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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.

\(19\dfrac{1}{7}\)

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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.


(4) Comment(s)


Saikrishna
 ()

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

ab/a+b=1200/70

answer: 120/7



Kavana
 ()

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
 ()

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
 ()

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.