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Hcf & Lcm


ngu2587

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Hey all,

 

I was stuck up with this problem..

 

Six bells commence tolling together and toll at intervals at 2,4,6,8,10 and 12 seconds respectively. In 30 min, how many times do they toll together

 

The answer for this problem is given as below

 

L.C.M of 2,4,6,8,10,12 is 120

So, the bells will toll together after every 120 seconds, i.e 2 min

In 30 minutes, they will toll together [(30/2)+1] = 16 times

 

Now over here in this problem why are they taking LCM to find when the bells will toll together

And the last line i.e in 30 min they will toll 16 times.. How did they arrive with that solution

Please explain me..

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"Now over here in this problem why are they taking LCM to find when the bells will toll together"

 

If two bells toll after every 3 secs and 4 secs respectively and if they commence tolling at the same time

 

then the first bell tolls after every 3, 6, 9, 12 secs...

the second bell tolls after every 4, 8, 12, ....

So they toll together again after 12 secs, which is the LCM

Henceforth they toll after every 12 seconds, i.e whenever the time is a common multiple of both 3 and 4.

 

 

"And the last line i.e in 30 min they will toll 16 times.. How did they arrive with that solution"

Since the bells start tolling together, the first toll also needs to be counted. therefore we need to add 1

 

I hope this makes it clear..

Thanx

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Hi GmatFundoo,

Thanks for the reply. But I could not understand why they added 1. How do you think this problem is different from this problem as below

 

A,B,C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 sec, B in 308 sec and C in 198 sec, all starting at the same point. After what time will they meet again at the starting point.

 

In this problem we just take the LCM, as per the logical explanation given by you..

 

Can you please explain me

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"A,B,C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 sec, B in 308 sec and C in 198 sec, all starting at the same point. After what time will they meet again at the starting point."

 

 

For the sake of simplicity lets take smaller numbers say 3 mins, 4 mins and 6 mins as the times taken by A, B and C respectively. Then as explained earlier they would meet again at the starting point after 12 mins, which is the LCM.

 

However, if the question is " How many times do they meet again at the starting point in 36 mins? ", the ans would be 36/12 = 3.

Here u don't need to add 1, since it asks u how many times do they meet AGAIN at the starting point.

 

But in the earlier question, it was how many times do the bells together in 30 mins. AFTER the first toll, the bells toll together 30/2 times, which is equal to 15. If we count the first toll, it becomes 15 + 1 = 16.

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