Vertical Multiplication Practice Problems & Worksheets

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

• Step 1: Multiply the unit digits.
• Step 2: Multiply the tens digits.
• Step 3: Add the results from Steps 1 and 2.

Let's execute these steps:
Step 1: Multiply the unit digit of 53, which is 3, by 3:
$3 \times 3 = 9$.

Step 2: Multiply the tens digit of 53, which is 5 (standing for 50), by 3:
$50 \times 3 = 150$.

Step 3: Add the results of Step 1 and Step 2:
$150 + 9 = 159$.

Therefore, the solution to the problem is $159$.

To solve this problem, we'll employ vertical multiplication.

Step 1: Set up the multiplication:
$62$
× $4$
---------

Step 2: Multiply each digit of 62 by 4. We start with the ones place, then the tens place.

• Multiply the ones digit: $2 \times 4 = 8$.
• Multiply the tens digit: $6 \times 4 = 24$.

Step 3: Consider the place value for each part of the calculation:
The result from multiplying the tens digit by 4 represents $240$ because it is $24 \times 10$.

Step 4: Add the two partial results:
8
+ 240
---------
248

Therefore, the solution to the problem is $248$.

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: $120$.

To solve this problem, we'll perform vertical multiplication of $82$ by $2$:

• Step 1: Multiply the ones place. Multiply $2$ (from 82) by $2$:

$2 \times 2 = 4$
This gives us $4$ in the ones place.

• Step 2: Multiply the tens place. Multiply $8$ (in the tens place of 82) by $2$:

$8 \times 2 = 16$
Since the result is $16$, we place $6$ in the tens place and carry over $1$ to the next higher place (hundreds place).

• Step 3: Add up the intermediate results.

The ones place has $4$, and the tens place has $6$ plus $1$ (carry-over), totaling to $7$ in the tens place. Thus, the full number now reads:

$164$

Therefore, the solution to the problem is $164$.

To solve this problem, we will multiply $91$ by $3$ using standard multiplication techniques:

• Step 1: Multiply the unit digit of $91$ by $3$:
  $1 \times 3 = 3$.
• Step 2: Multiply the tens digit of $91$ by $3$:
  $9 \times 3 = 27$.
• Step 3: Place the result of $27$ correctly one digit to the left (because it's actually $90 \times 3$), which gives $270$.
• Step 4: Add the results from Step 1 and Step 3:
  $270 + 3 = 273$.

Therefore, the product of $91 \times 3$ is $273$.