RS Aggarwal 2021-2022 for Class 6 Maths Chapter 6- Simplification
RS Aggarwal Class 6 Math Solution Chapter 6- Simplification Exercise 6A 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 6- Simplification
Exercise 6A
1. 21 − 12 ÷ 3 × 2
Solution:
21 – 12 ÷ 3 × 2
= 21 – 4 × 2
= 21 – 8
= 13. Ans
2. 16 + 8 ÷ 4 − 2 × 3
Solution:
16 + 8 ÷ 4 – 2 × 3
= 16 + 2 – 2 × 3
= 16 + 2 – 6
= 18 – 6
= 12. Ans.
3. 13 − (12 − 6 ÷ 3 )
Solution:
13 – (12 – 6 ÷ 3)
= 13 – (12 – 2)
= 13 – (10)
= 13 – 10
= 3 Ans.
4. 19 − [4 + {16 − (12 − 2)}]
Solution:
19 – [4 + {16 – (12 – 2)}]
= 19 – [4 + {16 – 10}]
= 19 – [4 + 6]
= 19 – 10
= 9. Ans
5. 36 − [18 − {14 − (15 − 4 ÷ 2 × 2)}]
Solution:
36 – [18 – {14 – (15 – 4 ÷ 2 × 2)}]
= 36 – [18 – {14 – (15 – 2 × 2)}]
= 36 – [18 – {14 – (15 – 4)}]
= 36 – [18 – {14 – 11}]
= 36 – [18 – 3]
= 36 – 15
= 21. Ans.
6. 27 − [18 − {16 − (5 − `bar(4 − 1)`}]
Solution:
27 − [18 − {16 − (5 − `bar(4 − 1)`}]
= 27 – [18 – {16 – (5 – 3)}]
= 27 – [18 – {16 – 2}]
= 27 – [18 – 14]
= 27 – 4
= 23. Ans.
7. 4`frac\{4}{5}` ÷ `frac\{3}{5}` of 5 + `frac\{4}{5}` × `frac\{3}{10}` − `frac\{1}{5}`
Solution:
4`frac\{4}{5}` ÷ `frac\{3}{5}` of 5 + `frac\{4}{5}` × `frac\{3}{10}` − `frac\{1}{5}`
= `frac\{24}{5}` ÷ `frac\{3}{cancel 5}` of `cancel 5` + `frac\{4}{5}` × `frac\{3}{10}` − `frac\{1}{5}`
= `frac\{24}{5}` ÷`frac\{3}{1}` + `frac\{4}{5}` × `frac\{3}{10}` − `frac\{1}{5}`
= `frac\{cancel 24^8}{5}` × `frac\{1}{cancel 3}` + `frac\{cancel 4^2}{5}` × `frac\{3}{cancel 10^5}` − `frac\{1}{5}`
= `frac\{8}{5}` + `frac\{6}{25}` − `frac\{1}{5}`
= `frac\{40 + 6 − 5}{25}`
= `frac\{46 − 5}{25}`
= `frac\{41}{25}`
= 1`frac\{16}{25}`
8. `(frac\{2}{3} + frac\{4}{9})` of `frac\{3}{5}` ÷ 1`frac\{2}{3}` × 1`frac\{1}{4}` − `frac\{1}{3})`
Solution:
`(frac\{2}{3} + frac\{4}{9})` of `frac\{3}{5}` ÷ 1`frac\{2}{3}` × 1`frac\{1}{4}` − `frac\{1}{3}`
= `(frac\{2}{3} + frac\{4}{9})` of `frac\{3}{5}` ÷ `frac\{5}{3}` × `frac\{5}{4}` − `frac\{1}{3}`
= `(frac\{6 + 4}{9})` of `frac\{3}{5}` ÷ `frac\{5}{3}` × `frac\{5}{4}` − `frac\{1}{3}`
= `frac\{10}{9}` of `frac\{3}{5}` ÷ `frac\{5}{3}` × `frac\{5}{4}` − `frac\{1}{3}`
= `frac\{cancel 10^2}{cancel 9^3}` of `frac\{cancel 3}{cancel 5}` ÷ `frac\{5}{3}` × `frac\{5}{4}` − `frac\{1}{3}`
