Question

Arrange the given distances (in miles) in ascending order.
$|-\frac{42}{5}|+2,|-5|+3,|-\frac{24}{5}|+5,\left|\frac{46}{5}\right|-3$

• A. $|-5|+3<\left|\frac{46}{5}\right|-3<|-\frac{24}{5}|+5<|-\frac{42}{5}|+2$
• B. $|-\frac{42}{5}|+2<\left|\frac{46}{5}\right|-3<|-5|+3<|-\frac{24}{5}|+5$
• C. $\left|\frac{46}{5}\right|-3<|-\frac{24}{5}|+5<|-5|+3|-\frac{42}{5}|+2$
• D. $\left|\frac{46}{5}\right|-3<|-5|+3<|-\frac{24}{5}|+5|<|-\frac{42}{5}|+2$

Answer 1

The correct option is D $\left|\frac{46}{5}\right|-3<|-5|+3<|-\frac{24}{5}|+5|<|-\frac{42}{5}|+2$

To compare the given numbers, they must have the same denominators
First number:
$|-\frac{42}{5}|+2=\frac{42}{5}+2=\frac{42+10}{5}=\frac{52}{5}$
Second number:
$|-5|+3=5+3=8=\frac{8\times 5}{5}=\frac{40}{5}$
Third number:
$|-\frac{24}{5}|+5=\frac{24}{5}+5=\frac{24+25}{5}=\frac{49}{5}$
Fourth number:
$\left|\frac{46}{5}\right|-3=\frac{46}{5}-3=\frac{46-15}{5}=\frac{31}{5}$
When arranged in ascending order, we get $\frac{31}{5}<\frac{40}{5}<\frac{49}{5}<\frac{52}{5}$.
Hence, the required order is $|-\frac{46}{5}|-3<|-5|+3<|-\frac{24}{5}|+5|<|-\frac{42}{5}|+2$