GATE2016 EC-1: GA-9

Question

If $\large q^{-a} = \dfrac{1}{r}$ and $\large r^{-b} = \dfrac{1}{s}$ and $\large s^{-c} = \dfrac{1}{q}$, the value of $abc$ is ________.

• A. $\left(rqs\right)^{-1}$
• B. $0$
• C. $1$
• D. $r+q+s$

Answer 1

Multiply all terms,we will get more faster than above method.

$q^{-a}\times r^{-b}\times s^{-c}=\frac{1}{r}\times\frac{1}{s}\times\frac{1}{q}$

$q^{-a}\times r^{-b}\times s^{-c}=q^{-1}\times r^{-1}\times s^{-1}$

Compare the terms

$a=1,b=1,c=1$

So,$abc=1$

----------------------------------------

Suppose they ask $a+b+c=?$

We can simply put the values of $a,b$ and $c$ and get the answer

$a+b+c=1+1+1=3$

Correct Answer: $C$

Comments

Option C

---

ans is 1

Answer 2

Given that,

$\large q^{-a} = \dfrac{1}{r},\;r^{-b}= \dfrac{1}{s},\; s^{-c} =\dfrac{1}{q}$

Now, $\large q^{-a} = \dfrac{1}{r}$

$\large\Rightarrow \dfrac{1}{q^{a}} =\dfrac{1}{r}$

$\large\Rightarrow q^{a} = r$

Similarly, $\large r^{b} = s$ and $\large s^{c} = q$

In $\large q^{a} = r$, we replace $q$ by $\large s^{c}$ and $r$ by $\large s^{\dfrac{1}{b}}$

We get, $\large s^{ca} = s^{\dfrac{1}{b}}$

Since, the bases are equal, we can work with the exponents.

So, we get, $ac$ = $\dfrac{1}{b}$ $\Rightarrow abc = 1$