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Common Core: 5th Grade Math : Number & Operations with Fractions

Study concepts, example questions & explanations for Common Core: 5th Grade Math

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All Common Core: 5th Grade Math Resources

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Example Questions

Example Question #981 : Common Core Math: Grade 5

Identify the division problem that equals the following:

$\frac{13}{15}$ $\displaystyle \frac{13}{15}$

Possible Answers:

$14\div15$ $\displaystyle 14\div15$

$13\div15$ $\displaystyle 13\div15$

$15\div14$ $\displaystyle 15\div14$

$15\div13$ $\displaystyle 15\div13$

Correct answer:

$13\div15$ $\displaystyle 13\div15$

Explanation:

The fraction line (-) $\displaystyle (-)$ means divided by; therefore, we can read $\frac{13}{15}$ $\displaystyle \frac{13}{15}$ as $\displaystyle 13$ divided by $\displaystyle 15$, or $13\div15$ $\displaystyle 13\div15$

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Example Question #102 : Solve Word Problems Involving Division Of Whole Numbers Leading To Answers In The Form Of Fractions Or Mixed Numbers: Ccss.Math.Content.5.Nf.B.3

Identify the division problem that equals the following:

$\frac{3}{15}$ $\displaystyle \frac{3}{15}$

Possible Answers:

$3\div15$ $\displaystyle 3\div15$

$2\div15$ $\displaystyle 2\div15$

$15\div3$ $\displaystyle 15\div3$

$15\div2$ $\displaystyle 15\div2$

Correct answer:

$3\div15$ $\displaystyle 3\div15$

Explanation:

The fraction line (-) $\displaystyle (-)$ means divided by; therefore, we can read $\frac{3}{15}$ $\displaystyle \frac{3}{15}$ as $\displaystyle 3$ divided by $\displaystyle 15$, or $3\div15$ $\displaystyle 3\div15$

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Example Question #132 : Number & Operations With Fractions

Identify the division problem that equals the following:

$\frac{12}{15}$ $\displaystyle \frac{12}{15}$

Possible Answers:

$15\div11$ $\displaystyle 15\div11$

$11\div15$ $\displaystyle 11\div15$

$12\div15$ $\displaystyle 12\div15$

$15\div12$ $\displaystyle 15\div12$

Correct answer:

$12\div15$ $\displaystyle 12\div15$

Explanation:

The fraction line (-) $\displaystyle (-)$ means divided by; therefore, we can read $\frac{12}{15}$ $\displaystyle \frac{12}{15}$ as $\displaystyle 12$ divided by $\displaystyle 15$, or $12\div15$ $\displaystyle 12\div15$

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Example Question #101 : Solve Word Problems Involving Division Of Whole Numbers Leading To Answers In The Form Of Fractions Or Mixed Numbers: Ccss.Math.Content.5.Nf.B.3

Identify the division problem that equals the following:

$\frac{4}{15}$ $\displaystyle \frac{4}{15}$

Possible Answers:

$15\div5$ $\displaystyle 15\div5$

$4\div15$ $\displaystyle 4\div15$

$5\div15$ $\displaystyle 5\div15$

$15\div4$ $\displaystyle 15\div4$

Correct answer:

$4\div15$ $\displaystyle 4\div15$

Explanation:

The fraction line (-) $\displaystyle (-)$ means divided by; therefore, we can read $\frac{4}{15}$ $\displaystyle \frac{4}{15}$ as $\displaystyle 4$ divided by $\displaystyle 15$, or $4\div15$ $\displaystyle 4\div15$

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Example Question #105 : Solve Word Problems Involving Division Of Whole Numbers Leading To Answers In The Form Of Fractions Or Mixed Numbers: Ccss.Math.Content.5.Nf.B.3

A baker has $\displaystyle 7$ cakes to make this weekend. If he has $25\textup{ pounds}$ $\displaystyle 25\textup{ pounds}$ of sugar, how much sugar does can he put in each cake, assuming each cake's recipe calls for an equal amount of sugar? Select the answer that contains the pair of numbers that the answer falls between.

