theperfectgentleman wrote:
A team of copper miners planned to mine 1800 tons of ore during certain number of days. Due to technical difficulties in the one third of the planned number of days, the team was able to achieve an output of 20 tons less than the planned output per day. To make up for this, the team overachieved for the rest of the days by 20 tons per day. The end result was the team completed the task one day ahead of time. How many tons of ore did the team initially plan to mine per day (Tons)?
A) 50
B) 200
C) 150
D) 100
E) 250
We can let x = the number of days originally planned for mining 1800 tons of ore. Thus, 1800/x tons of ore were originally planned to be mined per day.
For the first ⅓ of the x days, 20 fewer tons of ore per day were mined; thus, for the first ⅓ of the days, a total of (⅓x)(1800/x - 20) tons of ore were mined.
After the first ⅓ of the days, 20 more tons of ore per day were mined. However, since the 1800 tons of ore were completely mined 1 day ahead of schedule, for 1 day less than the last ⅔ of the x days, a total of (⅔x - 1)(1800/x + 20) tons of ore were mined.
Thus, we can create the following equation:
(⅓x)(1800/x - 20) + (⅔x - 1)(1800/x + 20) = 1800
600 - (20/3)x + 1200 + (40/3)x - 1800/x - 20 = 1800
(20/3)x - 1800/x - 20 = 0
Multiply both sides by 3x, we have:
20x^2 - 5400 - 60x = 0
x^2 - 3x - 270 = 0
(x - 18)(x + 15) = 0
x = 18 or x = -15
Since x, the number of days, can’t be negative, x = 18. Thus, the number of tons of ore planned to be mined per day is 1800/x = 1800/18 = 100.
Answer: D