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M16-12

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M16-12  [#permalink]

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New post 15 Sep 2014, 23:58
1
20
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

37% (01:07) correct 63% (01:19) wrong based on 165 sessions

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Re M16-12  [#permalink]

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New post 15 Sep 2014, 23:58
2
4
Official Solution:


Let \(P\) denote the original price.

Statement (1) by itself is insufficient. If \(x\) is much larger than \(y\), the new price is higher than the original. But if \(x\) is only marginally larger, the new price is lower. For example, if \(x = 20\) and \(y = 19\), the new price is \(P*1.2*0.81 \lt P\).

Statement (2) by itself is insufficient. Use the same reasoning. If \(y\) is large, the new price is small (if \(y = 100\), the new price is 0). If \(y\) is small, the new price is higher than the original (if \(y = 10\), the new price is \(P*1.12*0.9 = P*1.008 \gt P\)).

Statements (1) and (2) combined are insufficient. Adding S1 to S2 provides no new information.


Answer: E
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Re: M16-12  [#permalink]

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New post 16 Apr 2015, 06:44
Bunuel wrote:
Official Solution:


Let \(P\) denote the original price.

Statement (1) by itself is insufficient. If \(x\) is much larger than \(y\), the new price is higher than the original. But if \(x\) is only marginally larger, the new price is lower. For example, if \(x = 20\) and \(y = 19\), the new price is \(P*1.2*0.81 \lt P\).

Statement (2) by itself is insufficient. Use the same reasoning. If \(y\) is large, the new price is small (if \(y = 100\), the new price is 0). If \(y\) is small, the new price is higher than the original (if \(y = 10\), the new price is \(P*1.12*0.9 = P*1.008 \gt P\)).

Statements (1) and (2) combined are insufficient. Adding S1 to S2 provides no new information.


Answer: E


Hi Bunuel !

if x is only marginally larger, the new price is lower.
This doesn't hold true for x= 12 and y= 10

if we start with 100,
100 incr by 12% = 112, then decr by 10% = 112-11.2 = 100.8 (greater than 100)

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Re: M16-12  [#permalink]

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New post 07 Sep 2015, 15:25
Well E is correct.

Lets assume principal is P

P(1+x/100)(1-y/100) is the final price.

Now if we have x=1.2y

P(1+x/100)(1-x/120)
To find out if this will be <P

P(1+x/100)(1-x/120)<P

or (100+x)(120-x)<120*100
or 20x-x^2< 0

or x>20

So lets take Y = 30
X =42

So 100 becomes 142 and multiply 142 by 0.7 =99.4

So X=12, Y =10.. The final value is greater than P
but in X =42, Y=30.. the final value is less than P
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Re: M16-12  [#permalink]

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New post 08 Feb 2016, 06:06
lastminutemba 1.2x/100 is not equal to x/120
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Re: M16-12  [#permalink]

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New post 24 Jul 2016, 07:26
Hi! Is it fair to say that we can never know if the original price will be higher or lower when given only the percents?
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Re: M16-12  [#permalink]

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New post 27 Jul 2016, 05:46
There is not enough information to say/answer the question since we don't know x>y or x<y
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Re: M16-12  [#permalink]

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New post 13 Sep 2016, 17:34
lastminutemba
As per (2) x=1.2y, If y=30 then x=36 not 42, so P becomes 136 and then 95.2.
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Re: M16-12  [#permalink]

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New post 10 Oct 2016, 11:37
1
Bunuel wrote:
If a price was increased by \(x\)% and then decreased by \(y\)%, is the new price higher than the original?


(1) \(x \gt y\)

(2) \(x = 1.2y\)



Lets assume that New price is higher than the old price and see if it always holds true.

\(P*\frac{100+X}{100}*\frac{100-Y}{100} - P > 0\)

Solving, 100*(X-Y) > XY

1. Given, \(x \gt y\)
Now when X is infinitesimally larger then y , then L.H.S of the equality tends to 0 and RHS is X squared or Y squared, which can be large depending on the value of the X
When X is much larger than Y, say X = 100000, Y = 0.0001, then LHS is much larger then RHS
Hence the inequality flips sign based on the values of X and Y so not sufficient.

2. Putting \(x = 1.2y\) in our derived inequality
20> 1.2 Y
hence again whether this always holds true will depend on the value of Y. Not sufficient.

The condition given in point 2 also satisfies point 1 hence together also they are not sufficient.
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M16-12  [#permalink]

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New post 20 Feb 2017, 08:48
LanDrobnack
IMO it's not correct to say that one cannot determine whether the new price will be higher/lower than the original price if no percents are given.
If the percent change is given one should always be able to determine the relationship between the two prices; the thing is that in this case you are only given the relationship between the two percent changes!
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Re: M16-12  [#permalink]

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New post 06 Mar 2017, 10:50
Hi. New to the forum.

I think I must be doing something wrong but I cannot seem to find the answer choices anywhere. For the verbal question of the day the answers were displayed underneath the question but they seem to be absent in this case. What am I missing?
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Re: M16-12  [#permalink]

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New post 06 Mar 2017, 10:54
1
brettbrettbrettcom wrote:
Hi. New to the forum.

I think I must be doing something wrong but I cannot seem to find the answer choices anywhere. For the verbal question of the day the answers were displayed underneath the question but they seem to be absent in this case. What am I missing?


This is a data sufficiency question. Options for DS questions are always the same.

The data sufficiency problem consists of a question and two statements, labeled (1) and (2), in which certain data are given. You have to decide whether the data given in the statements are sufficient for answering the question. Using the data given in the statements, plus your knowledge of mathematics and everyday facts (such as the number of days in July or the meaning of the word counterclockwise), you must indicate whether—

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
D. EACH statement ALONE is sufficient to answer the question asked.
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

I suggest you to go through the following post ALL YOU NEED FOR QUANT.

Hope this helps.
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Re: M16-12  [#permalink]

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New post 15 Jul 2018, 05:22
Great question, almost got tricked into selecting C, but then decided to continue solving and plugging in values to test.

My solution below (a bit too time-consuming for GMAT, but helps me with understanding).
Let P = original price.
New price = P(1+ x/100)(1- y/100) = P(1+ x/100 - y/100 -xy/1000)
Test if (1+ x/100 - y/100 -xy/1000) > 1
x-y- xy/100 > 0
100x - 100y - xy > 0 --- (1)

Statement (1) gives the same information as statement (2), which has more details.

Sub x = 1.2y into (1),
120y - 100y - (1.2y)*y > 0
20y - 1.2y^2 > 0
y (20 - 1.2y) > 0

Since y is a %, it has to be > 0
20 - 1.2y > 0
1.2y < 20
y < 20/1.2 = 16

The new price is only higher for y < 16
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Re: M16-12 &nbs [#permalink] 15 Jul 2018, 05:22
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