Use >, <, or = appropriately, using the property of cross products.
(a) $\dfrac{3}{8}....\dfrac{4}{10}$
(b) $\dfrac{23}{40}....\dfrac{12}{30}$
(c) $\dfrac{-6}{11}....\dfrac{-11}{20}$
(d) $\dfrac{18}{40}....\dfrac{9}{20}$
(e) $\dfrac{72}{80}....\dfrac{9}{10}$
(f) $\dfrac{-3}{7}....\dfrac{-4}{10}$

Hint: Consider the expressions in each of the subparts of the question one by one. Now, multiply the numerator of the left fraction with the denominator of the right fraction to get the L.H.S. similarly multiply the numerator of the right fraction with the denominator of the left fraction to get the R.H.S. Check which side is greater in numerical value and according include the symbols >, <; or =. If L.H.S is greater then use the symbol ‘>’, if R.H.S is greater use the symbol ‘>’ and if L.H.S and R.H.S are equal then use ‘=’.

Complete step-by-step solution:
Here we have been provided with fractions and we are asked to compare them using the symbols >, < or = by considering their cross products. That means we need to determine which fraction is greater or if they are equal. Let us consider each subpart one by one.
(a) Here we have the fraction $\dfrac{3}{8}....\dfrac{4}{10}$. Now, we have to take the cross product. In cross product we multiply the numerator of the left fraction with the denominator of the right fraction to get the L.H.S and the numerator of the right fraction with the denominator of the left fraction to get the R.H.S. So we get,
$\Rightarrow L.H.S=3\times 10=30$ and $R.H.S=4\times 8=32$
Clearly we can see that R.H.S is greater than L.H.S, so we will use the symbol ‘<’.
$\therefore \dfrac{3}{8}<\dfrac{4}{10}$
(b) Here we have the fraction $\dfrac{23}{40}....\dfrac{12}{30}$. Considering the cross product we get,
$\Rightarrow L.H.S=23\times 30=690$ and $R.H.S=40\times 12=480$
Clearly we can see that L.H.S is greater than R.H.S, so we will use the symbol ‘>’.
$\therefore \dfrac{23}{40}>\dfrac{12}{30}$
(c) Here we have the fraction $\dfrac{-6}{11}....\dfrac{-11}{20}$, so considering the cross product we get,
$\Rightarrow L.H.S=-6\times 20=-120$ and \[R.H.S=-11\times 11=-121\]
We know that 121 is greater than 120 so the negative form of 121 will be less than the negative form of 120. Therefore, we can see that L.H.S is greater than R.H.S, so we will use the symbol ‘>’.
$\therefore \dfrac{-6}{11}>\dfrac{-11}{20}$
(d) Here we have the fraction $\dfrac{18}{40}....\dfrac{9}{20}$ so considering the cross product we get,
$\Rightarrow L.H.S=18\times 20=360$ and $R.H.S=40\times 9=360$
Clearly we can see that the R.H.S is equal to the L.H.S, so we will use the symbol ‘=’.
$\therefore \dfrac{18}{40}=\dfrac{9}{20}$
(g) Here we have the fraction $\dfrac{72}{80}....\dfrac{9}{10}$ so considering the cross product we get,
$\Rightarrow L.H.S=72\times 10=720$ and $R.H.S=80\times 9=720$
Clearly we can see that the R.H.S is equal to the L.H.S, so we will use the symbol ‘=’.
$\therefore \dfrac{72}{80}=\dfrac{9}{10}$
(f) Here we have the fraction $\dfrac{-3}{7}....\dfrac{-4}{10}$, so considering the cross product we get,
$\Rightarrow L.H.S=-3\times 10=-30$ and $R.H.S=-4\times 7=-28$
We know that 30 is greater than 28 so the negative form of 30 will be less than the negative form of 28. Therefore, we can see that R.H.S is greater than L.H.S, so we will use the symbol ‘<’.
$\therefore \dfrac{-3}{7}<\dfrac{-4}{10}$

Note: Always remember that the expression towards the pointed part of the symbol ‘<’ represents the smaller value and the expression towards the open part represents the greater value. Note that greater the absolute value of a number smaller will be its negative form. This can be seen by the representation of numbers on a number line.