Aptitude Discussion

Q. |
$A$ can do piece of work in 30 days while $B$ alone can do it in 40 days. In how many days can $A$ and $B$ working together do it? |

✖ A. |
\(16\dfrac{1}{7}\) |

✔ B. |
\(17\dfrac{1}{7}\) |

✖ C. |
\(18\dfrac{1}{7}\) |

✖ D. |
\(19\dfrac{1}{7}\) |

**Solution:**

Option(**B**) is correct

Therefore $A'$s one day’s work = \(\dfrac{1}{30}\)

$B’$ s one day’s work = \(\dfrac{1}{40}\)

$(A+B)’$s one day’s work = \(\dfrac{1}{30}+\dfrac{1}{40}\)

\(\dfrac{4+3}{120}=\dfrac{7}{120}\)

Number of days required for $A$ and $B$ to finish the work = \(\dfrac{1}{7/120}=17\dfrac{1}{7}\text{days}\)

**Edit:** For an alternate way of solving this, check **Arijit GanaiB**'s comment.

**Edit 2:** For yet another alternative solution, check comment by **Vandana.**

**PULOK KHAN**

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**Saikrishna**

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A=30 B=40 then ab work together so ab/a+b

ab/a+b=1200/70

answer: 120/7

**Kavana**

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A can do a work in x days

B can do in y days

A and B can together in xy/(x+y)

x=30,y=40

==>30*40/(30+40)

=1200/70

=120/7

=17.15

**Vandana**

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LCM of 30 & 40 is 120

Hence $\dfrac{120}{30} = 4$ and $\dfrac{120}{40} = 3$

Therefore,

$\dfrac{\text{Total work}}{\text{Work done in one day}}$

Now, total work done in a day $= 3+4 = 7$

Hence, Total work done together,

$=\dfrac{120}{7} = 17.15$

**Arijit Ganai**

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Eficiency of $A=\dfrac{100}{30}=3.33\%$ & efficiency of $B=\dfrac{100}{40}=2.5\%$.

Total efficiency,

$=\dfrac{100}{(3.33+2.5)}=\dfrac{100}{5.83}=17.15$.

Hence option B is correct.

we know, ab/a+b=total time

so,40*30/40+30

=>120/7

=> 17 and 1/7 days