Bunuel wrote:
Official Solution:
In the month of November, a company sold 3000 items of model A and 1000 items of model B. Items of model A accounted for 60% of the company's monthly sales in dollars while items of model B accounted for 40% of the monthly sales in dollars. If the company had sold 1000 items of model A fewer than it actually did, what percent of the total monthly sales in dollars would have been attributed to model A?
A. 48
B. 50
C. 52
D. 54
E. 55
Let \(S\) denote the total November sales of the company. The price of an item of model A = \(0.6*\frac{S}{3000}\); the price of an item of model B = \(0.4*\frac{S}{1000}\). If the company had sold 2000 items of model A, the revenue from sales of model A would have amounted to \(0.6*\frac{S}{3000}*2000 = 0.6S*\frac{2}{3} = 0.4S\) which is equal to the revenue from sales of model B. So, in the hypothetical case described in the stem, the two models would have accounted for 50% of the monthly sales each.
Alternative explanation - picking numbers
Let the total revenue be \($10,000\); then revenue from Model A is \($6,000\) for 3,000 units or \($2\) per unit and Model B's revenue is \($4,000\) for 1,000 units or \($4\) per unit.
If we had sold only 2,000 units of model A, the total revenue would have been \($2*2,000 +$4*1,000 =$8,000\). At this point it is clear that model A and B bring equal amounts of revenue (\($4,000\) each) and half of monthly sales are attributed to model A.
Answer: B
Hi Bunuel,
How did you arrive at price of A in 1st explanation..
I arrived at a different equation using the following approach -
Sales of A account for 60% of total sales.
So S = 0.6*3000*A.
I am unable to understand why is this wrong?
Posted from my mobile device\(S\) is the total November sales.
Items of model A accounted for 60% of the company's monthly sales in dollars --> revenue from A = 0.6S.
In the month of November, a company sold 3000 items of model A --> price of 1 unit of A = 0.6S/3000.