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saby007

· started a discussion

· 1 Months ago

sol (Question: 200720.0 )

totally wrong explanation.
A left 5 dyas ago..that menas..B+C together do the work
now..B left 2days after A.
so..B+C work for 2days= 9*2=18 unit
so last three dyas C done=4*3=12 unit
total work done in 5 dyas is=30 unit...reamaining 30 unit..which is done by A+B+C=30/15=2 dyas.
total dyas..5+2=7 days.

Question:
A, B and C complete a work in 10, 12 and 15 days respectively. They started the work together. But A left the work before 5 days of its completion. B also left the work 2 days after A left. In how many days will the work completed ? 
Options:
A) 4 days 
B) 5 days 
C) 7 days 
D) 8 days 
Solution:
Ans: (c)


If A work 5 days more than total units of work = 60 + 5 × 6 = 90 units

If B work 3 days more than total units of work = 90 + 3 × 5 = 105 units

So, time required to complete the work = \(\cfrac{105}{15}\)  = 7 days.

Pramendra Singh Samota

· commented

· 1 Months ago

Yet to be approved!

BRAJESH KUMAR GUPTA

· commented

· 1 Months ago

Yet to be approved!

Manik

· commented

· 1 Months ago

Good explanation thanks

Yogesh Kumar

· commented

· 1 Months ago

Bhai 9 kha s aaya.............?
main thoda slow hu...

Abhishek Chaudhary

· commented

· 1 Months ago

Yet to be approved!

alok ranjan

· commented

· 1 Months ago

We r assuming that 60 units of work is required.
Lets say that actual work completed in "x" days as per above given situation. Now accd to question, A left 5 days before work could be completed. So if A would have continued to work till x days then total no of units produced will be 60 + 5*6 (5 days A would have worked more*6 units/day A's capability) = 90
Similarly if B also would have continued to work till x days, then total units produced would be 90 + 3*5 = 105 units (B left 2 days after A, i.e., 3 days before completion).

Now A, B, C all worked for x days and produced 105 units in total. They are able to produced 15 units per day. So x = 105/15 = 7
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