Problem of the Week
These provide you with opportunities to solve problems that will help you build your logical, creative and mathematical thinking. This means, we want to know ‘how’ you think! When submitting your entry, be sure to explain how you executed the problem solving strategy you used or when you can, show more than one way of solving the problem!
Week 2 Marbles https://youtu.be/p6Bw3rZRiaA https://youtu.be/2Xik6XxGREw
The focus strategy is ‘Make a Table’. It is a problem solving strategy that you can use to write the information in a more organised way. It helps you to show your mathematical thinking in a logical way enabling you to look critically at the data to find patterns and develop a solution.
|
A dealer packages marbles in two different box sizes. One size holds 5 marbles and the other size holds 12 marbles. If the dealer packaged 99 marbles and used more than 10 boxes, how many boxes of each size did he use? |
After some marbles are packaged in boxes of 12, the remaining marbles must be completely packaged in boxes of 5. The following table shows what happens when some marbles are packaged in boxes of 12.
|
Number of boxes of 12 |
Number of extra marbles |
Number of boxes of 5 |
Total number |
|
1 |
99 - 12 = 87 |
17; 2 marbles leftover |
-- |
|
2 |
99 - 24 = 75 |
15 |
17 |
|
3 |
99 - 36 = 63 |
12; 3 marbles leftover |
-- |
|
. . . |
. . . |
. . . |
. . . |
|
7 |
99 - 84 = 15 |
3 |
10 |
Only two cases (highlighted) can all 99 marbles be completely packaged in 12-marble and 5-marble boxes. However, only the first of these two cases will satisfy the condition that more than 10 boxes must be used. Therefore 2 of the large boxes and 15 of the small boxes were used.
Week 3 Coins
The focus strategy is 'Test all possible combinations'. Have you ever been given a bunch of keys and have no clue which key opens the lock? You would have to test all the keys until you get that one key (or maybe more than one key) that opens it.
https://youtu.be/FaVyO_uqATU https://youtu.be/lGT0_5FZTPc https://youtu.be/QikFsKGa_e4
From a pile of 100 × 1c coins (A), 100 × 5c coins (B), and 100 × 10c coins (C), select 21 coins which have a total value of exactly $1.00. In your selection, you
must also use at least one coin of each type.
Since 1¢ coins must be used in the selection, the number of 1¢ coins used must be a multiple of 5.
Case 1: Suppose 5 × 1¢ coins are selected.
Trial 1 5 × 1¢ = 5¢
16 × 5¢ = 80¢
85¢ ← 15¢ short
Trial 2 5 × 1¢ = 5¢
13 × 5¢ = 65¢
3 × 10¢ = 30¢
$1 ✔
Case 2: Suppose 10 × 1¢ coins are selected.
Trial 1 10 × 1¢ = 10¢
11 × 5¢ = 55¢
65¢ ← 35¢ short
Trial 2 10 × 1¢ = 10¢
4 × 5¢ = 20¢
7 × 10¢ = 70¢
$1 ✔
Case 3: Suppose 15 × 1¢ coins are selected.
Trial 1 15 × 1¢ = 15¢
6 × 10¢ = 60¢
75¢ ← 25¢ short (IMPOSSIBLE!)
You can access the POW's for term 2
here:https://sites.google.com/parra.catholic.edu.au/numeracy/term-2