LCM Questions and Answers

Let the numbers be 13X and 15X

Their LCM will be (13*5)X

Now (13*15)X= 39780

Therefore X= 204

Hence the numbers are (13*204, 15*204) or 2652 and 3060

The lowest number which is exactly divisible by numbers 15 27 35 and 42 will be the LCM of these numbers which are mentioned below:

= 3*5*7*9*2= 1890

Therefore the required number is 1890+9= 1899

The required number will be=

= {LCM of (30   45   65   78) – 8}

LCM of 30   45   65   78 =

5 30    45    65  78
3 6   9   13    78
13  2   3   13   26
2 2   3   1   2

5*3*13*2*3 = 1170

Hence the required number = 1170-8= 1162

Clearly the required which when reduced by 5 will be  LCM of 20   28   35   105

Hence the required number = {LCM of 20   28   35   105) + 5}

LCM of 20   28   35   105 =

= 5*7*4*3 = 420

420+5= 425

Here the difference between the divisor and its corresponding remainder = 17 (35-18= 17 similarly 45-28= 17 and 55-38= 17)

Therefore the required number will be = { (LCM of 35   45   55) – 17}

= (5*7*9*11) – 17= 3465-17

= 3448

Therefore option B is the correct one.

The number of times the bells will toll together = LCM of 6, 7, 8, 9, and 12 = 504

Let the number be 4y and 5y

L.C.M = 20 y

20y = 40

Y = 2

So the numbers will be 4*2 = 8

5*2 = 10

L.C.M of 8 and 10 is 40

Let the given numbers be = 3y and 2y

LCM = 6Y

6Y = 30

Y = 5

The two numbers will be 3*5 = 15 and 2*5 = 10

Their sum will be = 10+15 = 25

LCM=\frac{product of  two numbers}{HCF} = \frac{2025}{15}=135

Let the numbers be 12a and 12b

Then 12a*12b= 2160 or ab=15

Therefore the values of co-primes a and b are (1,15) and (3,5)

Hence the two-digit numbers are (3*12, 5*12), i.e. 36, 60