RS Aggarwal 2021-2022 for Class 6 Maths Chapter 5- Fractions
RS Aggarwal Class 6 Math Solution Chapter 5- Fractions Exercise 5F 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 5- Fractions
Exercise 5F
Find the difference:
1. `frac\{5}{8}` −`frac\{1}{8}`
Solution:
`frac\{5}{8}` −`frac\{1}{8}`
= `frac\{5 − 1}{8}`
= `frac\{cancel 4^1}{cancel 8^2}`
= `frac\{1}{2}`
2. `frac\{7}{12}` −`frac\{5}{12}`
Solution:
`frac\{7}{12}` −`frac\{5}{12}`
= `frac\{7 − 5}{12}`
= `frac\{cancel 2^1}{cancel 12^6}`
= `frac\{1}{6}`
3. 4`frac\{3}{7}` − 2`frac\{4}{7}`
Solution:
4`frac\{3}{7}` − 2`frac\{4}{7}`
= `frac\{31}{7}` − `frac\{18}{7}`
= `frac\{31 − 18}{7}`
= `frac\{13}{7}`
= 1`frac\{6}{7}`
4. `frac\{5}{6}` −`frac\{4}{9}`
Solution:
`frac\{5}{6}` −`frac\{4}{9}`
L.C.M of 6 and 9 = 18
= `frac\{15 − 8}{18}`
= `frac\{7}{18}`
5. `frac\{1}{2}` −`frac\{3}{8}`
Solution:
`frac\{1}{2}` −`frac\{3}{8}`
L.C.M of 2 and 8 = 8
= `frac\{4 − 3}{8}`
= `frac\{1}{8}`
6. `frac\{5}{8}` −`frac\{7}{12}`
Solution:
`frac\{5}{8}` −`frac\{7}{12}`
L.C.M of 8 and 12 = 24
= `frac\{15 − 14}{24}`
= `frac\{1}{24}`
7. 2`frac\{7}{9}` − 1`frac\{8}{15}`
Solution:
2`frac\{7}{9}` − 1`frac\{8}{15}`
= `frac\{25}{9}` − `frac\{23}{15}`
L.C.M of 9 and 15 = 45
= `frac\{125 − 69}{45}`
= `frac\{56}{45}`
= 1`frac\{11}{45}`
8. 3`frac\{5}{8}` − 2`frac\{5}{12}`
Solution:
3`frac\{5}{8}` − 2`frac\{5}{12}`
= `frac\{29}{8}` − `frac\{29}{12}`
L.C.M of 8 and 12 = 24
= `frac\{87 − 58}{24}`
= `frac\{29}{24}`
= 1`frac\{5}{24}`
9. 2`frac\{3}{10}` − 1`frac\{7}{15}`
Solution:
2`frac\{3}{10}` − 1`frac\{7}{15}`
= `frac\{23}{10}` − `frac\{22}{15}`
L.C.M of 10 and 15 = 30
= `frac\{69 − 44}{30}`
= `frac\{cancel 25^5}{cancel 30^6}`
= `frac\{5}{6}`
10. 6`frac\{2}{3}` − 3`frac\{3}{4}`
Solution:
6`frac\{2}{3}` − 3`frac\{3}{4}`
= `frac\{20}{3}` − `frac\{15}{4}`
L.C.M of 3 and 4 = 12
= `frac\{80 − 45}{12}`
= `frac\{35}{12}`
11. 7 − 5`frac\{2}{3}`
Solution:
7 − 5`frac\{2}{3}`
= `frac\{7}{1}` − `frac\{17}{3}`
L.C.M of 1 and 3 = 3
= `frac\{21 − 17}{3}`
= `frac\{4}{3}`
= 1`frac\{1}{3}`
12. 10 − 6`frac\{3}{8}`
Solution:
10 − 6`frac\{3}{8}`
= `frac\{10}{1}` − `frac\{51}{8}`
L.C.M of 1 and 8 = 8
= `frac\{80 − 51}{8}`
= `frac\{29}{8}`
= 3`frac\{5}{8}`