= `frac\{cancel 2}{cancel 3}` × `frac\{cancel 3}{cancel 5}` × `frac\{cancel 5}{cancel 4^2}` − `frac\{1}{3}`
= `frac\{1}{2}` − `frac\{1}{3}`
= `frac\{3 − 2}{6}`
= `frac\{1}{6}`
9. 7`frac\{1}{3}` ÷ `frac\{2}{3}` of 2`frac\{1}{5}` + 1`frac\{3}{8}` ÷ 2`frac\{3}{4}` − 1`frac\{1}{2}`
Solution:
7`frac\{1}{3}` ÷ `frac\{2}{3}` of 2`frac\{1}{5}` + 1`frac\{3}{8}` ÷ 2`frac\{3}{4}` − 1`frac\{1}{2}`
= `frac\{22}{3}` ÷ `frac\{2}{3}` of `frac\{11}{5}` + `frac\{11}{8}` ÷ `frac\{11}{4}` − `frac\{3}{2}`
= `frac\{22}{3}` ÷ `frac\{22}{15}` + `frac\{11}{8}` ÷ `frac\{11}{4}` − `frac\{3}{2}`
= `frac\{22}{3}` × `frac\{15}{22}` + `frac\{11}{8}` × `frac\{4}{11}` − `frac\{2}{2}`
= `frac\{cancel 22}{cancel 3}` × `frac\{cancel 15^5}{cancel 22}` + `frac\{cancel 11}{cancel 8^2}` × `frac\{cancel 4}{cancel 11}` − `frac\{3}{2}`
= `frac\{5}{1}` + `frac\{1}{2}` − `frac\{3}{2}`
= `frac\{10 + 1 − 3}{2}`
= `frac\{11 − 3}{2}`
= `frac\{cancel 8^4}{cancel 2}`
= 4
10. 5`frac\{1}{7}` − `{3 frac\{3}{10}` ÷ `(2 frac\{4}{5} − frac\{7}{10})}`
Solution:
5`frac\{1}{7}` − `{3 frac\{3}{10}` ÷ `(2 frac\{4}{5} − frac\{7}{10})}`
= `frac\{36}{7}` − `{frac\{33}{10}` ÷ `( frac\{14}{5} − frac\{7}{10})}`
= `frac\{36}{7}` − `{frac\{33}{10}` ÷ `( frac\{28 − 7}{10})}`
= `frac\{36}{7}` − `{frac\{33}{10}` ÷ `frac\{21}{10}}`
= `frac\{36}{7}` − `{frac\{cancel 33^11}{cancel 10}` × `frac\{cancel 10}{cancel 21^7}}`
= `frac\{36}{7}` − `frac\{11}{7}`
= `frac\{36 − 11 }{7}`
= `frac\{25}{7}`
= 3`frac\{4}{7}`
11. 9`frac\{3}{4}` ÷ `[2 frac\{1}{6}` + `{4 frac\{1}{3} − (1 frac\{1}{2} + 1 frac\{3}{4})}]`
Solution:
`frac\{39}{4}` ÷ `[frac\{13}{6}` + `{ frac\{13}{3} − ( frac\{3}{2} + frac\{7}{4})}]`
= `frac\{39}{4}` ÷ `[frac\{13}{6}` + `{ frac\{13}{3} − ( frac\{6 + 7}{4})}]`
= `frac\{39}{4}` ÷ `[frac\{13}{6}` + `{ frac\{13}{3} − frac\{13}{4}}]`
= `frac\{39}{4}` ÷ `[frac\{13}{6}` + `{ frac\{52 − 39}{12}}]`
= `frac\{39}{4}` ÷ `[frac\{13}{6}` + `frac\{13}{12}]`
= `frac\{39}{4}` ÷ `[frac\{26 + 13}{12}]`
= `frac\{39}{4}` ÷ `frac\{39}{12}`
= `frac\{cancel 39}{cancel 4}` × `frac\{cancel 12^3}{cancel 39}`
= 3
12. 4`frac\{1}{10}` − `[2 frac\{1}{2}` −`{frac\{5}{6} − (frac\{2}{5} + frac\{3}{10} − frac{4}{15})}]`
Solution:
4`frac\{1}{10}` − `[2 frac\{1}{2}` −`{frac\{5}{6} − (frac\{2}{5} + frac\{3}{10} − frac{4}{15})}]`
