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Ex 10.2, 17 - The owner of a milk store finds that sell 980

Ex 10.2, 17 - Chapter 10 Class 11 Straight Lines - Part 2
Ex 10.2, 17 - Chapter 10 Class 11 Straight Lines - Part 3
Ex 10.2, 17 - Chapter 10 Class 11 Straight Lines - Part 4
Ex 10.2, 17 - Chapter 10 Class 11 Straight Lines - Part 5

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Transcript

Ex 10.2, 17 The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs 17/litre? Let selling price be P along x-axis & demand of milk be D along y-axis We know that the equation of line is y = mx + c Here, P is along x-axis and D is along y-axis So, our equation becomes D = mP + c Now, Owner sells 980 litre milk at Rs 14 /litre So, D = 980 & P = 14 satisfies the equation Putting values in (1) 980 = 14m + c Owner sells 1220 litre milk at Rs 16/litre So, D = 1220 & P = 16 satisfies the equation Putting values in (1) 1220 = 16m + c So, our equations are 980 = 14m + c 1220 = 16m + c From (A) 980 = 14m + c 980 − 14m = c Putting value of c in (B) 1220 = 16m + 980 − 14m 1220 – 980 = 16m − 14m 240 = 2m 240/2 = 2 120 = m m = 120 Putting m = 120 in (A) 980 = 14m + c 980 − 14m = c 980 − 14(120) = c 980 − 1680 = c −700 = c c = –700 Putting value of m & c in (1) D = mP + c D = 120P − 700 Hence, the required equation is D = 120P − 700 We need to find how many litres could he sell weekly at Rs 17/litre i.e. we need to find D when P = 17 Putting P = 17 in the equation D = 120P − 700 D = 120(17) − 700 D = 2040 − 700 D = 1340 Hence when price is Rs 17/ litre, 1340 litres milk could be sold

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.