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
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