Formulate the following problem as a pair of equations, and hence find their solutions<\/p>\n
(i)<\/strong>\u00a0 Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours, find her speed of rowing in still water and the speed of the current.<\/p>\n (ii)<\/strong>\u00a0 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 women alone to finish the work, and also that time taken by 1 man alone.<\/p>\n (iii)<\/strong>\u00a0 Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours, if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she taken 10 minutes longer. Find the speed of the train and the bus seperately.<\/p>\n (i)\u00a0 Let x km\/hr be the speed of boat in still water and y km\/hr be the speed of the current.<\/p>\n Speed upstream = (x – y) km\/hr<\/p>\n and Speed downstream = (x + y) km\/hr<\/p>\n According to question,<\/p>\n Time take to cover 20 km downstream = 2 hrs<\/p>\n \\(\\implies\\)\u00a0 \\(20\\over x + y\\) = 2<\/p>\n \\(\\implies\\)\u00a0 \u00a0x + y = 10\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 …….(1)<\/p>\n Time take to cover 4 km upstream = 2 hrs<\/p>\n \\(\\implies\\)\u00a0 \\(4\\over x – y\\) = 2<\/p>\n \\(\\implies\\)\u00a0 x – y = 2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………(2)<\/p>\n Now, Add equations (1) and (2), we get<\/p>\n 2x = 12\u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0x = 6<\/p>\n Now, Put the value of x = 6 in equation (1), we get<\/p>\n 6 + y = 10\u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0y = 4<\/p>\n Hence, the speed of current is 4 km\/hr and speed boat is 6 km\/hr.<\/strong><\/p>\n (ii)\u00a0 Let x days be the time taken by 1 woman to finish the embroidery and y days be the time taken by 1 man to finish the embroidery.<\/p>\n Then, 1 day work of 1 woman = \\(1\\over x\\)<\/p>\n and 1 day work of 1 man = \\(1\\over y\\)<\/p>\n According to Question,<\/p>\n \\(2\\over x\\) + \\(5\\over y\\) = \\(1\\over 4\\)<\/p>\n and \\(3\\over x\\) + \\(6\\over y\\) = \\(1\\over 3\\)<\/p>\n Put u = \\(1\\over x\\) and v = \\(1\\over y\\), then equation becomes<\/p>\n 2u + 5v = \\(1\\over 4\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………(1)<\/p>\n and 3u + 6v = \\(1\\over 3\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0……..(2)<\/p>\n Multiply equation (1) by 3 and equation (2) by 2 and then subtract equation (2) from equation (1), we get<\/p>\n 3v = \\(1\\over 12\\)\u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0v = \\(1\\over 36\\)<\/p>\n Put the value of v in equation (1), we get<\/p>\n 2u = \\(1\\over 9\\)\u00a0 \\(\\implies\\)\u00a0 u = \\(1\\over 18\\)<\/p>\n Now, u = \\(1\\over x\\)\u00a0 \u00a0\\(\\implies\\)\u00a0 x = 18<\/p>\n and v = \\(1\\over y\\)\u00a0 \\(\\implies\\)\u00a0 y = 36<\/p>\n Hence, 18\u00a0days<\/strong> be the time taken by 1 woman to finish the embroidery and 36 days<\/strong> be the time taken by 1 man to finish the embroidery.<\/p>\n (iii)\u00a0 Let x km\/hr be the speed of train and y km\/hr be the speed of the bus.<\/p>\n Then,\u00a0 \\(60\\over x\\) + \\(240\\over y\\) = 4<\/p>\n and\u00a0 \\(100\\over x\\) + \\(200\\over y\\) = 4 + \\(10\\over 60\\) = \\(25\\over 6\\)<\/p>\n Put u = \\(1\\over x\\) and v = \\(1\\over y\\), then equation becomes<\/p>\n 60u + 240v = 4\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………(1)<\/p>\n 100u + 200v = \\(25\\over 6\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0…….(2)<\/p>\n Multiply equation (1) by 5 and equation (2) by 6 and subtract equation (2), from equation (1), we get<\/p>\n -300u = – 5\u00a0 \u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0u = \\(1\\over 60\\)<\/p>\n Put the value of u in equation (1), we get<\/p>\n 240v = 3\u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0v = \\(1\\over 80\\)<\/p>\n Now, u = \\(1\\over x\\)\u00a0 \u00a0\\(\\implies\\)\u00a0 x = 60<\/p>\n and v = \\(1\\over y\\)\u00a0 \\(\\implies\\)\u00a0 y = 80<\/p>\nSolution :<\/h2>\n