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$A$ does half as much work as $B$ in three-fourths of the time. If together they take $18$ days to complete a work, how much time shall $A$ & $B$ take to do it individually ?

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A does half as much work as B = $\dfrac{1}{2}$ B of work in three-fourths of the time

∴ A does $\dfrac{1}{2}$B / $\dfrac{3}{4}$T = $\dfrac{2}3{B}$ of work in t time.

Therefore, work done by A = $\dfrac{2}3{B}$

Together they take 18 days to complete a work,

in 18 days they have done 1 job

in 1 day they have done $\dfrac{1}{18}$ job

A + B = $\dfrac{1}{18}$

or, $\dfrac{2}3{B}$ + B = $\dfrac{1}{18}$

or, $\dfrac{5}3{B}$ = $\dfrac{1}{18}$

or, B = $\dfrac{3}{5*18}$

or, B = $\dfrac{1}{30}$ job/day

so,if B completes $\dfrac{1}{30}$ of a job in a day,then B takes 30 days to complete the job.

Therefore, A = $\dfrac{2}3{B}$

= $\dfrac{2}{3}$ * $\dfrac{1}{30}$

= $\dfrac{2}{3*30}$

= $\dfrac{1}{45}$ job/day

so,if A completes $\dfrac{1}{45}$ of a job in a day,then A takes 45 days to complete the job.

by Boss (15.4k points)
selected

### In 18 day work done by both will be (A+B)'s work in 18 days = (1/x)18+(2/3x)18= 30/x. now time taken by B to do same 30/x work will be (30/x)x= 30 days, similarly time taken by A to do (30/x) work will be (30/x)(3/2)x= 45 days.

by Active (2.3k points)