Operations and Place Value

1. (a) The digit $7$ is in the thousands place, so its value is:

$7 \times 1{,}000 = 7{,}000$

(b) A digit in the ten-thousands place is always $\mathbf{10}$ times greater than the same digit in the thousands place. For example, if the digit $7$ were in the ten-thousands place it would represent $70{,}000$, while in the thousands place it represents $7{,}000$:

$\frac{70{,}000}{7{,}000} = 10$

This is because each place to the left is $10$ times the place to its right.

2. (a) $36{,}000 = 36 \times 1{,}000 = 36 \times 10^3$

(b)

$0.125 \times 10^3 = 0.125 \times 1{,}000 = 125$

This means $1{,}000$ muffins weigh $125$ kilograms.

3. (a) Using the standard algorithm:

$238 \times 14$

Step 1 — Multiply $238 \times 4$:

$238 \times 4 = 952$

Step 2 — Multiply $238 \times 10$:

$238 \times 10 = 2{,}380$

Step 3 — Add the partial products:

$952 + 2{,}380 = 3{,}332$

(b) $3{,}332 < 3{,}500$, so the school did not reach the discount threshold. They are $3{,}500 - 3{,}332 = 168$ notebooks short.

4. (a) Divide $2{,}784 \div 6$:

Step 1 — $6 \times 400 = 2{,}400$. Remainder: $2{,}784 - 2{,}400 = 384$.

Step 2 — $6 \times 60 = 360$. Remainder: $384 - 360 = 24$.

Step 3 — $6 \times 4 = 24$. Remainder: $0$.

$2{,}784 \div 6 = 464$

They will drive $\mathbf{464}$ miles each day.

(b) The remainder is $\mathbf{0}$ miles — the distance divides evenly.

5. (a) Divide $4{,}368 \div 12$:

Step 1 — $12 \times 300 = 3{,}600$. Remainder: $4{,}368 - 3{,}600 = 768$.

Step 2 — $12 \times 60 = 720$. Remainder: $768 - 720 = 48$.

Step 3 — $12 \times 4 = 48$. Remainder: $0$.

$4{,}368 \div 12 = 364$

They can fill $\mathbf{364}$ full boxes.

(b) The remainder is $\mathbf{0}$ — no granola bars are left over.

(c) Divide $364 \div 30$:

$30 \times 12 = 360$

$364 - 360 = 4 \text{ boxes remaining}$

Since there are $4$ leftover boxes that still need a shelf, they need $12 + 1 = \mathbf{13}$ shelves.