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. 5858 −1818
Solution:
5858 −1818
= 5−185−18
= 4182
= 12
2. 712 −512
Solution:
712 −512
= 7−512
= 21126
= 16
3. 437 − 247
Solution:
437 − 247
= 317 − 187
= 31−187
= 137
= 167
4. 56 −49
Solution:
56 −49
L.C.M of 6 and 9 = 18
= 15−818
= 718
5. 12 −38
Solution:
12 −38
L.C.M of 2 and 8 = 8
= 4−38
= 18
6. 58 −712
Solution:
58 −712
L.C.M of 8 and 12 = 24
= 15−1424
= 124
7. 279 − 1815
Solution:
279 − 1815
= 259 − 2315
L.C.M of 9 and 15 = 45
= 125−6945
= 5645
= 11145
8. 358 − 2512
Solution:
358 − 2512
= 298 − 2912
L.C.M of 8 and 12 = 24
= 87−5824
= 2924
= 1524
9. 2310 − 1715
Solution:
2310 − 1715
= 2310 − 2215
L.C.M of 10 and 15 = 30
= 69−4430
= 255306
= 56
10. 623 − 334
Solution:
623 − 334
= 203 − 154
L.C.M of 3 and 4 = 12
= 80−4512
= 3512
11. 7 − 523
Solution:
7 − 523
= 71 − 173
L.C.M of 1 and 3 = 3
= 21−173
= 43
= 113
12. 10 − 638
Solution:
10 − 638
= 101 − 518
L.C.M of 1 and 8 = 8
= 80−518
= 298
= 358
13. 56 − 49 + 23
Solution:
56 − 49 + 23
L.C.M of 6, 9 and 3 = 18
= 15−8+1218
= 27−8 18
= 1918
= 1118
14. 58 + 34 − 712
Solution:
58 + 34 − 712
L.C.M of 8, 4 and 12 = 24
= 15+18− 1424
= 33−14 24
= 1924
15. 2 + 1115 − 59
Solution:
2 + 1115 − 59
= 21 + 1115 − 59
L.C.M of 1, 15 and 9 = 45
= 90+33− 2545
= 123−2545
= 9845
= 2845
16. 534 − 4512 + 316
Solution:
534 − 4512 + 316
= 234 − 5312 + 196
L.C.M of 4, 12 and 6 = 12
= 69−53+3812
= 107−5312
= 549122
= 92
= 412
17. 2 + 5710 − 31415
Solution:
2 + 5710 − 31415
= 21 + 5710 − 5915
L.C.M of 1, 10 and 15 = 30
= 60+171− 11830
= 231−11830
= 11330
= 32330
18. 8 − 312 − 214
Solution:
8 − 312 − 214
= 81 − 72 − 94
L.C.M of 1, 2 and 4 = 4
= 32− 14−94
= 32−234
= 94
= 214
19. 856 − 338 + 2712
Solution:
534 − 4512 + 316
= 234 − 5312 + 196
L.C.M of 4, 12 and 6 = 12
= 69−53+3812
= 107−5312
= 549122
= 92
= 412
20. 616 − 515 + 313
Solution:
616 − 515 + 313
= 376 − 265 + 103
L.C.M of 6, 5 and 3 = 30
= 185−156+10030
= 285−15630
= 129433010
= 4310
= 4310
21. 3 + 115 + 23 − 715
Solution:
3 + 115 + 23 − 715
= 31 + 65 + 23 − 715
L.C.M of 1, 5, 3 and 15 = 15
= 45+18+10−715
= 73−715
= 6622155
= 225
= 425
22. What should be added to 923 to get 19?
Solution:
Let the required number be x
Now, according to question
923 + x = 19
⇒ 293 + x = 19
⇒ x = 19 − 293
⇒ x = 191 − 293
⇒ x = 57− 293
⇒ x = 283
⇒ x = 913
Hence, the required number is 913
23. What should be added to 6715 to get 815?
Solution:
Let the required number be x
Now, according to question
6715 + x = 815
⇒ 9715 + x = 415
⇒ x = 415 − 9715
⇒ x = 123− 9715
⇒ x = 2615
⇒ x = 11115
Hence, the required number is 11115
24. Subtract the sum of 359 and 313 from the sum of 556 and 419.
Solution:
(556 + 419) − (359 + 313)
= (356 + 379) − (329 + 103)
= (105+7418) − (32+309)
= 17918 − 629
= 179−12418
= 5518
= 3 118
25. Of 34 and 57, which is greater and by how much ?
Solution:
First we compare 34 and 57
By cross multiply
3 × 7 = 21 and 4 × 5 = 20
Clearly, 21 > 20
∴ 34 > 57
Now, Required difference:
= 34 − 57
L.C.M of 4 and 7 = 28
= 21−2028
= 128
Hence, 34 is greater than 57 by 128
26. Mrs. Soni bought 712 liters of milk. Out of this milk, 534 liters was consumed. How much milk is left with her?
Solution:
Milk bought by Mrs. Soni = 712 liters
Milk consumed by her = 534 litres
Milk left with her = 712 − 534
= 152 − 234
= 30−234
= 74 litres
= 134 litres
27. A film show last for 313 hours. Out of this time, 134 hours are spent on advertisements. What was the actual duration of the film?
Solution:
Total duration of a film show = 313 hours.
Time spent on advertisement = 134 hours
The actual duration of the film = 313 − 134
= 103 − 74
= 40−2112
= 1912
= 1712 hours
28. In one day, a rickshaw puller earned Rs. 13712. Out of this money, he spent Rs. 5634 on food. How much money is left with him?
Solution:
Rickshaw puller earned in a day = Rs. 13712 = Rs.2752
He spent on food = Rs. 5634 = Rs.2274
Money left with him = 2752 − 2274
= 550−2274
= 3234
= Rs. 8034
29. A piece of wire, 234 metres long, broken into two pieces. One piece is 58 metre long. How long is the other piece?
Solution:
Total length of wire = 234m = 114m
Length of one piece of wire = 58m
Length of other piece of wire = 114 − 58
= 22−58
= 178
= 218m
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