Hi there!

Beautiful (high-level) problem... let me use it as a nice example to show some of the features of my method, ok?!

(i) Let us put the question stem into Math....

Given: m (1-x/100)(1-y/100) = n

Asked (therefore our "focus"): (I will use the fact that m is positive... where?)

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n/m = (1-x/100)(1-y/100) < 1 ?
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(1) Focusing in the question above, it is easy to see that sttm (1) is equivalent to:

y. (1-x/100) = x then (y/100).(1-x/100) = x/100 then (-y/100).(1-x/100) = -x/100 then the last one, a bit more tricky...

then (1-y/100).(1-x/100) = -x/100 + (1-x/100) , therefore sttm (1) tells us that (1-x/100)(1-y/100) = 1-x/50 correct ?!

(Please note that there is no room here for "random manipulations"... I am 100% focus on the question above!!!)

Therefore we must ask ourselves: is n/m = 1-x/50 less than 1? Answer: YES, because x/50 is positive (we assume x is not zero by the context) therefore 1 - positive is less than 1.

We are done for sttm(1), that DECIDES about the question asked, so it is sufficient.


(2) Let us go back to the focus and let us use the proportion y : x = 3 : 4 using my "k technique", I mean, y = 3k and x = 4k, where (in this context) k is certainly positive. For a reason that deals ONLY with easiness in calculations, we MAY and we will change y = 3k to y = 100.3k and x = 4k to x = 100.4k .... no generallity was lost, but the calculations will be much easier... observe:


n/m = (1-100.4k/100)(1-100.3k/100) = (1-4k)(1-3k) < 1 ?

Now it´s just a matter of studying an easy 2o degree inequality... trivial!

Let´s do it: (1-4k)(1-3k) < 1 is equivalent to 12 k^2 - 7k < 0 therefore k(12k-7) < 0 is equivalent to k < 7/12 ? (because I know k is positive)

Now we have to be careful... y = 300k and x = 400k must be between 0 and 100, correct?

If k is greater than 7/12, that means that 400k is greater than 400.7/12 but this is more than 100, therefore we guarantee that k is less than 7/12... therefore sttm (2) is also sufficient.

The answer is D!

Regards,
Fabio.

P.S.: although I may have done something wrong (I´m in a hurry, therefore I typed directly here...sorry), the "philosophy" I use is my method, I mean, we are not randomly "shooting" for numbers, for smart tricks, it is really applied elementary math, but with FOCUS, with care and with techniques. I really hope you see the difference...

Hi there!

Beautiful (high-level) problem... let me use it as a nice example to show some of the features of my method, ok?!

(i) Let us put the question stem into Math....

Given: m (1-x/100)(1-y/100) = n

Asked (therefore our "focus"): (I will use the fact that m is positive... where?)

Fabio,

m (1-x/100)(1+y/100) = n

Please check it.

selango wrote:

Hi there!

Beautiful (high-level) problem... let me use it as a nice example to show some of the features of my method, ok?!

(i) Let us put the question stem into Math....

Given: m (1-x/100)(1-y/100) = n

Asked (therefore our "focus"): (I will use the fact that m is positive... where?)

Fabio,

m (1-x/100)(1+y/100) = n

Please check it.

You are absolutely right, selango. Thanks for the polite way of saying that I am wrong! Now I simply have no time to make the small modifications needed, could you have a try and do it for me? I guess it would be a really good exercise... have a try! (I will come back later and see/help what you did/tried... deal?)

Regards,
Fabio.

Rahul@gurome wrote:

replayyyy wrote:The price of a car was m dollars. It then depreciated by x%. Later, it appreciated by y% to n dollars.
If there are no other changes in the price, is m > n ?

(1) y = x/(1-x/100)

(2) y = 3/4*x

Given: Price was m dollars. Decreased by x%. New price say, p = m - (mx/100) = m[1 - (x/100)]
Later increased by y%. Final price, n = p + (py/100) = p[1 + (y/100)] = m[(1 - (x/100)][1 + (y/100)]

Statement 1: y = x/[1 - (x/100)] = (100x)/(100 - x) => y/100 = x/(100 - x) => [1 + (y/100)] = 100/(100 - x)
Therefore, n = m[(1 - (x/100)][1 + (y/100)] = m[(100 - x)/100][100/(100 - x)] = m

Sufficient.

Statement 2: y = 3x/4. Therefore,
# [(1 - (x/100)][1 + (y/100)]
= [(1 - (x/100)][1 + (3x/400)]
= [1 - (x/100) + (3x/400) - (3x^2/40000)]
= [1 - (x/400) - (3x^2/40000)] < 1

Therefore, n = m[(1 - (x/100)][1 + (y/100)] < m

Sufficient.

The correct answer is D.

Rahul,

I followed the same steps.

But in stmt1 n=m and stmt2 n<m

Both statements contradict?

The official answer, though, is A. I do not agree with it and that is why I have posted this question here. You can see the explanation provided in the attached file.

I´m back.

I guess nobody had interested in modifying my (pseudo-) solution, therefore I will just say that it is really quick and the inequalities modified are even easier than the ones I shown above. In this sense, I guess the (misunderstood) problem (the one I solved) is even nicer/harder/richer in details than the original one.

As far as the "contradiction" in the (D) alternative is concerned (that really occurs in the problem as originally posted by replayyyy), I´ve already (lengthly) mentioned in another BTG DS post that this is not really an issue (it does NOT invalidade the question stem), but I will not discuss it further.

Regards,
Fabio.

@Rahul´s comment: I also believe/used that the percentages were positive...

If you do not translate the problem into any equations or inequalities for Statement 2 and try to solve it with just a pure logic, it is sufficient. Since the stem says "depreciated by x%" and not "changed by x%", x cannot be negative ( depreciated by -10%, for example, doesn`t make sense - it either depreciates by 10% or changes by -10%). If this is true, the fact that y = 3/4x means that y IS smaller than x (because it appreciated, again cannot be negative). If you decrease a number with no matter what % and then increase the new value with smaller %, in every case the result will be smaller than the original value. Is there a flaw in the above argument?

replayyyy wrote:If you do not translate the problem into any equations or inequalities for Statement 2 and try to solve it with just a pure logic, it is sufficient. Since the stem says "depreciated by x%" and not "changed by x%", x cannot be negative ( depreciated by -10%, for example, doesn`t make sense - it either depreciates by 10% or changes by -10%). If this is true, the fact that y = 3/4x means that y IS smaller than x (because it appreciated, again cannot be negative). If you decrease a number with no matter what % and then increase the new value with smaller %, in every case the result will be smaller than the original value. Is there a flaw in the above argument?

NO, everything you wrote is PERFECT, and that means that:

(a) we do not have any more doubts that the percentages could NOT be negative ;
(b) sttm (2) could have been done without calculations.

Congrats,
Fabio.

replayyyy wrote:If you do not translate the problem into any equations or inequalities for Statement 2 and try to solve it with just a pure logic, it is sufficient. Since the stem says "depreciated by x%" and not "changed by x%", x cannot be negative ( depreciated by -10%, for example, doesn`t make sense - it either depreciates by 10% or changes by -10%). If this is true, the fact that y = 3/4x means that y IS smaller than x (because it appreciated, again cannot be negative). If you decrease a number with no matter what % and then increase the new value with smaller %, in every case the result will be smaller than the original value. Is there a flaw in the above argument?

Very true Buddy, Has to be D,
Nice explanation though