From the pair of linear equations in the following problem and find their solutions (if they exist) by any algebraic method :<\/p>\n
(i)<\/strong>\u00a0 A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student, A takes food for 20 days, A has to pay Rs 1000 as hostel charges, whereas a student B, who takes food for 26 days pays Rs 1180 as hostel charges. Find the fixed charge and cost of food per day.<\/p>\n (ii)<\/strong>\u00a0 A fraction becomes \\(1\\over 3\\)\u00a0 when 1 is subtracted from the numerator and it becomes \\(1\\over 4\\) when 8 is added to its denominator. Find the fraction.<\/p>\n (iii)<\/strong>\u00a0 Yash scored 40 marks in a test, receiving 3 marks for each correct answer and loosing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test ?<\/p>\n (iv)<\/strong>\u00a0 Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the car travel in the same direction at a different speed, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speed of the two cars ?<\/p>\n (v)<\/strong>\u00a0 The area of a rectangle gets reduced by 9 square units if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.<\/p>\n (i)<\/strong>\u00a0 Let Rs x be the fixed hostel charges<\/p>\n And Rs y be the cost of food per day.<\/p>\n ATQ,<\/p>\n A’s hostel charges are<\/p>\n Fixed hostel charges + cost of food for 20 days = Rs 1000<\/p>\n x + 20y = 1000\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ……..(1)<\/p>\n B’s hostel charges are<\/p>\n Fixed hostel charges + cost of food for 26 days = Rs 1180<\/p>\n x + 26y = 1180\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……….(2)<\/p>\n Now, subtract equation (1) from equation (2), we get<\/p>\n 6y = 180\u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0y = 30<\/p>\n Put y = 30 in equation (1),<\/p>\n x + 20(30) = 1000\u00a0 \u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0 x + 600 = 1000<\/p>\n \\(\\implies\\)\u00a0 \u00a0 x = 400<\/p>\n Hence, x = 400 and y = 30.<\/strong><\/p>\n (ii)<\/strong>\u00a0 Let \\(x\\over y\\) be the fraction,<\/p>\n According to Question,<\/p>\n \\(x – 1\\over y\\) = \\(1\\over 3\\)\u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a03x – 3 = y\u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a03x – y – 3 = 0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………(1)<\/p>\n and\u00a0 \\(x\\over y + 8\\)\u00a0 \\(\\implies\\)\u00a0 \u00a0\\(1\\over 4\\)\u00a0 \\(\\implies\\)\u00a0 \u00a04x = y + 8\u00a0 \\(\\implies\\)\u00a0 \u00a04x – y – 8 = 0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……….(2)<\/p>\n Now, subtracting equation (1) from equation (2), we get<\/p>\n x – 5 = 0\u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0x = 5<\/p>\n Put the value of x in equation (1), we get<\/p>\n y = 12<\/p>\n Hence, the fraction is \\(5\\over 12\\).<\/strong><\/p>\n (iii)<\/strong>\u00a0 Let x be the number of correct answers and y be the number of wrong answers given by yash.<\/p>\n Then, ATQ,<\/p>\n Case 1<\/strong> :\u00a0 He got 40 marks if 3 marks are given for correct answer and 1 mark is deducted for every incorrect answers.<\/p>\n 3x – y = 40\u00a0 \u00a0 \u00a0 \u00a0 ……..(1)<\/p>\n Case 2<\/strong> : He gets 50 marks if 4 marks are given for correct answer and 2 marks are deducted for every incorrect answers.<\/p>\n 4x – 2y = 50\u00a0 \u00a0 \u00a0 \u00a0 \u00a0…….(2)<\/p>\n Multiplying equation (1) by (2) and subtracting from equation (2), we get<\/p>\n -2x = -30\u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0 x = 15<\/p>\n Putting x = 15 in equation (1), we get<\/p>\n 45 – y = 40\u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0y = 5<\/p>\n Hence, the total number of questions = 15 + 5 = 20<\/strong>.<\/p>\n (iv)<\/strong>\u00a0 Let x km\/hr\u00a0 be the speed of the first car, starting from A.<\/p>\n and y km\/hr is the speed of second car, starting from B.<\/p>\n Distance travelled by first car in 5 hours = 5x km<\/p>\n Distance travelled by second car in 5 hours = 5y km<\/p>\n ATQ,<\/p>\n When they are moving in the same direction,<\/p>\n 5x = 100 + 5y<\/p>\n \\(\\implies\\)\u00a0 x = 20 + y<\/p>\n \\(\\implies\\)\u00a0 x – y = 20\u00a0 \u00a0 \u00a0 \u00a0 \u00a0……..(1)<\/p>\n When they are moving in the opposite direction,<\/p>\n Distance travelled by first car in 1 hour = x km<\/p>\n Distance travelled by second car in 1 hour = y km<\/p>\n \\(\\implies\\)\u00a0 \u00a0x + y = 100\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………..(2)<\/p>\n Adding equation (1) and equation (2), we get<\/p>\n 2x = 120\u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0x = 60<\/p>\n Put the value of in equation (1), we get<\/p>\n 60 – y = 20\u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0y = 40<\/p>\nSolution :<\/h2>\n