Vincen 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
The OA is D.
I got lost. I need an expert to explain this PS question step by step. Please.
Planned output in 1/3rd of planned days = 600 tons
Sine the planned output in 1/3rd of planned days was short by 20 tons per day, and the team increased the output by 20 tons per day, the loss of output in the first 1/3rd days will be offset by the output in the next 1/3rd of days.
Actual output in the first 1/3rd days = (600 - some amount)
Actual output in the second 1/3rd days = (600 + the same some amount)
Thus, the actual output in 2/3rd days = (600 - some amount) + (600 - some amount) = 1200 tons
Thus, we have 1800 - 1200 = 600 tons to be mined in the last 1/3rd of days.
We know that the rate of output is increased by 20 tons per day over the planned output per day.
Say the planned days = x days
Thus, the planned output per day = 1800/x tons per day
Increased output per day = (1800/x + 20) tons per day
Since the target was completed in one day less, the remaining 600 tons was mined in (x/3 - 1) days.
=> 600 = (1800/x + 20)*(x/3 - 1)
Upon simplifying, we get x = 18 days
Thus, the planned output was [spoiler]1800/18 = 100[/spoiler] tons per day
The correct answer:
D
Hope this helps!
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