[Maths Class Notes] on Partial Quotient Pdf for Exam

Division results in a different number as a result of dividing a number by another number. Dividends are numbers or integers that get divided, and the integer that divides a given number is the divisor. The remainder of a divisor is that number that does not divide the number entirely. Division symbols are represented by a ÷  or a /. Therefore, we can represent the division method as follows;

Dividend = Quotient × Divisor + Remainder

If the remainder is zero, then; Dividend = Quotient × Divisor

Therefore, Quotient = Dividend ÷ Divisor

Definition of a Partial Quotient

An approach to solving large division problems by using partial quotients is called a partial fraction. By taking a more logical approach to the problem, the student is able to see it less abstractly.

If you want to try this technique in your classroom, you might want to start with the Box Model/Area Model. Partial Quotients and the Box Method are similar in approach, but the Box Method is structured differently and is a good introduction to the quantitative method.

Using Partial Quotients as a Division Strategy

In the division of repeated subtraction and partial quotients, there is a strong correlation. This correlation makes dividing long divisions easy to comprehend and apply. You can easily divide partial quotients according to the three simple steps outlined here. The final quotient can be calculated by adding up each multiplier in the repeated subtraction division method. Long division of numbers can be easily performed by dividing using partial quotients. A fraction of a second after the partial quotient division of the numbers has been performed, the quotient and remainder will be displayed.

How to Divide Using the Partial Quotients Method?

To solve basic division problems, the partial quotients approach (also known as chunking) employs repetitive subtraction. When dividing a big number by a small number.

• Step 1– Find out an easy multiple of the divisor and deduct from the dividend (for example 100 ×, 10 ×, 5 × 2 ×, etc.) 

• Step 2 – Continue subtracting until the large number is reduced to zero or the remainder equals the divisor.

• Step 3 – To find the division answer, add up the multipliers of the divisor that were used in the repeated subtraction.

The partial quotient approach is shown in the diagram below. More explanations and solutions can be found further down the list

Using Partial Quotients to Divide One-Digit Numbers

Method of Partial Division

The use of partial quotients is useful for dividing large numbers. Using partial quotients you can divide the problem into smaller pieces and simplify division. All of the bits are then added back together to get the total.

 Let’s try it with 654 ÷ 3.

Step 1 – Start by subtracting multiples of 3 until you reach 0.

A multiple of 3 that goes into 654 is 600, because 3×200=600. Subtract 600 from 654.

You have 54 left. Now subtract another multiple of 3. You can use 30, since 3×10=30.

You have 24 left. Keep going! Subtract 24, since 3×8=24.

You’ve reached 0, so move to step 2.

Step 2 – Now, see how many times of ‘3’ it took to reach 654.

You broke 654 into 600, 30, and 24. Add the number of times it took 3 to reach each of those numbers.

200 + 10 + 8 = 218

So, 654 ÷ 3 = 218!

Divide Decimals with Partial Quotients 

Using Partial Quotient, You can divide Decimals with Remainders. This will help you find the problem. Make your answer as precise as possible.

Partial Quotients for Dividing by 2-Digit Numbers

Using partial quotients is also an option for dividing large numbers.

Let’s try it with 5,520 ÷ 23.

Step 1 – Start by subtracting multiples of 23 until you reach 0.

A multiple of 23 that goes into 5,520 is 4,600, because 23 × 200 = 4,600. Subtract 4,600 from 5,520.

You have 920 left. Now subtract another multiple of 23. You can use 690, since 23 × 30 = 690.

You have 230 left. Subtract 230, since 23 × 10 = 230.

You’ve reached 0, so move to step 2.

Step 2 – Now, see how many times 23 went into 5,520.

You broke 5,520 into 4,600, 690, and 230. Add the number of times it took 23  to reach each of these –

200 + 30 + 10 = 240

So, 5,520 ÷ 23 = 240!