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Given that $\frac{\log P}{y-z} = \frac{\log Q}{z-x} = \frac{\log R}{x-y} = 10$ for $x \neq y \neq z$, what is the value of the product $PQR$?

  1. 0
  2. 1
  3. $xyz$
  4. $10^{xyz}$
in Quantitative Aptitude recategorized by
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Migrated from GO Civil 3 years ago by Arjun

1 Answer

3 votes
3 votes
Best answer

$\frac{\log P}{y-z} = \frac{\log Q}{z-x} = \frac{\log R}{x-y} = 10$   for $x \neq y \neq z$

  • $\frac{\log P}{y-z} =10\implies\log P=10(y-z)\qquad \Rightarrow (1)$
  • $\frac{\log Q}{z-x} = 10\implies\log Q=10(z-x)\qquad \Rightarrow(2)$
  • $\frac{\log R}{x-y} = 10\implies\log R=10(x-y)\qquad \Rightarrow (3)$

Adding $(1),(2)$ and $(3)$ we get

$\log P+\log Q+\log R=10(y-z)+10(z-x)+10(x-y)$

$\implies \log P+\log Q+\log R=10\left(y-z+z-x+x-y\right)$

$\implies\log P+\log Q+\log R=10\times 0$

$\implies\log P+\log Q+\log R=0$

$\implies\log PQR=0$

$\implies PQR={10}^{0} = 1.$

Hence, (B) is the correct choice.

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1 comment

$\log_{10}PQR = 0$

$\implies\log_{10}PQR = \log_{10}1$

$\implies PQR = 1$
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