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Exercisethreepointthreequestionthirdkafourth

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Exercisethreepointthreequestionthirdkafourth
  • 2 answers

Priya Swain 5 years, 8 months ago

Let the fixed charge be x and the charge for distance covered be y. So ATQ, x+10y=105 x+15y=155 By elimination method we get 5y=50 So y=10 Now x=105-100=5

Aadya Singh 5 years, 8 months ago

Bhai space key bhi hoti h mobile me ??
http://mycbseguide.com/examin8/

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