Answer
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Hint: First remove the decimal by multiplying all the numbers with 1000. To balance, divide them with 1000. Leave the number 1000 in the denominator as it is and consider the numbers obtained in the numerator, find their L.C.M. by prime factorization method. Once the L.C.M. is found, divide it with 1000 to get the answer.
Complete step-by-step answer:
Here, we have been provided with the numbers 2.5, 0.5 and 0.175 and we are asked to find their L.C.M. But first let us make the numbers free form decimal. Now, on observing the three numbers we can say that we have to multiply them with 1000 so that all of them become free from the decimal. To balance, we need to divide them with 1000. So, we get,
\[\begin{align}
& \Rightarrow 2.5,0.5,0.175=\dfrac{2500}{1000},\dfrac{500}{1000},\dfrac{175}{1000} \\
& \Rightarrow 2.5,0.5,0.175=\dfrac{1}{1000}\left( 2500,500,175 \right) \\
\end{align}\]
Since, 1000 was common in all the three numbers in the denominator so we have taken it aside. Now, what we will do is we will find the L.C.M. of 2500, 500 and 175 and divide it with 1000 to get the answer.
Now, let us use the method of prime factorization to get the L.C.M. of three numbers. The L.C.M. of these numbers will be the product of the highest power of each prime factor appearing. So, writing 2500, 500 and 175 as the product of its primes, we have,
\[\begin{align}
& \Rightarrow 2500={{2}^{2}}\times {{5}^{4}} \\
& \Rightarrow 500={{2}^{2}}\times {{5}^{3}} \\
& \Rightarrow 175={{5}^{2}}\times 7 \\
\end{align}\]
Clearly, we can see that the highest power of the prime factor 2 is 2, 5 is 4 and 7 is 1, so the L.C.M. of the above three numbers can be given as: -
\[\Rightarrow L.C.M.={{2}^{2}}\times {{5}^{4}}\times 7\]
Now, we have to divide this L.C.M. with 1000 to get the L.C.M. of initial three numbers 2.5, 0.5 and 0.175, so we get,
\[\begin{align}
& \Rightarrow L.C.M.=\dfrac{{{2}^{2}}\times {{5}^{4}}\times 7}{1000} \\
& \Rightarrow L.C.M.=17.5 \\
\end{align}\]
Hence, 17.5 is our answer.
Note: One may note that the numbers 2.5 and 0.5 will get free form the decimal even after multiplying with 10, but the third number 0.175 will have to be multiplied with 1000 to get free from decimal and that is why we choose this number. You can note that the general method of finding the L.C.M. of two fractions \[\dfrac{a}{b}\] and \[\dfrac{c}{d}\] is that, first we find the L.C.M. of a and c, i.e., the numerators, then in the next step we find the H.C.F. of b and d, i.e., the denominators and finally we use the relation: - L.C.M. required = \[\dfrac{L.C.M.\left( a,c \right)}{H.C.F.\left( b,d \right)}\], to get the answer.
Complete step-by-step answer:
Here, we have been provided with the numbers 2.5, 0.5 and 0.175 and we are asked to find their L.C.M. But first let us make the numbers free form decimal. Now, on observing the three numbers we can say that we have to multiply them with 1000 so that all of them become free from the decimal. To balance, we need to divide them with 1000. So, we get,
\[\begin{align}
& \Rightarrow 2.5,0.5,0.175=\dfrac{2500}{1000},\dfrac{500}{1000},\dfrac{175}{1000} \\
& \Rightarrow 2.5,0.5,0.175=\dfrac{1}{1000}\left( 2500,500,175 \right) \\
\end{align}\]
Since, 1000 was common in all the three numbers in the denominator so we have taken it aside. Now, what we will do is we will find the L.C.M. of 2500, 500 and 175 and divide it with 1000 to get the answer.
Now, let us use the method of prime factorization to get the L.C.M. of three numbers. The L.C.M. of these numbers will be the product of the highest power of each prime factor appearing. So, writing 2500, 500 and 175 as the product of its primes, we have,
\[\begin{align}
& \Rightarrow 2500={{2}^{2}}\times {{5}^{4}} \\
& \Rightarrow 500={{2}^{2}}\times {{5}^{3}} \\
& \Rightarrow 175={{5}^{2}}\times 7 \\
\end{align}\]
Clearly, we can see that the highest power of the prime factor 2 is 2, 5 is 4 and 7 is 1, so the L.C.M. of the above three numbers can be given as: -
\[\Rightarrow L.C.M.={{2}^{2}}\times {{5}^{4}}\times 7\]
Now, we have to divide this L.C.M. with 1000 to get the L.C.M. of initial three numbers 2.5, 0.5 and 0.175, so we get,
\[\begin{align}
& \Rightarrow L.C.M.=\dfrac{{{2}^{2}}\times {{5}^{4}}\times 7}{1000} \\
& \Rightarrow L.C.M.=17.5 \\
\end{align}\]
Hence, 17.5 is our answer.
Note: One may note that the numbers 2.5 and 0.5 will get free form the decimal even after multiplying with 10, but the third number 0.175 will have to be multiplied with 1000 to get free from decimal and that is why we choose this number. You can note that the general method of finding the L.C.M. of two fractions \[\dfrac{a}{b}\] and \[\dfrac{c}{d}\] is that, first we find the L.C.M. of a and c, i.e., the numerators, then in the next step we find the H.C.F. of b and d, i.e., the denominators and finally we use the relation: - L.C.M. required = \[\dfrac{L.C.M.\left( a,c \right)}{H.C.F.\left( b,d \right)}\], to get the answer.
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