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A team of copper miners planned to mine 1800 tons of ore during

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A team of copper miners planned to mine 1800 tons of ore during  [#permalink]

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New post 17 Jun 2017, 02:37
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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
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Re: A team of copper miners planned to mine 1800 tons of ore during  [#permalink]

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New post 17 Jun 2017, 03:09
I went by options for this one!
1800 = number of days(let it be x) * units of work per day(let it be u)

For 1/3rd of the total days, the rate is u - 20
For the remaining 2/3rd of the days is u + 20
Since, the finish the work in 1 day less, u units of work must be reduced from the total amount of work
We are given the answer options for the initial amount of units, u.

If u=50, x=1800/50 = 36
First 1/3rd days = x/3*(u-20) = 12*30 = 360
Next 2/3rd days - 1day(since work completes before time) = (2x/3 - 1)*(u+20) = 25*70 = 1750
Sum is greater than 1800

If u=200, x=1800/200 = 9
First 1/3rd days = x/3*(u-20) = 3*180 = 540
Next 2/3rd days - 1day(since work completes before time) = (2x/3 - 1)*(u+20) = 8*220 = 1780
Sum is greater than 1800

If u=150, x=1800/150 = 12
First 1/3rd days = x/3*(u-20) = 4*130 = 520
Next 2/3rd days - 1day(since work completes before time) = (2x/3 - 1)*(u+20) = 7*170 = 1190
Sum is lesser than 1800

If u=100, x=1800/100 = 18
First 1/3rd days = x/3*(u-20) = 6*80 = 480
Next 2/3rd days - 1day(since work completes before time) = (2x/3 - 1)*(u+20) = 11*120 = 1320
Sum is equal to 1800. Hence, that's our answer. (Option D)

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A team of copper miners planned to mine 1800 tons of ore during  [#permalink]

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New post Updated on: 19 Jun 2017, 05:45
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First we need the total number of "planned days" to be a multiple of 3. So,
A. 1800/50 = 36
B. 1800/200 = 9
C. 1800/150 = 16 [ eliminate]
D. 1800/100 = 18
E. 1800/250 = not an integer [ eliminate]

Between A,B and D - applying the condition given,

A: 12 X 30 + 23 X 70 = not equal to 1800... eliminate
B : 3 X 180 + 5 X 220 = not equal to 1800..eliminate
D. 6 X 80 + 11 X 120 = 480 + 1320 = 1800 - and we are done.

To note, the answer choices are not in ascending or descending order, so theperfectgentleman.. whats the source of this problem ?

Cheers !! :-D
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Originally posted by godot53 on 19 Jun 2017, 04:48.
Last edited by godot53 on 19 Jun 2017, 05:45, edited 1 time in total.
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Re: A team of copper miners planned to mine 1800 tons of ore during  [#permalink]

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New post 19 Jun 2017, 05:34
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godot53 wrote:
First we need the total number of "planned days" to be a multiple of 3. So,
A. 1800/50 = 36
B. 1800/200 = 9
C. 1800/150 = 16 [ eliminate]
D. 1800/100 = 18
E. 1800/250 = not an integer [ eliminate]

Between A,B and D - applying the condition given,

A: 12 X 80 + 22 X 120 = much above 1800... eliminate
B : 3 X 80 + 5 X 120 = 240 + 600 = 840..eliminate
D. 6 X 80 + 11 X 120 = 480 + 1320 = 1800 - and we are done.

To note, the answer choices are not in ascending or descending order, so theperfectgentleman.. whats the source of this problem ?

Cheers !! :-D

Any reason why you chose 80 & 120 as rate across all choices? Shouldn't it be

A: 12* 30 + 23* 70
B: 3 * 180 + 5*220
D: 6*80 + 11*120 ?
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A team of copper miners planned to mine 1800 tons of ore during  [#permalink]

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New post 19 Jun 2017, 05:43
roadrunner wrote:
Any reason why you chose 80 & 120 as rate across all choices? Shouldn't it be

A: 12* 30 + 23* 70
B: 3 * 180 + 5*220
D: 6*80 + 11*120 ?


Thank you, I considered 100 for A and B also...
Edited my post.
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Re: A team of copper miners planned to mine 1800 tons of ore during  [#permalink]

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New post 19 Jun 2017, 08:18
godot53 wrote:
First we need the total number of "planned days" to be a multiple of 3. So,
A. 1800/50 = 36
B. 1800/200 = 9
C. 1800/150 = 16 [ eliminate]
D. 1800/100 = 18
E. 1800/250 = not an integer [ eliminate]



How do you know that the number of planned days needs to be a multiple of 3?
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Re: A team of copper miners planned to mine 1800 tons of ore during  [#permalink]

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New post 19 Jun 2017, 11:08
Has anyone tried solving the question by making equations?
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A team of copper miners planned to mine 1800 tons of ore during  [#permalink]

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New post 19 Jun 2017, 14:30
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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


let 3d=planned work days
t=1800/3d=planned tons per day
d(t-20)+(2d-1)(t+20)=1800
t(3d-1)+20d-20=1800
substituting 1800/3d for t,
d^2-d-30=0
d=6
3d=18
t=1800/18=100 tons per day
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Re: A team of copper miners planned to mine 1800 tons of ore during  [#permalink]

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New post 20 Jun 2017, 18:42
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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
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A team of copper miners planned to mine 1800 tons of ore during  [#permalink]

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New post 29 Oct 2017, 23:23
Let '3n' be number of days planned.

so planned ores /day = 1800/(3n) = 600/n

1/3 of planned days i.e. for n days => 20 less than estimated per day => total ores = n * ((600/n) - 20)
now by extra 20 ores/day, team was able to extract one day ahead => (2n - 1) days => total ores => (2n - 1) ( (600/n) + 20)

So, n * ((600/n) - 20) + (2n - 1)((600/n) + 20) = 1800

Solving above quadratic eqn,n = 6 0r -5, as n cant't be -ve, n = 6

so ore extracted per day = (600/n) = 100

Answer (D)
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A team of copper miners planned to mine 1800 tons of ore during &nbs [#permalink] 29 Oct 2017, 23:23
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