L.C.M
Question
Q1.
Three persons step off together , their steps cover 60 cm, 55 cm and 58 cm respectively . What is the minimum distance , so that
each can cover the distance in complete steps ?
Solution.
Required distance is LCM of 60, 55 & 58
= 19140 cm
= 191.4 M
Solution
Q2.
Find the LCM of
4/5
,
6/11
,
7/15
Solution.
LCM of fraction =
LCM of Numerators/
HCF of Denominators
LCM of 4,6,7 = 84
HCF of 5,11,15 = 1
LCM of fraction = 84
Solution
Q3.
Find the prime factors in the product 42 x 94 x 77 .
(a) 21 (b) 10 (c) 19 (d) 20
Solution.
= (22)2 x (32)4 x 77
= 24 x 38 x 77
= 4 + 8 + 7
= 19
Solution
Q4.
A circular field has a circumference of 210 km . Three cyclists start together and can travel 30 km,42 km and 70 km an hour, on circular
track. when they will meet again ?
Solution.
Find the LCM of
210/30
,
210/42
,
210/70
find the LCM of 7, 5 , 3
= 105 hrs
they will after 105 hrs.
Or
210/HCF of 30, 70, 42
=
210/2
= 105 hrs
Solution
Q5.
Two numbers are in ratio of 4:5 , and LCM is 180 , find both the numbers .
Solution.
(4x) x (5x) = (x) x 180
x = 9
Numbers are 4x = 36 and 5x = 45
Note : Let numbers are 4x and 5x , their LCM is 180 and HCF is x .
find number x second number =
HCF x LCM
Solution
Q6.
Three sportsman A,B and C run on a circular track , having speed of 4 hrs, 3 hrs and 9 hrs respectively . Circumference of circular
track is 27 km. When they will at the starting point.
Solution.
First find time to complete one revolution .
Time =
distance/
speed
LCM of
27/4
,
27/3
and
27/9
=
27/4
,
27/3
,
3/1
LCM of
27/4
,
27/3
,
3/1
=
LCM of 27,27,3/HCF of 4,3,1
=
27/1
= 27
All will meet after 27 hrs.
Solution
Q7.
Two numbers are in ratio of 4:5 , and LCM is 180 , find both the numbers .
Solution.
(4x) x (5x) = (x) x 180
x = 9
Numbers are 4x = 36 and 5x = 45
Note : Let numbers are 4x and 5x , their LCM is 180 and HCF is x .
find number x second number =
HCF x LCM
Solution
Q8.
Two numbers are in ratio 3:4 and their LCM is 48. Find the difference of these two numbers.
Solution.
Let numbers are 3x and 4x , their HCF is x
first number x second number = HCF x LCM
(3x) x (4x) =
(x) x 48
x = 4
So numbers are 3x = 3 x 4 = 12 and 4x = 4 x 4 = 16
difference of the
two numbers = 16 - 12 = 4
Solution
Q9.
Sum of two numbers is 40 and difference is 1/5 of the sum . find LCM and HCF .
Solution.
Let numbers are x and y
x + y = 40
x - y = 8
x = 24, y = 16
LCM = 48
HCF = 8
Solution
Q10.
Find the least number which is divided by 2,3,4,5,6 leaves 2 as remainder in each case(except 2) but it is exactly divisible by 7 and 2 .
Solution.
The LCM of 2,3,4,5,6 is 60
So the desired number is 60k + 2 (k is any positive integer)
= (7 x 8 + 4)k + 2 = (7 x 8)k + (4k + 2)
Hence , choose the least value of k for which 4k + 2 is divisible by 7.
(that is k = 3)
∴ The desired number = 182
Solution
