The document RD Sharma Solutions -Page No.5.21, Negative Numbers And Integers, Class 6, Maths Class 6 Notes | EduRev is a part of the Class 6 Course RD Sharma Solutions for Class 6 Mathematics.

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**Question 16: Evaluate 7 Ã—Ã— |âˆ’ 15| âˆ’ |âˆ’ 9| Ã—Ã— 8.**

**ANSWER: **âˆµ |âˆ’ 15| = 15 and |âˆ’ 9| = 9

âˆ´ 7 Ã—Ã— |âˆ’ 15| âˆ’ |âˆ’ 9| Ã—Ã— 8 = 7 Ã—Ã— 15 âˆ’ 9 Ã—Ã— 8

= 105 âˆ’ 72

= 33

Hence, 7 Ã—Ã— |âˆ’ 15| âˆ’ |âˆ’ 9| Ã—Ã— 8 = 33.

**Question 17: Find the value of 38 âˆ’ (âˆ’ 25) âˆ’ 58 + (âˆ’ 15) + 23 âˆ’ (âˆ’ 8).**

**ANSWER: **38 âˆ’ (âˆ’ 25) âˆ’ 58 + (âˆ’ 15) + 23 âˆ’ (âˆ’ 8)

= 38 âˆ’ (âˆ’ 25) âˆ’ 58 + (âˆ’ 15) + 23 +8

= 38 âˆ’ (âˆ’ 25) âˆ’ 58 + (âˆ’ 15) + 31

= 38 âˆ’ (âˆ’ 25) âˆ’ 58 + (31 âˆ’ 15)

= 38 âˆ’ (âˆ’ 25) âˆ’ 58 + 16

= 38 âˆ’ (âˆ’ 25) âˆ’ (58 âˆ’ 16)

= 38 âˆ’ (âˆ’ 25) âˆ’ 42

= 38 + 25 âˆ’ 42

= 38 âˆ’ (42 âˆ’ 25)

= 38 âˆ’ 17

= 21

Hence, 38 âˆ’ (âˆ’ 25) âˆ’ 58 + (âˆ’ 15) + 23 âˆ’ (âˆ’ 8) = 21.

**Question 18: Simplify: 5 + (âˆ’ 5) + 5 + (âˆ’ 5) + ...(i) When the number of terms is 20 (ii) When the number of terms is 25**

**ANSWER: **(i) When the number of terms is 20

5 + (âˆ’ 5) + 5 + (âˆ’ 5) + ... = (5 + 5 + 5 + ... + 10 terms) + (âˆ’ 5) + 5 + (âˆ’ 5) + ... + 10 terms

= 10 Ã—Ã— 5 + 10 Ã—Ã— (âˆ’ 5)

= 50 âˆ’ 50

= 0

(ii) When the number of terms is 25

5 + (âˆ’ 5) + 5 + (âˆ’ 5) + ... = (5 + 5 + 5 + ... + 12 terms) + (âˆ’ 5) + 5 + (âˆ’ 5) + ... + 13 terms

= 12 Ã—Ã— 5 + 13 Ã—Ã— (âˆ’ 5)

= 60 âˆ’ 65

= âˆ’ (65 âˆ’ 60)

= âˆ’ 5

**Question 19: If âˆ† is an operation on integers such that for integers a and b, a âˆ† b = a âˆ’ b âˆ’ (âˆ’ 5). Find the values of**

(ii) (âˆ’ 9) âˆ† (âˆ’ 4)

(iii) 2 âˆ† 5

(iv) 4 âˆ† (âˆ’ 5)

**ANSWER:**(i)

âˆµ a âˆ† b = a âˆ’ b âˆ’ (âˆ’ 5)

âˆ´ (âˆ’ 7) âˆ† 3 = (âˆ’ 7) âˆ’ 3 âˆ’ (âˆ’ 5)

= âˆ’ 7 âˆ’ 3 + 5

= âˆ’ (7 + 3) + 5

= âˆ’ 10 + 5

= âˆ’ (10 âˆ’ 5)

