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4 men and 6boys can finish a |
| Answer» Let the man finishes the work in x days and that the boy finishes in y days.One day\'s work of a man =\xa0{tex}\\frac{1}{x}{/tex}and one day\'s work of a boy ={tex}\\frac{1}{y}{/tex}Since, 4 men and 6 boys finish a piece of work in 5 days.{tex}\\therefore{/tex}\xa0One day\'s work of 4 men and 6 boys =\xa0{tex}\\frac{1}{5}{/tex} of the work{tex} \\Rightarrow \\frac{4}{x} + \\frac{6}{y} = \\frac{1}{5}{/tex}Similarly, in second case,\xa0One day\'s work of 3 men and 4 boys =\xa0{tex}\\frac{1}{7}{/tex} part of the work{tex}\\therefore \\frac{3}{x} + \\frac{4}{y} = \\frac{1}{7}{/tex}Thus, we have the following equations{tex}\\frac{4}{x} + \\frac{6}{y} = \\frac{1}{5}{/tex}\xa0......(i)and{tex}\\frac{3}{x} + \\frac{4}{y} = \\frac{1}{7}{/tex}\xa0.....(ii)Here , Eqs. (i) and (ii) are not in linear form , so we reduce them in linear form by putting\xa0{tex}\\begin{array}{l}\\frac1x\\;=\\;u\\;and\\;\\frac1y\\;=\\;v\\\\\\end{array}{/tex}Now, Eq. (i) becomes\xa0{tex}4u + 6v = \\frac{1}{5}{/tex} .....(iii)and Eq(ii) becomes\xa0{tex}3u + 4v = \\frac{1}{7}{/tex}\xa0.....(iv)On multiplying Eq(iii) by 3 and Eq(iv) by 4 and then subtract Eq , we get{tex}18v - 16v = \\frac{3}{5} - \\frac{4}{7}{/tex}{tex}\\Rightarrow 2v = \\frac{{21 - 20}}{{35}} \\Rightarrow 2v = \\frac{1}{{35}} \\Rightarrow v = \\frac{1}{{70}}{/tex}Put\xa0{tex}v=\\frac{1}{70}{/tex} in Eq(iv), we get\xa0{tex}3u + \\frac{4}{{70}} = \\frac{1}{7}{/tex}{tex} \\Rightarrow 3u = \\frac{1}{7} - \\frac{4}{{70}}{/tex}{tex} \\Rightarrow 3u = \\frac{6}{{70}}{/tex}{tex} \\Rightarrow u = \\frac{1}{{35}}{/tex}Thus\xa0{tex}u=\\frac{1}{35}{/tex}\xa0and\xa0{tex}v=\\frac{1}{70}{/tex}{tex} \\Rightarrow \\frac{1}{x} = \\frac{1}{{35}}{/tex}and\xa0{tex}\\frac{1}{y} = \\frac{1}{{70}}{/tex}{tex} \\Rightarrow x = 35{/tex}\xa0and\xa0{tex}y=70{/tex}Hence, 1 man alone and 1 boy alone finishes the work in 35 and 70 days, respectively. | |