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Ex 4.3, 4 - Sum of reciprocals of Rehman ages, 3 years ago - Ex 4.3

  1. Chapter 4 Class 10 Quadratic Equations
  2. Serial order wise
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Ex 4.3 ,4 The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1/3 . Find his present age. Let Rehman's current age = x Rehman’s age 3 years ago = x – 3 Rehman’s age 5 years from now = x + 5 Given that Sum of reciprocal of Rehman’s ages 3 yr. ago and 5 yr. from now = 1/3 1/((𝐴𝑔𝑒 𝑜𝑓 𝑅𝑒ℎ𝑚𝑎𝑛 3 𝑦𝑟 𝑎𝑔𝑜)) + 1/((𝐴𝑔𝑒 𝑜𝑓 𝑟𝑒ℎ𝑚𝑎𝑛 5 𝑦𝑟 𝑓𝑟𝑜𝑚 𝑛𝑜𝑤)) = 1/3 1/(x − 3)+1/(x + 5)=1/3 (x + 5 + x − 3)/((x − 3)(x + 5))=1/3 (2x + 2)/((x − 3)(x + 5))=1/3 (2x + 2) "×" 3=1" ×" (x−3)(x+5) 6x + 6 = x(x + 5) – 3 (x + 5) 6x + 6 = x2 + 5x – 3x – 15 0 = x2 + 5x – 3x – 15 – 6x – 6 0 = x2 + 5x – 3x – 6x – 15 0 = x2 – 4x – 21 x2 – 4x – 21 = 0 We factorize by splitting the middle term method x2 + 3x – 7x – 21 = 0 x (x + 3) –7 (x + 3) = 0 (x – 7) (x + 3) = 0 So, x = 7 and x = -3 But x cannot be negative, as x is Rehman’s current age So, x = 7 ∴ Rehman’s current age = x = 7 years

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