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Hint: It is a word problem related to the money exchange, and the only thing that you need to focus on for solving this problem is the percentage calculation. Remember that Bargained discount is calculated using the marked price while profit is calculated with respect to cost price.

Complete step-by-step answer:

To start with the solution, we let the cost price of a dozen of the article to be x and the marked price of a dozen of the article to be y.

Therefore, the cost price of 16 articles is $16\times \dfrac{x}{12}=\dfrac{4x}{3}$ and marked price = $\dfrac{4}{3}y$

From the question, we can deduce that the dealer gave a discount in two ways: he had to give a 20% discount on the marked price, and the other was to give him 16 articles for the price of 12. So, the loss he faced can be mathematically represented as:

Discount = 20% of the marked price of a dozen + the marked of the extra four articles he gave.$Discount=\dfrac{20}{100}\times

y+(16-12)\dfrac{y}{12}=\dfrac{y}{5}+\dfrac{4y}{12}=\dfrac{y}{5}+\dfrac{y}{3}$

The question also states that he made an overall profit of 20% of the cost price by cracking the deal.

So, selling price = 20% of cost price + the cost price

$\text{selling price=}\dfrac{20}{100}\times \dfrac{4}{3}x+\dfrac{4}{3}x=1.2\times \dfrac{4}{3}x=1.6x$

Now, if we see after subtracting the discount on the marked price from the marked price, we get the selling price, and from the above equation, we know that the selling price is equal to 1.6x. So, writing it mathematically, we get

$\text{selling price=marked price - discount}$

\[\Rightarrow \text{1}\text{.6x=}\dfrac{4}{3}y-\dfrac{y}{5}-\dfrac{y}{3}\text{ }\]

$\Rightarrow \dfrac{\text{16x}}{10}\text{= y}\left( \dfrac{4}{3}-\dfrac{1}{3}-\dfrac{1}{5}

\right)$

$\Rightarrow \dfrac{16x}{10}\text{= }\dfrac{12y}{15}$

$\Rightarrow \dfrac{240x}{120}\text{= y}..............\text{(i)}$

Now we percent above the cost price which the dealer took on the marked price is given by:

$\dfrac{\text{marked price - cost price}}{\text{cost price }}\times 100$

$=\dfrac{\dfrac{4}{3}y-\dfrac{4x}{3}}{\dfrac{4x}{3}\text{ }}\times 100$

Now we will substitute the value of y from equation (i).

$=\dfrac{\dfrac{4}{3}\left( \dfrac{240}{120}-1 \right)}{\dfrac{\text{4 }}{3}}\times 100$

$=\left( 2-1 \right)\times 100$

$=100%$

So, 100% is the price marked by the dealer over his cost price.

Note: Don’t get confused and take the percentages with respect to selling price. At the same time you should be very clear that the percentage loss or profit are terms related to the actual pricing not to the price for which you crack the deal. Also remember that bargained discount is with respect to marked price while profit and loss are with respect to cost price.

Complete step-by-step answer:

To start with the solution, we let the cost price of a dozen of the article to be x and the marked price of a dozen of the article to be y.

Therefore, the cost price of 16 articles is $16\times \dfrac{x}{12}=\dfrac{4x}{3}$ and marked price = $\dfrac{4}{3}y$

From the question, we can deduce that the dealer gave a discount in two ways: he had to give a 20% discount on the marked price, and the other was to give him 16 articles for the price of 12. So, the loss he faced can be mathematically represented as:

Discount = 20% of the marked price of a dozen + the marked of the extra four articles he gave.$Discount=\dfrac{20}{100}\times

y+(16-12)\dfrac{y}{12}=\dfrac{y}{5}+\dfrac{4y}{12}=\dfrac{y}{5}+\dfrac{y}{3}$

The question also states that he made an overall profit of 20% of the cost price by cracking the deal.

So, selling price = 20% of cost price + the cost price

$\text{selling price=}\dfrac{20}{100}\times \dfrac{4}{3}x+\dfrac{4}{3}x=1.2\times \dfrac{4}{3}x=1.6x$

Now, if we see after subtracting the discount on the marked price from the marked price, we get the selling price, and from the above equation, we know that the selling price is equal to 1.6x. So, writing it mathematically, we get

$\text{selling price=marked price - discount}$

\[\Rightarrow \text{1}\text{.6x=}\dfrac{4}{3}y-\dfrac{y}{5}-\dfrac{y}{3}\text{ }\]

$\Rightarrow \dfrac{\text{16x}}{10}\text{= y}\left( \dfrac{4}{3}-\dfrac{1}{3}-\dfrac{1}{5}

\right)$

$\Rightarrow \dfrac{16x}{10}\text{= }\dfrac{12y}{15}$

$\Rightarrow \dfrac{240x}{120}\text{= y}..............\text{(i)}$

Now we percent above the cost price which the dealer took on the marked price is given by:

$\dfrac{\text{marked price - cost price}}{\text{cost price }}\times 100$

$=\dfrac{\dfrac{4}{3}y-\dfrac{4x}{3}}{\dfrac{4x}{3}\text{ }}\times 100$

Now we will substitute the value of y from equation (i).

$=\dfrac{\dfrac{4}{3}\left( \dfrac{240}{120}-1 \right)}{\dfrac{\text{4 }}{3}}\times 100$

$=\left( 2-1 \right)\times 100$

$=100%$

So, 100% is the price marked by the dealer over his cost price.

Note: Don’t get confused and take the percentages with respect to selling price. At the same time you should be very clear that the percentage loss or profit are terms related to the actual pricing not to the price for which you crack the deal. Also remember that bargained discount is with respect to marked price while profit and loss are with respect to cost price.

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