Question

Let $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ be 2 sets of solution satisfying the following equations:
${\log }_{10}\left(2xy\right)=4+({\log }_{10}x-1)({\log }_{10}y-2)$
${\log }_{10}\left(2yz\right)=4+({\log }_{10}y-2)({\log }_{10}z-1)$
${\log }_{10}\left(zx\right)=2+({\log }_{10}z-1)({\log }_{10}x-1)$
such that $(x_1>x_2)$,
then match the elements of List - I with the correct answer in List -II.

$\begin{array}{cc}\text{List -I}& \text{List -II}\\ \left(I\right)\frac{y_1}{x_1}& \left(P\right)2\\ \left(II\right)\frac{z_1}{x_2}& \left(Q\right)100\\ \left(III\right)\frac{z_1{x}_2}{z_2}& \left(R\right)1000\\ \left(IV\right)\frac{y_2+z_1}{x_2}& \left(S\right)150\end{array}$

Which of the following is the only 'INCORRECT' combination?

• A. $\left(I\right)\to \left(P\right)$
• B. $\left(II\right)\to \left(Q\right)$
• C. $\left(III\right)\to \left(S\right)$
• D. $\left(IV\right)\to \left(S\right)$

Answer 1

The correct option is C $\left(III\right)\to \left(S\right)$

Let ${\log }_{10}x-1=A$
${\log }_{10}y-2=B$
${\log }_{10}z-1=C$
So $(A-1)(B-1)={\log }_{10}2$
$(B-1)(C-1)={\log }_{10}2$
$(A-1)(C-1)=1$
$\Rightarrow (A-1)(B-1)(C-1)=\pm {\log }_{10}2$
$\begin{array}{c}A-1=\pm 1\\ C-1=\pm 1\\ B-1=\pm lo{g}_{10}2\end{array}\}$
$(A,B,C)=(2,1+lo{g}_{10}2,2)or(0,1-lo{g}_{10}2,0)$
$(x,y,z)=(\underset{x_1}{1000},\underset{y_1}{2000},\underset{z_1}{1000}),(\underset{x_2}{10},\underset{y_2}{500},\underset{z_2}{10})$
Now match correct combinations