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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Ex 3.2, 3 Form the pair of linear equations for the following problems and find their solution by substitution method. (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km? Let Fixed charge = Rs x & Charge per km = Rs y Given that Charge paid for 10 km is Rs 105 Fixed charge + 10 × (Charge per km) = Rs 105 x + 10y = 105 Also, Given that Charge paid for 15 km is ₹ 155 Fixed charge + 15 × (Charge per km) = ₹ 155 x + 15y = 155 Our equations are x + 10y = 105 …(1) x + 15y = 155 …(2) From (1) x + 10y = 105 x = 105 – 10y Putting value x in (2) x + 15y = 155 (105 – 10y) + 15y = 155 105 – 10y + 15y = 155 –10y + 15y = 155 – 105 5y = 50 y = 50/5 y = 10 Putting y = 10 in (1) x + 10y = 105 x + 10(10) = 105 x + 100 = 105 x = 105 – 100 x = 5 Hence, x = 5, y = 10 Hence, Fixed charges = x = Rs 5 Change per km = y = Rs 10 We also need to find How much does a person have to pay for travelling 25 km Amount paid by person travelling 25 km = Fixed charges + (Charges per km) × 25 = 5 + 10 × 25 = 5 + 250 = Rs 255

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.