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 |

**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**

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.