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TANCET 2013 Quant Qn 3: Indices

Question

If ( 0.004 * 10a )( 0.32 * 10b ) = 128 * 103, a + b = ?

  1. 3
  2. 5
  3. 6
  4. 8
  5. -2

  Correct Answer    Choice 4. a + b = 8


Explanatory Answer

  • Step by step solution

    1. (0.004 * 10a)(0.32 * 10b) = 128 * 103
    2. We can write 0.004 as 4/1000

      Similarly, we can write 0.32 as 32/100

    3. So, the expression can be rewritten as (4/1000 * 10a)(32/100 * 10b) = 128 * 103

    4. Or (128/10^5)*10^a*10^b = 128 * 103

    5. i.e., (128/10^5)*10^(a+b) = 128 * 103

    6. i.e., 128 * 10a + b - 5 = 128 * 103

    7. Dividing both sides by 128, we get 10a + b - 5 = 103

    8. As the bases are equal, we can equate the indices.
      a + b - 5 = 3
      Or a + b = 8

      The correct answer is Choice 4. (a + b) = 8

  • Remember these Rules of Indices

    1. ax * ay = ax + y

    2. (ax)y = axy

    3. a^x / a^y = a^(x - y) = ax - y

    4. (a^x)^(1/y)  = a^(x/y)

      e.g., (a^2)^(1/3)  = a^(2/3)

    5. a-x = 1/(a^x)

    6. ax * bx = (a*b)x

Video Explanation

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