in Quantitative Aptitude edited by
5,136 views
11 votes
11 votes

 If $pqr \ne 0$ and $p^{-x}=\dfrac{1}{q},q^{-y}=\dfrac{1}{r},r^{-z}=\dfrac{1}{p},$ what is the value of the product $xyz$ ?

  1. $-1$
  2. $\dfrac{1}{pqr}$
  3. $1$
  4. $pqr$
in Quantitative Aptitude edited by
by
5.1k views

4 Comments

got 1
0
0
same here ans should be 1
0
0
you have type wrong question

p−x=1/q

q−y=1/r

r−z=1/p
0
0
pqr !=0, p^-x = 1/q, q^-y = 1/r, r^-z = 1/p

Taking log both sides,

-x(log p) = -log q,

-y(log q) = -log r,

-z(log r) = -log p

=> x = log q/ log p

=> y = log r/ log q

=> z = log p/ log r

multiplying x,y and z,

(x)(y)(z) = 1
7
7

9 Answers

10 votes
10 votes
Best answer
$\color{maroon}{pqr \neq 0}$

$\color{maroon}{p^{-x} = \dfrac{1}{q}}$

Or,$\log(p^{-x}) = \log(\dfrac{1}{q})$

Or, $-x \log (p) = \log(1)-\log(q)$  $\qquad\Big[∵\color{blue}{\log_b(m^n)= n. \log_b(m)\\\qquad\qquad \log_b(\dfrac{m}{n})= \log_b(m)-\log_b(n)} \Big]$

Or, $x\log(p) = \log (q)$ $\qquad[∵\color{blue}{\log(1)=0}]$

Or, $\color{green}{x= \dfrac{\log(q)}{\log(p)}}$

$\color{maroon}{q^{-y} = \dfrac{1}{r}}$

Or,$\log(q^{-y}) = \log(\dfrac{1}{r})$

Or, $-y \log (q) = \log(1)-\log(r)$  $\qquad\Big[∵\color{blue}{\log_b(m^n)= n. \log_b(m)\\\qquad\qquad \log_b(\dfrac{m}{n})= \log_b(m)-\log_b(n)}\Big]$

Or, $y\log(q) = \log (r)$ $\qquad[∵\color{blue}{\log(1)=0}]$

Or, $\color{green}{y= \dfrac{\log(r)}{\log(q)}}$

$\color{maroon}{r^{-z} = \dfrac{1}{p}}$

Or,$\log(r^{-z}) = \log(\dfrac{1}{p})$

Or, $-z \log (r) = \log(1)-\log(p)$  $\qquad\Big[∵\color{blue}{\log_b(m^n)= n. \log_b(m)\\\qquad\qquad \log_b(\dfrac{m}{n})= \log_b(m)-\log_b(n)}\Big]$

Or, $z\log(r) = \log (p)$ $\qquad[∵\color{blue}{\log(1)=0}]$

Or, $\color{green}{z= \dfrac{\log(p)}{\log(r)}}$

∴ $\color{black}{x \times y \times z}$ = $ \dfrac{\log(q)}{\log(p)} \times  \dfrac{\log(r)}{\log(q)} \times \dfrac{\log(p)}{\log(r)}$

$\qquad \qquad = \color{black}{1}$

Correct Answer: $C$
edited by

1 comment

$p^{-x}=\dfrac{1}{q},q^{-y}=\dfrac{1}{r},r^{-z}=\dfrac{1}{p}$

We can write like this $\dfrac{1}{p^{x}}=\dfrac{1}{q},\dfrac{1}{q^{y}}=\dfrac{1}{r},\dfrac{1}{r^{z}}=\dfrac{1}{p}$

$\implies \dfrac{1}{p^{x}}\times \dfrac{1}{q^{y}}\times \dfrac{1}{r^{z}} = \dfrac{1}{q}\times\dfrac{1}{r} \times \dfrac{1}{p}$

$\implies \dfrac{1}{p^{x}}\times \dfrac{1}{q^{y}}\times \dfrac{1}{r^{z}} = \dfrac{1}{p}\times\dfrac{1}{q} \times \dfrac{1}{r}$

Compare the coefficients of both sides, and we get,

$\implies x=y=z=1$

$\therefore xyz = 1\times 1 \times 1 = 1$

So, the correct answer is $(C).$
7
7
13 votes
13 votes
Given: $pqr ≠ 0$

$p^{-x}= 1/q$

$q^{-y}= 1/r$

$r{-z}= 1/p$

Take $\log$ on both side

$x \log p= \log q$

$\implies x= \frac{\log q}{\log p}$

$y \log q= \log r$

$\implies y = \frac{\log r}{ \log q}$

$z \log r= \log p$

$\implies z= \frac{\log p}{\log r}$

So, $xyz= 1.$
edited by
7 votes
7 votes

Here we can write as follows:

px=q --(i)

qy=r --(ii)

rz=p --(iii)

from (ii) and (iii)

(qy)z=p => qyz=p --(iv)

from (i) and (iv)

(px)yz=p => pxyz=p1   

              => xyz=1

 

 

6 votes
6 votes
multiply each equation

 

p^-x *q^-y*r^-z=1/pqr

on comparing

x=-1

y=-1

z=-1

so xyz=1

2 Comments

yes got same
0
0
if p=q=r=1 then pqr!=0 it can take any  positive value of x,y,z.
0
0
Answer:

Related questions