The price of an automobile decreased m percent between 2010

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?

Posting official solution of this problem.

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!

CrackVerbal Academics Team

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.

Answer: B.

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.

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

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

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

\(10000<10000 +100n-100m-mn\)

\(100n-100m>mn\)

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