A and B together can complete a piece of work in 10 2/7days white B and C together can complete the same work in 13 1/3 days. B is 25% more efficient than C. In how many days will A and C together complete the same work ?
A and B together can complete a piece of work in 10 2/7days white B and C together can complete the same work in 13 1/3 days. B is 25% more efficient than C. In how many days will A and C together complete the same work ?
1). $$11\frac{1}{4}$$
2). $$12\frac{1}{4}$$
3). $$11\frac{1}{3}$$
4). $$12\frac{1}{3}$$
3 answers
B is 25% more efficient than C
=> Ratio of time taken by B and C = 4 : 5
Let time taken by B and C respectively be 4x and 5x days
=> $\frac{1}{4x} + \frac{1}{5x} = \frac{1}{\frac{40}{3}}$
=> $\frac{9}{20x} = \frac{3}{40}$
=> $\frac{3}{x} = \frac{1}{2}$
=> $x = 6$
Thus, time taken by B = 24 days and time taken by C = 30 days
Let time taken by A be y days
=> $\frac{1}{y} + \frac{1}{24} = \frac{7}{72}$
=> $\frac{1}{y} = \frac{7}{72} - \frac{1}{24} = \frac{1}{18}$
=> $y = 18$
Thus, (A + C)'s 1 day's work = $\frac{1}{18} + \frac{1}{30} = \frac{4}{45}$
$\therefore$ Time taken by A & C together
= $\frac{45}{4} = 11\frac{1}{4}$ days
B is 25% more efficient than C
=> Ratio of time taken by B and C = 4 : 5
Let time taken by B and C respectively be 4x and 5x days
=> $\frac{1}{4x} + \frac{1}{5x} = \frac{1}{\frac{40}{3}}$
=> $\frac{9}{20x} = \frac{3}{40}$
=> $\frac{3}{x} = \frac{1}{2}$
=> $x = 6$
Thus, time taken by B = 24 days and time taken by C = 30 days
Let time taken by A be y days
=> $\frac{1}{y} + \frac{1}{24} = \frac{7}{72}$
=> $\frac{1}{y} = \frac{7}{72} - \frac{1}{24} = \frac{1}{18}$
=> $y = 18$
Thus, (A + C)'s 1 day's work = $\frac{1}{18} + \frac{1}{30} = \frac{4}{45}$
$\therefore$ Time taken by A & C together
= $\frac{45}{4} = 11\frac{1}{4}$ days
B is 25% more efficient than C
=> Ratio of time taken by B and C = 4 : 5
Let time taken by B and C respectively be 4x and 5x days
=> $\frac{1}{4x} + \frac{1}{5x} = \frac{1}{\frac{40}{3}}$
=> $\frac{9}{20x} = \frac{3}{40}$
=> $\frac{3}{x} = \frac{1}{2}$
=> $x = 6$
Thus, time taken by B = 24 days and time taken by C = 30 days
Let time taken by A be y days
=> $\frac{1}{y} + \frac{1}{24} = \frac{7}{72}$
=> $\frac{1}{y} = \frac{7}{72} - \frac{1}{24} = \frac{1}{18}$
=> $y = 18$
Thus, (A + C)'s 1 day's work = $\frac{1}{18} + \frac{1}{30} = \frac{4}{45}$
$\therefore$ Time taken by A & C together
= $\frac{45}{4} = 11\frac{1}{4}$ days