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I was preparing for a standardized test and came across the following problem:

During a closeout sale 30 suits are sold at \$60 each.The price is the reduced to \$50 each and 20 suits are sold. At what average price must the remaining 10 suits be sold in order to obtain an average of \55 for the 60 suits?

The answer is supposed to be \50. Could anyone tell me how they got that? A little working shown would be appreciated..

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    Up to now 50 suits are sold for \$2800 alltogether. Now find $x$ with $(2800 + 10x)/60 = 55$.2012-05-22

1 Answers 1

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The first and most important step is to introduce a variable:

$\ \ \ \ \ \ $Let $x$ be the common price of each of the remaining $10$ suits.


The second step is to find an equation in $x$. Of course, we may obtain one by knowing that the average sale price of the twenty suits is $55$ dollars. Once we have our equation, we'll solve it for $x$.

The average sale price of the suits is obtained by computing the total revenue from the $60$ suits and dividing by $60$.

Let's find the total revenue first. Thirty suits were sold at $60$ dollars and twenty suits were sold at $50$ dollars. The revenue from these is $30\cdot60+20\cdot 50$ dollars. The remaining ten suits are sold at $x$ dollars and the revenue from these is $x\cdot10$ dollars. So, adding these together, the total revenue from the $60$ suits is $30\cdot 60+ 20 \cdot 50 +x\cdot 10$ dollars.

Thus, the average sale price of the twenty suits is $\tag{1}{\text{total revenue}\over 60}={30\cdot 60+ 20 \cdot 50 +x\cdot 10\over 60}.$

Now we find our equation. You want $(1)$ to turn out to be $55$. So set $\tag{2} {30\cdot 60+ 20 \cdot 50 +x\cdot 10\over 60}=55. $


Now the third and final step: we solve equation $(2)$ for $x$. Towards that end, first multiply both sides of $(2)$ by $60$: $ {30\cdot 60+ 20 \cdot 50 +x\cdot 10 }=55\cdot 60. $ Simplify the arithmetic: $\tag{3} 2800+10x=3300, $ or, subtracting $2800$ from both sides of $(3)$, $\tag{4} 10x=500. $ Finally, divide both sides of $(4)$ by $10$ to obtain $x=50$ dollars.

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    @Rajeshwar You're welcome; glad to help.2012-05-22