Question

Write 'T' for true and 'F' for false for each of the following:

(i) $\frac{-15}{-11}$ lies to the left of 0 on the number line.
(ii) $\frac{1}{3}\mathrm{and}\frac{-3}{2}$ lie on opposite side of 0 on the number line.
(iii) $\frac{-8}{13}$ lies to the left of 0 on the number line.
(iv) $\frac{-4}{5}>\frac{-2}{3}$.
(v) $\frac{-3}{5}$ is the largest among $\frac{-3}{7},\frac{-7}{10}\mathrm{and}\frac{-5}{6}$.

Answer 1

$\begin{array}{l}\left(\text{i}\right)\text{False. This is because}\frac{-15}{-11}=\frac{15}{11},\text{which lies to the right of 0}\text{.}\\ \left(\text{ii}\right)\text{True}\\ \left(\text{iii}\right)\text{True}\\ \left(\text{iv}\right)\text{False}\\ L.C.M.\mathrm{of}5\mathrm{and}3\mathrm{is}15.\\ \begin{array}{l}\text{We know:}\frac{-4\times 3}{5\times 3}=\frac{-12}{15}\text{and}\frac{-2\times 5}{5\times 3}=\frac{-10}{15}\\ \therefore \frac{-4}{5}<\frac{-2}{5}\\ \left(\text{v}\right)\text{True}\\ \text{L.C.M. of 5},\text{1}0and\text{6 is 3}0.\\ \therefore \frac{-3\times 6}{5\times 6}=\frac{-18}{30},\frac{-7\times 3}{10\times 3}=\frac{-21}{30}\text{and}\frac{-5\times 5}{6\times 5}=\frac{-25}{30}\end{array}\\ \therefore \frac{-3}{5}\text{is the largest among the given fractions.}\end{array}$