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
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
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.
L.C.M of given fractions = L.C.M.
of 2,8,16,10 = 80_
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
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_______
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.
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