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Joonas has an unlimited supply of €5 notes and €2 coins - Junior Cycle Mathematics - Question 7 - 2018
Question 7
Joonas has an unlimited supply of €5 notes and €2 coins. (a) Fill in the table to show three different ways in which he can use these to make exactly €27. One way i... show full transcript
Worked Solution & Example Answer:Joonas has an unlimited supply of €5 notes and €2 coins - Junior Cycle Mathematics - Question 7 - 2018
Step 1
Way 2
Answer
To calculate another way to create €27, we can use 3 €5 notes and a certain number of €2 coins. Since 3 €5 notes give €15, we need an additional €12 to reach €27. The equation for the number of €2 coins needed is:
Thus, for Way 2, we can fill in the table:
- Number of €5 notes: 3
- Number of €2 coins: 6
Step 2
Way 3
Answer
For the last way (Way 3), we will use 5 €5 notes. This contributes €25 to our total. Therefore, to reach €27, we need an additional €2:
- Number of €5 notes: 5
- Number of €2 coins: 1
Now the table is complete:
| Number of €5 notes | Number of €2 coins | Total amount of money |
|---|---|---|
| Way 1 | 1 | 11 |
| Way 2 | 3 | 6 |
| Way 3 | 5 | 1 |
Step 3
Explain how he could use his supply of €5 notes and €2 coins to make every whole number value of money greater than €3
Answer
To create every whole number value greater than €3, Joonas can utilize his unlimited supply of €5 notes and €2 coins in a systematic way.
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Using €2 coins: Every even number amount can be obtained using €2 coins alone. Every odd number can be crafted by pairing an odd quantity of €2 coins with a single €5 note. For example:
- No €2 coins leads to €5— all odd amounts ending in 0 or 5.
- One €2 coin gives: €2, €7; all odd amounts ending in 2 or 7.
- Two €2 coins give: €4, €9; all odd amounts ending in 4 or 9.
- Three €2 coins yield: €6, €11; all odd amounts ending in 6 or 1.
- Four €2 coins provide: €8, €13; all odd amounts ending in 8 or 3.
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General Method: By using 5 + 2k (where k is any non-negative integer), Joonas can achieve every even number. Adding one €5 note to these combinations, he can generate every odd number greater than €3.
In summary, with this combination, all values greater than €3 can be achieved.