Forumlas with Unspecified Amounts

[Videoorchard]

Hi Tommy,

I have a quick question related to Formulas with Unspecified Amounts found on algebra book.

Question:
If the length of the side of a cube decreases by two-thirds its original value, by what percentage will the volume of the cube decrease?

My Question:
1. How did the author arrive at 3 - (2/3)(3) =1 unit?

2. How do go about to pick "Smart" Numbers(3 in this case) in these kind of problem?

Question PG No:
Algebra Strategy guide.
Pg. 105

[n00bpron00bpron00b]

While picking numbers for testing cases, you could pick any number (keeping in mind the additional constraints given, if any). => you will always arrive at the same answer.

But instead of picking any random number, it's always ideal to pick numbers which are small as well as easy for computations. (ex. 1,0,100,10,500,-10 etc)

Percents - multiples of 10 preferred
Fractions - multiples of the fraction divisors preferred (or multiples of LCM in case of multiple fractions)

Now,

In this case the side of the cube decreases by 2/3rd's. If you notice if we pick side length of cube as "3", the 3 in the fractional denominator and length of cube gets cancelled out easily leaving behind 2. Instead if you picked say "37" or sqrt of 5 ; you end with tedious calculations i.e. (2/3)(37)

Regarding the solution to the problem,

Let's pick the original length of side of the cube as 9
Algebraic translation of "decreases by 2/3 its original length" = (2/3)*9 = 6 (this is not the new length ; this is the value by which the original length decreases)

9 - 6 = 3 (new length)

Original Volume of cube = (side)^3 = (9)^3 = 729
New Volume = 27

% decrease = 96.2 %

[tommywallach]

Agreed with noob 99% here (I'll get to the 1% disagreement in a moment).

The thing about smart numbers is that the only thing that makes them smart is that they make your calculations easier. But even if you pick "dumb" numbers, you will still get the right answer. The whole point with a question like this is that it doesn't matter what numbers you pick (otherwise, you wouldn't be able to throw the "Smart" number in there).

The only disagreement I have is to be a teensy bit careful on what we call VICs (Variables in Answer Choices). In these cases, you almost never want to pick 0 or 1 (As Noob suggested), for a complicated set of reasons there's no need to get into right now.

Otherwise, the numbers you pick will always work! : )

-t

[n00bpron00bpron00b]

Oh yes, Tommy is right (for VIC'S the strategy needs to be tweaked a bit :P)

[Videoorchard]

Hey,

Thank you for giving such indept explanation Noob.It helped me a ton.

Tommy, thank you for your take on the problem. much appreciated! :)


Regards,
Manish A.

[tommywallach]

Glad to help, and double thanks to Noob, who's killing it on these forums!

-t