Buckets of Thinking

1169 /www/nrich/html/content/03/05/penta1/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> There are three buckets: one red, one blue and one yellow. They each hold a maximum of $5$ litres. <mdo:image height="91" width="235" alt="three buckets" src="3buckets.gif"></mdo:image> Liquid is measured carefully in whole litres and poured into the buckets, a different number of litres in each one. If the liquid in the red bucket was poured into the blue bucket, it would then contain the same amount of liquid as the yellow bucket. Half the contents of the yellow bucket is the same as twice that in the red bucket. How much liquid is there in each bucket? This problem is also available in French: <a href="http://nrich.maths.org/7121&part=">Question de chaudières</a>. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> There were a number of solutions that came in and here are those that had some extra information about how they went about it. Shendy First, I used letters to show the amount of liquid in each bucket. R= The amount of liquid in the red basket. B= The amount of liquid in the blue basket. Y= The amount of liquid in the yellow basket. One of the sentences states that: Y$\div2 = 2$R Multiply both sides by $2$. Y$=4$R As Y must be smaller or equal to $5$, R can only be $1$ So, the red bucket contains $1$ litre of liquid. Y$=4$R Y$=4\times1$ Y$=4$ The yellow bucket contains $4$ litres of liquid. Also, $x$ is the amount of liquid poured from bucket A As the amount of liquid in bucket B plus the amount of liquid poured from bucket A equals to the amount of water in bucket C, B$+x=5$ As the red bucket contains only $1$ litre of water, $x$ can only equal to $1$ So, B$+1=4$ B$=4-1 =3$ The blue bucket contains $3$ litres of liquid. Jemima, Jasmine, Stephen and Olly used a trial and improvement approach: red $1$ litre, blue $3$ litres, yellow $4$ litres We tried yellow as $5$ litres and that didn't work because you're not allowed half litres. But if yellow was $4$ and red was $1$ plus blue as $3$ it would work because if red was poured into blue it would make $4$ like yellow and twice the amount in red is $2$ and half of $4$ (yellow) is $2$. From Ania: Let the volume of liquid in the red bucket be N. Let the volume of liquid in the blue bucket be P. Let the volume of liquid in the yellow bucket be Q. N, P and Q are given to be whole numbers. Each volume is less than $5$ litres, therefore N, P, Q are less or equal to $5$ (litres). We are also given: N+P=Q and $\frac{1}{2}\times$Q$=2\times$N Re-arranging the last equation we get Q$=4\times$N As Q cannot be bigger than $5$ and both Q and N are whole numbers we must take N$=1$ (if N$=2$ than Q$=8$, which is too much!) Therefore Q$=4\times$N$=4\times1=4$ As P=Q-N we get P$=4-1=3$ So, the red bucket contains $1$ litre, the blue bucket contains $3$ litres and the yellow bucket contains $4$ litres. From Nur: <div>Because half the liquid in the yellow bucket is the same as twice that in the red bucket, this means that there is a quarter of the amount in the red compared to the yellow.</div> <div>Since they must be whole numbers and they can't be bigger than $5$, there must be $1$ litre in the red and $4$ in the yellow.</div> <div>So then the blue bucket must have $3$ litres.</div> Finally, a solution that came in right at the end of the month from Ollie I was working systematically. First I did the red bucket as $1L$, the blue bucket as $2L$ and the yellow bucket as $3L$. $1L$ add $2L$ equals $3L$ so that's ok but half of $3L$ equals $1.5L$ and $1$ times $2$ isn't $1.5L$ so I knew that was wrong so I went on to the next one. The red bucket is $1L$, the blue bucket is $3L$ and the yellow bucket is $4L$. I saw that $1L$ add $3L$ equals $4L$. And $4L$ divided by $2$ equals $2L$ and $1L$ add $1L$ equals $2L$, so I knew it was right. That's how I solved thiis problem. Well done all of you and others who sent in the correct solution as well. I hope that those of you who did not send anything in but worked on it enjoyed the thinking that was necessary. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3>Why do this problem?</h3> <a href="http://nrich.maths.org/public/viewer.php?obj_id=1169&part=index">This problem</a> is a nice, simple activity which stimulates discussion and some real thinking. It can also be opened out in ways that will appeal to older or higher-attaining learners. It uses some basic arithmetic and encourages trial and improvement. It can also be used as an introduction to algebra because of the unknowns in each bucket. <h3>Possible approach</h3> <div>You could start by showing the group the picture of the three buckets and asking for suggestions as to the maximum amount each might hold. Use this discussion to inform the class that each is $5$ litres. Then, reveal each clue in turn and invite pairs of children to talk about possible conclusions.</div> Give learners time to work in pairs on the problem, warning them that you will be focusing on how they worked out their solution. Learners could be asked to find some other arrangements of buckets along with two or three statements that would challenge someone else to work out the amount of water in each. They could keep to the rules that there is a different amount in each bucket, measurements are in whole litres and $5$ litres is the maximum. (See <a href="/content/03/05/penta1/ExtensionBucketsThinking.doc">this sheet</a> for further ideas.) <div>At the end of the lesson when various solutions and methods of reaching them have been discussed, it might be appropriate to model how the problem could be written algebraically.</div> <h3>Key questions</h3> <div>What is the maximum amount of water that the bucket can hold?</div> What do you know about the amount of water in this bucket? <div>Can you think of a way of writing this down or doing a drawing to help?</div> <h3>Possible extension</h3> <a href="/content/03/05/penta1/ExtensionBucketsThinking.doc">This sheet</a> gives more detail about extending the task by encouraging children to make up their own problems, firstly by sticking to the same 'rules', then by varying the constraints. <h3>Extension for the exceptionally mathematically able</h3> Go to <a href="http://nrich.maths.org.uk/6850">More and More buckets</a> <h3>Possible support</h3> Suggest working with counters representing each litre and pictures of the buckets. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Can you work out a relationship between the amount of liquid in the blue bucket compared with the amount in the red bucket? Now you have relationships between red compared with yellow, and red compared with blue, which you can use to suggest some answers. </mdoxml> 2 4 0 1 0 0 0 Buckets of Thinking There are three buckets each of which holds a maximum of 5 litres. Use the clues to work out how much liquid there is in each bucket. Calculations and Numerical Methods Addition & subtraction Calculations and Numerical Methods Multiplication & division Using, Applying and Reasoning about Mathematics Trial and improvement Algebra Introducing algebra