Arrange $\dfrac{5}{8}, -\dfrac{3}{16}, -\dfrac{1}{4} \text{ and } \dfrac{17}{32}$ in the descending order of their magnitudes.

Also, find the sum of the lowest and the largest of these rational numbers. Express the result obtained as a decimal fraction correct to two decimal places.

L.C.M. of 4, 8, 16 and 32 is 32.

So, converting denominator of each fraction $\dfrac{5}{8}, -\dfrac{3}{16}, -\dfrac{1}{4} \text{ and } \dfrac{17}{32}$ into 32.

$\Rightarrow \dfrac{5}{8} \times \dfrac{4}{4} = \dfrac{20}{32} \\[1em] \Rightarrow -\dfrac{3}{16} \times \dfrac{2}{2} = -\dfrac{6}{32} \\[1em] \Rightarrow -\dfrac{1}{4} \times \dfrac{8}{8} = -\dfrac{8}{32} \\[1em] \Rightarrow \dfrac{17}{32} \times \dfrac{1}{1} = \dfrac{17}{32}.$

Since, -8 < -6 < 17 < 20.

$\therefore -\dfrac{8}{32} \lt -\dfrac{6}{32} \lt \dfrac{17}{32} \lt \dfrac{20}{32}$

$\therefore -\dfrac{1}{4} \lt -\dfrac{3}{16} \lt \dfrac{17}{32} \lt \dfrac{5}{8}$

So, in descending order.

$\Rightarrow \dfrac{5}{8} \gt \dfrac{17}{32} \gt -\dfrac{3}{16} \gt -\dfrac{1}{4}$

Sum of largest and lowest :

$= \dfrac{5}{8} + \Big(-\dfrac{1}{4}\Big) \\[1em] = \dfrac{5}{8} - \dfrac{1}{4} \\[1em] = \dfrac{5 \times 4}{32} - \dfrac{1 \times 8}{32} \\[1em] = \dfrac{20 - 8}{32} \\[1em] = \dfrac{12}{32} \\[1em] = \dfrac{3}{8} \\[1em] = 0.38$

Hence, fractions in descending order are : $\dfrac{5}{8} \gt \dfrac{17}{32} \gt -\dfrac{3}{16} \gt -\dfrac{1}{4}$ and required sum = 0.38