1 |

Chapter Name | Integers |

Exercise | Ex 1.1 |

Table of Contents

**RD Sharma Solutions for Class 7 Chapter 1 Integers Ex 1.1 Download PDF**

**Question: 1**

Determine each of the following products:

(i) 12 × 7

(ii) (-15) × 8

(iii) (- 25) × (- 9)

(iv) (125) × (- 8)

**Solution:**

(i) We have,

12 × 7 = 84 [The product of two integers of like signs is equal to the product of their absolute value]

(ii) We have,

(- 15) × 8 [The product of two integers of opposite

= (- 15 × 8) signs is equal to the additive inverse of the

= –120 [product of their absolute values]

(iii) We have,

(-25) × (-9)

= + (25 × 9)

= 225

(iv) We have,

(125) × (- 8)

= – (125 × 8)

= –1000

**Question: 2**

Find each of the following products:

(i) 3 × (- 8) × 5

(ii) 9 × (- 3) × (- 6)

(iii) (- 2) × 36 × (- 5)

(iv) (- 2) × (- 4) × (- 6) × (- 8)

**Solution:**

(i) We have,

3 × (- 8) × 5

= – (3 × 8) × 5

= (- 24) × 5

= – (24 × 5)

= – 120

(ii) We have,

9 × (-3) × (- 6)

= – (9 × 3) × (- 6)

= (- 27) × (- 6)

= + (27 × 6)

= 162

(iii) We have,

(-2) × 36 × (- 5)

= – (2 × 36) × (- 5)

= (- 72) × (- 5)

= (72 × 5)

= 360

(iv) We have,

(- 2) × (- 4) × (- 6) × (- 8)

= (2 × 4) × (6 × 8)

= (8 × 48)

= 384

**Question: 3**

Find the value of:

(i) 1487 × 327 + (- 487) × 327

(ii) 28945 × 99 – (- 28945)

**Solution:**

(i) We have,

1487 × 327 + (- 487) × 327

= 486249 – 159249

= 327000

(ii) We have,

28945 × 99 – (- 28945)

= 2865555 – 28945

= 2894500

**Question: 4**

Complete the following multiplication table:

Second Number

X | – 4 | -3 | – 2 | -1 | 0 | 1 | 2 | 3 | 4 |

– 4 | |||||||||

First Number | |||||||||

– 2 | |||||||||

-1 | |||||||||

0 | |||||||||

1 | |||||||||

2 | |||||||||

3 | |||||||||

4 |

Is the multiplication table symmetrical about the diagonal joining the upper left corner to the lower right corner?

**Solution:**

Second number

X | – 4 | – 3 | – 2 | – 1 | 0 | 1 | 2 | 3 | 4 |

– 4 | 16 | 12 | 8 | 4 | 0 | -4 | -8 | -12 | -16 |

First number | 12 | 9 | 6 | 3 | 0 | -3 | -6 | -9 | -12 |

-3 | |||||||||

-2 | 8 | 6 | 4 | 2 | 0 | -2 | -4 | -6 | -8 |

-1 | 4 | 3 | 2 | 1 | 0 | -1 | -2 | -3 | -4 |

0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

1 | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |

2 | -8 | -6 | -4 | -2 | 0 | 2 | 4 | 6 | 8 |

3 | -12 | -9 | -6 | -3 | 0 | 3 | 6 | 9 | 12 |

4 | -16 | -12 | -8 | -4 | 0 | 4 | 8 | 12 | 16 |

**Question: 5**

Determine the integer whose product with ‘-1’ is

(i) 58

(ii) 0

(iii) – 225

**Solution:**

(i) 58 x (–1) = – (58 x 1)

= – 58

(ii) 0 x (–1) = 0

(iii) (–225) x (–1) = + (225 x 1)

= 225

**Question: 6**

What will be the sign of the product if we multiply together

(i) 8 negative integers and 1 positive integer?

(ii) 21 negative integers and 3 positive integers?

(iii) 199 negative integers and 10 positive integers?

**Solution:**

(i) Positive ∵ [- ve × – ve = + ve]

(ii) Negative ∵ [- ve × + ve = – ve]

(iii) Negative

**Question: 7**

State which is greater:

(i) (8 + 9) × 10 and 8 + 9 × 10

(ii) (8 – 9) × 10 and 8 – 9 × 10

(iii) ((-2) – 5) × – 6 and (-2) – 5 × (- 6)

**Solution:**

(i) (8 + 9) × 10 = 17 × 10

= 170

8 + 9 × 10 = 8 + 90 = 98

(8 + 9) × 10 > 8 + 9 × 10

(ii) (8 – 9) × 10 = – 1 × 10

= – 10

8 – 9 × 10 = 8 – 90 = – 82

(8 – 9) × 10 > 8 – 9 × 10

(iii) ((-2) – 5) × – 6 = (- 7) × (- 6)

= (7 x 6)

= 42

(– 2) – 5 x (– 6) = – 2 + (5 x 6)

= 30 – 2

= 28

Therefore, ((-2) – 5×(- 6)) > (- 2) – 5 × (- 6)

**Question: 8**

(i) If a× (-1) = – 30, is the integer a positive or negative?

(ii) If a × (-1) = 30, is the integer a positive or negative?

**Solution:**

(i) When multiplied by ‘a’ negative integer, a gives a negative integer. Hence, ‘a’ should be

a positive integer.

a = 30

(ii) When multiplied by ‘a’ negative integer, a gives a positive integer. Hence, ‘a’ should be

a negative integer.

a = – 30

**Question: 9**

Verify the following:

(i) 19 × (7 + (-3)) = 19 × 7 + 19 × (-3)

(ii) (-23)[(-5)+ (+19)] = (-23) × (- 5) + (- 23) × (+19)

**Solution:**

(i) L.H.S = 19 × (7+ (-3))

= 19 × (7-3)

= 19 × 4

= 76

R.H.S = 19 × 7 + 19 × (-3)

= 133 – 57

= 76

Therefore, L.H.S = R.H.S

(ii) L.H.S = (-23)[(-5) + (+19)]

= (-23)[-5 + 19]

= (-23)[14]

= – 322

R.H.S = (-23) × (-5) + (-23) × (+19)

= 115 – 437

= –322

Therefore, L.H.S = R.H.S

**Question: 10**

Which of the following statements are true?

(i) The product of a positive and a negative integer is negative.

(ii) The product of three negative integers is a negative integer.

(iii) Of the two integers, if one is negative, then their product must be positive.

(iv) For all non-zero integers a and b, a × b is always greater than

either a or b.

(v) The product of a negative and a positive integer may be zero.

(vi) There does not exist an integer b such that for a >1, a × b = b × a = b. <

**Solution:**

(i) True

(ii) True

(iii) False

(iv) False

(v) False

(vi) True

**All Chapter RD Sharma Solutions For Class 7 Maths**

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