A shopkeeper increases the price of an article by X%

Hi Bunuel,

If you increase an amount by x% and then decrease the amount by x%, is the net decrease always equal to a constant a%? You seem to assume that in your reasoning.

Thanks!
Eric

Yes, it does not depend on the amount. I mean that whether initial amount is $10 or $20, if you increase both by x% and then decrease both by x%, you'll get the net decrease of the same percentage:

For example if you increase both 10 and 100 by 10% and then decrease both by 10%, you'll get the net decrease of 1% for both amounts.

Hope it's clear.

Hi,
Is there any other method to solve this problem....if someone dont get the logic in the solution provided by Bunuel?

Hi Bunuel,
I have a doubt,U have equated the changed price value to 180 rather than equating the change in price.

pa is the final value after reduction in price,so to represent the change it should be p-pa=180.isnt it?

Please Correct me if i m wrong

We are told that after a% decrease the price of the article is reduced by $180, which means that a% of p is $180 --> pa=180.

hi, i was confused by the question:
A shopkeeper increases the price of an article by X% and then decreases it by X%. As a result the price of the article is reduced by $180. After one more such change the price is further reduced by $153 Find the original price of the article.

I think the question run in the following logic:
1. Price increase by x%
2. Price decrease by x %
3. After step 1 and step 2, price is reduced by $180 from the original price
4. Price increase by x%
5. Price decrease by x%
6. After step 4 and 5, price is reduced by $153 further.

Is it incorrect??

As usual Bunuel answer is best

But when i solved it , i did it this way (which is quite similar to bunuel , just more verbose)

Since p is first increased by x% and then decreased by x% , we will get
p(1 + x/100)(1 - x/100) = p-180 ---1st eq

Again the remaining amount( which is p-180) is first increased by x% and then decreased by x% , we will get
(p-180) (1 + x/100)(1 - x/100) = (p-180)-153 --- 2nd eq

Divide 1st eq by 2nd we get

p/(p-180) = (p-180)/(p-333)
(the equation is quite easy to solve once you cross multiply it)
on solving we will get p = 1200

When an article is increased and decreased by the same percentage the net value of the article decreases because :
a(1+x)(1-x) = a(1-\(x^2\)) since \( x^2\) is always positive this is less than the original value a.

Given that the when by X% and decreased by X% the net value reduced by $180 and hence considering the net percentage reduction to be p%.
\(a\cdot\left(1-\frac{p}{100}\right)\ =\ a\ -\ 180\)

Similarly using the second condition :

\(\left(a-180\right)\cdot\left(1-\frac{p}{100}\right)\ =\ a-180-153\ =\ a-333\)

Hence \(1-\frac{p}{100}=\ \frac{\left(a-180\right)}{a}=\frac{\left(a-323\right)}{\left(a-180\right)}\)

Solving this we have :

\(27a\ =\ 180\cdot180\ =\ 32400\)

a = $1200

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