= `frac\{41}{10}` − `[ frac\{5}{2}` −`{frac\{5}{6} − (frac\{2}{5} + frac\{3}{10} − frac{4}{15})}]`
= `frac\{41}{10}` − `[ frac\{5}{2}` −`{frac\{5}{6} − (frac\{12 + 9 − 8}{30})}]`
= `frac\{41}{10}` − `[ frac\{5}{2}` −`{frac\{5}{6} − (frac\{21 − 8}{30})}]`
= `frac\{41}{10}` − `[ frac\{5}{2}` −`{frac\{5}{6} − frac\{13}{30}}]`
= `frac\{41}{10}` − `[ frac\{5}{2}` −`{frac\{25 − 13}{30}}]`
= `frac\{41}{10}` − `[ frac\{5}{2}` − `frac\{12}{30}]`
= `frac\{41}{10}` − `[ frac\{75 − 12}{30}]`
= `frac\{41}{10}` − `frac\{63}{30}`
= `frac\{123 − 63}{30}`
= `frac\{cancel 60^2}{cancel 30}`
= 2
13. 1`frac\{5}{6}` + `[2 frac\{2}{3}` − `{3 frac\{3}{4} (3 frac\{4}{5} ÷ 9 frac\{1}{2} )}]`
Solution:
1`frac\{5}{6}` + `[2 frac\{2}{3}` − `{3 frac\{3}{4} (3 frac\{4}{5} ÷ 9 frac\{1}{2} )}]`
= `frac\{11}{6}` + `[ frac\{8}{3}` − `{ frac\{15}{4} ( frac\{19}{5} ÷ frac\{19}{2} )}]`
= `frac\{11}{6}` + `[ frac\{8}{3}` − `{ frac\{15}{4} ( frac\cancel {19}{5} × frac\{2}{cancel 19} )}]`
= `frac\{11}{6}` + `[ frac\{8}{3}` − `{ frac\{cancel 15^3}{cancel 4^2} × frac\{cancel 2}{cancel 5}}]`
= `frac\{11}{6}` + `[ frac\{8}{3}` − `frac\{3}{2}]`
= `frac\{11}{6}` + `[ frac\{16 − 9}{6}]`
= `frac\{11}{6}` + `frac\{7}{6}`
= `frac\{11 + 7}{6}`
= `frac\{cancel 18^3}{cancel 6}`
= 3
14. 4`frac\{4}{5}` ÷ `{2 frac\{1}{5}` − `frac\{1}{2} (1 frac\{1}{4} − bar (frac\{1}{4} − frac\{1}{5}))}`
Solution:
`frac\{24}{5}` ÷ `{ frac\{11}{5}` − `frac\{1}{2} ( frac\{5}{4} − bar (frac\{1}{4} − frac\{1}{5}))}`
= `frac\{24}{5}` ÷ `{ frac\{11}{5}` − `frac\{1}{2} ( frac\{5}{4} − frac\{5 − 4}{20})}`
= `frac\{24}{5}` ÷ `{ frac\{11}{5}` − `frac\{1}{2} ( frac\{5}{4} − frac\{1}{20})}`
= `frac\{24}{5}` ÷ `{ frac\{11}{5}` − `frac\{1}{2} ( frac\{25 − 1}{20})}`
= `frac\{24}{5}` ÷ `{ frac\{11}{5}` − `frac\{1}{2} × frac\{cancel 24^6}{cancel 20^5}}`
= `frac\{24}{5}` ÷ `{ frac\{11}{5}` − `frac\{1}{cancel 2} × frac\{cancel 6^3}{5}}`
= `frac\{24}{5}` ÷ `{ frac\{11}{5}` − `frac\{3}{5}}`
= `frac\{24}{5}` ÷ `{ frac\{11 − 3}{5}}`
= `frac\{24}{5}` ÷ ` frac\{8}{5}`
= `frac\{cancel 24^3}{cancel 5}` × ` frac\{cancel 5}{cancel 8}`
= 3
15. 7`frac\{1}{2}` − `[2 frac\{1}{4}` ÷ `{1 frac\{1}{4}` − `frac\{1}{2} ( frac\{3}{2} − bar (frac\{1}{3} − frac\{1}{6}))}]`
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