Vertical Multiplication Practice Problems & Worksheets

To solve this problem, we'll follow these steps:

• Step 1: Identify the numbers to be multiplied: $15$ and $2$.
• Step 2: Calculate the product of these two numbers.

Now, let's work through each step:
Step 1: We are given the numbers $15$ and $2$.
Step 2: We will multiply these numbers together: $15 \times 2$ = $30$.

Therefore, the solution to the problem is $30$.

To solve this problem, we'll follow these steps:

• Step 1: Multiply the ones digit of the two-digit number by the single-digit number.
• Step 2: Multiply the tens digit of the two-digit number by the single-digit number.
• Step 3: Add the two products from the above steps to find the final result.

Now, let's work through each step:
Step 1: Multiply $5 \times 9$. This results in $45$, which includes the 5 in the ones place, and we carry over 4.
Step 2: Next, multiply $2 \times 9$ (from the tens place), which equals $18$. Add the carried-over 4 to get $18 + 4 = 22$. This 22 represents 220 when taking place value into account.
Step 3: Combining steps 1 and 2, we put the $5$ from step 1 in the ones digit and the result from step 2 as tens (which corresponds to $220 + 5 = 225$).

Therefore, the solution to the problem is $x = 225$, aligning with choice 4.

To solve this multiplication problem, follow these clear steps:

• Step 1: Align the numbers vertically (place 26 above 6), ensuring the digits are properly arranged by place value.
• Step 2: Begin multiplication with the unit digit of the bottom number (6). Multiply 6 by each digit in 26, starting from the right.

Now, let's perform the calculations:

Step 1: Multiply the units digit of 6 with the number 26:
- $6 \times 6 = 36$. Write 6 in the units place of the answer, and carry over the 3.
- Next, multiply $6 \times 2 = 12$. Then, add the carryover (3) to 12, resulting in 15.

Step 2: Write 15 next to the 6 in the result. Thus, the complete multiplication gives 156.

Therefore, the solution to the problem is $\boxed{156}$.

To solve this problem, we'll use long multiplication. Here's how to proceed step-by-step:

• Step 1: Begin with the multiplication of the units digit of 28, which is 8, by 5.
• Step 1 Calculation: $8 \times 5 = 40$. We write 0 in the units place and carry over 4.
• Step 2: Multiply the tens digit of 28, which is 2, by 5.
• Step 2 Calculation: $20 \times 5 = 100$.
• Step 3: Add the products from Step 1 and Step 2.
• Addition: $100 + 40 = 140$.

Therefore, the product of 28 and 5 is $140$.

We will solve the problem using direct multiplication of the two numbers, 30 and 4.

Steps:

• First, multiply the one's place of 30 by 4:
  $0 \times 4 = 0$

• Second, multiply the ten's place of 30 by 4:
  $3 \times 4 = 12$

• The result from the tens multiplication is over the magnitude of the number 30 (since it's in the tens place), so we already account for place by multiplying 3 by 4 directly forming a product 12, no tens digit carries from one's digit.

• Combine these results to get the total product:
  $0 + 120 = 120$

Therefore, the product of $30$ and $4$ is $\mathbf{120}$.

By referencing the multiple-choice options provided, the correct choice matches the calculation we performed and is choice 3: $120$.