RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev

Mathematics (Maths) Class 6

Class 6 : RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev

The document RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev is a part of the Class 6 Course Mathematics (Maths) Class 6.
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Find the L.C.M. of the numbers given below :
Q.1. 42, 63
Ans. We have
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
So, 42 = 2 × 3 × 7
63 = 3 × 3 × 7
= 32 × 7
∴ L.C.M. of 42 and 63 = 2 × 32 × 7
= 2 × 9 × 7
= 18 × 7 = 126

Q.2. 60, 75
Ans. We have
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
So, 60 = 2 × 2 × 3 × 5 = 22 × 3 × 5
75 = 3 × 5 × 5 = 3 × 52
L.C.M. of 60 and 75 = 22 × 3 × 52
= 4 × 3 × 25
= 4 × 75 = 300

Q.3. 12. 18, 20
Ans. We have
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
So, 12 = 2 × 2 × 3 = 22 × 3
18 = 2 × 3 × 3 = 2 × 32
20 = 2 × 2 × 5 = 22 × 5
∴ L.C.M. of 12, 18 and 20 = 22 × 32 × 5
= 4 × 9 × 5
= 20 × 9 = 180

Q.4. 36, 60, 72
Ans. We have
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ 36 = 2 × 2 × 3 × 3 = 22 × 32
60 = 2 × 2 × 3 × 5 = 22 × 3 × 5
72 = 2 × 2 × 2 × 3 × 3 = 23 × 32
∴ L.C.M. of 36, 60 and 72 = 23 × 32 × 5

= 8 × 9 × 5
= 40 × 9 = 360

Q.5. 36, 40, 126
Ans. We have
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
So, 36 = 2 × 2 × 3 × 3 = 22 × 32
40 = 2 × 2 × 2 × 5 = 23 × 5
126 = 2 × 3 × 3 × 7 = 2 × 32 × 7
L.C.M. of 36, 40 and 126
= 2× 32 × 5 × 7
= 8 × 9 × 5 × 7

= 72 × 35 = 2520

Q.6. 16, 28, 40, 77
Ans.
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ L.C.M. of given numbers
= 2 × 2 × 2 × 7 × 2 × 5 × 11

= 8 × 14 × 55
= 112 × 55 = 6160

Q.7. 28, 36, 45, 60
Ans.

RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ L.C.M. of given numbers
= 2 × 2 × 3 × 3 × 5 × 7
= 36 × 35 = 1260

Q.8. 144, 180, 384
Ans.
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev

∴ L.C.M. of given numbers
= 2 × 2 × 2 × 2 × 3 × 3 × 5 × 8

= 16 × 9 × 40 = 144 × 40 = 5760

Q.9. 48, 64, 72, 96, 108
Ans.
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ L.C.M. of given numbers
= 2 × 2 × 2 × 2 × 2 × 3 × 3 × 2 × 3
= 32 × 54 = 1728

Find the H.C.F. and L.C.M. of :
Q.10. 117, 221
Ans. First we find the H.C.F. of the given numbers as under :
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ H.C.F. of 117 and 221 = 13
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev

∴ H.C.F. = 13 and L.C.M. = 1989

Q.11. 234, 572
Ans. First we find the H.C.F. of 234 and 572 as under :
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ H.C.F. of 234 and 572 = 26
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev

Q.12. 693, 1078
Ans. First we find the H.C.F. of 693 and 1078 as under :
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ H.C.F. of 693 and 1078 = 77

RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ H.C.F. = 77 and L.C.M. = 9702

Q.13. 145, 232
Ans. First we find the H.C.F. of 145 and 232 as under :
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ H.C.F. of 145 and 232 = 29
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ H.C.F. = 29 and L.C.M. = 1160

Q.14. 861, 1353
Ans. First we find the H.C.F. of 861 and 1353 as under :

RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ H.C.F. of 861 and 1353 = 123

RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
= 7 × 1353 = 9471

∴ H.C.F. = 123 and L.C.M. = 9471

Q.15. 2923, 3239
Ans. First we find the H.C.F. of 2923 and 3239 as under :
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ H.C.F. of 2923 and 3239 = 79
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
= 37 × 3239 = 119843
∴ H.C.F. = 79 and L.C.M. = 119843

Q.16. For each pair of numbers, verify that their product = (H.C.F. × L.C.M.)
(i) 87, 145
(ii) 186, 403
(iii) 490, 1155
Ans.
(i) We have
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
So, 87 = 3 × 29
145 = 5 × 29
∴ H.C.F. of 87 and 145 = 29
L.C.M. of 87 and 145 = 3 × 5 × 29
= 15 × 29 = 435
Now product of 87 and 145 = 87 × 145

= 12615
Their H.C.F. × L.C.M. = 29 × 435
= 12615
Hence verified.
(ii) We have
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
So, 186 = 2 × 3 × 31
403 = 13 × 31
∴ H.C.F. of 186 and 403 = 31
L.C.M. of 186 and 403
= 2 × 3 × 13 × 31 = 2418
Now product of 186 and 403
= 186 × 403 = 74958
Their H.C.F. × L.C.M. = 31 × 2418
= 74958

Hence verified.
(iii) 490, 1155
We have
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
So, 490 = 2 × 5 × 7 ×7 = 2 × 5 × 72
1155 = 3 × 5 × 7 × 11
H.C.F. of 490 and 1155 = 5 × 7 = 35
L.C.M. of 490 and 1155

= 2 × 3 × 5 × 72 × 11

= 2 × 3 × 5 × 49 × 11
= 30 × 539 = 16170
Now, product of 490 and 1155
= 490 × 1155 = 565950
Their H.C.F. × L.C.M. = 35 × 16170
= 565950
Hence verified.

