Answers can be found at the end of the book.
1 15 × 4 =
2 23 × 6 =
3 35 × 9 =
4 26 × 22 =
5 72 × 18 =
6 124 × 12 =
7 161 × 13 =
8 148 × 17 =
9 257 × 14 =
10 321 × 67 =
11 You need to order an inhaler to last a patient for the next 28 days. The patient takes four inhalations (doses) of the inhaler daily. How many inhalations does the patient take over 28 days and will an inhaler containing 200 doses be sufficient for that period?
12 A patient is due to go home for the weekend. They take 5 millilitres of an oral solution four times each day. Over the weekend they will need: one dose for Friday night, four doses for Saturday, four doses for Sunday and one dose for Monday morning. How many millilitres need to be in the bottle to last until the patient returns to hospital?
Multiplying decimals
Clinical calculations will sometimes involve multiplying decimals together. Treat this as you would any other multiplication and then use the technique below to get the decimal point in exactly the right place.
Imagine that you had to calculate the annual leave entitlement for a member of staff. Nurse Williams is a Practice nurse and is entitled to 2.33 days annual leave for each month that she works. She has worked at the surgery for 4.5 months. How much time off is she owed?
To calculate the amount of annual leave that she can take means multiplying 2.33 (days) by 4.5 (months):
Method
As before, the multiplication is calculated vertically from right to left, starting under the ‘hundredths’ column (h) and involves three individual calculations, one for the ‘hundredths’ column (h), one for the ‘tenths’ (t) column and a final calculation for the ‘ones’ column. The results of these three individual calculations are then added together.
Process
Starting with the bottom number at the ‘tenths’ column:
The digit 5 from the number 4.5 is used to multiply the digit 3 in the ‘hundredths’ column from the 2.33. It is then used to multiply the digit 3 in the ‘tenths’ column and then the digit 2 from the number 2.33.
5 × 3 = 15
Similar to the earlier multiplication examples, the 5 from the 15 is recorded below the line beneath the hundredths column and the 1 from the 15 is carried over to the ‘tenths’ column (written as a small ‘1’ in this column).
The digit 5 from the number 4.5 is then used to multiply the digit 3 in the ‘tenths’ column from the number 2.33
5 × 3 = 15
and the 1 carried over is added
15 + 1 = 16
The 6 ‘ones’ in this number are written below the line beneath the 5 of the number 4.5 under the ‘tenths’ column and the 1 ‘ten’ from this number is carried over to the ‘ones’ column
The digit 5 from the number 4.5 is then used to multiply the digit 2 in the ‘ones’ column from the number 2.33
5 × 2 = 10
and the 1 carried over is added
10 + 1 = 11
The 1 ‘one’ in this number is written below the line beneath the 4 of the number 4.5 under the ‘ones’ column and the 1 ‘ten’ from this number is written below the line in the ‘tens’ column
Moving left to the ‘ones’ column:
The first action is to write a zero in the ‘hundredths’ column below the 5 in the first part of the answer. The overall answer to this calculation is calculated by adding up three short answers and this zero acts as a placeholder within the short answer, maintaining the value of the digits.
The digit 4 from the number 4.5 is used to multiply the digit 3 in the ‘hundredths’ column from the 2.33. It is then used to multiply the digit 3 in the ‘tenths’ column and then the digit 2 from the number 2.33
4 × 3 = 12
Similar to the earlier multiplication examples, the 2 from the 12 is recorded below the line beneath the ‘tenths’ column and the 1 from the 12 is carried over to the ‘ones’ column (written as a small ‘1’ in this column).
The digit 4 from the number 4.5 is then used to multiply the digit 3 in the ‘tenths’ column from the number 2.33
4 × 3 = 12
and the 1 carried over is added
12 + 1 = 13
The 3 ‘ones’ in this number are written below the line beneath the 4 of the number 4.5 under the ‘ones’ column and the 1 ‘ten’ from this number is carried over to the ‘tens’ column (written as a small ‘1’ in this column).
The digit 4 from the number 4.5 is then used to multiply the digit 2 in the ‘ones’ column from the number 2.33
4 × 2 = 8
and the 1 carried over is added
8 + 1 = 9
This number is written below the line in the ‘tens’ column.
The next stage of the calculation involves adding up the individual answer numbers that are written below the original sum.
The final stage of the process is identifying the correct location for the decimal point. This is obtained by adding up the number of digits to the right of the decimal point in the numbers being multiplied together. The upper number 2.33 has two digits after the decimal
