I want the proper solution of this question...if you need 100 points...
Along with this I also need the answer of the following question...
This one is little bit easy. but I need the solution and properly.
Karan and Pavan can complete a job in 40 days and 50 days , respectively. They worked on alternative days to complete it . Find the minimum possible time in which they could have completed it.(in days)
(it can be in fractions also)
Answers
Answer:
Option(b) - 2(1/2) hours
Step-by-step explanation:
A and B can fill a tank in 6 hours and 9 hours.
A's 1 - hour work = (1/6)
B's 1 - hour work = (1/9)
Also said that, C can empty the tank in 7.5 hours.
C's 1 - hour work = (1/7.5)
Let it be 'x' hours tap C to be kept open.
Given that,
Tap B is opened 3 hours after tap A is opened.
⇒ 6 * (1/6) + 3 * (1/9) - (x/7.5) = 1
⇒ 1 + (1/3) - (x/7.5) = 1
⇒ (4/3) - (x/7.5) = 1
⇒ -(x/7.5) = -1/3
⇒ -x = 2.5
⇒ x = 2.5
⇒ x = 2 (1/2) hours.
Therefore,
Tap C to be kept open for 2(1/2) hours.
Hope it helps!
Answer:
A can do in = 40 days
B can do in = 50 days
both can do in = 40×50÷40+50
= 2000/90
= 200/9 days
= 200/9×24 hours
= 533.333333......... hours
This Will be the minimum time working as repeatedly