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(B) Let $A$ alone do the work in $x$ days and $B$ alone do the work in $y$ days.According to the problem,the work done by $A$ and $B$ in one day is $1

Vedclass · Accepted Answer

$6 \frac{1}{4}$

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(B) Let $A$ alone do the work in $x$ days and $B$ alone do the work in $y$ days.According to the problem,the work done by $A$ and $B$ in one day is $1/x$ and $1/y$ respectively.Given that $A$ and $B$ together complete the job in $5$ days,we have: $1/x + 1/y = 1/5$ .....$(1)$If $A$ works with double efficiency $(2/x)$ and $B$ works with $1/3$ efficiency $(1/(3y))$,they complete the job in $3$ days: $2/x + 1/(3y) = 1/3$ .....$(2)$To solve for $x$,multiply equation $(2)$ by $3$: $6/x + 1/y = 1$ .....$(3)$Subtract equation $(1)$ from equation $(3)$: $(6/x + 1/y) - (1/x + 1/y) = 1 - 1/5$$5/x = 4/5$$x = 25/4 = 6 \frac{1}{4}$ days.Thus,$A$ alone would require $6 \frac{1}{4}$ days to complete the job.

Two workers $A$ and $B$ working together completed a job in $5$ days. If $A$ worked twice as efficiently as he actually did and $B$ worked $1/3$ as efficiently as he actually did,the work would have been completed in $3$ days. To complete the job alone,$A$ would require (in days):

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