The first question was fairly straightforward. Following on from our work on fractions and percentages last week, the children found a method to find 20% of £60:
In this way, we could prove that Jim made £120 when he sold the cat and the dog, as the profit and loss cancelled each other out. This prompted a valuable discussion about how 20% of one number is different to 20% of another. Finding the same percentage doesn't automatically mean you will get the same answer!
The second question provided a greater challenge.
After some independent conversations and calculations, we came together as a class to find the best way to solve the problem. Our working out is shown below.
It took some time to get our heads around the fact that £60 was 120% of the value that Jim had paid for the dog, as we were told that this included the 20% profit that he made. We then had to work out from this point what 100% was (i.e. the value that he paid originally).
We then looked at the cat and realised that as he had lost 20%, the final £60 that he received for the cat was only 80% of what he paid! From here we could work out what 20% was (guided expertly by Thomas) and then what the full 100% was that he paid.
Bradley K then correctly observed, "He paid £50 + £75, which is £125, but he only received £120 from the sales, so he made a loss of £5 overall."
Relying on trial and error to begin with, the children quickly realised that the best place to put the 0 was in the centre of the square, so that it could be involved in the most calculations. They then realised that they could balance out the square with positive and negative numbers on either side of the 0. I.e. -3 could be balanced out by 3 on the other side of the 0.
We then tried putting the 0 in other squares to see how the outcome was affected. Each time, one of the diagonal calculations let our solution down!
This week's problem was very wordy, so Class 3 first had to identify which information was unnecessary, and which would help them to solve the problem. Once they had done this, it became clear that the Maths expectations of the question were not dissimilar some of our previous Maths Challenges.
Most groups used 'Guess and Check', which is fine, but they realised that with a little more structure, this method could become even more effective.
Both totals are multiples of 10, so it made perfect sense to start by presuming that the prices of the buns and lollies would also be multiples of 10. We did this in the form of a table, so that we could keep track of the prices we had tried and how they affected the totals. By changing the price of only the buns each time, and keeping the amount paid by Multi the same, we could see how the resulting price of the lollies affected the total that Divvy had to pay.
As as you can see, it only took three attempts to match all of our columns with the criteria in the question.
Coins on the table
Anna put some 10p coins on the table.
One half of them were tails up.
Anna turned over two of the coins, and then one third of them were tails up.
How many coins did Anna put on the table?
An explanation by Kirsty and Leah:
First we thought about the different strategies, and the strategy we used was making a list because we wrote out the 3's and the 2's, and saw which two or more matched up.
When we matched up the numbers, we found that 12 was in the 2's and the 3's so we did half of 12 which is 6 and then we took off 2 which gave us 4. Four is a third of 12.
12 was our answer! ( This was the right answer )
We found that the ‘guess and check’ strategy worked the best here. Starting by finding combinations which fitted the first equation (c + t = 4), we applied the numbers that we found to the following two equations.
E.g. if c = 2 and t = 2, (2 x 2) + 2p would have to be 9.
In this instance, the numbers we have chosen do not work. Therefore, we would have to go back to the original equation, reflect on our chosen numbers and try a new combination.
We were lucky that the early pairs that we tried were successful. We know that this won’t always be the case!
Our final answer:
We tried lots of different methods for this problem.
Some of us tried repeated addition of each number to eventually make a total of £5.00.
Others started with 10 of each bar and added/subtracted until we found a total of £5.00. This worked very successfully!
Another pair tried starting at £5.00 rather than 0 and tried taking different combinations away, rather than adding them on. This was also successful but a little less efficient than the final method.
As a class, we decided to write out the 18 and 26 times tables. We then looked for pairs of numbers (one from each column), which would give us a total of £5.00. This was the quickest method and definitely the easiest to explain!
Our final answer was that Anil bought 12 choc bars and 11 fruit bars.
How to solve the problem:
1. Read the question carefully so that you can use what you know.
2. Calculate the value of the circle because you know the total and there are only circles in the calculation. 20 ÷ 4 = 5.
