Aptitude - Time and Work - Discussion

Discussion Forum : Time and Work - General Questions (Q.No. 9)
9.
A does 80% of a work in 20 days. He then calls in B and they together finish the remaining work in 3 days. How long B alone would take to do the whole work?
23 days
37 days
371/2
40 days
Answer: Option
Explanation:

Whole work is done by A in ( 20 x 5 ) = 25 days.
4

Now, ( 1 - 4 ) i.e., 1 work is done by A and B in 3 days.
5 5

Whole work will be done by A and B in (3 x 5) = 15 days.

A's 1 day's work = 1 , (A + B)'s 1 day's work = 1 .
25 15

Therefore B's 1 day's work = ( 1 - 1 ) = 4 = 2 .
15 25 150 75

So, B alone would do the work in 75 = 37 1 days.
2 2

Discussion:
166 comments Page 16 of 17.

Abica said:   6 years ago
Nice explanation, Thanks @Vikram Sihag.

Siddhant said:   6 years ago
80% work done by A -------->20 days (given)
1 work ie. 100% work done by A ------->20*100/80 = 25 days
therefore, total work is done by A = 25 days

Now,
20% work done by (A+B) implies,

20/100 work done --------> 3 days.
1 work ie. 100% work done in ------>3*100/20.= 15 days.

Now we have,
A's 1 days work = 1/25 ----------------------> 1
and, (A+B)'s 1 days work = 1/15 -------------------------> 2

Putting A's work in equation 2.
B's one day work = 2/75.

Therefrom total work done by B = 75/2.
= 37 1/2.

Devika said:   6 years ago
Solution by LCM Method:

A does 80% of the work in 20 days => 4/5th of the work in 20 days
=> (4/5)x = 20 => x=25 days (No of days A takes to complete the work)

Remaining work = 1/5 which is completed by A+B in 3 days.
(1/5)x = 3 => x=15 days (No of days A+B takes to complete the work).

Therefore,
A - 25 days - 3 units.
A+B - 15 days - 5 units.
LCM - 75.
A+B = 5 => B=2 units.
Therefore, total work done by B = 75/2.
= 37 1/2.

Surekha Merla said:   5 years ago
Let's take 5 parts total.
A finishes 80% of it in 20 days, A can finish whole work in 25 days.
A one day work is 1/25.
A= 4/5 work in 20 days.
Remaining, 1-(4/5)= 1/5 part.
(A+B)= 1/5 work in 3 days.
A's 3 days work = 3/25.
Therefore,
For 3 days, A+B= 1/5.
(3/25)+B= 1/5.
B= 2/25 in 3 days.
B's one day work, 2/75 work.
Whole work by B= 75/2.
B = 37 1/2 days.

Sahil Negi said:   5 years ago
Thanks @Ganesh.

Mr pai said:   5 years ago
Thank you so much @Baskar.

Ravi said:   5 years ago
A do 80%work for 20days.

(Simplify it remaining 20%do in 5days).

Now work for 20%is 1/5 and work are done by both a+b is 1/3. Do lcm we get 15 and subtract 5-3/15=2/15.
Now it is 20% of work done by B is 2/15.
Now 20%*5=100% i.e,2/15*5=2/75 ans is 37 1/2 days.

Pavani Durga said:   5 years ago
Thank you @Rajendra.

Rashmi said:   5 years ago
Thank you @Baskar.

S. Kalyan said:   5 years ago
80% = (80/100).

so, A's (80/100) work done in 20 days.

Total Work of A = (100/80 * 20) = 25 //
Now if we consider total Work as 1.
4/5 part was already done by A remaining 1/5 he completed with the help of B.

So, (1/5) of work done by (A+B) in 3 days,
by calculating we get (A+B) = 3*(5/1) = 15 days,

Now to find B's Work we have A+B=15.
substitute A value and in above equation i.e (A+B = 15).

so
1/25 + 1/B = 1/15.
1/B = (1/15) - (1/25) ---- (LCM = 75).
1/B = (5-3) / 75,
1/B = 2/75,
B = 75 / 2 = 37 1/2.


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