Time and Work
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Q.

Jay and Anup can do a job, each working alone in 30 and 15 days respectively. Jay started the work and after a few days, Anup joined him. They completed the work in 18 days from the start. After how many days did Anup join A?

 A.

6

 B.

10

 C.

12

 D.

14

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Solution:
Option(C) is correct

Since Jay does \(\left(\dfrac{1}{30}\right)^{th}\)of the work in one day, total work done by Jay in 18 days

\(\dfrac{18}{30}=\dfrac{3}{5}\)

Remaining \(\left(\dfrac{2}{5}\right)^{th}\)work was done by Anup

Since Anup does \(\left(\dfrac{1}{15}\right)^{th}\) of the work = \(\dfrac{2/5}{1/15}\)  = 6 days.

So, Anup joined Jay after $(18−6) = \textbf{12 days}$

Edit: For an alternative solution, check comment by Saif Pathan.


(2) Comment(s)


Ashutosh Upadhyay
 ()

total no of unit will be LCM of 30 and 15 is 30

jay per day unit work 30/30=1 unit

anup per day unit work 30/15=2 unit

jay let x days jay work =x*1=x unit

now anup has done 18-x days but these 18-x both have done work so

(18-x)*3

3 is total no. of unit per day

now total should be equal to 30 unit

x*1+(18-x)3=30

x=12



Saif Pathan
 ()

Jay $\frac{1}{30}$ Anup $\frac{1}{15}$

Jay worked for 18 days while Anup worked for $18-x$ days

Work done by jay $= \dfrac{18}{30}$ and anup $= \dfrac{(18-x)}{15}$

Equating,

$\dfrac{18}{30} + \dfrac{(18-x)}{15} = 1$

Solving the equation,

We get,

$x = 12$