9. A can do a piece of work in 10 days and B in 20 days. They begin together but A leaves 2 days before the
completion of the work. In how many days will the whole work be completed?
Answers
A can do the work in 10 days
A can do the work in 10 daysB can do the work in 20 days
A can do the work in 10 daysB can do the work in 20 daysTherefore, their efficiencies
A can do the work in 10 daysB can do the work in 20 daysTherefore, their efficienciesTotal work = LCM(10,20)= 20 units
A can do the work in 10 daysB can do the work in 20 daysTherefore, their efficienciesTotal work = LCM(10,20)= 20 unitsA's efficiency= 20/10= 2 units of work/day
A can do the work in 10 daysB can do the work in 20 daysTherefore, their efficienciesTotal work = LCM(10,20)= 20 unitsA's efficiency= 20/10= 2 units of work/dayB's efficiency= 20/20= 1 unit of work / day
A can do the work in 10 daysB can do the work in 20 daysTherefore, their efficienciesTotal work = LCM(10,20)= 20 unitsA's efficiency= 20/10= 2 units of work/dayB's efficiency= 20/20= 1 unit of work / daySo, according to the question
A can do the work in 10 daysB can do the work in 20 daysTherefore, their efficienciesTotal work = LCM(10,20)= 20 unitsA's efficiency= 20/10= 2 units of work/dayB's efficiency= 20/20= 1 unit of work / daySo, according to the questionIt takes n days to complete the total job and A leaves 2 days early. So A works for n-2 days and B works for n days
A can do the work in 10 daysB can do the work in 20 daysTherefore, their efficienciesTotal work = LCM(10,20)= 20 unitsA's efficiency= 20/10= 2 units of work/dayB's efficiency= 20/20= 1 unit of work / daySo, according to the questionIt takes n days to complete the total job and A leaves 2 days early. So A works for n-2 days and B works for n daysSo, A's work in n-2 days + B's work in n days = 20 units
A can do the work in 10 daysB can do the work in 20 daysTherefore, their efficienciesTotal work = LCM(10,20)= 20 unitsA's efficiency= 20/10= 2 units of work/dayB's efficiency= 20/20= 1 unit of work / daySo, according to the questionIt takes n days to complete the total job and A leaves 2 days early. So A works for n-2 days and B works for n daysSo, A's work in n-2 days + B's work in n days = 20 unitsOr, 2(n-2) + 1(n)= 20
A can do the work in 10 daysB can do the work in 20 daysTherefore, their efficienciesTotal work = LCM(10,20)= 20 unitsA's efficiency= 20/10= 2 units of work/dayB's efficiency= 20/20= 1 unit of work / daySo, according to the questionIt takes n days to complete the total job and A leaves 2 days early. So A works for n-2 days and B works for n daysSo, A's work in n-2 days + B's work in n days = 20 unitsOr, 2(n-2) + 1(n)= 20Or, n= 8 days
A can do the work in 10 daysB can do the work in 20 daysTherefore, their efficienciesTotal work = LCM(10,20)= 20 unitsA's efficiency= 20/10= 2 units of work/dayB's efficiency= 20/20= 1 unit of work / daySo, according to the questionIt takes n days to complete the total job and A leaves 2 days early. So A works for n-2 days and B works for n daysSo, A's work in n-2 days + B's work in n days = 20 unitsOr, 2(n-2) + 1(n)= 20Or, n= 8 daysSo, the work gets completed in 8 days.
Step-by-step explanation:
refer to the given attachment Hope it helps
