My Math Forum Number of steps to form 1 fluid oz

 Probability and Statistics Basic Probability and Statistics Math Forum

 February 18th, 2019, 02:29 PM #1 Newbie   Joined: Feb 2019 From: Scotland Posts: 6 Thanks: 0 Number of steps to form 1 fluid oz Jane wants to have two separate 1 fluid ounce measures of water at the same time. However the only measures she has are for six, ten and fifteen fl oz. Show how this could be done in the smallest number of steps without marking the measures or using any container other than the original large beaker of water. The only steps allowed are filling or emptying a measure or transferring water from one measure to another.
 February 18th, 2019, 04:27 PM #2 Senior Member   Joined: Aug 2012 Posts: 2,212 Thanks: 654 You can fill the ten, pour it into the fifteen, then fill the six and pour it into the fifteen. Now you have 1 ounce in the six. Now you need a second ounce, but you can no longer use the six? Or if you can, pour the contents of the six (the 1 ounce) into the empty beaker, if you're allowed to empty the beaker. Now repeat the same procedure and you end up with 1 ounce in the beaker and 1 ounce in the six. How one shows that's the minimum solution, I don't know. Can we empty the beaker? You said we can empty the measures, did you mean to exclude emptying the beaker? If you can't empty the beaker you have no place to put that one ounce that won't interfere with getting the second ounce. Unless there's some other way. Thanks from Estermont Last edited by Maschke; February 18th, 2019 at 04:30 PM.
February 19th, 2019, 02:16 AM   #3
Newbie

Joined: Feb 2019
From: Scotland

Posts: 6
Thanks: 0

Quote:
 Originally Posted by Maschke You can fill the ten, pour it into the fifteen, then fill the six and pour it into the fifteen. Now you have 1 ounce in the six. Now you need a second ounce, but you can no longer use the six? Or if you can, pour the contents of the six (the 1 ounce) into the empty beaker, if you're allowed to empty the beaker. Now repeat the same procedure and you end up with 1 ounce in the beaker and 1 ounce in the six. How one shows that's the minimum solution, I don't know. Can we empty the beaker? You said we can empty the measures, did you mean to exclude emptying the beaker? If you can't empty the beaker you have no place to put that one ounce that won't interfere with getting the second ounce. Unless there's some other way.
No I don't think it's allowed since they specifically mentioned what the operations were... You can use the measures again though.

February 19th, 2019, 06:22 AM   #4
Math Team

Joined: Oct 2011

Posts: 14,146
Thanks: 1003

Quote:
 Originally Posted by RachelGreen No I don't think it's allowed .....
Why not? Problem states:
"...or using any container other than the original large beaker of water."
So you can USE it, right?
Perhaps you can ask your teacher..."I don't think" is not allowed in maths!!

 February 19th, 2019, 07:42 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,386 Thanks: 2012 The rules permit using the beaker, but emptying the beaker isn't a valid step. However, Maschke's procedure can be modified to work, though requiring more steps. After Maschke's first 4 steps, transfer 10 from the 15 to the 10, then empty the 15, transfer all the water in the 10 to the 15, refill the 10 from the beaker, transfer 5 from the 10 to the 15 (filling it), and then empty the 15. You now have 1 in the 6, 5 in the 10 and the 15 is empty. Transfer the 1 in the 6 to the 15, fill the 6, and then transfer 5 from the 6 to the 10 (filling it). There is now 1 in the 6, 10 in the 10 and 1 in the 15, achieving the desired objective after a total of 13 steps. I don't know whether a shorter method is possible. Thanks from Maschke and RachelGreen
 February 19th, 2019, 08:01 AM #6 Newbie   Joined: Jan 2017 From: London Posts: 11 Thanks: 0 I found a similar question but with 2 "jugs" instead of 3... https://math.stackexchange.com/quest...-of-operations I don't really understand everything but maybe it could be applied for this question?
February 19th, 2019, 08:11 AM   #7
Newbie

Joined: Feb 2019
From: Scotland

Posts: 6
Thanks: 0

Quote:
 Originally Posted by Estermont I found a similar question but with 2 "jugs" instead of 3... https://math.stackexchange.com/quest...-of-operations I don't really understand everything but maybe it could be applied for this question?
Thanks! So.. What they're saying is that there can be a general algorithm for this? Wouldn't it be a lot more complex with 3 measures though?

February 19th, 2019, 09:52 AM   #8
Math Team

Joined: Oct 2011

Posts: 14,146
Thanks: 1003

Quote:
 Originally Posted by skipjack The rules permit using the beaker, but emptying the beaker isn't a valid step.
What meanest thou?
If the rules did NOT permit using the beaker, then where
would the water come from?
You HAVE to use the beaker: it's the source of the water...no?

If beaker is used only as a water source, then there is no need
to mention it, and the problem would be CLEAR this way:

Jane wants to have 2 separate 1 ounce measures of water at the same time.
However the only measures she has are for 6, 10 and 15 ounces.
So she needs to end up with 2 of these 3 measures containing 1 ounce each.
Show how this could be done in the smallest number of steps,
without marking the measures or using any other container.
The only steps allowed are filling or emptying a measure or transferring
water from one measure to another.
Assume the kitchen water tap is used as the source of water!

 February 19th, 2019, 12:05 PM #9 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,146 Thanks: 1003 I "nervously!" have a 10 step solution (water from kitchen tap!): Code: STEP [15] [10] [6] 0 0 0 1 0 10 0 : fill [10] 2 10 0 0 : [10] to [15] 3 10 0 6 : fill [6] 4 15 0 1 : fill [15] from [6] 5 5 10 1 : fill [10] from [15] 6 0 10 1 : empty [15] 7 10 0 1 : dump [10] in [15] 8 10 1 0 : dump [6] in [10] 9 10 1 6 : fill [6] 10 15 1 1 : fill [15] from [6] Am I ok or did I goof?
 February 19th, 2019, 06:38 PM #10 Global Moderator   Joined: Dec 2006 Posts: 20,386 Thanks: 2012 That works. There's another solution in 10 steps, and 10 seems to be the minimum number of steps.

 Tags fluid, form, number, steps

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Punch Algebra 2 January 29th, 2012 04:35 AM kent162 Complex Analysis 1 October 29th, 2011 05:38 AM sivela Math Events 3 February 23rd, 2011 12:31 AM zeroman89 Advanced Statistics 3 October 7th, 2010 02:35 PM BlackOps Algebra 2 June 22nd, 2008 04:01 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top