13. `frac\{5}{6}` − `frac\{4}{9}` + `frac\{2}{3}`
Solution:
`frac\{5}{6}` − `frac\{4}{9}` + `frac\{2}{3}`
L.C.M of 6, 9 and 3 = 18
= `frac\{15 − 8 + 12 }{18}`
= `frac\{27 − 8 }{18}`
= `frac\{19}{18}`
= 1`frac\{1}{18}`
14. `frac\{5}{8}` + `frac\{3}{4}` − `frac\{7}{12}`
Solution:
`frac\{5}{8}` + `frac\{3}{4}` − `frac\{7}{12}`
L.C.M of 8, 4 and 12 = 24
= `frac\{15 + 18 − 14}{24}`
= `frac\{33 − 14 }{24}`
= `frac\{19}{24}`
15. 2 + `frac\{11}{15}` − `frac\{5}{9}`
Solution:
2 + `frac\{11}{15}` − `frac\{5}{9}`
= `frac\{2}{1}` + `frac\{11}{15}` − `frac\{5}{9}`
L.C.M of 1, 15 and 9 = 45
= `frac\{90 + 33 − 25}{45}`
= `frac\{123 − 25}{45}`
= `frac\{98}{45}`
= 2`frac\{8}{45}`
16. 5`frac\{3}{4}` − 4`frac\{5}{12}` + 3`frac\{1}{6}`
Solution:
5`frac\{3}{4}` − 4`frac\{5}{12}` + 3`frac\{1}{6}`
= `frac\{23}{4}` − `frac\{53}{12}` + `frac\{19}{6}`
L.C.M of 4, 12 and 6 = 12
= `frac\{69 − 53 + 38}{12}`
= `frac\{107 − 53}{12}`
= `frac\{cancel 54^9}{cancel 12^2}`
= `frac\{9}{2}`
= 4`frac\{1}{2}`
17. 2 + 5`frac\{7}{10}` − 3`frac\{14}{15}`
Solution:
2 + 5`frac\{7}{10}` − 3`frac\{14}{15}`
= `frac\{2}{1}` + `frac\{57}{10}` − `frac\{59}{15}`
L.C.M of 1, 10 and 15 = 30
= `frac\{60 + 171 − 118}{30}`
= `frac\{231 − 118}{30}`
= `frac\{113}{30}`
= 3`frac\{23}{30}`
18. 8 − 3`frac\{1}{2}` − 2`frac\{1}{4}`
Solution:
8 − 3`frac\{1}{2}` − 2`frac\{1}{4}`
= `frac\{8}{1}` − `frac\{7}{2}` − `frac\{9}{4}`
L.C.M of 1, 2 and 4 = 4
= `frac\{32 − 14 − 9}{4}`
= `frac\{32 − 23}{4}`
= `frac\{9}{4}`
= 2`frac\{1}{4}`
19. 8`frac\{5}{6}` − 3`frac\{3}{8}` + 2`frac\{7}{12}`
Solution:
5`frac\{3}{4}` − 4`frac\{5}{12}` + 3`frac\{1}{6}`
= `frac\{23}{4}` − `frac\{53}{12}` + `frac\{19}{6}`
L.C.M of 4, 12 and 6 = 12
= `frac\{69 − 53 + 38}{12}`
= `frac\{107 − 53}{12}`
= `frac\{cancel 54^9}{cancel 12^2}`
= `frac\{9}{2}`
= 4`frac\{1}{2}`
20. 6`frac\{1}{6}` − 5`frac\{1}{5}` + 3`frac\{1}{3}`
Solution:
6`frac\{1}{6}` − 5`frac\{1}{5}` + 3`frac\{1}{3}`
= `frac\{37}{6}` − `frac\{26}{5}` + `frac\{10}{3}`
L.C.M of 6, 5 and 3 = 30
= `frac\{185 − 156 + 100}{30}`
= `frac\{285 − 156}{30}`
= `frac\{cancel 129^43}{cancel 30^10}`
= `frac\{43}{10}`
= 4`frac\{3}{10}`
21. 3 + 1`frac\{1}{5}` + `frac\{2}{3}` − `frac\{7}{15}`
Solution:
