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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

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 $

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