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02 HCF and LCM Solved Examples
Ex. 1. Find the H.C.F. of 23 X 32 X 5 X 74, 22 X 35 X 52 X 73,23 X 53 X 72
Sol. The prime numbers common to given numbers are 2,5 and 7.
H.C.F. = 22 x 5 x72 = 980.
Ex. 2. Find the H.C.F. of 108, 288 and 360.
Sol. 108 = 22 x 33, 288 = 25 x 32 and 360 = 23 x 5 x 32.
H.C.F. = 22 x 32 = 36.
Ex. 3. Find the H.C.F. of 513, 1134 and 1215.
Sol.
1134 ) 1215 ( 1
1134
81 ) 1134 ( 14
81
324
324
x
\H.C.F. of 1134 and 1215 is 81.
So, Required H.C.F. = H.C.F. of 513 and 81.
______
81 ) 513 ( 6
__486____
27) 81 ( 3
81
0
H.C.F. of given numbers = 27.
Ex. 4. Reduce 391 to lowest terms .
667
to lowest terms.
Sol. H.C.F. of 391 and 667 is 23.
On dividing the numerator and denominator by 23, we get :
391 = 391 ¸ 23 = 17
667 667¸ 23 29
Ex.5.Find the L.C.M. of 22 x 33 x 5 x 72 , 23 x 32 x 52 x 74 , 2 x 3 x 53 x 7 x 11.
Sol. L.C.M. = Product of highest powers of 2, 3, 5, 7 and 11 = 23 x 33 x 53 x 74 x 11
Ex.6. Find the L.C.M. of 72, 108 and 2100.
Sol. 72 = 23 x 32, 108 = 33 x 22, 2100 = 22 x 52 x 3 x 7.
L.C.M. = 23 x 33 x 52 x 7 = 37800.
Ex.7.Find the L.C.M. of 16, 24, 36 and 54.
Sol.
|
2 |
16 |
- 24 |
- 36 |
- 54 |
|
2 |
8 |
- 12 |
- 18 |
- 27 |
|
2 |
4 |
- 6 |
- 9 |
- 27 |
|
3 |
2 |
- 3 |
- 9 |
- 27 |
|
3 |
2 |
- 1 |
- 3 |
- 9 |
|
|
2 |
- 1 |
- 1 |
- 3 |
|
|
|
|
|
|
\ L.C.M. = 2 x 2 x 2 x 3 x 3 x 2 x 3 = 432.
Ex. 8. Find the H.C.F. and L.C.M. of 2 , 8 , 16 and 10.
3 9 81 27
Sol. H.C.F. of given fractions = H.C.F. of 2,8,16,10 = 2_
L.C.M. of 3,9,81,27 81
L.C.M of given fractions = L.C.M. of 2,8,16,10 = 80_
H.C.F. of 3,9,81,27 3
Ex. 9. Find the H.C.F. and L.C.M. of 0.63, 1.05 and 2.1.
Sol. Making the same number of decimal places, the given numbers are 0.63, 1.05 and 2.10.
Without decimal places, these numbers are 63, 105 and 210.
Now, H.C.F. of 63, 105 and 210 is 21.
H.C.F. of 0.63, 1.05 and 2.1 is 0.21.
L.C.M. of 63, 105 and 210 is 630.
L.C.M. of 0.63, 1.05 and 2.1 is 6.30.
Ex. 10. Two numbers are in the ratio of 15:11. If their H.C.F. is 13, find the numbers.
Sol. Let the required numbers be 15.x and llx.
Then, their H.C.F. is x. So, x = 13.
The numbers are (15 x 13 and 11 x 13) i.e., 195 and 143.
Ex. 11. TheH.C.F. of two numbers is 11 and their L.C.M. is 693. If one of the numbers is 77,find the other.
Sol. Other number = 11 X 693 = 99
77
Ex. 12. Find the greatest possible length which can be used to measure exactly the lengths 4 m 95 cm, 9 m and 16 m 65 cm.
Sol. Required length = H.C.F. of 495 cm, 900 cm and 1665 cm.
