SOLUTION: A takes 16 Days less than B to do a piece of work. If both working together can do it in 15 days ,in how many days will A alone complete the work.
My solution:
Let time taken b
Question 1085310: A takes 16 Days less than B to do a piece of work. If both working together can do it in 15 days ,in how many days will A alone complete the work.
My solution:
Let time taken by B be x
Time taken by A= x-16
One day work of B =1/x
One day work of A =1/x-16
They together complete the work in 15 days
A+B one day work = 1/15
According to question
1/(x-16) -1/x =1/15
X-(x-16)/x^2-16x= 1/15
16/x^2-16x= 1/15
15×16=x^2-16x
x^2-16x-240=0
On solving this eq my ans is not coming. Please give me the correct solution. Found 2 solutions by josgarithmetic, ikleyn:Answer by josgarithmetic(39613) (Show Source):
You made a mistake in your way. The correct equation is
+ = . (I am sure you are able to identify your error on your own !)
Now, to solve it, multiply both sides by 15*x*(x-16). You will get
15x + 15(x-16) = x*(x-16),
15x + 15x - 240 = ,
= 0. (1)
Factor left side polynomial:
(x-40)*(x-6) = 0
So, the equation (1) has two solutions: x= 40 and x= 6.
Let us check both:
1) x = 40 means B makes the job in 40 days. Then A makes it in 40-16 = 24 days.
= = = 1. Correct !!
2) x= 6 means B makes the job in 6 days. Then A makes it in 6-16 = -10 days.
It doesn't make sense.
So, the only solution is this:
Answer. B can do the job in 40 days. A can do the job in 40 - 16 = 24 days.
Solved.
It is a typical joint work problem.
There is a wide variety of similar solved joint-work problems with detailed explanations in this site. See the lessons