Question 3 - Case Based Questions (MCQ) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables

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Question It is common that Governments revise travel fares from time to time based on various factors such as inflation ( a general increase in prices and fall in the purchasing value of money) on different types of vehicles like auto, Rickshaws, taxis, Radio cab etc. The auto charges in a city comprise of a fixed charge together with the charge for the distance covered. Study the following situations Situation 1: In city A, for a journey of 10 km, the charge paid is Rs 75 and for a journey of 15 km, the charge paid is Rs 110. Situation 2: In a city B, for a journey of 8km, the charge paid is Rs 91 and for a journey of 14 km, the charge paid is Rs 145.Refer Situation 1: In city A, for a journey of 10 km, the charge paid is Rs 75 and for a journey of 15 km, the charge paid is Rs 110. Question 1 If the fixed charges of auto rickshaw be Rs x and the running charges be Rs y km/hr, the pair of linear equations representing situation is (a) x + 10y = 110, x + 15y = 75 (b) x + 10y = 75, x + 15y = 110 (c) 10x + y = 110, 15x + y = 75 (d) 10x + y = 75, 15x + y =110 Given that Fixed charges = Rs x Running charges = Rs y For a journey of 10 km, the charge paid is Rs 75 x + 10y = 75 For a journey of 15 km, the charge paid is Rs 110. x + 15y = 110 So, the correct answer is (b) Let’s solve (1) and (2) Subtracting (2) and (1) (x + 15y) − (x + 10y) = 110 − 75 (x + 15y) − (x + 10y) = 110 − 75 5y = 35 y = 35/5 y = 7 Putting y = 7 in (1) x + 10y = 75 x + 10(7) = 75 x + 70 = 75 x = 75 − 70 x = 5 Refer Situation 1: In city A, for a journey of 10 km, the charge paid is Rs 75 and for a journey of 15 km, the charge paid is Rs 110. Question 2 A person travels a distance of 50 km. The amount he has to pay is (a) Rs 155 (b) Rs 255 (c) Rs 355 (d) Rs 455 If a person Travels 50 km Total Charges = Fixed Charges + Running charges × 50 = x + 50y = 5 + 50 × 7 = 5 + 350 = 355 ∴ Total charges is Rs 355 So, correct answer is (c) Refer Situation 2: In a city B, for a journey of 8km, the charge paid is Rs 91 and for a journey of 14 km, the charge paid is Rs 145 Question 3 What will a person have to pay for travelling a distance of 30 km? (a) Rs 185 (b) Rs 289 (c) Rs 275 (d) Rs 305 Given that Fixed charges = Rs p Running charges = Rs q For a journey of 8 km, the charge paid is Rs 91 p + 8q = 91 For a journey of 14 km, the charge paid is Rs 145 p + 14q = 145 Let’s solve (3) and (4) Subtracting (4) and (2) (p + 14q) − (p + 8q) = 145 − 91 6q = 54 q = 54/6 q = 9 Putting q = 9 in (1) p + 8q = 91 p + 8(9) = 91 p + 72 = 91 p = 91 − 72 p = 19 Thus, Fixed charges = Rs p = Rs 19 Running charges = Rs q = Rs 9 Now, we have to find what will a person have to pay for travelling a distance of 30 km If a person Travels 30 km Total Charges = Fixed Charges + Running charges × 30 = p + 30q = 19 + 30 × 9 = 19 + 270 = 289 ∴ Total charges is Rs 289 So, correct answer is (b) Question 4 The graph of lines representing the conditions are: (situation 2) For situation 2, the lines are p + 8q = 91 p + 14q = 145 There is not graph which matches both lines No option is correct In graph (iii), only one line matches But not the other line