Possible Answers:

$\displaystyle 5$ and $\displaystyle 6$

$\displaystyle 3$ and $\displaystyle 4$

$\displaystyle 6$ and $\displaystyle 7$

$\displaystyle 2$ and $\displaystyle 3$

$\displaystyle 4$ and $\displaystyle 5$

Correct answer:

$\displaystyle 3$ and $\displaystyle 4$

Explanation:

We can think of this problem as an improper fraction and solve for the mixed number by placing the amount of sugar over the number of cakes to be made. We get the following:

$\small \frac{25\textup{ pounds of sugar}}{7\textup{ cakes}}$ $\displaystyle \small \frac{25\textup{ pounds of sugar}}{7\textup{ cakes}}$

$\displaystyle 7$ can go into $\displaystyle 25$ only $\displaystyle 3$ 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:

$7\times 3=21$ $\displaystyle 7\times 3=21$

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

25-21=4 $\displaystyle 25-21=4$

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:

$\frac{4}{7}$ $\displaystyle \frac{4}{7}$

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

$\small \frac{25}{7}=3\frac{4}{7}$ $\displaystyle \small \frac{25}{7}=3\frac{4}{7}$

Last, we know that $3\frac{4}{7}$ $\displaystyle 3\frac{4}{7}$ is between the numbers $\displaystyle 3$ and $\displaystyle 4$; therefore, the correct answer is:

$\displaystyle 3$ and $\displaystyle 4$

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Example Question #110 : Solve Word Problems Involving Division Of Whole Numbers Leading To Answers In The Form Of Fractions Or Mixed Numbers: Ccss.Math.Content.5.Nf.B.3

A baker has $\displaystyle 5$ cakes to make this weekend. If he has $24\textup{ pounds}$ $\displaystyle 24\textup{ pounds}$ of sugar, how much sugar does can he put in each cake, assuming each cake's recipe calls for an equal amount of sugar? Select the answer that contains the pair of numbers that the answer falls between.

Possible Answers:

$\displaystyle 3$ and $\displaystyle 4$

$\displaystyle 4$ and $\displaystyle 5$

$\displaystyle 1$ and $\displaystyle 2$

$\displaystyle 5$ and $\displaystyle 6$

$\displaystyle 2$ and $\displaystyle 3$

Correct answer:

$\displaystyle 4$ and $\displaystyle 5$

Explanation:

We can think of this problem as an improper fraction and solve for the mixed number by placing the amount of sugar over the number of cakes to be made. We get the following:

$\small \frac{24\textup{ pounds of sugar}}{5\textup{ cakes}}$ $\displaystyle \small \frac{24\textup{ pounds of sugar}}{5\textup{ cakes}}$

$\displaystyle 5$ can go into $\displaystyle 24$ only $\displaystyle 4$ 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:

$5\times 4=20$ $\displaystyle 5\times 4=20$

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

24-20=4 $\displaystyle 24-20=4$

Then, we put the difference over the denominator:

$\frac{4}{5}$ $\displaystyle \frac{4}{5}$

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

$\small \frac{24}{5}=4\frac{4}{5}$ $\displaystyle \small \frac{24}{5}=4\frac{4}{5}$

Last, we know that $4\frac{4}{5}$ $\displaystyle 4\frac{4}{5}$ is between the numbers $\displaystyle 4$ and $\displaystyle 5$; therefore, the correct answer is:

$\displaystyle 4$ and $\displaystyle 5$

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Example Question #131 : Number & Operations With Fractions

A baker has $\displaystyle 4$ cakes to make this weekend. If he has $27\textup{ pounds}$ $\displaystyle 27\textup{ pounds}$ of sugar, how much sugar does can he put in each cake, assuming each cake's recipe calls for an equal amount of sugar? Select the answer that contains the pair of numbers that the answer falls between.

