Free Rational Numbers 02 Practice Test - 7th grade 

Let the number to be added be x.

$\frac{-3}{8}+x=\frac{5}{16}$

$x=\frac{5}{16}+\frac{3}{8}=\frac{5}{16}+\frac{6}{16}=\frac{11}{16}$

The space between 0 and 1 is divided into 10 equal parts. The ball lies in sixth place.

Therefore, the location ball lies in is, $\frac{6}{10}=\frac{3}{5}$

$-\frac{16}{3}÷-\frac{4}{3}=-\frac{16}{3}\times -\frac{3}{4}=4$

Reducing  $\frac{39}{26}$   to its lowest form ,  we get, 

    $\frac{39÷13}{26÷13}$ = $\frac{3}{2}$( 13 is the common factor)

Multiplying numerator and denominator by 2 we get

$\frac{3\times 2}{2\times 2}$ = $\frac{6}{4}$

Hence, $\frac{39}{26}$ is also equal to  $\frac{6}{4}$

So, both the fraction are equal

Hence,  $\frac{39}{26}=\frac{6}{4}$
                   

Consider the given rational number $\frac{4}{7}$.

To make the numerator as 16, we have to multiply it by 4.

Multiplying both numerator and denominator by 4, we get

$\frac{4\times 4}{7\times 4}=\frac{16}{28}$

Therefore, $\frac{16}{28}$ is an equivalent rational number of $\frac{4}{7}$ with numerator equal to 16.

Reciprocal of $\frac{-26}{12}$ is $\frac{-12}{26}$.

$\frac{2}{13}+\frac{-12}{26}$      [Taking LCM]

$=\frac{4}{26}-\frac{12}{26}=\frac{4-12}{26}=\frac{-8}{26}=\frac{-4}{13}$

$\frac{-7}{19}+\frac{-2}{57}=\frac{(-7\times 3)}{(19\times 3)}+\frac{-2}{57}$

$=\frac{-21}{57}+\frac{-2}{57}=\frac{-23}{57}$

$\frac{10}{2}=\frac{2\times 5}{2}$

$=5$

Any positive number is greater than 0. 

A negative number cannot be greater than a positive number.

so,$\frac{-4}{3}<\frac{24}{36}$