Weighing Time
Solve simultaneous equations
Devise and use problem solving strategies to explore situations mathematically (be systematic, use equipment, draw a diagram, use smaller cases, use algebra).
This problem can be done in at least three ways though the methods are all essentially the same. If you have spreadsheets handy then that would give a fourth method that relies on the same basic idea.
If your students have been solving two linear equations in two unknowns, then this problem could be set up as a system of three equations in two unknowns. Hence their algebra abilities could be extended. But it is possible to do the problem without any knowledge of algebra – if you think carefully about it.
The Problem
Mr and Mrs Watson and their son work on the dairy farm. Mrs Watson decided that they were all getting a little overweight so she wanted to weigh the family. Unfortunately the only scales they had were some old ones that had once been used to weigh cows and these couldn’t weigh anything less than 100kg. So Mr Watson said that they should get on the scales two at a time.
When they weighed Mr and Mrs Watson the scales showed 124 kg. When they weighed Mrs Watson and her son they showed 140 kg. When they weighed Mr Watson and his son they showed 152 kg. How heavy was each member of the family?
Teaching sequence
- This problem is relatively straightforward and you can probably give it directly to the class to start on without any whole class discussion.
- Help the groups that need it though don’t be too quick to intervene. You might suggest that they try guess and check or algebra, depending on their ability. Encourage the quicker groups to try the Extension.
- Let a few groups report back to the whole class.
- Leave time for students to write up their solutions (or let them do this as homework).
Extension to the problem
Is there an easier way to find the weights of the three Watsons? Can you devise a system of weighing them that is a bit easier to solve, given that the scales can’t weigh anything less than 100kg?
Can their weights be found in less than three weighings?
Solution
Method 1: Guess 50 kg for Mrs Watson. Then Mr Watson weighs 124 – 50 = 74 and the son weighs 140 – 50 = 90. But 90 + 74 = 164 > 152.
What information do we get from this? Can we use guess and improve now rather than just make another wild guess? By choosing 50 kg for Mrs Watson we have got too big an answer for the other two. So we have to increase the weight of Mrs Watson.
Suppose that Mrs Watson was 60 kg. Then Mr Watson would be 64 and their son 80. Now 64 + 80 = 144 < 152. So Mrs Watson weighs something between 50 and 60 kg. As the error in weights was less with the 60 kg guess than maybe we should try our next guess closer to 60 then 50.
Keeping going like this we will eventually get the right answer.
Method 2: If we add all the three weighings together we will find twice the weight of the whole family. So the weight of the whole family is 416/2 = 208. Now Mr and Mrs Watson together weigh 124 kg so their son must weigh 208 – 124 = 84 kg. It’s straightforward then to see that Mrs Watson weighs 140 – 84 = 56 kg and Mr Watson weighs 140 – 84 = 68 kg.
Method 3: Set up three equations. Let Mrs Watson weigh f (for female) kg, Mr Watson weigh m kg and their son weigh s kg. Then
f + m = 124 (1)
f + s = 140 (2) and
s + m = 152 (3).
These equations can be solved by guess and check or by using a table in a spreadsheet. The guess and check method is essentially the same as Method 1, so we will solve the equations formally.
Adding them all together we get 2(f + m + s) = 416, so f + m + s = 208 (4). (Does this ring any bells?) Subtracting (1) from (4) gives s = 208 – 124 = 84. Substituting for s in (2) gives f = 140 – 84 = 56. Substituting for s in (3) gives m = 152 – 84 = 68.
It can also be solved algebraically by rearranging to find:
f = 124 -4 and s = 152 - m. Then we substitute these into f + s = 140 to get
124 - m + 152 - m = 140
276 - 2m - 140
m = 68.
Then f = 56 and s = 84.
Mrs Watson weighs 56kg, Mr Watson weighs 68 kg and their son weighs 84 kg. Do they sound overweight to you?
Solution to the extension
If all three of them were weighed together and then two of the original weighings were made, then the calculations would be a little easier.
There are three people and so three unknowns. As we need three equations to find three unknowns then we need three weighings here. (We have no other information of an informal nature that we can use here. It’s not like some Diophantine equations where we can use the fact that the numbers involved are integers.)
| Attachment | Size |
|---|---|
| WeighingTime.pdf | 37.02 KB |
| WeighingTimeMaori.pdf | 41.24 KB |
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