Q.17. The product of two numbers is 2160 and their H.C.F. is 12. Find their L.C.M.
Ans. We know that
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev

Q.18. The product of two numbers is 2560 and their L.C.M. is 320. Find their H.C.F.
Ans. We know that
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev

Q.19. The H.C.F. of two numbers is 145 and their L.C.M. is 2175. If one of the numbers is 725, find the other.
Ans.
We know that
One number × The other number
= H.C.F. × L.C.M.
∴ The other number
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ Required number = 435

Q.20. The H.C.F. and L.C.M. of two numbers are 131 and 8253 respectively. If one of the numbers is 917, find the other.
Ans.
We know that
One number × The other number
= H.C.F. × L.C.M.
∴ The other number

RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ Required number = 1179

Q.21. Find the least number divisible by 15, 20, 24, 32 and 36.
Ans. Required least number = L.C.M. of 15, 20, 24, 32 and 36
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ L.C.M. = 3 × 2 × 2 × 2 × 5 × 4 × 3
= 24 × 60 = 1440
Hence, required least number = 1440

Q.22. Find the least number which when divided by 25, 40 and 60 leaves 9 as the remainder in each case.
Ans. Clearly, required least number = (L.C.M. of the given numbers + 9)
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ L.C.M. of the given numbers
= 4 × 5 × 5 × 2 × 3
= 600
∴ Required least number = 600 + 9
= 609

Q.23. Find the least number of five digits that is exactly divisible by 16, 18, 24 and 30.
Ans.
First we find the L.C.M. of the given numbers as under :
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ L.C.M. of the given numbers
= 2 × 2 × 2 × 3 × 2 × 3 × 5
= 24 × 30 = 720
Now least number of five digits = 10000
Dividing 10000 by 720, we get
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
Clearly if we add 80 to 640, it will become 720 which is exactly divisible by 720.
∴ Required least number of five digits
= 10000 + 80 = 10080

Q. 24. Find the greatest number of five digits exactly divisible by 9, 12, 15, 18 and 24.
Ans. The greatest number of five digits exactly divisible by the given numbers = The greatest number of five digits exactly divisible by the L.C.M. of given numbers.
Now
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ L.C.M. of given numbers
= 2 × 2 × 3 × 3 × 5 × 2
= 360
Now greatest number of five digits
= 99999
Dividing 99999 by 360, we get
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ Required greatest number of five digits
= 99999 – 279
= 99720

Q.25. Three bells toll at intervals of 9, 12, 15 minutes respectively. If they start tolling together, after what time will they next toll together ?
Ans.
Three bells will again toll together after an interval of time which is exactly divisible by 9, 12, 15 minutes.
∴ Required time = L.C.M. of 9, 12, 15 minutes
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ L.C.M. of 9, 12, 15 minutes
= 3 × 3 × 4 × 5 minutes
= 9 × 20 minutes
= 180 minutes
∴ Required time = 180 minutes
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev

Q. 26. Three boys step off together from the same place. If their steps measure 36 cm, 48 cm and 54 cm, at what distance from the starting point will they again step together ?
Ans. Required distance = L.C.M. of 36 cm, 48 cm and 54 cm
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ L.C.M. of 36 cm, 48 cm, 54 cm
= 2 × 2 × 3 × 3 × 4 × 3 cm
= 36 × 12 cm
= 432 cm
= 4 m 32 cm
∴ Required distance = 4 m 32 cm.

Q. 27. The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 secondsrespectively. If they start changing simultaneously at 8 a.m., after how much time will they change again simultaneously ?
Ans. Required time = L.C.M. of 48 seconds, 72 seconds and 108 seconds
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ L.C.M. of 48 sec., 72 sec. and 108 sec.

= 2 × 2 × 2 × 3 × 3 × 2 × 3 sec.
= 24 × 18 sec.
= 432 sec.
∴ Required time = 432 sec.
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev

Q.28. Three measuring rods are 45 cm, 50 cm and 75 cm in length. What is the least length (in metres) of a rope that can be measured by the full length of each of these three rods ?
Ans.
Lengths of three rods = 45 cm, 50 cm and 75 cm
Required least length of the rope = L.C.M. of 45 cm, 50 cm, 75 cm
We have
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ L.C.M. of 45 cm, 50 cm and 75 cm
= 3 × 5 × 5 × 3 × 2 cm
= 15 × 30 cm
= 450 cm
∴ Required length of the rope = 450 cm
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev

Q.29. An electronic device makes a beep after every 15 minutes. Another device makes a beep after every 20 minutes. They beeped together at 6 a.m. At what time will they next beep together ?
Ans.
The time after which both the devices will beep together = L.C.M. of 15 minutes and 20 minutes
Now,
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ L.C.M. of 15 minutes and 20 minutes
= 5 × 3 × 4
= 60 minutes
= 1 hour
∴ Both the devices will beep together after 1 hour from 6 a.m.
∴  Required time = 6 + 1
= 7 a.m.

Q.30. The circumferences of four wheels are 50 cm, 60 cm, 75 cm and 100 cm respectively. They start moving simultaneously. What least distance should they cover so that each wheel makes a complete number of revolutions ?
Ans.
The circumferences of four wheels = 50 cm, 60 cm, 75 cm and 100 cm
Required least distance = L.C.M. of 50 cm, 60 cm, 75 cm and 100 cm

Now,
RS Aggarwal Solutions: Factors and Multiples Exercise - 2E Notes | EduRev
∴ L.C.M. of 50 cm, 60 cm, 75 cm, 100 cm
= 2 × 2 × 3 × 5 × 5 cm
= 300 cm = 3 m
∴ Required least distance = 3 m.

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