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A team of copper miners

This topic has 2 expert replies and 0 member replies

A team of copper miners

Post Mon Sep 25, 2017 6:42 pm
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.

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Post Mon Sep 25, 2017 10:24 pm
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 1800/18 = 100 tons per day

The correct answer: D

Hope this helps!

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Matt@VeritasPrep GMAT Instructor
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Post Tue Sep 26, 2017 5:00 pm
If you find yourself getting lost on these, take them one chunk at a time rather than all at once.

First chunk:

The miners wanted to mine 1800 tons of ore in d days, or 1800/d per day.

Second chunk:

For the first d/3 days, the miners were only able to mine (1800/d) - 20 tons per day.

Third chunk:

For the remaining ((2/3)d - 1) days, the miners were able to mine (1800/d) + 20 tons per day. (The -1 is important: the team was able to finish one day ahead of the planned schedule of d days.)

We're told that this was all of the work needed to be done, so

(d/3) * (1800/d - 20) + ((2/3)d - 1) * (1800/d + 20) = 1800

Then solve for d, and you too will be done Smile

d = 18, making the original plan for 1800/d => 1800/18 => 100 tons per day

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