Problem Solving Task Cards - Drawing a Table: Three math word problems for Level 2 learners, illustrated with a dog eating nuts, a bouncing ball, and a decorated biscuit.

Let’s solve each problem one by one, using logical reasoning and, where needed, drawing a table to organize our thinking — as the title suggests: “Drawing a Table”.

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## 🔢 Problem 7: Wild Dog Eating Nuts

> A wild dog swallowed a total of 105 nuts in five days. Each day he managed to eat eight more than he had on the previous day. How many did he eat on each day?

🧠 Strategy:

This is an arithmetic sequence problem.

Let the number of nuts eaten on Day 1 = `x`
Then:
- Day 2 = x + 8
- Day 3 = x + 16
- Day 4 = x + 24
- Day 5 = x + 32

Total = x + (x+8) + (x+16) + (x+24) + (x+32) = 105

Combine like terms:
→ 5x + (8 + 16 + 24 + 32) = 105
→ 5x + 80 = 105
→ 5x = 25
→ x = 5

✔ Answer:

- Day 1: 5 nuts
- Day 2: 13 nuts
- Day 3: 21 nuts
- Day 4: 29 nuts
- Day 5: 37 nuts

✔️ Check: 5 + 13 + 21 + 29 + 37 = 105 → Correct!

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## 📏 Problem 8: Bouncy Ball Bouncing

> Each time a bouncy ball bounces, it bounces to half the height from which it falls. If Jack drops a ball from a building and it rises to a height of 24 metres, how high will it bounce on the fifth bounce?

🧠 Strategy:

The ball is dropped from 24m → first bounce reaches half of that → 12m
Second bounce → half of 12m → 6m
Third → 3m
Fourth → 1.5m
Fifth → 0.75m

We can make a table:

| Bounce Number | Height (metres) |
|---------------|------------------|
| 1 | 24 ÷ 2 = 12 |
| 2 | 12 ÷ 2 = 6 |
| 3 | 6 ÷ 2 = 3 |
| 4 | 3 ÷ 2 = 1.5 |
| 5 | 1.5 ÷ 2 = 0.75 |

> Note: The problem says “it rises to a height of 24 metres” — this is ambiguous. But since it says “Jack drops a ball from a building and it rises to a height of 24 metres”, we interpret that as: the first bounce reaches 24m, meaning the drop was from 48m? But wait — let’s read carefully.

Actually, re-reading:
> *“If Jack drops a ball from a building and it rises to a height of 24 metres...”*

This likely means: he drops it from some height, and after the first bounce, it rises to 24m. But then the question asks for the fifth bounce.

But if the first bounce is 24m, then:

- Bounce 1: 24m
- Bounce 2: 12m
- Bounce 3: 6m
- Bounce 4: 3m
- Bounce 5: 1.5m

Wait — now I’m confused. Let’s clarify:

In physics, when you *drop* a ball, the first bounce is the first time it rebounds. So if it’s dropped from height H, the first bounce reaches H/2, second bounce H/4, etc.

But the problem says:
> “it rises to a height of 24 metres” — this is probably the height after the first bounce.

So if first bounce = 24m, then:

- Second bounce = 24 / 2 = 12m
- Third = 6m
- Fourth = 3m
- Fifth = 1.5m

✔ So answer is 1.5 metres

BUT — let’s check the wording again:
> “Jack drops a ball from a building and it rises to a height of 24 metres”

This could mean the ball was dropped from 24m, and then the first bounce is 12m. That’s also possible.

However, in standard problem phrasing, if it says “rises to 24m”, that usually refers to the height reached after the first bounce — especially since it’s followed by “how high on the fifth bounce”.

To resolve ambiguity, let’s consider both:

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Option A: Drop from 24m → First bounce = 12m → Fifth bounce = 24 × (1/2)^5 = 24 / 32 = 0.75m

Option B: First bounce = 24m → Fifth bounce = 24 × (1/2)^4 = 24 / 16 = 1.5m

Which is correct?

Look at the language:
> “Jack drops a ball from a building and it rises to a height of 24 metres”

This implies: after being dropped, the ball rises to 24m — meaning first bounce height = 24m.

Therefore, bounce 1 = 24m, so bounce 5 = 24 × (1/2)^4 = 24 / 16 = 1.5m

✔ Final Answer: 1.5 metres

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## 🍪 Problem 9: Nicole’s Biscuits

> Nicole is baking 100 biscuits. She tops every third biscuit with a nut, ices every fourth biscuit, and places a cherry on every fifth biscuit. When she has finished all 100, how many biscuits will have all three decorations?

🧠 Strategy:

We need to find how many biscuits are multiples of 3, 4, and 5 — because:

- Nut: every 3rd → positions: 3, 6, 9, 12, 15, ...
- Ice: every 4th → 4, 8, 12, 16, 20, ...
- Cherry: every 5th → 5, 10, 15, 20, 25, ...

A biscuit gets all three if its position is divisible by 3, 4, and 5 → i.e., divisible by the LCM (Least Common Multiple) of 3, 4, and 5.

Step 1: Find LCM of 3, 4, 5

- Prime factors:
- 3 = 3
- 4 = 2²
- 5 = 5
→ LCM = 2² × 3 × 5 = 60

So, biscuits at positions: 60, 120, 180, ... get all three decorations.

Within 1 to 100, only 60 is ≤ 100.

✔ So only 1 biscuit (biscuit #60) has all three decorations.

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## 📋 Summary of Answers:

| Problem | Answer |
|---------|--------|
| Problem 7 | Day 1: 5, Day 2: 13, Day 3: 21, Day 4: 29, Day 5: 37 |
| Problem 8 | 1.5 metres |
| Problem 9 | 1 biscuit |

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✔ All problems solved with clear reasoning and tables where helpful!