Aptitude Discussion

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 |

**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.**

**Ashutosh Upadhyay**

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**Saif Pathan**

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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,

$x = 12$

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