Question

If the rational numbers $\frac{2}{5},\frac{5}{8},\frac{4}{7}$ and $\frac{7}{10}$ are arranged in the descending order what will be the square of the second last fraction?

• A. $\frac{4}{25}$
• B. $\frac{16}{49}$
• C. $\frac{49}{100}$
• D. $\frac{25}{64}$

Answer 1

Answer: (B)

We have,

$\frac{2}{5},\frac{5}{8},\frac{4}{7}$ and $\frac{7}{10}$

$LCM$ of $5,8,7,$ and $10=280$

Hence,

$\frac{2}{5}=\frac{2\times 56}{5\times 56}=\frac{112}{280}$

$\frac{5}{8}=\frac{5\times 35}{8\times 35}=\frac{175}{280}$

$\frac{4}{7}=\frac{4\times 40}{7\times 40}=\frac{160}{280}$

$\frac{7}{10}=\frac{7\times 28}{10\times 28}=\frac{196}{280}$

Now, it is clear that,

$\frac{196}{280}>\frac{175}{280}>\frac{160}{280}>\frac{112}{280}$

Or $\frac{7}{10}>\frac{5}{8}>\frac{4}{7}>\frac{2}{5}$

and square of second last fraction is ${\left(\frac{4}{7}\right)}^2=\frac{16}{49}$

Hence, this is the answer.