topspin360 wrote:what's a methodological way of solving this problem that can be generalized and applied to other problems. solution showed plugging in numbers.
A car dealership sold two cars: the first car at a 10% profit and the second car at a 10% loss, which gave them an overall profit of 5% from these two sales. If the dealership's total profit was $1000, what was the cost price of each car?
$5,000 and $1,000
$9,000 and $5,000
$11,000 and $9,000
$15,000 and $5,000
$20,000 and $10,000
Thanks.
This is a MIXTURE problem: a 10% profit and a 10% loss are being combined to form a MIXTURE with a 5% overall profit.
The following approach is called
alligation.
It's a very good way to handle MIXTURE PROBLEMS.
Let F = the first car and S = the second car.
Step 1: Plot the 3 percentages on a number line, with the two starting percentages (+10% and -10%) on the ends and the goal percentage (+5%) in the middle.
F(10%)--------------------(+5%)--------S(-10%)
Step 2: Calculate the distances between the percentages.
F(10%)----------
5---------(+5)----
15----S(-10%)
Step 3: Determine the ratio in the mixture.
The ratio of F to S in the mixture is the RECIPROCAL of the distances in red.
F : S = 15 : 5 = 3:1.
Only answer choice
D has the same ratio:
15000:5000 = 3:1.
The correct answer is
D.
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