Skip to content

CA Foundation · Quantitative Aptitude · Ratio and Proportion, Indices and Logarithms

If 2^x = 3^y = 6^(−z), then the value of 1/x + 1/y + 1/z is:

Set 2^x, 3^y and 6^(-z) all equal to k, so 2 = k^(1/x), 3 = k^(1/y) and 6 = k^(-1/z). Since 6 equals 2 times 3, the exponents add: -1/z = 1/x + 1/y. Therefore 1/x + 1/y + 1/z equals 0.

  1. A1
  2. B−1
  3. C0Correct
  4. D6

Explanation

Let 2^x = 3^y = 6^(−z) = k. Then 2 = k^(1/x), 3 = k^(1/y) and 6 = k^(−1/z). Since 2 × 3 = 6, k^(1/x + 1/y) = k^(−1/z), so 1/x + 1/y = −1/z. Therefore 1/x + 1/y + 1/z = 0. The answer 1 would arise if the negative sign in the power of 6 were ignored.

Did you get it right without looking?

One question tells you little. A timed set on Ratio and Proportion, Indices and Logarithms shows your real accuracy, how long you take and where you lose marks.

Practise Ratio and Proportion, Indices and LogarithmsFree diagnostic mock

More Ratio and Proportion, Indices and Logarithms questions