Question

Three bells ring at intervals of 5, 15 and 25 seconds respectively. After how many seconds do all the bells ring together?​

Answer 1

Question :-

Three bells ring at intervals of 5, 15 and 25 seconds respectively. After how many seconds do all the bells ring together?

$\large\underline{\sf{Solution-}}$

♣ $\huge \tt \red{\underline{\bold{Given \: that}}}$

Three bells ring at intervals of 5, 15 and 25 seconds respectively.

The three bells ring together after how many seconds is equals to LCM of 5, 15, 25 seconds respectively.

In order to find LCM of 5, 15 and 25 seconds respectively, we use Method of Prime factorization.

So,

$\rm :\longmapsto\:Prime \: factors \: of \: 5 \: = \: 5$

$\rm :\longmapsto\:Prime \: factors \: of \: 15 \: = \: 5 \times 3$

$\rm :\longmapsto\:Prime \: factors \: of \: 25 \: = \: 5 \times 5 = {5}^{2}$

Hence,

$\sf\implies \:LCM(5,15,25) = 3 \times {5}^{2} = 75$

Hence,

• Three bells ring together exactly after 75 seconds.

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Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

Three bells ring at intervals of 5, 15 and 25 seconds respectively.

The three bells ring together after how many seconds is equals to LCM of 5, 15, 25 seconds respectively.

In order to find LCM of 5, 15 and 25 seconds respectively, we use Method of Prime factorization.

So,

$\rm :\longmapsto\:Prime \: factors \: of \: 5 \: = \: 5$

$\rm :\longmapsto\:Prime \: factors \: of \: 15 \: = \: 5 \times 3$

$\rm :\longmapsto\:Prime \: factors \: of \: 25 \: = \: 5 \times 5 = {5}^{2}$

Hence,

$\rm\implies \:LCM(5,15,25) = 3 \times {5}^{2} = 75$

Hence,

• Three bells ring together exactly after 75 seconds.

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Additional information

If a and b are two natural numbers then,

• HCH(a, b) × LCM(a, b) = a × b

• HCF always divides a, b and LCM

• LCM is always divisible by HCF, a and b

• HCF of two prime numbers is always 1.