Calculate the HCF of $$\frac{1.75}{9}, \frac{5.6}{12}, \frac{7}{18}$$​​

Correct option is B

Given:

Number given are $$\frac{1.75}{9}, \frac{5.6}{12}, \frac{7}{18}$$​​

Formula Used:

HCF of fractions =$$\frac{HCF\ of\ Numerators}{LCM\ of\ denominators}$$​​

Solution:

Numbers in fraction are:

​$$\frac{1.75}{9} =\frac{175}{900}= \frac{7}{36}$$​ 

​$$\frac{5.6}{12}=\frac{56}{120}= \frac{7}{15}$$​ 

HCF of $$(\frac{7}{36},\frac{7}{15}, \frac{7}{18}) = \frac{HCF\ of\ 7,7,7}{LCM\ of\ 36,15,18}$$​​

HCF of 7,7,7 is 7

LCM of 36, 15 and 18

36 =$$2^2 \times 3^2$$​​

15 = $$3 \times 5$$​​

18 = $$2 \times 3^2$$​​

LCM =$$2^2 \times 3^2 \times 5 = 180$$​​

HCF of $$(\frac{1.75}{9}, \frac{5.6}{12}, \frac{7}{18}) = \frac{7}{180}$$​​