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TANCET 2013 Quant Qn 1: GP

Question

If a rubber ball consistently bounces back ⅔ of the height from which it is dropped, what fraction of its original height will the ball bounce after being dropped and bounced four times without being stopped?

  1. 16/81
  2. 16/27
  3. 4/9
  4. 37/81
  5. 9/12

  Correct Answer    Choice 1. 16/81


Explanatory Answer

  • Step by step solution

    1. Each time the ball is dropped and it bounces back, it reaches ⅔ of the height it was dropped from.
    2. After the first bounce, the ball will reach ⅔ of the height from which it was dropped - let us call it the original height.
    3. After the second bounce, the ball will reach ⅔ of the height it would have reached after the first bounce.

      So, at the end of the second bounce, the ball would have reached ⅔ * ⅔ of the original height = 4/9th of the original height.
    4. After the third bounce, the ball will reach ⅔ of the height it would have reached after the second bounce.

      So, at the end of the third bounce, the ball would have reached ⅔ * ⅔ * ⅔ = 8/27th of the original height.
    5. After the fourth and last bounce, the ball will reach ⅔ of the height it would have reached after the third bounce.

      So, at the end of the last bounce, the ball would have reached ⅔ * ⅔ * ⅔ * ⅔ of the original height = 16/81 of the original height.

      The correct answer is Choice 1.

  • Formulae to remember in Geometric Progression

    A geometric progression is such a sequence in which every subsequent term of the sequence is obtained by multiplying the preceding term with a constant number. This constant number is called the common ratio (r) of the sequence.

    e.g., 2, 4, 8, 16, 32 is a geometric sequence.

    The second term is obtained by multiplying the first term by 2. The third term is obtained by multiplying the second term by 2 and so on.

    Formula 1: nth term of a Geometric Progression an = arn-1, where a is the first term of the sequence and r is the common ratio.

    Formula 2: Sum of the first n terms of a geometric progression = (ar^n - a)/(r - 1)

    Formula 3: Sum up to infinite terms of a geometric progression whose common ratio r lies between 0 and 1 i.e., S = a/(1 - r)

Video Explanation

More Questions on Arithmetic, Geometric Progression

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