3 + 1`frac\{1}{5}` + `frac\{2}{3}` − `frac\{7}{15}`
= `frac\{3}{1}` + `frac\{6}{5}` + `frac\{2}{3}` − `frac\{7}{15}`
L.C.M of 1, 5, 3 and 15 = 15
= `frac\{45 + 18 + 10 − 7}{15}`
= `frac\{73 − 7}{15}`
= `frac\{cancel 66^22}{cancel 15^5}`
= `frac\{22}{5}`
= 4`frac\{2}{5}`
22. What should be added to 9`frac\{2}{3}` to get 19?
Solution:
Let the required number be x
Now, according to question
9`frac\{2}{3}` + x = 19
⇒ `frac\{29}{3}` + x = 19
⇒ x = 19 − `frac\{29}{3}`
⇒ x = `frac\{19}{1}` − `frac\{29}{3}`
⇒ x = `frac\{57 − 29}{3}`
⇒ x = `frac\{28}{3}`
⇒ x = 9`frac\{1}{3}`
Hence, the required number is 9`frac\{1}{3}`
23. What should be added to 6`frac\{7}{15}` to get 8`frac\{1}{5}`?
Solution:
Let the required number be x
Now, according to question
6`frac\{7}{15}` + x = 8`frac\{1}{5}`
⇒ `frac\{97}{15}` + x = `frac\{41}{5}`
⇒ x = `frac\{41}{5}` − `frac\{97}{15}`
⇒ x = `frac\{123 − 97}{15}`
⇒ x = `frac\{26}{15}`
⇒ x = 1`frac\{11}{15}`
Hence, the required number is 1`frac\{11}{15}`
24. Subtract the sum of 3`frac\{5}{9}` and 3`frac\{1}{3}` from the sum of 5`frac\{5}{6}` and 4`frac\{1}{9}`.
Solution:
`(5 frac\{5}{6}` + 4`frac\{1}{9})` − `(3 frac\{5}{9}` + 3`frac\{1}{3})`
= `(frac\{35}{6}` + `frac\{37}{9})` − `(frac\{32}{9}` + `frac\{10}{3})`
= `(frac\{105 + 74}{18})` − `(frac\{32 + 30}{9})`
= `frac\{179}{18}` − `frac\{62}{9}`
= `frac\{179 − 124}{18}`
= `frac\{55}{18}`
= 3 `frac\{1}{18}`
25. Of `frac\{3}{4}` and `frac\{5}{7}`, which is greater and by how much ?
Solution:
First we compare `frac\{3}{4}` and `frac\{5}{7}`
By cross multiply
3 × 7 = 21 and 4 × 5 = 20
Clearly, 21 > 20
∴ `frac\{3}{4}` > `frac\{5}{7}`
Now, Required difference:
= `frac\{3}{4}` − `frac\{5}{7}`
L.C.M of 4 and 7 = 28
= `frac\{21 − 20 }{28}`
= `frac\{1}{28}`
Hence, `frac\{3}{4}` is greater than `frac\{5}{7}` by `frac\{1}{28}`
26. Mrs. Soni bought 7`frac\{1}{2}` liters of milk. Out of this milk, 5`frac\{3}{4}` liters was consumed. How much milk is left with her?
Solution:
Milk bought by Mrs. Soni = 7`frac\{1}{2}` liters
Milk consumed by her = 5`frac\{3}{4}` litres
Milk left with her = 7`frac\{1}{2}` − 5`frac\{3}{4}`
= `frac\{15}{2}` − `frac\{23}{4}`
= `frac\{30 − 23}{4}`
= `frac\{7}{4}` litres
= 1`frac\{3}{4}` litres
27. A film show last for 3`frac\{1}{3}` hours. Out of this time, 1`frac\{3}{4}` hours are spent on advertisements. What was the actual duration of the film?
Solution:
Total duration of a film show = 3`frac\{1}{3}` hours.
Time spent on advertisement = 1`frac\{3}{4}` hours
The actual duration of the film = 3`frac\{1}{3}` − 1`frac\{3}{4}`
= `frac\{10}{3}` − `frac\{7}{4}`
= `frac\{40 − 21}{12}`
= `frac\{19}{12}`
= 1`frac\{7}{12}` hours
28. In one day, a rickshaw puller earned Rs. 137`frac\{1}{2}`. Out of this money, he spent Rs. 56`frac\{3}{4}` on food. How much money is left with him?
Solution:
Rickshaw puller earned in a day = Rs. 137`frac\{1}{2}` = Rs.`frac\{275}{2}`
He spent on food = Rs. 56`frac\{3}{4}` = Rs.`frac\{227}{4}`
Money left with him = `frac\{275}{2}` − `frac\{227}{4}`
= `frac\{550 − 227}{4}`
= `frac\{323}{4}`
= Rs. 80`frac\{3}{4}`
29. A piece of wire, 2`frac\{3}{4}` metres long, broken into two pieces. One piece is `frac\{5}{8}` metre long. How long is the other piece?
Solution:
Total length of wire = 2`frac\{3}{4}`m = `frac\{11}{4}`m
Length of one piece of wire = `frac\{5}{8}`m
Length of other piece of wire = `frac\{11}{4}` − `frac\{5}{8}`
= `frac\{22 − 5}{8}`
= `frac\{17}{8}`
= 2`frac\{1}{8}`m
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