Given:
- Working alone at its own constant rate, a machine seals k cartons in 8 hours. (say, Machine A)
- Working alone at its own constant rate, a second machine seals k cartons in 4 hours. (say, Machine B)
- The two machines, each working at its own rate and for the same time, together sealed some cartons.
To find: What percentage of the cartons were sealed by the machine working at the faster rate?
Solution:
We are interested in the machine that works at the faster rate. First, let’s find out which one this machine is.
- Machine A: k cartons in 8 hours
- Machine B: k cartons in 4 hours
Since machine B seals k cartons in less time than A, we can be sure that machine B is faster.
So, we need to find what percentage of the cartons A and B sealed together did B seal.
- We know that for a given Total Amount of Work, the Number of Workers is inversely proportional to the total Time spent doing that work.
- That is, for fixed Total Time, Rate of Work (R) ∝ Work (W).
- Hence, \(\frac{Rate-B}{Rate-A}\) = \(\frac{Work-B}{Work-A}\)
- Since Machine B takes half the time that machine A takes in doing the same work, \(\frac{Rate-B}{Rate-A}\) = \(\frac{2}{1}\).
- So, \(\frac{2}{1}\) = \(\frac{Work-B}{Work-A}\)
- If Total work = 3 units, then Work-B = 2 units and Work-A = 1 unit.
- Hence, Required percentage =\( \frac{2}{3}\) × 100 = \(66 \frac{2}{3}\%\)
Correct Answer: D
study
Working alone at its own constant rate, a machine seals k cartons in 8 hours, and working alone at its own constant rate, a second machine seals k cartons in 4 hours. If the two machines, each working at its own constant rate and for the same period of time, together sealed a certain number of cartons, what percent of the cartons were sealed by the machine working at the faster rate?
A. \(25\%\)
B. \(33\frac{1}{3}\%\)
C. \(50\%\)
D. \(66\frac{2}{3}\%\)
E. \(75\%\)