# Difficult Percentages Solved QuestionAptitude Discussion

 Q. A shopkeeper gives a discount of $12%$, whereas a customer makes cash payment. Let '$p$' denotes the percentage, above the cost price, that the shopkeeper must mark up the price of the articles ['$p$' is an integer] in order to make a profit of $x% (x < 100)$. Which of the following is the possible value(s) of $x$?
 ✖ A. 54 ✖ B. 76 ✔ C. 96 ✖ D. 32

Solution:
Option(C) is correct

Let the cost price of one article is Rs $r$ then:

Marked price of the article

$=r\left(1+\dfrac{p}{100}\right)$

Selling price

$=r\left(1+\dfrac{p}{100}\right)\left(1-\dfrac{12}{100}\right)$

Profit percentage $= x\%$

$=r\left(1+\dfrac{p}{100}\right)\left(1-\dfrac{12}{100}\right)=r\left(1+\dfrac{x}{100}\right)$

$\Rightarrow x = \dfrac{25(12+x)}{22}$

As, $p$ is an integer, $x$ must be multiple of 22.

All the possible values $x$, less than 100 are $k=10,32,54,76,98$

Except option 'C' all the options are possible.

## (4) Comment(s)

Gideon
()

The question has a lot of faws. First it should be 12%, P% and X%. And the question must be which of the following cannot be the value of 'X'.

And for the the solution.

Take all the values that are given in the options. (54,76,96 and 32)

Say when X = 54%

Suppose cost price = ₹100

Then profit would be ₹54. And the selling price ₹154

The selling price is basically 12% less than the marked price. That is the selling price is 88% of the marked price.

Thus, 154*100/88 (to find the marked price)

Marked price is ₹175.

According to the question 'P is the percentage above the cost price and must be an integer'. For this case it is P = 75%. (75 is an integer)

It satisfies the given requirement. Hence 54 can be one of the solutions.

Similarly solve for other options as well.

Only X = 96 would give a non integer value for P.

Gideon
()

Another imp. Fact...this is true for any value other than 100 as well...so it can be verified by using other numbers as well..i tried with 90 and 80... Just keep in mind that it is.. X% and P%.

Areeba
()

I am also confused about this question. The forum needs to review this solution once and should give proper explanation of the steps specified in the solution.

Nishant
()

Can someone please explain how?

$r\left(1+\dfrac{p}{100}\right)\left(1-\dfrac{12}{100}\right)=r\left(1+\dfrac{x}{100}\right)$

becomes $p=\dfrac{25(12+x)}{22}$

I am getting $p=\dfrac{x+8700}{88}$