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A alone can do a work in 20 days. B is 25% more efficient than A. A and B started working and worked for 4 days. If C alone completed the remaining job in 22 days. How many days C alone takes to complete the entire job?
A alone can do a work in 20 days. B is 25% more efficient than A. A and B started working and worked for 4 days. If C alone completed the remaining job in 22 days. How many days C alone takes to complete the entire job?
1). 24 days
2). 36 days
3). 12 days
4). 20 days
1). 24 days
2). 36 days
3). 12 days
4). 20 days
This Question has 1 answers.
$ A_{\text{efficiency}} = \frac{1}{20} $
$ B_{\text{efficiency}} = \frac{125}{100} \times \frac{1}{20} = \frac{5}{4} \times \frac{1}{20} = \frac{1}{16} $
$ \text{Work done by A and B in 4 days} = 4 \times \left( \frac{1}{20} + \frac{1}{16} \right) $
$ = 4 \times \left( \frac{4}{80} + \frac{5}{80} \right) = 4 \times \frac{9}{80} = \frac{36}{80} = \frac{9}{20} $
$ \text{Remaining work} = 1 - \frac{9}{20} = \frac{11}{20} $
$ C_{\text{efficiency}} = \frac{\frac{11}{20}}{22} = \frac{11}{20} \times \frac{1}{22} = \frac{11}{440} = \frac{1}{40} $
$ \text{Total days for C to complete full work} = \frac{1}{\frac{1}{40}} = 40 $
$ B_{\text{efficiency}} = \frac{125}{100} \times \frac{1}{20} = \frac{5}{4} \times \frac{1}{20} = \frac{1}{16} $
$ \text{Work done by A and B in 4 days} = 4 \times \left( \frac{1}{20} + \frac{1}{16} \right) $
$ = 4 \times \left( \frac{4}{80} + \frac{5}{80} \right) = 4 \times \frac{9}{80} = \frac{36}{80} = \frac{9}{20} $
$ \text{Remaining work} = 1 - \frac{9}{20} = \frac{11}{20} $
$ C_{\text{efficiency}} = \frac{\frac{11}{20}}{22} = \frac{11}{20} \times \frac{1}{22} = \frac{11}{440} = \frac{1}{40} $
$ \text{Total days for C to complete full work} = \frac{1}{\frac{1}{40}} = 40 $
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