The Distributive Property: Additional Practice (Set 1)

Split $47$ as $40 + 7$ and distribute the $9$ to both parts.

$9 \times 47 = 9(40 + 7) = 9 \times 40 + 9 \times 7 = 360 + 63 = 423$

The form $9 \times 40 + 7$ multiplies only the first part, and $9 \times 4 + 9 \times 7$ uses $4$ instead of $40$, so both are wrong.

Since $95$ is $5$ below the round $100$, write it as $100 - 5$ and distribute the $8$.

$8 \times 95 = 8(100 - 5) = 8 \times 100 - 8 \times 5 = 800 - 40 = 760$

The correction $8 \times 5 = 40$ is small because $95$ is close to $100$.

The outside $6$ multiplies both terms inside, and the minus sign between them is kept.

$6(20 - 3) = 6 \times 20 - 6 \times 3 = 120 - 18 = 102$

The line $6 \times 20 - 3$ multiplies only the first term, and $6 \times 20 + 6 \times 3$ flips the subtraction to addition, so both are wrong.

Write $197$ as $200 - 3$ and distribute the $5$ to each part.

$5 \times 197 = 5(200 - 3) = 5 \times 200 - 5 \times 3 = 1000 - 15 = 985$

Subtracting only $3$ leaves $997$, and adding $5 \times 3$ gives $1015$, so both of those are wrong.

Distribute the $9$ across the difference, keeping the minus sign.

$9(100 - 4) = 9 \times 100 - 9 \times 4 = 900 - 36 = 864$

Check the inside first: $100 - 4 = 96$, and $9 \times 96 = 864$.

All four splits are true, and all four reach the same answer, so the question is only which leaves the least work.

$6 \times 148 = 6(150 - 2) = 900 - 12 = 888$

Both of those products are ones you can do at a glance. The other splits still leave a two-digit product to work out, $6 \times 48$, $6 \times 74$, or $6 \times 28$, so none of them is as light as $150 - 2$.

The greatest common factor of $45$ and $27$ is $9$, since $45 = 9 \times 5$ and $27 = 9 \times 3$.

$45 + 27 = 9 \times 5 + 9 \times 3 = 9(5 + 3)$

Check: $9(5 + 3) = 9 \times 8 = 72$, and $45 + 27 = 72$. The form $3(15 + 9)$ pulls out only a $3$, so it is not fully factored, and the others give wrong totals.

Split $45$ as $40 + 5$ and distribute the $12$ to both parts.

$12 \times 45 = 12(40 + 5) = 12 \times 40 + 12 \times 5 = 480 + 60 = 540$

Forgetting the $12 \times 5 = 60$ would leave $480$, which is too small.

Both parts of $50 + 3$ must be multiplied by $9$, but the student forgot to multiply the $3$.

$9 \times 53 = 9(50 + 3) = 9 \times 50 + 9 \times 3 = 450 + 27 = 477$

The student added the unmultiplied $3$ instead of $9 \times 3 = 27$.

Split $88$ as $80 + 8$ and distribute the $7$ to both parts.

$7 \times 88 = 7(80 + 8) = 7 \times 80 + 7 \times 8 = 560 + 56 = 616$

The form $7 \times 80 + 2$ multiplies only the first part, so it is wrong.

Multiply each of the three terms by $4$, then add the products.

$4(12 + 9 + 6) = 4 \times 12 + 4 \times 9 + 4 \times 6 = 48 + 36 + 24 = 108$

Check the inside first: $12 + 9 + 6 = 27$, and $4 \times 27 = 108$.

One box yields $12 - 2 = 10$ sellable mugs, and there are $7$ identical boxes.

$7(12 - 2) = 7 \times 10 = 70$

Distributing gives the same count: $7 \times 12 - 7 \times 2 = 84 - 14 = 70$. The form $7 \times 12 - 2$ removes only $2$ chipped mugs in total instead of $2$ from every box, and $12(7 - 2)$ subtracts boxes rather than mugs.