Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce 5/4 w widgets in 3 days, how many days would it take machine X alone to produce 2w widgets?
A. 4
B. 6
C. 8
D. 10
E. 12
E
E
Let machine Y takes U days to complete the work W.
So work done by Y in one day = W/U
work done by X in one day = W/(U+2)
Total work done in one day= W(1/U + 1/U+2) = W 5/4 * 1/3
solve it
Sorry it’s Option E only
E
Option C
Ans. e
12
E : 12 days
e
e
E
E
e
e
Days M/c Y takes to make W widgets = D
Days M/c X takes to make W widgets = D+2
Rate of work done by X = W/(D+2)
Rate of work done by Y = W/(D)
Combined rate of work = 1/(D+2) + 1/D = 2((D+1)W)/(D(D+2))
5W/4 widgets made in 3 days by the two m/cs, so rate = 5W/12
5W/12 = 2((D+1)W)/(D(D+2)
On solving gives D=4 ==> rate of X = 4+2 = 6. So to make 2W widgets it needs 12 days.
Hi, can somebody please explain this step by step. No clue (feel stupid). thx
E. 12
apply W/y and W/y+2 and calculate combined rate at 2W(2y+2)/y(y+2) and solve using 3 days and 5/4W to get y =4, and x=y+2=6, thus 2x=6*2=12
e
e