We can think of this problem as an improper fraction and solve for the mixed number by placing the size of the sub over the number of pieces to be cut. We get the following:

 can go into  only  times. In other words, we are looking for the number that when multiplied by the denominator gets us as close to the numerator as possible, without going over the value of the numerator.

Simple multiplication reveals the following:

In order to find out what is left over, we must subtract this number from the numerator.

The difference will always be less than the denominator from the original improper fraction. If it is not, then check your work in the previous step because this means that the denominator could have been divided into the numerator at least one more time.

Then, we put the difference over the denominator:

Let's solve for the entire improper fraction by putting these values together and forming a mixed number:

Last, we know that  is between the numbers  and ; therefore, the correct answer is:

 and 

We can think of this problem as an improper fraction and solve for the mixed number by placing the size of the sub over the number of pieces to be cut. We get the following:

 can go into  only  times. In other words, we are looking for the number that when multiplied by the denominator gets us as close to the numerator as possible, without going over the value of the numerator.

Simple multiplication reveals the following:

In order to find out what is left over, we must subtract this number from the numerator.

The difference will always be less than the denominator from the original improper fraction. If it is not, then check your work in the previous step because this means that the denominator could have been divided into the numerator at least one more time.

Then, we put the difference over the denominator:

Let's solve for the entire improper fraction by putting these values together and forming a mixed number:

Last, we know that  is between the numbers  and ; therefore, the correct answer is:

 and 

We can think of this problem as an improper fraction and solve for the mixed number by placing the size of the sub over the number of pieces to be cut. We get the following:

 can go into  only  times. In other words, we are looking for the number that when multiplied by the denominator gets us as close to the numerator as possible, without going over the value of the numerator.

Simple multiplication reveals the following:

In order to find out what is left over, we must subtract this number from the numerator.

The difference will always be less than the denominator from the original improper fraction. If it is not, then check your work in the previous step because this means that the denominator could have been divided into the numerator at least one more time.

Then, we put the difference over the denominator:

Let's solve for the entire improper fraction by putting these values together and forming a mixed number:

Last, we know that  is between the numbers  and ; therefore, the correct answer is:

 and 

We can think of this problem as an improper fraction and solve for the mixed number by placing the size of the sub over the number of pieces to be cut. We get the following:

 can go into  only  times. In other words, we are looking for the number that when multiplied by the denominator gets us as close to the numerator as possible, without going over the value of the numerator.

Simple multiplication reveals the following:

In order to find out what is left over, we must subtract this number from the numerator.

The difference will always be less than the denominator from the original improper fraction. If it is not, then check your work in the previous step because this means that the denominator could have been divided into the numerator at least one more time.

Then, we put the difference over the denominator:

Let's solve for the entire improper fraction by putting these values together and forming a mixed number:

Last, we know that  is between the numbers  and ; therefore, the correct answer is:

 and 

We can think of this problem as an improper fraction and solve for the mixed number by placing the size of the sub over the number of pieces to be cut. We get the following:

 can go into  only  times. In other words, we are looking for the number that when multiplied by the denominator gets us as close to the numerator as possible, without going over the value of the numerator.

Simple multiplication reveals the following:

In order to find out what is left over, we must subtract this number from the numerator.

The difference will always be less than the denominator from the original improper fraction. If it is not, then check your work in the previous step because this means that the denominator could have been divided into the numerator at least one more time.

Then, we put the difference over the denominator:

Let's solve for the entire improper fraction by putting these values together and forming a mixed number:

Last, we know that  is between the numbers  and ; therefore, the correct answer is:

 and 

We can think of this problem as an improper fraction and solve for the mixed number by placing the size of the sub over the number of pieces to be cut. We get the following:

 can go into  only  times. In other words, we are looking for the number that when multiplied by the denominator gets us as close to the numerator as possible, without going over the value of the numerator.

Simple multiplication reveals the following:

In order to find out what is left over, we must subtract this number from the numerator.

The difference will always be less than the denominator from the original improper fraction. If it is not, then check your work in the previous step because this means that the denominator could have been divided into the numerator at least one more time.

Then, we put the difference over the denominator:

Let's solve for the entire improper fraction by putting these values together and forming a mixed number:

Last, we know that  is between the numbers  and ; therefore, the correct answer is:

 and 

We can think of this problem as an improper fraction and solve for the mixed number by placing the size of the sub over the number of pieces to be cut. We get the following:

 can go into  only  times. In other words, we are looking for the number that when multiplied by the denominator gets us as close to the numerator as possible, without going over the value of the numerator.

Simple multiplication reveals the following:

In order to find out what is left over, we must subtract this number from the numerator.

The difference will always be less than the denominator from the original improper fraction. If it is not, then check your work in the previous step because this means that the denominator could have been divided into the numerator at least one more time.

Then, we put the difference over the denominator:

Let's solve for the entire improper fraction by putting these values together and forming a mixed number:

Last, we know that  is between the numbers  and ; therefore, the correct answer is:

 and 

We can think of this problem as an improper fraction and solve for the mixed number by placing the size of the sub over the number of pieces to be cut. We get the following:

 can go into  only  times. In other words, we are looking for the number that when multiplied by the denominator gets us as close to the numerator as possible, without going over the value of the numerator.

Simple multiplication reveals the following:

In order to find out what is left over, we must subtract this number from the numerator.

The difference will always be less than the denominator from the original improper fraction. If it is not, then check your work in the previous step because this means that the denominator could have been divided into the numerator at least one more time.

Then, we put the difference over the denominator:

Let's solve for the entire improper fraction by putting these values together and forming a mixed number:

Last, we know that  is between the numbers  and ; therefore, the correct answer is:

 and 

We can think of this problem as an improper fraction and solve for the mixed number. 

 can go into  only  times with  left over.