= âˆ’ 5

(ii)(âˆ’ 9) âˆ† (âˆ’ 4) = (âˆ’ 9) âˆ’ (âˆ’ 4) âˆ’ (âˆ’ 5)

= âˆ’ 9 + 4 + 5

= âˆ’ 9 + 9

= 0

(iii) 2 âˆ† 5 = 2 âˆ’ 5 âˆ’ (âˆ’ 5)

= 2 âˆ’ 5 + 5

= 2

(iv)

4 âˆ† (âˆ’ 5) = 4 âˆ’ (âˆ’ 5) âˆ’ (âˆ’ 5)

= 4 + 5 + 5

= 14

**Question 20: Evaluate: âˆ’ 36 âˆ’ 40 + 43 âˆ’ (âˆ’ 29) + 18 âˆ’ (âˆ’ 74).**

**ANSWER: **âˆ’ 36 âˆ’ 40 + 43 âˆ’ (âˆ’ 29) + 18 âˆ’ (âˆ’ 74)

= âˆ’ 36 âˆ’ 40 + 43 âˆ’ (âˆ’ 29) + 18 + 74

= âˆ’ 36 âˆ’ 40 + 43 âˆ’ (âˆ’ 29) + 92

= âˆ’ 36 âˆ’ 40 + 43 + 29 + 92

= âˆ’ 36 âˆ’ 40 + 43 + 121

= âˆ’ 36 âˆ’ 40 + 164

= âˆ’ (36 + 40) + 164

= âˆ’ 76 + 164

= 164 âˆ’ 76

= 88

**Question 21: The largest negative integer is ...............**

**ANSWER:** The numbers ... âˆ’ 5, âˆ’ 4, âˆ’ 3, âˆ’ 2, âˆ’ 1, 0, 1, 2, 3, 4, 5, ... form the set of integers.

Thus, the largest negative integer is âˆ’ 1.

**Question 22: The smallest positive integer is .................**

**ANSWER: **The collection ... âˆ’ 4, âˆ’ 3, âˆ’ 2, âˆ’ 1, 0, 1, 2, 3, 4, ... forms the set of integers.

Thus, the smallest positive integer is 1.

**Question 23: (âˆ’ 22) + 21 + (âˆ’ 22) + 21 + ... + 20 terms is equal to .............**

**ANSWER: **âˆµ (âˆ’ 22) + 21 = âˆ’ (22 âˆ’ 21) = âˆ’ 1

âˆ´ (âˆ’ 22) + 21 + (âˆ’ 22) + 21 + ... + 20 terms = (âˆ’ 1) + (âˆ’ 1) + (âˆ’ 1) + ... + 10 terms

= 10 Ã—Ã— (âˆ’ 1)

= âˆ’ 10

**Question 24: (âˆ’ 3) (âˆ’ 4) (12) (âˆ’ 1) = ............**

**ANSWER:** (âˆ’ 3) (âˆ’ 4) (12) (âˆ’ 1) = (âˆ’ 3) (âˆ’ 4) (âˆ’ 12) [ âˆµ (12) (âˆ’ 1) = (âˆ’ 12)]

= (âˆ’ 3) (48) [âˆµ (âˆ’ 4) (âˆ’ 12) = 48]

= âˆ’ (3 Ã—Ã— 48)

= âˆ’ 144

Hence, (âˆ’ 3) (âˆ’ 4) (12) (âˆ’ 1) = âˆ’ 144.

**Question 25: (âˆ’ 1) (âˆ’ 1) (âˆ’ 1) (âˆ’ 1) (âˆ’ 1) = ...............**

**ANSWER: **Here, âˆ’ 1 is multiplied 5 times and since, 5 is an odd integer, therefore

(âˆ’ 1) (âˆ’ 1) (âˆ’ 1) (âˆ’ 1) (âˆ’ 1) = âˆ’ 1