Xavier Aptitude Test 2013 Paper

(c) Statement I : Prime factors of 1000027 are 7, 19, 73, 103.
Hence, the statement is false.
Statement II :
$6!=6\times 5!=\mathrm{}\frac{6^6}{6^5}=5!$
or $\mathrm{}\sqrt[6]{6!}=6\times \sqrt[6]{\mathrm{}\frac{5!}{6^5}}$ ........ (1)
$7!=7\times 6\times 5!=\mathrm{}\frac{6^7}{6^6}\times 7\times 5!$
or $7\sqrt[7]{7!}=6\times \sqrt[7]{\mathrm{}\frac{7\times 5!}{6^6}}$
or $\sqrt[7]{7!}=6\times \sqrt[7]{\mathrm{}\frac{7}{6}}\times \sqrt[7]{\mathrm{}\frac{5!}{6^5}}$
$\therefore \sqrt[7]{7!}>\sqrt[6]{6!}$
Statement II is false.
Statement III :
One half of the journey is completed x km/hr. Let theother half is completed at an average speed of km/hr. If the total distance is taken as ‘d’, then the overall average
$\mathit{s}\mathit{p}\mathit{e}\mathit{e}\mathit{d}=\frac{\mathit{T}\mathit{o}\mathit{t}\mathit{a}\mathit{l}\mathrm{}\mathit{d}\mathit{i}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{c}\mathit{e}}{\mathit{T}\mathit{o}\mathit{t}\mathit{a}\mathit{l}\mathrm{}\mathit{t}\mathit{i}\mathit{m}\mathit{e}\mathrm{taken}}$
$=\mathrm{}\frac{\mathrm{}\mathit{d}}{\frac{\mathrm{}\mathit{d}}{2\mathrm{}\mathit{x}}+\mathrm{}\frac{\mathrm{}\mathit{d}}{2\mathrm{}\mathit{y}}}=\mathrm{}\frac{2\mathrm{}\mathit{x}\mathit{y}}{\left(\mathrm{}\mathit{x}+\mathit{y}\right)}$
Let us check whether this could be equal to 2x.
$\mathrm{}\frac{2\mathit{x}\mathit{y}}{\left(\mathit{x}+\mathit{y}\right)}=2\mathit{x}$
$\Rightarrow \mathit{y}=\mathit{x}+\mathit{y}$
Hence, $\mathit{x}=0$ which is wrong.
$\therefore$ Statement III is correct.