495 = 32 x 5 x 11, 900 = 22 x 32 x 52, 1665 = 32 x 5 x 37.
\H.C.F. = 32 x 5 = 45.
Hence, required length = 45 cm.
Ex. 13. Find the greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.
Sol. Required number = H.C.F. of (1657 - 6) and (2037 - 5) = H.C.F. of 1651 and 2032
_______
1651 ) 2032 ( 1 1651
1651_______
381 ) 1651 ( 4
1524_________
127 ) 381 ( 3
381
0
Required number = 127.
Ex. 14. Find the largest number which divides 62, 132 and 237 to leave the same remainder in each case.
Sol . Required number = H.C.F. of (132 - 62), (237 - 132) and (237 - 62)
= H.C.F. of 70, 105 and 175 = 35.
Ex.15.Find the least number exactly divisible by 12,15,20,27.
Sol.
|
3 |
12 |
- 15 |
- 20 |
- 27 |
|
4 |
4 |
- 5 |
- 20 |
- 9 |
|
5 |
1 |
- 5 |
- 5 |
- 9 |
|
|
1 |
- 1 |
- 1 |
- 9 |
Ex.16.Find the least number which when divided by 6,7,8,9, and 12 leave the same remainder 1 each case
Sol. Required number = (L.C.M OF 6,7,8,9,12) + 1
|
3 |
6 |
- 7 |
- 8 |
- 9 - 12 |
|
4 |
2 |
- 7 |
- 8 |
- 3 - 4 |
|
5 |
1 |
- 7 |
- 4 |
- 3 - 2 |
|
|
1 |
- 7 |
- 2 |
- 3 - 1 |
\L.C.M = 3 X 2 X 2 X 7 X 2 X 3 = 504.
Hence required number = (504 +1) = 505.
Ex.17. Find the largest number of four digits exactly divisible by 12,15,18 and 27.
Sol. The Largest number of four digits is 9999.
Required number must be divisible by L.C.M. of 12,15,18,27 i.e. 540.
On dividing 9999 by 540,we get 279 as remainder .
\Required number = (9999-279) = 9720.
Ex.18.Find the smallest number of five digits exactly divisible by 16,24,36 and 54.
Sol. Smallest number of five digits is 10000.
Required number must be divisible by L.C.M. of 16,24,36,54 i.e 432,
On dividing 10000 by 432,we get 64 as remainder.
\Required number = 10000 +( 432 – 64 ) = 10368.
Ex.19.Find the least number which when divided by 20,25,35 and 40 leaves remainders 14,19,29 and 34 respectively.
Sol. Here,(20-14) = 6,(25 – 19)=6,(35-29)=6 and (40-34)=6.
\Required number = (L.C.M. of 20,25,35,40) – 6 =1394.
Ex.20.Find the least number which when divided by 5,6,7, and 8 leaves a remainder 3, but when divided by 9 leaves no remainder .
Sol. L.C.M. of 5,6,7,8 = 840.
\ Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
\Required number = (840 X 2 + 3)=1683
Ex.21.The traffic lights at three different road crossings change after every 48 sec., 72 sec and 108 sec.respectively .If they all change simultaneously at 8:20:00 hours,then at what time they again change simultaneously .
Sol. Interval of change = (L.C.M of 48,72,108)sec.=432sec.
So, the lights will agin change simultaneously after every 432 seconds i.e,7 min.12sec
Hence , next simultaneous change will take place at 8:27:12 hrs.
Ex.22.Arrange the fractions 17 , 31, 43, 59 in the ascending order.
18 36 45 60
Sol.L.C.M. of 18,36,45 and 60 = 180.
Now, 17 = 17 X 10 = 170 ; 31 = 31 X 5 = 155 ;
18 18 X 10 180 36 36 X 5 180
43 = 43 X 4 = 172 ; 59 = 59 X 3 = 177 ;
45 45 X 4 180 60 60 X 3 180
Since, 155<170<172<177, so, 155 < 170 < 172 < 177
180 180 180 180
Hence, 31 < 17 < 43 < 59
36 18 45 60
