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Wall Area Percentage Change

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

This is an easy 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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The length, breadth and height of a room are in the ratio 3:2:13 : 2 : 1. If the breadth and height are halved while the length is doubled, then the total area of the four walls of the room will:

  1. A.

    remain the same

  2. B.

    decrease by 13.64%

  3. C.

    decrease by 15%

  4. D.

    decrease by 18.75%

  5. E.

    decrease by 30%

Answer

decrease by 30%

Explanation

Let l=3x,b=2x,h=xl = 3x, b = 2x, h = x. Initial area of 4 walls =2(l+b)h=2(3x+2x)x=10x2= 2(l + b)h = 2(3x + 2x)x = 10x^2. New dimensions: l′=6x,b′=x,h′=x/2l' = 6x, b' = x, h' = x/2. New area of 4 walls =2(6x+x)(x/2)=7x2= 2(6x + x)(x/2) = 7x^2. Percentage decrease =10x2−7x210x2×100%=30%= \frac{10x^2 - 7x^2}{10x^2} \times 100\% = 30\%.

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