Possible Answers:

$\displaystyle 2$ and $\displaystyle 3$

$\displaystyle 4$ and $\displaystyle 5$

$\displaystyle 3$ and $\displaystyle 4$

$\displaystyle 6$ and $\displaystyle 7$

$\displaystyle 5$ and $\displaystyle 6$

Correct answer:

$\displaystyle 6$ and $\displaystyle 7$

Explanation:

We can think of this problem as an improper fraction and solve for the mixed number by placing the amount of sugar over the number of cakes to be made. We get the following:

$\small \frac{27\textup{ pounds of sugar}}{4\textup{ cakes}}$ $\displaystyle \small \frac{27\textup{ pounds of sugar}}{4\textup{ cakes}}$

$\displaystyle 4$ can go into $\displaystyle 27$ only $\displaystyle 6$ 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:

$4\times 6=24$ $\displaystyle 4\times 6=24$

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

27-24=3 $\displaystyle 27-24=3$

Then, we put the difference over the denominator:

$\frac{3}{4}$ $\displaystyle \frac{3}{4}$

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

$\small \frac{27}{4}=6\frac{3}{4}$ $\displaystyle \small \frac{27}{4}=6\frac{3}{4}$

Last, we know that $6\frac{3}{4}$ $\displaystyle 6\frac{3}{4}$ is between the numbers $\displaystyle 6$ and $\displaystyle 7$; therefore, the correct answer is:

$\displaystyle 6$ and $\displaystyle 7$

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Example Question #988 : Common Core Math: Grade 5

A baker has $\displaystyle 7$ cakes to make this weekend. If he has $30\textup{ pounds}$ $\displaystyle 30\textup{ pounds}$ of sugar, how much sugar does can he put in each cake, assuming each cake's recipe calls for an equal amount of sugar? Select the answer that contains the pair of numbers that the answer falls between.

Possible Answers:

$\displaystyle 7$ and $\displaystyle 8$

$\displaystyle 6$ and $\displaystyle 7$

$\displaystyle 5$ and $\displaystyle 6$

$\displaystyle 4$ and $\displaystyle 5$

$\displaystyle 8$ and $\displaystyle 9$

Correct answer:

$\displaystyle 4$ and $\displaystyle 5$

Explanation:

We can think of this problem as an improper fraction and solve for the mixed number by placing the amount of sugar over the number of cakes to be made. We get the following:

$\small \frac{30\textup{ pounds of sugar}}{7\textup{ cakes}}$ $\displaystyle \small \frac{30\textup{ pounds of sugar}}{7\textup{ cakes}}$

$\displaystyle 7$ can go into $\displaystyle 30$ only $\displaystyle 4$ 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:

$7\times 4=28$ $\displaystyle 7\times 4=28$

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

30-28=2 $\displaystyle 30-28=2$

Then, we put the difference over the denominator:

$\frac{2}{7}$ $\displaystyle \frac{2}{7}$

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

$\small \frac{30}{7}=4\frac{2}{7}$ $\displaystyle \small \frac{30}{7}=4\frac{2}{7}$

Last, we know that $4\frac{2}{7}$ $\displaystyle 4\frac{2}{7}$ is between the numbers $\displaystyle 4$ and $\displaystyle 5$; therefore, the correct answer is:

$\displaystyle 4$ and $\displaystyle 5$

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Example Question #138 : Number & Operations With Fractions

A baker has $\displaystyle 9$ cakes to make this weekend. If he has $26\textup{ pounds}$ $\displaystyle 26\textup{ pounds}$ of sugar, how much sugar does can he put in each cake, assuming each cake's recipe calls for an equal amount of sugar? Select the answer that contains the pair of numbers that the answer falls between.

Possible Answers:

$\displaystyle 3$ and $\displaystyle 4$

$\displaystyle 4$ and $\displaystyle 5$

$\displaystyle 2$ and $\displaystyle 3$

$\displaystyle 1$ and $\displaystyle 2$

$\displaystyle 5$ and $\displaystyle 6$

Correct answer:

$\displaystyle 2$ and $\displaystyle 3$

Explanation:

We can think of this problem as an improper fraction and solve for the mixed number by placing the amount of sugar over the number of cakes to be made. We get the following:

$\small \frac{26\textup{ pounds of sugar}}{9\textup{ cakes}}$ $\displaystyle \small \frac{26\textup{ pounds of sugar}}{9\textup{ cakes}}$

$\displaystyle 9$ can go into $\displaystyle 26$ only $\displaystyle 2$ 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:

