Free Fractions and Decimals 02 Practice Test - 7th grade 

$\frac{1}{3}-(\frac{5}{6}\times \frac{9}{18})+\frac{2}{9}$

$=\frac{1}{3}-\left(\frac{15}{36}\right)+\frac{2}{9}$

$=\frac{1}{3}-\frac{15}{36}+\frac{2}{9}$

Taking L.C.M of denominators, $\frac{12-15+8}{36}=\frac{5}{36}$

$\frac{1}{2}\times (1+\frac{1}{4})\times (1+\frac{2}{4})\times (1+\frac{3}{4})\times (1+\frac{5}{4}$)

= $\frac{1}{2}\times \frac{5}{4}\times \frac{6}{4}\times \frac{7}{4}\times \frac{9}{4}=\frac{945}{256}$

$(1\frac{2}{4}$ -$\frac{1}{2}$ +3$\frac{6}{16}$ -$\frac{1}{16}$ +$\frac{2}{8}$)

=$\frac{6}{4}-\frac{1}{2}+\frac{54}{16}-\frac{1}{16}+\frac{2}{8}$

Taking LCM of denominators and solving,

$\frac{24-8+54-1+4}{16}$=$\frac{73}{16}$

Now, If we multiply $\frac{73}{16}$ by $\frac{16}{1}$, it becomes 73, which is an integer
 

Hence, $\frac{16}{1}$ is the required fraction.

$\frac{\left(\frac{1}{2}\right)\times \left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)}{\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)\times \left(\frac{3}{4}\right)}$ = $\frac{\left(\frac{1}{4}\right)+\left(\frac{1}{2}\right)}{1\times \left(\frac{3}{4}\right)}$ = $\frac{\left(\frac{3}{4}\right)}{\left(\frac{3}{4}\right)}=1$

Given $[5\frac{1}{4}-\frac{2}{5}\times 5\frac{5}{9}+(\frac{1}{2}+\frac{3}{4})]$

Apply BODMAS rule to solve

= $[\frac{21}{4}-(\frac{2}{5}\times \frac{50}{9})+\frac{5}{4}]$

= $[\frac{21}{4}-(\frac{20}{9})+\frac{5}{4}]$

Taking LCM of denominators, we have

$\frac{189-80+45}{36}=\frac{154}{36}=\frac{77}{18}$

$[5\frac{1}{4}-\frac{2}{5}\times 5\frac{5}{9}+(\frac{1}{2}+\frac{3}{4})]=\frac{77}{18}$

First, let us do the conversion of mixed fractions into improper fractions.

i.e  $3\frac{1}{2}=\frac{(2\times 3)+1}{2}=\frac{7}{2}$
     
$2\frac{2}{3}=\frac{(3\times 2)+2}{3}=\frac{8}{3}$, and 
     
$1\frac{1}{3}=\frac{(3\times 1)+1}{3}=\frac{4}{3}$

$\text{Total quantity of fruits}=(\frac{7}{2}+\frac{8}{3}+\frac{4}{3})kg$
                                     
                                      $=\frac{7}{2}+4$

                                      $=\frac{7}{2}+\frac{8}{2}$ kg
   
                                      $=\frac{15}{2}$ kg

When $\frac{15}{2}kg$ is divided among 4 people,

each person's share $=\frac{\frac{15}{2}}{4}kg$
                                 
                                  $=(\frac{15}{2}\times \frac{1}{4})kg$

                                  $=\frac{15}{8}kg$

                                  $=1\frac{7}{8}kg$

Hence, each person gets $\left(1\frac{7}{8}\right)kg$ of fruits. 

$\frac{\frac{5}{2}+\frac{7}{3}+\frac{9}{4}}{\frac{1}{10}\times \frac{2}{5}\times \frac{125}{6}}$

$=\frac{\frac{85}{12}}{\frac{5}{6}}$

$=\frac{85}{12}\times \frac{6}{5}$

$=\frac{17}{2}=8.5$

Let the total distance to be covered $=dkm$

Then, $\frac{1}{8}\times d$ = $56km$

$\text{d = (56}\times \text{8)km = 448km}$

$\text{Remaining distance = (448 - 56)km = 392km}$

Concentrate on numerator. The expression has '$0$' multiplied to it, and any number multiplied by '$0$' is '$0$'.

And 0 divided by any number always '$0$'.

Hence the answer is '$0$'.