All Common Core: 5th Grade Math Resources
Example Questions
Example Question #1041 : Common Core Math: Grade 5
The cafe has of coffee to last days. How many liters of coffee does the cafe have per day, assuming that each day has an equal amount of coffee? Select the answer that contains the pair of numbers that the answer falls between.
and
and
and
and
and
and
We can think of this problem as an improper fraction and solve for the mixed number by placing the amount of coffee over the number of days. 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
Example Question #1042 : Common Core Math: Grade 5
The cafe has of coffee to last days. How many liters of coffee does the cafe have per day, assuming that each day has an equal amount of coffee? Select the answer that contains the pair of numbers that the answer falls between.
and
and
and
and
and
and
We can think of this problem as an improper fraction and solve for the mixed number by placing the amount of coffee over the number of days. 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
Example Question #1051 : Common Core Math: Grade 5
The cafe has of coffee to last days. How many liters of coffee does the cafe have per day, assuming that each day has an equal amount of coffee? Select the answer that contains the pair of numbers that the answer falls between.
and
and
and
and
and
and
We can think of this problem as an improper fraction and solve for the mixed number by placing the amount of coffee over the number of days. 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
Example Question #1052 : Common Core Math: Grade 5
The cafe has of coffee to last days. How many liters of coffee does the cafe have per day, assuming that each day has an equal amount of coffee? Select the answer that contains the pair of numbers that the answer falls between.
and
and
and
and
and
and
We can think of this problem as an improper fraction and solve for the mixed number by placing the amount of coffee over the number of days. 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
Example Question #1051 : Common Core Math: Grade 5
The cafe has of coffee to last days. How many liters of coffee does the cafe have per day, assuming that each day has an equal amount of coffee? Select the answer that contains the pair of numbers that the answer falls between.
and
and
and
and
and
and
We can think of this problem as an improper fraction and solve for the mixed number by placing the amount of coffee over the number of days. 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
Example Question #1052 : Common Core Math: Grade 5
The cafe has of coffee to last days. How many liters of coffee does the cafe have per day, assuming that each day has an equal amount of coffee? Select the answer that contains the pair of numbers that the answer falls between.
and
and
and
and
and
and
We can think of this problem as an improper fraction and solve for the mixed number by placing the amount of coffee over the number of days. 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
Example Question #1053 : Common Core Math: Grade 5
The cafe has of coffee to last days. How many liters of coffee does the cafe have per day, assuming that each day has an equal amount of coffee? Select the answer that contains the pair of numbers that the answer falls between.
and
and
and
and
and
and
We can think of this problem as an improper fraction and solve for the mixed number by placing the amount of coffee over the number of days. 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
Example Question #1054 : Common Core Math: Grade 5
Ashley has pieces of candy that she wants to divide amongst her friends. How many pieces of candy will each friend get, assuming that each friend gets the same amount? Select the answer that contains the pair of numbers that the answer falls between.
and
and
and
and
and
and
We can think of this problem as an improper fraction and solve for the mixed number by placing the amount of candy over the number of Ashley's friends. 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
Example Question #1055 : Common Core Math: Grade 5
Melissa has pieces of candy that she wants to divide amongst her friends. How many pieces of candy will each friend get, assuming that each friend gets the same amount? Select the answer that contains the pair of numbers that the answer falls between.
and
and
and
and
and
and
We can think of this problem as an improper fraction and solve for the mixed number by placing the amount of candy over the number of Melissa's friends. 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
Example Question #1056 : Common Core Math: Grade 5
Melissa has pieces of candy that she wants to divide amongst her friends. How many pieces of candy will each friend get, assuming that each friend gets the same amount? Select the answer that contains the pair of numbers that the answer falls between.
and
and
and
and
and
and
We can think of this problem as an improper fraction and solve for the mixed number by placing the amount of candy over the number of Melissa's friends. 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