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System of Exponential Equations

CAT 2006 Slot 1 · Quantitative Ability · Medium · Quantitative Aptitude

This is a medium Quantitative Ability question from the CAT 2006 Slot 1 paper. It tests Quantitative Aptitude. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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What are the values of xx and yy that satisfy both equations? 20.7x⋅3−1.25y=86272^{0.7x} \cdot 3^{-1.25y} = \frac{8\sqrt{6}}{27} 40.3x⋅90.2y=8⋅(81)1/54^{0.3x} \cdot 9^{0.2y} = 8 \cdot (81)^{1/5}

  1. A.

    x = 2, y = 5

  2. B.

    x = 2.5, y = 6

  3. C.

    x = 3, y = 5

  4. D.

    x = 3, y = 4

  5. E.

    x = 5, y = 2

Answer

x = 5, y = 2

Explanation

From equation 2: 20.6x⋅30.4y=23⋅34/52^{0.6x} \cdot 3^{0.4y} = 2^3 \cdot 3^{4/5}. Equating powers of 2 and 3: 0.6x=3  ⟹  x=50.6x = 3 \implies x = 5. 0.4y=4/5=0.8  ⟹  y=20.4y = 4/5 = 0.8 \implies y = 2.

Verify in equation 1: 20.7(5)⋅3−1.25(2)=23.5⋅3−2.5=23⋅20.532.5=8293=86272^{0.7(5)} \cdot 3^{-1.25(2)} = 2^{3.5} \cdot 3^{-2.5} = \frac{2^3 \cdot 2^{0.5}}{3^{2.5}} = \frac{8\sqrt{2}}{9\sqrt{3}} = \frac{8\sqrt{6}}{27}. Correct!

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