document.write( "Question 1181832: A rectangular box has a volume of 200cm^3.
\n" ); document.write( "(a) find three possible whole number sets of dimensions that give the correct volume\r
\n" ); document.write( "\n" ); document.write( "(b) Find three sets of dimensions with at least one non-whole number that give the correct volume. \r
\n" ); document.write( "\n" ); document.write( "okay so for (a) I'm experienced enough to get all three sets in my head: 5*4*10, 2*10*10 and 20*2*5, which all equal to 200cm cube.\r
\n" ); document.write( "\n" ); document.write( "but for (b) how can I find non-whole numbers like decimals as the dimensions, are there any reliable algebraic calculations for this?\r
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\n" ); document.write( "\n" ); document.write( "My Ideas:
\n" ); document.write( "So for a decimal number right, its a whatever number divide by 10 or 10 to the power of something. So for example to find a decimal number we can randomly generate a number 825 in our brain. To make this number a decimal, we divide it by 10 which equals 82.5 or we can divide by another 10 which = 8.25. Okay, with this knowledge I create these equations n = a/10 and a/10 * b * h = 200. But the things is b and h can be anything we want. So I tried a/10 * 4 * 8 = 200, a = 62.5, therefore 6.25 is the non-whole number I need.
\n" ); document.write( " But I realize there is a flaw in my way, because if I try 4 and 2 as b and h, then a/10*4*2=200
\n" ); document.write( "a = 250
\n" ); document.write( "which makes my number n = 250/10 = 25 which is still a whole number. Do I just use trial and error on (b and h) until I get my non-whole number or are there different ways?\r
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Algebra.Com's Answer #811749 by greenestamps(13203)\"\" \"About 
You can put this solution on YOUR website!


\n" ); document.write( "A useful trick for performing mental multiplication is to multiply one of the numbers by some constant and divide the other by the same constant.

\n" ); document.write( "For example, to multiply 24 times 35, you could cut the 24 in half to 12 and double the 35 to 70; then 12 times 70 is easier than 24 times 35 because one of the numbers now ends in 0.

\n" ); document.write( "You can use the same kind of process to find as many combinations as you want of three dimensions that give the required volume of 200.

\n" ); document.write( "Start with any single set of three dimensions, like your first one: 5*4*10.

\n" ); document.write( "You can double the 5 and cut the 4 in half to get 10*2*10; that is your second one.

\n" ); document.write( "Next you can double one of the 10s and cut the other in half to get 20*2*5, which is your third.

\n" ); document.write( "Continue doing the same kind of thing. If you want a set of dimensions that is not whole numbers, divide one of the measurements by a number that doesn't give a whole number result.

\n" ); document.write( "Your third set was 20*2*5. You can divide the 5 by 2 and double the 2 to get 20*4*2.5.

\n" ); document.write( "Then you could take that and again divide the 2.5 by 2 and double the 20 to get 40*4*1.25.

\n" ); document.write( "And you can continue with that process to get as many sets as you want that all give a volume of 200.

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