<p>"In a certain parking lot that contains 200 cars, 50 percent of the cars are red, 60 percent are four-door cars and 70 percent have alloy rims. what is the greatest number of cars in the parking lot that could be green two-door cars with alloy rims?"</p>
<p>You want to maximize the number of alloy rim cars that are neither red nor 4-door.</p>
<p>There are 140 alloy rim cars. So minimize the number of red cars with alloy rims. There are 100 red cars. So the desired minimum is 40. That means 60 red cars do not have alloy rims – 60 + 140 = 200. Now minimize the number of 4-door cars with alloy rims. There are 120 4-door cars. That means 60 of the 4-door cars do not have alloy rims. The other 60 do. Of these 40 can be red. The other 20 have alloy rims but are not red. So of the 140 alloy rim cars 80 are neither red nor 4-door. That’s the answer: 80.</p>
<p>Draw a picture to make this clearer. Use circles – a big one for the 140 alloy rimmed cars. Then have the other circles overlap to show the number alloyed rimmed and the number not, and the number alloyed rimmed + red + 4-door, the number alloyed rimmed + 4-door + red, and the number not.</p>
<p>basically you want to make a hypothetical “best concept”. </p>
<p>if 50% are red… then its sure the other 50% is green
if 60% have four doors, then 40% would have 2 doors
the best concept here is that in that “60%” four door cars, are the “50%” red cars
…to sum up , until now we have 50% red and four door cars and 10% four door cars. That leaves us with a 40% of cars being both green and 2 doored. (hypothetically again, the 70% of cars with rims would be in the whole of the green-2doored cars) . So we have 40% of the total cars being green-having 2 doors and having rims. Thats as i said “the best concept” so 40% of 200 = 80…here’s ur answer</p>