The document RD Sharma Solutions - Ex-1.1 Integers, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.

All you need of Class 7 at this link: Class 7

**Determine each of the following products:(i) 12 â˜“ 7(ii) (âˆ’15) â˜“ 8(iii) (âˆ’25) â˜“ (âˆ’9)(iv) 125 â˜“ (âˆ’8)**

(i) 12 Ã— 7 = 84

(ii) (âˆ’15) Ã— 8 = âˆ’120

(iii) (âˆ’25) Ã— (âˆ’9) = 225

(iv) 125 Ã— (âˆ’8) = âˆ’1000

**Find each of the following products:(i) 3 â˜“ (âˆ’8) â˜“ 5(ii) 9 â˜“ (âˆ’3) â˜“ (âˆ’6)(iii) (âˆ’2) â˜“ 36 â˜“ (âˆ’5)(iv) (âˆ’2) â˜“ (âˆ’4) â˜“ (âˆ’6) â˜“ (âˆ’8)**

(i) 3 Ã— (âˆ’8) Ã— 5 = âˆ’3Ã—(8Ã—5)-3Ã—(8Ã—5) = âˆ’-120

(ii) 9 Ã— (âˆ’3) Ã— (âˆ’6) = 9Ã— (3Ã—6) 9Ã— (3Ã—6) = 162

(iii) (âˆ’2) Ã— 36 Ã— (âˆ’5) = 36 Ã— (2Ã—5)36 Ã— (2Ã—5) = 360

(iv) (âˆ’2) Ã— (âˆ’4) Ã— (âˆ’6) Ã— (âˆ’8) = (2Ã—4Ã—6Ã—8)(2Ã—4Ã—6Ã—8) = 384

**Find the value of:(i) 1487 Ã— 327 + (âˆ’487) Ã— 327(ii) 28945 Ã— 99 âˆ’ (âˆ’28945)**

(i) 1487 Ã— 327 + (âˆ’487) Ã— 327 = 327 (1487âˆ’487) = 327 Ã—1000 =327000327 (1487-487) = 327 Ã—1000 =327000

(ii) 28945 Ã— 99 âˆ’ (âˆ’28945) = 28945 (99 âˆ’(âˆ’1)) = 28945 (99+1) = 2894500

**Complete the following multiplication table:**

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

Yes, the table is symmetrical along the diagonal joining the upper left corner to the lower right corner.

**Determine the integer whose product with 'âˆ’1' is(i) 58(ii) 0(iii) âˆ’225**

The integer, whose product with âˆ’1 is the given number, can be found by multiplying the given number by âˆ’1.

Thus, we have:

(i) 58 Ã— (âˆ’1) = âˆ’58

(ii) 0 Ã— (âˆ’1) = âˆ’ (0Ã—1) - (0Ã—1) = 0

(iii) (âˆ’225) Ã— (âˆ’1) = 225

**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?**

Negative numbers, when multiplied even number of times, give a positive number. However, when multiplied odd number of times, they give a negative number. Therefore, we have:

(i) (negative) 8 times Ã— (positive) 1 time = positive Ã— positivepositive Ã— positive = positive integer

(ii) (negative) 21 times Ã— (positive) 3 times = negative Ã—positive negative Ã—positive = negative integer

(iii) (negative) 199 times Ã— (positive) 10 times = negative Ã— positive negative Ã— positive = negative integer

**State which is greater:(i) (8 + 9) Ã— 10 and 8 + 9 Ã— 10(ii) (8 âˆ’ 9) Ã— 10 and 8 âˆ’ 9 Ã— 10(iii) {(âˆ’2) âˆ’ 5} Ã— (âˆ’6) and (âˆ’2) âˆ’5 Ã— (âˆ’6)**

(i) ( 8 + 9) Ã— 10 = 170 > 8 + 90 = 98

(ii) (8 âˆ’ 9) Ã— 10 = âˆ’10 > 8 âˆ’ 90 = âˆ’ 82

(iii) {(âˆ’2) âˆ’ 5 } Ã— (âˆ’6) = âˆ’7 Ã— (âˆ’6) = 42 > (âˆ’2) âˆ’ 5 Ã— (âˆ’6) = ( âˆ’2 ) âˆ’ (âˆ’30) = âˆ’2 + 30 = 28

**(i) If a Ã— (âˆ’1) = âˆ’30, is the integer a positive or negative?(ii) If a Ã— (âˆ’1) = 30, is the integer a positive or negative?**

(i) *a *Ã— (âˆ’1) = âˆ’30

When multiplied by a negative integer, *a* gives a negative integer. Hence, *a* should be a positive integer.*a =* 30

(ii) *a *Ã— (âˆ’1) = 30

When multiplied by a negative integer, *a* gives a positive integer. Hence, *a* should be a negative integer.*a =* âˆ’30

**Verify the following:(i) 19 Ã— {7 + (âˆ’3)} = 19 Ã— 7 + 19 Ã— (âˆ’3)(ii) (âˆ’23) {(âˆ’5) + (+19)} = (âˆ’23) Ã— (âˆ’5) + (âˆ’23) Ã— (+19)**

(i)

LHS = 19 Ã— { 7 + (âˆ’3) } = 19 Ã— {4} = 76

RHS = 19 Ã— 7 + 19 Ã— (âˆ’3) = 133 + (âˆ’57) = 76

Because LHS is equal to RHS, the equation is verified.

(ii)

LHS = (âˆ’23) {(âˆ’5) + 19} = (âˆ’23) { 14} = âˆ’322

RHS = (âˆ’23) Ã— (âˆ’5) + (âˆ’23) Ã— 19 = 115 + (âˆ’437) = âˆ’322

Because LHS is equal to RHS, the equation is verified.

**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 **

(i) True. Product of two integers with opposite signs give a negative integer.

(ii) True. Negative integers, when multiplied odd number of times, give a negative integer.

(iii) False. Product of two integers, one of them being a negative integer, is not necessarily positive. For example, (âˆ’1) Ã— 2 = âˆ’2

(iv) False. For two non-zero integers *a* and *b*, their product is not necessarily greater than either *a* or *b*. For example, if *a = *2 and *b =* âˆ’2, then, *a *Ã—* b =* âˆ’4, which is less than both 2 and âˆ’2.

(v) False. Product of a negative integer and a positive integer can never be zero.

(vi) True. If a > 1, then, aÃ—b â‰ bÃ—a â‰ b