Divisor = Dividend – Remaining/Quotient when the remainder is 0 and Dividend – Remainder/Quotient when the remainder is non-zero. A divisor can be made into a quotient, and the quotient can be made into a divisor. For example, 10 5 = 2 or 10 2 = 5 ( 5 and 2 are the quotient and divisor in different situations.)
How do you find the divisor of a dividend and a quotient?
The divisor of 12 becomes the quotient, and the quotient becomes the divisor of 12 if we have 12 12. A divisor can be found by using a formula. Divisor = Dividend – Quotient = 0 if the remainder is zero. A non-zero divisor is one where the dividend is divided by the remainder.
How do you find the divisor?
With quotient and dividend, finding the divisor is as simple as 1, 2, or 3. Take a look at this: The quotient is 5, the dividend is 49, and the divisor is… Taking the Dividend, subtract the remainder and divide by the Quotient.
What is the formula to find quotient?
Formula: Dividend – Divisor = Quotient. A simple example of 12 x 4 = 3 can help us comprehend this.
How do you divide in thousands?
- Decimal places are moved three columns to the right when multiplying by 1000.
- A ‘0’ will be placed in front of the decimal point, indicating that it will be followed immediately by the tenths column.
- Dividing 9000 by 1000 has the same effect as eliminating the final three numbers, which are all ‘0’s.
How many divisors does 1 have?
A prime number is a whole number that has only two divisors, 1 and itself, and is therefore a prime number. A prime number cannot have more than one divisor. Because it has three divisors (1, 2 and 4), the number 4 is not prime, and the number 6 has four divisors, it is also not prime ( 1 , 2 , 3 , and 6 ).
What are the divisors of 1080?
Ten thousand eight hundred and eighty is divisible by 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 27, 30… and 1080 itself.
Is there any relation between dividend divisor quotient and remainder?
A closer look at each of these division-related GMAT questions will help you prepare for all kinds of quotient/remainder problems. If we divide 7 by 4, we’ll get 7/4 = 1 + 3/4. This basic example illustrates the importance of division terminology.
We’re referring to the dividend as the phrase we’re dividing by something else. The four-digit number that divides is known as the divisor. The quotient of a mixed fraction is 1, which is the whole number component. 3 is the final number. Even if you have to re-learn the terms, this should feel familiar to you.
The basic remainder formula is: Dividend/Divisor = Quotient + Remainder/Divisor. When we increase the Divisor by the Remainder formula, we get another useful variation: Addition: Dividend = Quotient*Divisor + Remainder
The following official GMAT question can be answered using only this vocabulary and the remainder equation:
In this case, the quotient is S, and the remainder is V, when N is divided by T. How do you get from this statement to the number N?
Divided by something else, our dividend is N; the divisor is T; the quotient is S; and the remainder is V in this problem. After plugging the variables into our residual equation, we get N = ST + V… and we’re done! C is correct.
If you’ve forgotten the remainder formula, you can still answer this problem by picking simple numbers. If N is 7 and T is 3, then 2 + 1/3 Equals 7/3. V = 1 since the Quotient is 2 and the remainder is 1. ‘ To get a N of seven, we’ll need to enter in three times for T, two times for S, and one time for the V. Because 2*3+1=7, we can assume that answer C is accurate.)
Often a statement will offer us information about the dividend in terms of the divisor and the remainder when we need to build a list of possible numbers to verify for data sufficiency.
With this question in mind, what is the answer to: x divided by 5 yields 4. x, 5, and 4 are the divisor and remainder, respectively, in this example. To avoid confusion, we’ll refer to the quotient as q. It will look like this: x = 5q + 4 in equation form. To generate x, we can pick values for q, keeping in mind that the quotient must be a positive integer.
