JIPMAT 2025 (QA) - Which of the following statements is/are correct? A. If 2^x = 3^y = 6^-z, then 1/x + 1/y + 1/z = 0 B. (243)^0.16 x (9)^0.1 = 0.3 C. If 3.105 x 10^P = 0.00239 + 0.000715, then P = -3 | PYQs + Solutions

JIPMAT 2025

Algebra

Linear Equations

Hard

Which of the following statements is/are correct? $\newline$ $\newline$ A. If $2^x = 3^y = 6^{-z}$ , then $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$ $\newline$ B. $(243)^{0.16} \times (9)^{0.1} = 0.3$ $\newline$ C. If $3.105 \times 10^P = 0.00239 + 0.000715$ , then P = -3

B only

B and C only

A and B only

A and C only

✅ Correct Option: 4

It is best to first check the (B) statement from the given options.

(243)^{0.16} \times (9)^{0.1} = 0.3

\newline

LHS =

(3)^{5 \times{0.16}} \times (3)^{2 \times {0.1}} = 3^{(0.80)} \times 3^{(0.20)} = 3^1 = 3

Thus, B is an incorrect statement.

\newline

Hence, the only possible option is Option 4.

A. If

2^x = 3^y = 6^{-z}

, then

\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0

\newline

Given:

2^{x}=3^{y}=6^{-z}

Let:

2^{x}=3^{y}=6^{-z}=k \quad \text { (say) }

Now take logs:

\newline

- From

2^{x}=k

, we get

x=\log _{2} k

\newline

- From

3^{y}=k

, we get

y=\log _{3} k

\newline

- From

6^{-z}=k

, take log:

-z=\log _{6} k \Rightarrow z=-\log _{6} k

Now evaluate:

\newline

\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{\log _{2} k}+\frac{1}{\log _{3} k}+\frac{1}{-\log _{6} k}

Use change of base:

\frac{1}{\log _{b} k}=\log _{k} b

So:

=\log _{k} 2+\log _{k} 3-\log _{k} 6=\log _{k}\left(\frac{2 \times 3}{6}\right)=\log _{k}(1)=0

3.105 \times 10^{P}=0.00239+0.000715=0.003105

Thus,

3.105 \times 10^{P}=0.003105 \Rightarrow 10^{P}=\frac{0.003105}{3.105}=10^{-3} \Rightarrow P=-3