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Manager  Status: Never ever give up on yourself.Period.
Joined: 23 Aug 2012
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The price of an automobile decreased m percent between 2010  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 63% (02:11) correct 38% (02:54) wrong based on 161 sessions

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The price of an automobile decreased m percent between 2010 and 2011 and then increased n percent between 2011 and 2012. Was the price of the automobile lower in 2010 than in 2012?

(1) m < n
(2) mn < 100n – 100m

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Re: The price of an automobile decreased m percent between 2010  [#permalink]

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The price of an automobile decreased m percent between 2010 and 2011 and then increased n percent between 2011 and 2012. Was the price of the automobile lower in 2010 than in 2012?

The price in 2010 - $$p$$;
The price in 2011 - $$p*(1-\frac{m}{100})$$;
The price in 2012 - $$p*(1-\frac{m}{100})*(1+\frac{n}{100})$$;

Question: is $$p<p*(1-\frac{m}{100})*(1+\frac{n}{100})$$? --> is $$1<(1-\frac{m}{100})*(1+\frac{n}{100})$$? --> is $$100n-100m>mn$$?

(1) m < n. Not sufficient.
(2) mn < 100n – 100m. Directly answers the question. Sufficient.

Identical question from OG to practice: the-annual-rent-collected-by-a-corporation-from-a-certain-89184.html
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Re: The price of an automobile decreased m percent between 2010  [#permalink]

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1 ) is clearly insufficient

2) $$MN$$ $$<$$ $$100 N - 100 M$$

here a faster approach is simply pickying a %. for instance: first we have a decrease so we could have 0.8 % (decrease by 20%) the an increse %, for instance 1.2 % (increase by 20 %).

Always the LHS is not smaller than the RHS

$$0.96 < 0.4$$ The answer, even with other numbers will be always NO. So is sufficient

B is the answer
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Re: The price of an automobile decreased m percent between 2010  [#permalink]

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Bunuel wrote:
The price of an automobile decreased m percent between 2010 and 2011 and then increased n percent between 2011 and 2012. Was the price of the automobile lower in 2010 than in 2012?

The price in 2010 - $$p$$;
The price in 2011 - $$p*(1-\frac{m}{100})$$;
The price in 2012 - $$p*(1-\frac{m}{100})*(1+\frac{n}{100})$$;

Question: is $$p<p*(1-\frac{m}{100})*(1+\frac{n}{100})$$? --> is $$1<(1-\frac{m}{100})*(1+\frac{n}{100})$$? --> is $$100n-100m>mn$$?

(1) m < n. Not sufficient.
(2) mn < 100n – 100m. Directly answers the question. Sufficient.

Identical question from OG to practice: the-annual-rent-collected-by-a-corporation-from-a-certain-89184.html

Hi Bunuel,
How we can say that Stat.1 is NOT sufficient even without plugging in values? I'm getting it clear by plugging values but what I'd like to know is there any faster way to determine the sufficiency of Stat.1 from the inequality?

Would you please explain?
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Re: The price of an automobile decreased m percent between 2010  [#permalink]

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Posting official solution of this problem.
Attachments official_6.PNG [ 78.81 KiB | Viewed 1268 times ]

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GMAT 1: 780 Q51 V46 Re: The price of an automobile decreased m percent between 2010  [#permalink]

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1
Hi Daviesj,

The question here tests us on the concept of successive percentages. When dealing with successive percentage questions the following formula comes in very handy.

Overall %age = a + b + ab/100

where a and b is the percentage increase and decrease respectively. If a and b are percentage increases then we use a and b as positive. If a and b are percentage decreases then we use a and b as negative.

If you are dealing with three percentages (a, b and c) then you find the overall percentage between the first two (a and b) and then use this overall percentage with the third (c).

The question here tells us that the price decreased by m% and then increased by n%. So the overall percentage will be

-m + n - mn/100

Now let us break down the question. Here we are asked, Was the price of the automobile lower in 2010 than in 2012?

The price of the automobile will be lower in 2010 than in 2012 only if the overall percentage is negative. So the question can be rephrased as

Is -m + n - mn/100 < 0

Evaluating the statements individually

Statement 1 : m < n

This does not give us any information whether -m + n - mn/100 < 0. If we consider m to be 10 and n to be 11 then we get a YES, but if we consider m to be 10 and n to be 20 then we get a NO. Insufficient.

Statement 2 : mn < 100n – 100m

Rearranging the terms we get 100n - 100m - mn > 0. Dividing throughout by 100 we get

n - m - mn/100 > 0

This gives us a definite NO. Sufficient.

Hope this helps!

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Re: The price of an automobile decreased m percent between 2010  [#permalink]

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Question: is $$p<p*(1-\frac{m}{100})*(1+\frac{n}{100})$$? --> is $$1<(1-\frac{m}{100})*(1+\frac{n}{100})$$? --> is $$100n-100m>mn$$?

(1) m < n. Not sufficient.
(2) mn < 100n – 100m. Directly answers the question. Sufficient.

Hi, Bunuel can you elaborate on the last step here, i am unable to understand.--> is $$1<(1-\frac{m}{100})*(1+\frac{n}{100})$$? --> is $$100n-100m>mn$$?
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Re: The price of an automobile decreased m percent between 2010  [#permalink]

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1
sarat0994 wrote:
Question: is $$p<p*(1-\frac{m}{100})*(1+\frac{n}{100})$$? --> is $$1<(1-\frac{m}{100})*(1+\frac{n}{100})$$? --> is $$100n-100m>mn$$?

(1) m < n. Not sufficient.
(2) mn < 100n – 100m. Directly answers the question. Sufficient.

Hi, Bunuel can you elaborate on the last step here, i am unable to understand.--> is $$1<(1-\frac{m}{100})*(1+\frac{n}{100})$$? --> is $$100n-100m>mn$$?

Sure.

$$1<(1-\frac{m}{100})*(1+\frac{n}{100})$$

$$1<(\frac{100 -m}{100})*(\frac{100+n}{100})$$

$$10000<(100 -m)*(100+n)$$

$$10000<10000 +100n-100m-mn$$

$$100n-100m>mn$$
_________________ Re: The price of an automobile decreased m percent between 2010   [#permalink] 27 Jan 2019, 04:36
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