Assuming q is 0, x is 4. x Equals 9 if q is 1. The value of x is equal to 14 if q is 2. Keep an eye out for a trend emerging when we use these x values: x = 4… The remainder is the first acceptable value of x. After that, all we have to do is keep adding the divisor, 5. You could go on like this for as long as you wanted: 4, 9, 14, 19, 24, 29, etc. If you’re dealing with complex data sufficiency issues like the following:
1) The remaining of x – y divided by 5 equals 1
The leftover of x + y equals 2 when x + y is divided by 5.
There is no need to include statement (2) in addition to statement (1).
In this case, (B), statement (2) is sufficient, but (A) statement (1) is not.
As a whole, both statements are enough, yet neither statement on its own is sufficient.
Statement 1 provides us with possible values for x – y in this problem. Recall that x–y must be bigger than or equal to a multiple of 5. Starting with the divisor (5), we can determine that x–y = 1 or 6 or 11, etc. by adding the divisor to the remainder. If x – y = 1, we can claim that x is 1 and y is nil. After x2 + y2 = 1 + 0 = 1, then the residual when 1 is divided by 5 is 1. If x – y = 6, then we can argue that x = 7 and y = 0 There is a residue of zero when the sum of x and y is divided by 5, hence x and y are 49 and 1 respectively. Statement 1 isn’t enough because the rest of the situation varies from one to the next.
As stated in the second paragraph, we have some possible values for the sum of x and y. For the sake of clarity, let’s use the formula: x + y must be a multiple of 5. x + y = 2 or 7 or 12, etc., if we start with the remainder and keep adding the divisor (5). In this case, we can claim that x and y are equal to one each. This means that x2 + y2 = 2 + 2 = 1, and the leftover when 2 is divided by 5 is 2. It’s possible to claim that the value of x is 7 and y is 0. 49 + 0 Equals 49, and the rest of 49 divided by 5 is 4. Statement 2 is likewise insufficient on its own because the rest of the situation varies.
Simply choose one from Statement 1 and one from Statement 2 and watch what happens. This is the typical C or E situation, so let’s try it out! Assume that (x + y) = 7 and (x-y) = 1 We get 2x = 8 or x = 4 by combining these equations. If x is 4, y is 3. There are now 16+9=25, therefore the remaining when 25 is divided by 5 is zero.
Here’s a quick refresher: To get a non-E result on Data Sufficiency, we must know that the value will remain constant under all possible scenarios. Let’s attempt a different scenario just to be safe. Assume the following values: x–y = 6; and x+y = 12. We get 2x = 18, or x = 9 by combining the equations. To get 90, you need to multiply x by 9 and y by 3. Once again, the resulting number is zero. Regardless of the variables we choose, we can be sure that the remainder is zero. C is the correct response based on the statements.
In this section, we’ll highlight some of the most crucial points about GMAT quotient/remainder issues and the formula for solving them. This is a topic that will almost certainly come up on the GMAT, so make sure you understand it thoroughly. Make sure you understand the formula for the remainder: Dividend is equal to the product of the divisor and the quotient, plus the remainder. Another way to make selections is to start with a numeric value and then add the divisor in successive iterations. Rest questions will become much more manageable if you can get your head around these two concepts.
The term “division” can be a little intimidating for adults, but remember that you were a master at it as an elementary school student.
If you become bogged down in jargon or abstraction on the GMAT, simply use simple numbers to jog your memory of how the procedure actually works.
Dividend = Divisor*Quotient + Remainder is a crucial equation, yet even if you forget it, you can reassemble the equation.
Begin by dividing seven by four, as we did at the beginning.
That yields a result of 1, with a leftover of 3.
Furthermore, the quotient is 1, as 4 is divided by 7 just once, leaving 3 as the remainder.
After all of this has been done, you may reassemble the equation as follows: 7 = 4(1) + 3: 7 = 4(1) + 3; 7 = 4(3) + 3.
The rest of quotient/remainder difficulties is a concept you’ve mastered over the course of your entire life.
Want extra practice with the remainder equations and division problems from the GMAT?
Use the Veritas Prep Question Bank or practice tests to work through some of our other articles on this frequently-tested GMAT topic.
