Gulls

The Problem

It has been raining. In the paddock near the school 273 worms have come to the surface. As you can imagine, the seagulls have started to hover overhead.

1. Now gulls are satisfied if they can eat 11 worms in one sitting. How many gulls can be satisfied by the worms in the paddock?
2. Actually gulls can ‘drum up’ worms by ‘running on the spot’. Have you seen them doing that? A flock of 34 gulls lands in the paddock with 273 worms. How many worms will they have to drum up so that they are all satisfied?
3. In another paddock there are 359 worms. After another flock lands there and has drummed up enough worms, each gull can be satisfied. What’s more, the leader of the flock manages to get two extra worms. How many gulls were there in this flock?

Teaching sequence

1. Introduce the problem by showing the students the pictures of gulls and worms and asking them to pose questions about them. (List these for early finishers to try.)
2. Pose the problem to the class.
3. Ask them to think about the first question and the number operation (addition, subtraction, multiplication or division) that they would use. Encourage them to explain their selection.
4. As the students work, in pairs or individually, ask questions that focus their thinking on their selection of number operations and the algorithms they are using.
   What number operation(s) are you using? Why did you select it?
   Show me how you are working out the calculation.
   Are you sure that you have solved the problem correctly? How do you know?
   Does your answer look reasonable? Why?
5. Ask that the students record their solutions so that they can be displayed and shared with others.
6. Display solutions. Let the students look at the other solutions.
7. Discuss the solutions.

Other Contexts

This problem could be put in the context of a school trip gathering leaves. There are so many leaves (worms) and so many students (gulls). Each student has to collect so many leaves.

Extension to the Problem

Consider part (b). Can you predict how many worms will have to be drummed up for 49 gulls? How about for 100 gulls? Try to find a pattern for any number of gulls.

In part (c), is there only one possible solution to the number of gulls that could be fed? How many extra worms are needed to feed these gulls?

Solutions

1. This is a straightforward division problem. How many times does 11 go into 273? Now 273/11 = 24 with remainder 9. So 24 gulls can get their fill. (Perhaps 9 worms escape to live another day.)
2. Now 34 gulls need 34 x 11 = 374 worms. There are 273 there already. So they have to drum up 374 – 273 = 101 worms. (That should get them fit.)
3. In this problem 359 + the number of drummed up worms = 11 times the number of gulls plus 2. So 357 + the number of drummed up worms = 11 times the number of gulls. This means that we have to find a number that, when added to 357 gives a multiple of 11. Now 357 = 32 x 11 + 5. To make this up to a multiple of 11 we need to add on 6. Then we have a total of 363, which is 33 x 11.

So this means that there could have been 33 gulls in the flock that attacked the other paddock.

Solution to Extension (b):

Patterns are often most easily found by making a table. So we produce the table below. The number of drummed up worms is the number of worms needed less 273, the number of worms already on the surface.

The table makes clear what should have been obvious. Every new gull needs 11 more worms. So for 49 gulls we have 15 more gulls than the original 34 and so we need 15 x 11 = 165 more worms. Altogether then they need to drum up 101 + 165 = 266 worms.

Alternatively, 49 gulls need to drum up 49 x 11 - 273 = 539 - 273 = 266 worms.

So 100 gulls then will want 100 x 11 - 273 = 1100 - 273 = 827 worms. Get the students to put this pattern into words.

Solution to Extension (c):

The crucial step in the original problem was to notice that we had to make 357 = 32 + 5 up to a multiple of 11. In part (c) we did this by adding 6. However, we could have added 6 + 11, 6 + 11 + 11 and so on. So we could have added 6 plus any multiple of 11. In this case we manage to get 1 more gull for the ‘6’and one more for each ‘11’. So any number of gulls can be fed. To do this exactly though, the number of worms needed is 6 plus an appropriate multiple of 11.