? about using #s to solve a problem w/ variables

[blumnday99]

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Here's the question

If a > b > c > d > 0, then
(a) ad > bc
(b) ac > b*b
(c) abd > abc
(d) a - c > b - d
(e) none of these

If I use one set of numbers (say a = 4, b = 3, c = 2, and d =1) and I find the correct answer, can I assume that that answer will be correct for all sets of numbers that satisfy the if criteria?

Does that make sense? After I have an answer should I try and find a set of numbers that does not satisfy the correct answer...that could be time consuming depending on how many sets I try.

Mark

 

[coodoo]

Here's the question

If a > b > c > d > 0, then
(a) ad > bc
(b) ac > b*b
(c) abd > abc
(d) a - c > b - d
(e) none of these

If I use one set of numbers (say a = 4, b = 3, c = 2, and d =1) and I find the correct answer, can I assume that that answer will be correct for all sets of numbers that satisfy the if criteria?

Plugging #'s doesnt always work. Imagine if a = 10 and b = 3 c =2 d=1
now look at (a) 10*1 > 2*3 yes, that's true

but if we had used your #'s we would have gotten 4*1 > 2*3, which is false
you have to be very careful when you decide to plug in #'s, although it can sometimes save you time.

The answer should be e btw

 

[JayMiranti]

Here's the question

If a > b > c > d > 0, then
(a) ad > bc
(b) ac > b*b
(c) abd > abc
(d) a - c > b - d
(e) none of these

If I use one set of numbers (say a = 4, b = 3, c = 2, and d =1) and I find the correct answer, can I assume that that answer will be correct for all sets of numbers that satisfy the if criteria?

Does that make sense? After I have an answer should I try and find a set of numbers that does not satisfy the correct answer...that could be time consuming depending on how many sets I try.

Mark

you can pick ANY numbers that fit the criteria of the "if" statement... that is the purpose of the statement.

you can pick your numbers, or the numbers the poster above me picked, it doesnt matter as long as they satisfy the initial conditions, they are fair game.

The only time you have to be careful is if you are picking numbers that DO work, coincidentally, but other numbers that satisfy the initial conditions ( other numbers someone could plug in and satisfy the if condition) would not work, and so the statement wouldnt be true for ANY number, which is required, which the letters represent. So yes, you can pick ANY number.

 

[blumnday99]

So if a set of numbers works does it mean that all sets of numbers will work,

but if a set of numbers does NOT work , it does not necessarily mean that all sets of numbers will not work?