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

Question

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

A) 50
B) 200
C) 150
D) 100
E) 250

godot53 · Accepted Answer

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

JeffTargetTestPrep · Answer

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

pushpitkc · Answer

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)

roadrunner · Answer

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 ?

godot53 · Answer

Thank you, I considered 100 for A and B also...
Edited my post.

Smokeybear00 · Answer

How do you know that the number of planned days needs to be a multiple of 3?

pawani · Answer

Has anyone tried solving the question by making equations?

gracie · Answer

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
D

hellosanthosh2k2 · Answer

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)

HarshavardhanR · Answer

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Symoki786 · Answer

The number of days will be a multiple of 3. Now, it is given that the one third of the days, 20 tonnes were less produced, and to compensate on other days 20 tonnes were produced. So, if we take triplets of 3 days, then for every 3 days, an extra 20 tonnes were produced. For eg: if the rate was 40, then it should have been 40,40,40(total= 120). but it is 20,60,60(total = 140). Now let us assume that there were x days. then total such triplets possible will (x/3 -1), because it takes 1 day less. and the extra work done in these triplets will be equal to a day's work.

(x/3 -1)*20 = 1800/x
(x/3 -1) = 90/x
x^2 - 3x = 270
x^2 -18x + 15x -270 =0
x(x-18) + 15(x-18)=0

this gives x =18. as x= -15, not possible.
therefore the earlier comes out to be 1800/18 = 100.

IMO option D.

bumpbot · Answer

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