3. Move onto the line with 2 circles and 2 triangles and the total of 26. We know that the circles are 5 each, making a total of 10. 26 - 10 = 16, meaning that each triangle is worth 8.
4. We then moved onto the line with circles (5), triangles (8) and 2 clubs in, with a total of 25. The circle and triangle total 13. 25 - 13 = 12, meaning that each club is worth 6.
5. We now knew the value of each symbol and could simply calculate the missing totals.
The key is to use what you know!
We started by recognising how many different combinations we could find, which gave us 100 eggs in total, and had two of the same quantity.
Sophie: "We found that we needed 10 large eggs, 10 medium eggs and 80 small eggs. We realised that since the total spent was £10, we were likely to have whole pounds spent on each egg. We then used trial and error from the combinations we had found, making 100 in total."
Lauren: "We knew that 80 large eggs would have been too expensive, so when we trialled the different combinations, we had to estimate how much it would cost."
Using mostly 'Guess and Check', we found two different routes to 100.
We also found a few different routes when trying to find our large numbers. Lauren found 391, after realising that she had made an error. She went back and checked, proving how important it is to check your work! Well done, Lauren.
The smallest number that we were able to make was 37, which was made by Faith and Ben.
This task was challenging from the very beginning!
We realised quite quickly that there were two different ways of looking at this problem:
1) Only using combinations of digits which did not exceed 9, e.g. 9+9+7=25, but when placed on the abacus they represent the number 997.
2) The other method was to make a list of all of the possible combinations of numbers which would make 25 when added together. Inevitably, this involved some two-digit numbers, e.g. 7+13+5=25. We know that we cannot have two digits in one column, but here, 11 tens are equivalent to 110. Therefore, when the values are added together, we have 700+130+5=835.
Some of us used the first approach, while others preferred to focus on the second one, as it opened the task up to more possibilities!
- Tables and lists of the different combinations,
- Pictures of the abacus,
- 'Guess and check'
We also realised afterwards that making beads to act out the problem would also be an efficient method.
Those of us working on approach 1 found that there were 6 possible combinations. We also realised that we could only use the digits 7, 8 and 9.
For the second approach, the possibilities were endless, but the highest combination of digits that we found was 35!
Some of us started by guessing, using our addition knowledge. We knew that to make £6 for A and B, we could only have combinations of £5 and £1, £3 and £3 or £4 and £2. We started there and when we found a combination that worked, we carried on through the list, making sure that all of our answers met the criteria we were given.
We agreed that the starting point was the hardest and once we had a clear strategy, we answered the question quite quickly. Most of us relied on the 'guess and check' strategy, although we organised our work in different ways; some of us made tables and others preferred to draw a diagram to show the different pairs. Some of us even decided to start at the end and work backwards to see if it was quicker. It turned out that there was no benefit to this in this particular task.
The children used a combination of drama and trial-and-error for today's task and answered it very quickly! They were able to explain their thinking too. A very successful problem-solving session!
In our groups, we used a variety of different methods, such as lists, tables and separate calculations. However, having checked our working together in the format of a table, we discovered that the queen was actually incorrect. She spent £1,048,575, meaning that this method cost her an extra £48,575. We used the table so that we could keep track of how much each gold ring would cost her.
This proves how important it is to ALWAYS check our working thoroughly, and a different method can sometimes make this even more accurate.
After comparing our strategies, it seemed that everyone had spotted that multiplying was always going to give us a larger answer. Lots of groups also spotted that by putting the larger digits in the positions of hundreds or tens, the answers were also guaranteed to be larger than if these digits were in the ones positions.
We then put ourselves into size order to give our place value (and team-work!) skills a final test!
Lots of trial and error in today's problem solving session!
The children spotted that they were relying most heavily on the 'guess and check' strategy, but some pairs also drew diagrams and made a smaller square first and added to it.
Lots of excitement in today's problem solving session! It was very rewarding when the hard work paid off and the solutions were discovered!
Final solutions to follow!