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2 Answers

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Given $\frac{a}{b}=\frac{1}{3}$                 $\frac{b}{c}={2}$                      $\frac{c}{d}=\frac{1}{2}$              

$\frac{d}{e}=3$                  $\frac{e}{f}=\frac{1}{4}$

 

b=3a=2c=d   => d=3a    

b=2c=d=3e => e =  $\frac{b}{3}$       

c = $\frac{d}{2} = \frac{3e}{2} = \frac{3f}{8}$  

 

$\frac{a}{d} *\frac{b}{e} * \frac{c}{f}$ =  $\frac{a}{3a} *\frac{b}{\frac{b}{3}} * \frac{c}{\frac{8c}{3}}$

=$\frac{3}{8}$
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Given that:

$a/b=1/3,b/c=2,c/d=1/2,d/e=3,e/f=1/4$

Now calculate;

$\frac{a}{d}=(a/b)*(b/c)*(c/d)\implies(1/3)*(2)*(1/2)=1/3$

$\frac{b}{e}=(b/c)*(c/d)*(d/e)\implies (2)*(1/2)*(3)=3$

$\frac{c}{f}=(c/d)*(d/e)*(e/f)=(1/2)*(3)*(1/4)=3/8$

so $\frac{abc}{def}=(1/3)*(3)*(3/8)=\frac{3}{8}$

Option (A) is correct.

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