$9\times 2=18$ $\displaystyle 9\times 2=18$

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

26-18=8 $\displaystyle 26-18=8$

Then, we put the difference over the denominator:

$\frac{8}{9}$ $\displaystyle \frac{8}{9}$

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

$\small \frac{26}{9}=2\frac{8}{9}$ $\displaystyle \small \frac{26}{9}=2\frac{8}{9}$

Last, we know that $2\frac{8}{9}$ $\displaystyle 2\frac{8}{9}$ is between the numbers $\displaystyle 2$ and $\displaystyle 3$; therefore, the correct answer is:

$\displaystyle 2$ and $\displaystyle 3$

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Example Question #114 : Solve Word Problems Involving Division Of Whole Numbers Leading To Answers In The Form Of Fractions Or Mixed Numbers: Ccss.Math.Content.5.Nf.B.3

A baker has $\displaystyle 5$ cakes to make this weekend. If he has $31\textup{ pounds}$ $\displaystyle 31\textup{ pounds}$ of sugar, how much sugar does can he put in each cake, assuming each cake's recipe calls for an equal amount of sugar? Select the answer that contains the pair of numbers that the answer falls between.

Possible Answers:

$\displaystyle 2$ and $\displaystyle 3$

$\displaystyle 5$ and $\displaystyle 6$

$\displaystyle 3$ and $\displaystyle 4$

$\displaystyle 6$ and $\displaystyle 7$

$\displaystyle 4$ and $\displaystyle 5$

Correct answer:

$\displaystyle 6$ and $\displaystyle 7$

Explanation:

We can think of this problem as an improper fraction and solve for the mixed number by placing the amount of sugar over the number of cakes to be made. We get the following:

$\small \frac{31\textup{ pounds of sugar}}{5\textup{ cakes}}$ $\displaystyle \small \frac{31\textup{ pounds of sugar}}{5\textup{ cakes}}$

$\displaystyle 5$ can go into $\displaystyle 31$ only $\displaystyle 6$ 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:

$5\times 6=30$ $\displaystyle 5\times 6=30$

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

31-30=1 $\displaystyle 31-30=1$

Then, we put the difference over the denominator:

$\frac{1}{5}$ $\displaystyle \frac{1}{5}$

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

$\small \frac{31}{5}=6\frac{1}{5}$ $\displaystyle \small \frac{31}{5}=6\frac{1}{5}$

Last, we know that $6\frac{1}{5}$ $\displaystyle 6\frac{1}{5}$ is between the numbers $\displaystyle 6$ and $\displaystyle 7$; therefore, the correct answer is:

$\displaystyle 6$ and $\displaystyle 7$

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Common Core: 5th Grade Math : Number & Operations with Fractions

Study concepts, example questions & explanations for Common Core: 5th Grade Math

All Common Core: 5th Grade Math Resources

Example Questions

Example Question #981 : Common Core Math: Grade 5

Example Question #102 : Solve Word Problems Involving Division Of Whole Numbers Leading To Answers In The Form Of Fractions Or Mixed Numbers: Ccss.Math.Content.5.Nf.B.3

Example Question #132 : Number & Operations With Fractions

Example Question #101 : Solve Word Problems Involving Division Of Whole Numbers Leading To Answers In The Form Of Fractions Or Mixed Numbers: Ccss.Math.Content.5.Nf.B.3

Example Question #105 : Solve Word Problems Involving Division Of Whole Numbers Leading To Answers In The Form Of Fractions Or Mixed Numbers: Ccss.Math.Content.5.Nf.B.3

Example Question #110 : Solve Word Problems Involving Division Of Whole Numbers Leading To Answers In The Form Of Fractions Or Mixed Numbers: Ccss.Math.Content.5.Nf.B.3

Example Question #131 : Number & Operations With Fractions

Example Question #988 : Common Core Math: Grade 5

Example Question #138 : Number & Operations With Fractions

Example Question #114 : Solve Word Problems Involving Division Of Whole Numbers Leading To Answers In The Form Of Fractions Or Mixed Numbers: Ccss.Math.Content.5.Nf.B.3

All Common Core: 5th Grade Math Resources

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