Math worksheet with a word problem involving a lizard named Frazer building a wall with bricks.

Let's solve both parts of the problem step by step.

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Problem 1: Frazer’s Wall

Given:
- Total bricks needed: 650
- Day 1: 243 bricks
- Each subsequent day, he lays ⅔ of the number of bricks laid the previous day.
- We need to find how many days it takes him to lay at least 650 bricks.

This is a geometric sequence where:
- First term $ a = 243 $
- Common ratio $ r = \frac{2}{3} $

The total number of bricks laid after $ n $ days is the sum of the geometric series:

$$
S_n = a \cdot \frac{1 - r^n}{1 - r}
$$

Plug in the values:

$$
S_n = 243 \cdot \frac{1 - \left(\frac{2}{3}\right)^n}{1 - \frac{2}{3}} = 243 \cdot \frac{1 - \left(\frac{2}{3}\right)^n}{\frac{1}{3}} = 243 \cdot 3 \cdot \left(1 - \left(\frac{2}{3}\right)^n\right)
$$

$$
S_n = 729 \left(1 - \left(\frac{2}{3}\right)^n\right)
$$

We want $ S_n \geq 650 $. So:

$$
729 \left(1 - \left(\frac{2}{3}\right)^n\right) \geq 650
$$

Divide both sides by 729:

$$
1 - \left(\frac{2}{3}\right)^n \geq \frac{650}{729}
$$

$$
\left(\frac{2}{3}\right)^n \leq 1 - \frac{650}{729} = \frac{79}{729}
$$

Now take logarithms (or try small values of $ n $):

Try $ n = 1 $:
$ S_1 = 243 $

Try $ n = 2 $:
$ 243 \times \frac{2}{3} = 162 $ → Total: $ 243 + 162 = 405 $

Try $ n = 3 $:
$ 162 \times \frac{2}{3} = 108 $ → Total: $ 405 + 108 = 513 $

Try $ n = 4 $:
$ 108 \times \frac{2}{3} = 72 $ → Total: $ 513 + 72 = 585 $

Try $ n = 5 $:
$ 72 \times \frac{2}{3} = 48 $ → Total: $ 585 + 48 = 633 $

Try $ n = 6 $:
$ 48 \times \frac{2}{3} = 32 $ → Total: $ 633 + 32 = 665 $

✔ After 6 days, he has laid 665 bricks, which is more than 650.

So, Frazer finishes on day 6.

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Problem 2: Captain Salamander’s Wall

Given:
- Total bricks: 650
- Day 1: 180 bricks
- Each day, he lays 22 fewer bricks than the day before → Arithmetic sequence
- First term $ a = 180 $
- Common difference $ d = -22 $

We want to find the smallest $ n $ such that the sum of the first $ n $ terms is at least 650.

Sum of arithmetic sequence:

$$
S_n = \frac{n}{2} \left[2a + (n - 1)d\right]
$$

Plug in:

$$
S_n = \frac{n}{2} \left[2(180) + (n - 1)(-22)\right] = \frac{n}{2} \left[360 - 22(n - 1)\right]
$$

$$
S_n = \frac{n}{2} \left[360 - 22n + 22\right] = \frac{n}{2} (382 - 22n)
$$

Set $ S_n \geq 650 $:

$$
\frac{n}{2} (382 - 22n) \geq 650
$$

Multiply both sides by 2:

$$
n(382 - 22n) \geq 1300
$$

$$
382n - 22n^2 \geq 1300
$$

Bring all terms to one side:

$$
-22n^2 + 382n - 1300 \geq 0
$$

Multiply by -1 (reverse inequality):

$$
22n^2 - 382n + 1300 \leq 0
$$

Now solve the quadratic equation:

$$
22n^2 - 382n + 1300 = 0
$$

Use quadratic formula:

$$
n = \frac{382 \pm \sqrt{(-382)^2 - 4 \cdot 22 \cdot 1300}}{2 \cdot 22}
$$

Calculate discriminant:

$$
D = 382^2 - 4 \cdot 22 \cdot 1300 = 146,004 - 114,400 = 31,604
$$

$$
\sqrt{31,604} \approx 177.77
$$

Now:

$$
n = \frac{382 \pm 177.77}{44}
$$

First root:

$$
n = \frac{382 - 177.77}{44} = \frac{204.23}{44} \approx 4.64
$$

Second root:

$$
n = \frac{382 + 177.77}{44} = \frac{559.77}{44} \approx 12.72
$$

So the inequality $ 22n^2 - 382n + 1300 \leq 0 $ holds for $ n $ between approx 4.64 and 12.72.

Since $ n $ must be an integer, we test values from $ n = 5 $ to $ n = 12 $.

But we want the smallest $ n $ such that $ S_n \geq 650 $.

Let’s compute cumulative sums:

- Day 1: 180 → Total: 180
- Day 2: 180 - 22 = 158 → Total: 180 + 158 = 338
- Day 3: 158 - 22 = 136 → Total: 338 + 136 = 474
- Day 4: 136 - 22 = 114 → Total: 474 + 114 = 588
- Day 5: 114 - 22 = 92 → Total: 588 + 92 = 680

✔ On day 5, total is 680 ≥ 650 → Captain Salamander finishes on day 5

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Compare Both Builders

- Frazer: Takes 6 days
- Captain Salamander: Takes 5 days

👉 Captain Salamander builds his wall first.

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✔ Final Answers:

1. Frazer took 6 days to build the wall.
2. Captain Salamander built his wall first (in 5 days).

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✔ Answer:
Frazer took 6 days. Captain Salamander built his wall first.