teal wrote:

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 while items of model B accounted for 40% of the monthly sales. If the company had sold 1000 items of model A less than it actually did, what percent of the total monthly sales would have been attributed to model A?

48

50

52

54

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.

In the above explanation, when the final calculation gives you 0.4S how is that 50% of S shouldn't that be 40%??

It is better to express total sale in terms of sales from both models.

Denote by A the price of one item of model A, and by B the price of an item of model B. Then we can write:

3000A / (3000A+1000B) = 0.6, from which we get that 2A = B.

So, if finally only 2000 items of model A were sold, then we get 2000A / (2000A + 1000B) = B / (2B) = 0.5.

When you change the number of items of model A, you automatically change the total sales, so you should not compare sales from model A to the initial total sale.

Initially, total sale was 3000A + 1000B, then the total sale is assumed to be 2000A + 1000B.

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PhD in Applied Mathematics

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