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September 10th, 2018, 06:43 AM  #1 
Member Joined: Jun 2017 From: Lima, Peru Posts: 97 Thanks: 1 Math Focus: Calculus  Is the method to solve this single variable equation used correctly?
The problem is as follows: A king decides to split an award between his three best archers after organizing an accuracy contest in his court. The first one gets $\frac{2}{5}$ parts of the total minus $\frac{1}{5}$ of a pound, the second gets $\frac{2}{5}$ of the rest minus $\frac{1}{5}$ of a pound. If the third gets the $23$ remaining pounds. What was the total quantity of the cash prize that the king distributed?The alternatives given in my book are the following:  $53$  $36$  $38$  $63$  $72$ In my attempt to solve this problem, I stated the situation as follows: The total would be: $x$ pounds The first archer: $\frac{2}{5}x  \frac{1}{5}$ Second archer: $\frac{2}{5}\left(x\left(\frac{2}{5}x  \frac{1}{5}\right ) \right)\frac{1}{5}$ Third archer: $23$ So assuming that all summed up should become into the original prize, that would be: $\frac{2}{5}x  \frac{1}{5} + \frac{2}{5}\left(x\left(\frac{2}{5}x  \frac{1}{5}\right ) \right)\frac{1}{5} + 23 =x$ Doing up a little bit of algebra, I rearranged it in this form: $ \left ( \frac{2}{5}x  \frac{1}{5} \right ) \left( 1  \frac {2}{5} \right ) + \frac{2}{5}x\frac{1}{5} = x  23$ $ \left ( \frac{2}{5}x  \frac{1}{5} \right ) \left( 1  \frac {2}{5} + 1 \right )= x  23$ $ \left (\frac{2}{5}x  \frac{1}{5} \right ) \left( 2  \frac {2}{5} \right ) = x  23$ $ \left (\frac{2}{5}x  \frac{1}{5} \right ) \left( \frac{8}{5} \right ) = x  23$ Multiplying by $25$ both sides: $ \left (\frac{2x1}{5} \right ) \left( \frac{8}{5} \right ) = x  23$ $ \frac{8\left (2x1 \right )}{25} \left ( 25 \right ) = 25x  575$ $ 16x8 = 25x  575$ $ 9x = 567$ $ x= 63$ Therefore the answer is $63$, which is what appears within the alternatives. But I was wondering whether this is the only way to solve this problem or could there be another way to solve it?. Last edited by skipjack; September 10th, 2018 at 01:18 PM. 
September 10th, 2018, 12:11 PM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 
The first one gets 2/5 parts of the total minus 1/5 of a pound, the second gets 2/5 of the rest minus 1/5 of a pound. Unless above means: The first one gets 2/5 parts of (the total minus 1/5 of a pound), the second gets 2/5 of (the rest minus 1/5 of a pound). There are 4 possibilities if prize is less than 100: 13: 5,3,5 38: 15,9,14 63: 25,15,23 (yahoo!) 88: 35,21,32 Last edited by Denis; September 10th, 2018 at 12:43 PM. 
September 10th, 2018, 02:31 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,757 Thanks: 2138 
Just work back twice, starting from 23. Each time, subtract 1/5 and then multiply by 5/3. (23  1/5)(5/3) = 38 and (38  1/5)(5/3) = 63. Or (equivalently) add 1, multiply by 5, divide by 3, and then subtract 2. Then repeat. (23 + 1) × 5/3  2 = 38 and (38 + 1) × 5/3  2 = 63. That's a bit longer to write, but quicker to do. For either way, dividing by 3 and then multiplying by 5 happens to be slightly easier than multiplying by 5 and then dividing by 3. 

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