A large tanker can be filled by two pipes a and b in 60 minutes and 40 minutes respectively. how many minutes will it take to fill the tanker from empty state if b is used for half the time and a and b fill it together for the other half?
a. 15 min
Answers
Time taken by pipe "a"= 60 minutes
So one minute work of pipe "a"= 1/60
Time taken by pipe "b"= 40 minutes
one minute work of pipe "b"= 1/40
Now first we open pipe b only suppose it is opened for x minutes
So work done ( tank fill) by pipe "b" in x minutes = x* ( 1/40 )
Now and b work together for another x minutes
let us first find the combined one minute work of a and b = ( 1/60) + ( 1/40)
when we simplify we get : ( 40 + 60 )/( 40*60) = 100/ 2400 = 1/24
Because they work together for x minutes so the work done by them in x minutes = x*( 1/24 )
This entire work should be equal to 1 ( complete work )
work done by a only + work done by a and b together = 1
x*(1/40 ) + x*( 1/24) = 1
( x/40 ) + ( x/24) = 1
Let us simplify it and solve for x
24x + 40x = 1
------------------
24*40
64x = 1
---------
960
64x = 960
x= 960/ 64 = 15
total time take =
first x minutes ( when only a worked) + next x minutes when a and b worked together
Answer
total time = 15 + 15 = 30 minutes
☆Question ☆
A large tanker can be filled by two pipes a and b in 60 minutes and 40 minutes respectively. how many minutes will it take to fill the tanker from empty state if b is used for half the time and a and b fill it together for the other half?
☆Answer ☆
In the attachment...
