Although many answer explanations have been given above, but since mine was different - thought to share this perspective with the group.
I am using a simple quadratic equation to solve.
It is given that A takes sometime more than 3 minutes to make 20 burgers, let us say that he takes 3+x minutes. ------ (1)
Also is given that together A, B and C take 8 minutes to make 80 burgers (therefore B and C together make 60 burgers in {8 - (3+x)} mins. or (5-x) mins) ------- (2)
Also given that A, B, C make 20 burgers per minute, all working together. ------ (3)
In 1 min, A alone can make 20/(3+x) burgers, ------ (4)
In 1 min, B+C can together make 60/(5-x) burgers ------ (5)
Combining statements (3), (4) and (5),
\(20/(3+x) + 60/(5-x) = 20\)
Simplifying, we get, \(1/(3+x) + 3/(5-x) = 1\)
or \(x^2 - 1 = 0\) => x = 1 or -1
Since A's time is more than 3, therefore it cannot be -1. Hence x = 1.
Therefore, A takes 3+x = 3+1 = 4 mins to make 20 burgers => 4*8 = 32 mins to make 160 burgers.
virtualanimosity wrote:
3 cooks have to make 80 burgers.They are known to make 20 pieces every minute working together.The 1st cook began workin alone and made 20 pieces having worked for sometime more than 3 mins.The remaining part of the work was done by second and 3rd cook working together.It took a total of 8 minutes to complete the 80 burgers.How many minutes would it take the 1st cook alone to cook 160 burgers?
A. 16 minutes
B. 24 mins
C. 32 mins
D. 40 mins
E. 30 mins