## Q1: A bell rings every 9 seconds. A second bell rings every 10 seconds and a third one rings every 18 seconds . If all the three ring at the same time at 8 O' clock in the morning , at what other time will they all ring together ?

Solution : First bell rings every 9 seconds.

Second bell rings every 10 seconds.

Third bell rings every 18 seconds.

To get time at which all the bells ring together, we need to find L.C.M.

Multiples of 9 are

9,18,27,36,45,54,63,72,81,90,99.................

Multiples of 10 are

10,20,30,40,50,60,70,80,90,100........

Multiples of 18 are

18,36,54,72,90,108........................

The least common multiple of 9,10,18 is 90

Therefore every 90 seconds all the three bells ring together.

After 8 O' Clock all the three bells ring at 8:03:30 a.m.

Q 2 : In morning walk, three persons step off together . Their steps measure 25 cm, 35 cm and 45 cm respectively. What is the minimum distance each should walk so that all can cover the same distance in complete steps ?

Solution : The distance covered by each one of them is required to be the same as well

as minimum.

The required minimum distance each should walk would be the least

common multiple of the measures of their steps.

Prime factorisation of

25 = 5 x 5.

35 = 5 x 7.

45 = 5 x 3 x 3.

Thus L.C.M. of 25, 35 and 45 = 5 x 5 x 7 x 3 x 3 = 1575.

Q 3 : The traffic lights ar three different road crossings change after every 48 seconds 72seconds and 108 seconds respectively . If they change simultaneously at 7 a.m. at what time wil they changes simultaneously again ?

Solution : We need to find L.C.M. of 48, 72, 108 seconds to get change

simultaneously again.

48 = 2 x 2 x 2 x 2 x 3

72 = 2 x 2 x 2 x 3 x 3

108 = 2 x 2 x 3 x 3 x 3

L.C.M. of 48, 72 and 108 = 2 x 2 x 2 x 2 x 3 x 3 x 3 = 432 seconds

432 seconds = 7 minutes 12 seconds.

Therefore the three lights change simultaneously at 7:07:12 a.m.

Q4 : Determine the smallest 3-digit number which is exactly divisible by 4, 9 and 20 ?

Solution: First we need to find the L.C.M. of 4, 9, 20

Prime factorisation of 4 = 2 x 2

9 = 3 x 3

20 = 2 x 2 x 5

L.C.M. of 4, 9, and 20 = 2 x 2 x 3 x 3 x 5 = 160.

But we need to get 3-digit least number.

So, we need to take the multiple of 160 which is least 3-digit number.

multiples of 160 are 160, 320, ....

Least 3-digit number is 160.

Therefore the smallest 3-digit number which is exactly divisible by 4, 9, 20 is 160.