RS Aggarwal 2021-2022 for Class 6 Maths Chapter 4 - Integer
RS Aggarwal Class 6 Math Solution Chapter 4- Integer Exercise 4E is available here. RS Aggarwal Class 6 Math Solutions are solved by expert teachers in step by step, which help the students to understand easily. RS Aggarwal textbooks are responsible for a strong foundation in Maths. These textbook solutions help students in exams as well as their daily homework routine. The solutions included are easy to understand, and each step in the solution is described to match the students’ understanding.
Rs Aggarwal Class 6 Math Solution Chapter 4- Integer
Exercise 4E
1: Divide
(i) 85 by –17
(ii) – 72 by 18
(iii) – 80 by 16
(iv) – 121 by 11
(v) 108 by –12
(vi) – 161 by 23
(vii) – 76 by – 19
(viii) 147 by – 21
(ix) – 639 by – 71
(x) – 15625 by – 125
(xi) 2067 by – 1
(xii) 1765 by 1765
(xiii) 0 by – 278
(xiv) 3000 by – 100
Solution:
(i) 85 ÷ ( –17) = `\frac\{85}{–17}` = –5
(ii) –72 ÷ 18 = `\frac\{–72}{18}` = –4
(iii) –80 ÷ 16 = `\frac\{–80}{16}` = –5
(iv) –121 ÷ 11 = `\frac\{–121}{11}` = –11
(v) 108 ÷ (–12) = `\frac\{108}{–12}` = –9
(vi) –161 ÷ 23 = `\frac\{–161}{23}` = –7
(vii) –76 ÷ (–19) = `\frac\{–76}{–19}` = 4
(viii) 147 ÷ (–21) = `\frac\{147}{–21}` = –7
(ix) –639 ÷ (–71) = `\frac\{–639}{–71}` = 9
(x) –15625 ÷ (–125) = `\frac\{–15625}{–125}` = 125
(xi) 2067 ÷ (–1) = `\frac\{2067}{–1}` = –2067
(xii) 1765 ÷ 1765 = `\frac\{1765}{1765}` = 1
(xiii) 0 ÷ (–278) = `\frac\{0}{–278}` = 0
(xiv) 3000 ÷ (–100) = `\frac\{3000}{–100}` = –30
2: Fill in the blanks -
(i) 80 ÷ (…..) = – 5
(ii) – 84 ÷ (…..) = – 7
(iii) (….) ÷ ( – 5) = 25
(iv) (……) ÷ 372 = 0
(v) (….) ÷ 1 = – 186
(vi) (…..) ÷ 17 = – 2
(vii) (….) ÷ 165 = – 1
(viii) (….) ÷ ( – 1) = 73
(ix) 1 ÷ (…..) = – 1
Solution:
(i) 80 ÷ (…..) = – 5
Let 80 ÷ Z = – 5
then, Z = 80 ÷ ( – 5) = – 16
∴ 80 ÷ ( – 16) = – 5
(ii) – 84 ÷ (…..) = – 7
Let – 84 ÷ Z = – 7
then, Z = (−84) ÷ (−7) = 12
∴ – 84 ÷ 12 = – 7
(iii) (….) ÷ ( – 5) = 25
Let Z ÷ ( – 5) = 25
Z = 25 x ( – 5) = – 125
∴ ( – 125) ÷ ( – 5) = 25
(iv) (……) ÷ 372 = 0
Let Z ÷ 372 = 0
Then, Z = 0 x 372 = 0
∴ (0) ÷ 372 = 0
(v) (….) ÷ 1 = – 186
Let Z ÷ 1 = – 186
Then, Z = – 186 x 1 = – 186
∴ ( – 186) ÷ 1 = – 186
(vi) (…..) ÷ 17 = – 2
Let Z ÷ 17 = – 2
Then, Z = – 2 x 17 = – 34
∴ ( – 34) ÷ 17 = – 2
(vii) (….) ÷ 165 = – 1
Let Z ÷ 165 = – 1
Then, Z = – 1 x 165 = – 165
∴ ( – 165) ÷ 165 = – 1
(viii) (….) ÷ ( – 1) = 73
Let Z ÷ ( – 1) = 73
Then, Z = 73 x ( – 1) = – 73
∴ ( – 73) ÷ ( – 1) = 73
(ix) 1 ÷ (…..) = – 1
Let 1 ÷ Z = – 1
Then, Z = 1 ÷ (– 1) = – 1
∴ 1 ÷ ( – 1) = – 1 Ans.
3: Write (T) for true and (F) for false for each of the following statements
(i) 0 ÷ (– 6) = 0
(ii) ( – 8) ÷ 0 = 0
(iii) 15 ÷ (– 1)= – 15
(iv) (– 16) ÷ (– 4) = – 4
(v) (– 7) ÷ (– 1) = – 7
(vi) (– 18) ÷ 9 = – 2
(vii) 20 ÷ (– 5) = – 4
(viii) (– 10) ÷ 1 = – 10
(ix) (– 1) ÷ (– 1) = – 1
Solution:
(i) True: as if zero is divided by any non-zero integer, then quotient is always zero
(ii) False: As division by zero is not admissible
(iii) True: As dividing by one integer by another having opposite signs in negative.
(iv) False: As dividing one integer by another having the same signs is positive not negative.
(v) True: As dividing one istager by another with same sign is always positive
(vi) True: As dividing one integer by another having opposite signs is always negative.
(vii) True: As dividing one integer by another having opposite signs is always negative
(vii) True: As dividing one integer by another having opposite signs is always negative.
(ix) False: As dividing one integer by another having same signs is always positive not negative
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