The easiest way to do this is to backsolve. Let's start with the middle answer choice.
If car B's speed is 40 mph, car A's is 50 mph. Thus, it took car A 8 hours to go 400 miles, and it took car B 10 hours.
There's the answer (C) - since the times are two hours apart.
I just happened to pick the right answer first (it's always a good idea to start with the middle choice). If it hadn't worked out, we'd have known whether to move higher or lower in the answers.
A lot faster!
complicated distance rate problem...should i use quadratic F
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Ian Stewart
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A couple of comments here:
-you did indeed arrive at the correct quadratic equation here;
-if you ever have a factoring problem like this:
2x^2 + 20x - 4000 = 0
the first thing you should do is divide both sides by 2; as long as everything remains an integer, that will always simplify matters:
x^2 + 10x - 2000 = 0
Two numbers whose product is -2000, and which are 10 apart in absolute value: either 40 and -50 or -40 and 50. We need them to add to 10, so the factorization is:
(x - 40)(x+ 50) = 0
x = 40 or -50 --> x = 40, since x can't be negative.
-Of course, there are faster ways to arrive at an answer here. Backsolving is fairly quick if you land on the right answer early. The algebra can also be done much more quickly, if you use t = d/s. We know d = 400; let s be the speed of the slower car. Their times are 400/s and 400/(s+10). We know these times differ by 2 hours:
400/s - 400/(s+10) = 2
That's a single equation, single unknown, and leads to the factoring problem above if you find a common denominator.
Knowing how to do the algebra will definitely serve you well on the harder questions on the test- where you see fewer numbers and more letters.
-you did indeed arrive at the correct quadratic equation here;
-if you ever have a factoring problem like this:
2x^2 + 20x - 4000 = 0
the first thing you should do is divide both sides by 2; as long as everything remains an integer, that will always simplify matters:
x^2 + 10x - 2000 = 0
Two numbers whose product is -2000, and which are 10 apart in absolute value: either 40 and -50 or -40 and 50. We need them to add to 10, so the factorization is:
(x - 40)(x+ 50) = 0
x = 40 or -50 --> x = 40, since x can't be negative.
-Of course, there are faster ways to arrive at an answer here. Backsolving is fairly quick if you land on the right answer early. The algebra can also be done much more quickly, if you use t = d/s. We know d = 400; let s be the speed of the slower car. Their times are 400/s and 400/(s+10). We know these times differ by 2 hours:
400/s - 400/(s+10) = 2
That's a single equation, single unknown, and leads to the factoring problem above if you find a common denominator.
Knowing how to do the algebra will definitely serve you well on the harder questions on the test- where you see fewer numbers and more letters.
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Ian Stewart
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If you are backsolving, and you will know, by testing an answer, whether to go up or down, it's certainly better to start in the middle than to start at A or at E. But it's not the best choice, and certainly not 'always a good idea'. B or D are better first choices.VP_Jim wrote: I just happened to pick the right answer first (it's always a good idea to start with the middle choice). If it hadn't worked out, we'd have known whether to move higher or lower in the answers.
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chidcguy
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IMO, plugging in and testing values is a good idea but one needs to be very adept at doing it and should have used it extensively during prep. In the heat of the exam, its difficult to pick numbers and back solve. It gets worse if we have to back solve 3 times and probably equals the time to solve it directly.
my 2 cents
my 2 cents
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gmataspirant
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How did you arrive at the following equation, I am just trying to figure it out.
[/quote]2x^2 + 20x - 4000 = 0
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