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Bunuel
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I think the stem says that the value depreciates by the end of each year, so after the end of year 1 itself we have Original * 0.6 IMO. So by the end of 3rd year it becomes 0.216x IMO.
Statement A says, it lost more than 12k in 2nd year, meaning 0.6x was at the start of year 2 and it became 0.36x, thus decrease of 0.24x.
Thus 0.24x = 12000 then we get x = 50000 and if x = 50000 then,
50000*0.216 (end of 3rd year) = 50*216 = 10,800. Thus no lesser than 10,800.

Sufficient.
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A design studio bought a commercial 3D printer. If at the end of each year the printer was worth 60% of what it was worth at the beginning of that year, was the printer worth less than $10,800 at the end of 3 years?

(1) The printer lost more than $12,000 in value during its second year.


If at the end of the second year the printer was worth 60% of its value at the beginning of the year, then it lost 40% of its value.

So, 40% was more than $12,000.

We thus know that, at the beginning of the second year, the printer was worth more than $12,000/0.40 = $30,000.

So, at the end of the second year, the printer was worth more than $30,000 - $12,000 = $18,000.

Further, at the end of the third year, the printer was worth more than $18,000 × 0.60 = $10,800.

So, using this statement, we can determine that the answer to the question is No, the computer was not worth less than $10,800.

Sufficient.

(2) The printer’s starting value was more than $49,000.

Clearly, if the starting value was greater than $49,000, the value at the end of 3 years could be greater than $10,800.

So, the key question here is that of whether the value at the end of 3 years could be less than $10,800.

It's easier to use $50,000 as a starting point. So, let's do that and see what we get.

50,000 × 0.6 × 0.6 × 0.6 = 50 × 216 = $10,800.

So, if the starting value was between $49,000 and $50,000, the final value was less than $10,800.

Thus, given this statement, the final value could be greater or less than $10,000.

Insufficient.

Correct answer: A
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Since it is worth 60% of its value at the beginning of each year:
Value after 3 years=(0.6)*(0.6)*(0.6)V=0.216V
0.216V<10,800
V<50000
So the question is equivalent to asking: Was the starting value less than $50,000?
S1:
The loss during the second year was more than $12,000.
Value at start of second year =0.6V.
Loss during second year:
0.4(0.6V)=0.24V
0.24V>12,000
V>50,000
Sufficient

S2:
V>49,000 is very vague and many values can be assumed. So it is insufficient.
Answer is A
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