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Common Core: 6th Grade Math : Grade 6

Study concepts, example questions & explanations for Common Core: 6th Grade Math

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All Common Core: 6th Grade Math Resources

6 Diagnostic Tests 186 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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Example Question #2 : Solving Word Problems With One Unit Conversions

A carpenter is making a model house and he buys $8\textup{ feet}$ of crown moulding to use as accent pieces. He needs $24\textup{ inches}$ of the moulding for the house. How many feet of the material does he need to finish the model?

Possible Answers:

$2 \textup{ feet}$

$2 \tfrac{2}{3}\textup{ feet}$

$1 \tfrac{1}{2}\textup{ feet}$

$2 \tfrac{3}{4}\textup{ feet}$

$2 \tfrac{1}{2}\textup{ feet}$

Correct answer:

$2 \textup{ feet}$

Explanation:

We can solve this problem using ratios. There are $12\ inches$ in $1\ foot$ . We can write this relationship as the following ratio:

$12\ inches:1\ foot\rightarrow \frac{12\ inches}{1\ foot}$

We know that the carpenter needs $24\ inches$ of material to finish the house. We can write this as a ratio using the variable to substitute the amount of feet.

$24\ inches:x\ feet\rightarrow \frac{24\ inches}{x\ feet}$

Now, we can solve for by creating a proportion using our two ratios.

$\frac{12\ inches}{1\ foot}=\frac{24\ inches}{x\ feet}$

Cross multiply and solve for .

$12\ inches \times (x\ feet)=24\ inches \times (1\ foot)$

Simplify.

12x=24

Divide both sides by .

$\frac{12x}{12}=\frac{24}{12}$

Solve.

$x=2 \ feet$

The carpenter needs $2 \ feet$ of material.

Report an Error

Example Question #82 : Grade 6

A carpenter is making a model house and he buys $8\textup{ feet}$ of crown moulding to use as accent pieces. He needs $38\textup{ inches}$ of the moulding for the house. How many feet of the material does he need to finish the model?

Possible Answers:

$3 \tfrac{1}{2} \textup{ feet}$

$3 \tfrac{5}{6}\textup{ feet}$

$3 \tfrac{1}{6}\textup{ feet}$

$2 \tfrac{5}{6}\textup{ feet}$

$3 \tfrac{3}{4}\textup{ feet}$

Correct answer:

$3 \tfrac{1}{6}\textup{ feet}$

Explanation:

We can solve this problem using ratios. There are $12\ inches$ in $1\ foot$ . We can write this relationship as the following ratio:

$12\ inches:1\ foot\rightarrow \frac{12\ inches}{1\ foot}$

We know that the carpenter needs $38\ inches$ of material to finish the house. We can write this as a ratio using the variable to substitute the amount of feet.

$38\ inches:x\ feet\rightarrow \frac{38\ inches}{x\ feet}$

Now, we can solve for by creating a proportion using our two ratios.

$\frac{12\ inches}{1\ foot}=\frac{38\ inches}{x\ feet}$

Cross multiply and solve for .

$12\ inches \times (x\ feet)=38\ inches \times (1\ foot)$

Simplify.

12x=38

Divide both sides by .

$\frac{12x}{12}=\frac{38}{12}$

Solve.

$x=3 \tfrac{2}{12} \ feet$

Reduce.

$x=3 \tfrac{1}{6} \ feet$

The carpenter needs $3 \tfrac{1}{6} \ feet$ of material.

Report an Error

Example Question #83 : Grade 6

A carpenter is making a model house and he buys $8\textup{ feet}$ of crown moulding to use as accent pieces. He needs $64\textup{ inches}$ of the moulding for the house. How many feet of the material does he need to finish the model?

Possible Answers:

$6 \tfrac{1}{3} \textup{ feet}$

$5 \tfrac{1}{3} \textup{ feet}$

$5 \tfrac{2}{3} \textup{ feet}$

$4 \tfrac{1}{3} \textup{ feet}$

$4 \tfrac{2}{3} \textup{ feet}$

Correct answer:

$5 \tfrac{1}{3} \textup{ feet}$

Explanation:

We can solve this problem using ratios. There are $12\ inches$ in $1\ foot$ . We can write this relationship as the following ratio:

$12\ inches:1\ foot\rightarrow \frac{12\ inches}{1\ foot}$

We know that the carpenter needs $64\ inches$ of material to finish the house. We can write this as a ratio using the variable to substitute the amount of feet.

$64\ inches:x\ feet\rightarrow \frac{64\ inches}{x\ feet}$

Now, we can solve for by creating a proportion using our two ratios.

$\frac{12\ inches}{1\ foot}=\frac{64\ inches}{x\ feet}$

Cross multiply and solve for .

$12\ inches \times (x\ feet)=64\ inches \times (1\ foot)$

Simplify.

12x=64

Divide both sides by .

$\frac{12x}{12}=\frac{64}{12}$

Solve.

$x=5 \tfrac{4}{12} \ feet$

Reduce.

$x=5 \tfrac{1}{3} \ feet$

The carpenter needs $5 \tfrac{1}{3} \ feet$ of material.

Report an Error

Example Question #481 : New Sat

A carpenter is making a model house and he buys $8\textup{ feet}$ of crown moulding to use as accent pieces. He needs $72\textup{ inches}$ of the moulding for the house. How many feet of the material does he need to finish the model?

Possible Answers:

$4 \textup{ feet}$

$5\textup{ feet}$

$6 \tfrac{1}{3} \textup{ feet}$

$6 \textup{ feet}$

$7 \textup{ feet}$

Correct answer:

$6 \textup{ feet}$

Explanation:

We can solve this problem using ratios. There are $12\ inches$ in $1\ foot$ . We can write this relationship as the following ratio:

$12\ inches:1\ foot\rightarrow \frac{12\ inches}{1\ foot}$

We know that the carpenter needs $72\ inches$ of material to finish the house. We can write this as a ratio using the variable to substitute the amount of feet.

$72\ inches:x\ feet\rightarrow \frac{72\ inches}{x\ feet}$

Now, we can solve for by creating a proportion using our two ratios.

$\frac{12\ inches}{1\ foot}=\frac{72\ inches}{x\ feet}$

Cross multiply and solve for .

$12\ inches \times (x\ feet)=72\ inches \times (1\ foot)$

Simplify.

12x=72

Divide both sides by .

$\frac{12x}{12}=\frac{72}{12}$

Solve.

$x=6 \ feet$

The carpenter needs $6 \ feet$ of material.

Report an Error

Example Question #61 : How To Divide Fractions

Hydraulic fracturing is a process used by gas companies to rupture and collects pockets of gas trapped within pockets of shale rock. A particular shale fracking site is $\frac{1}{5}\ miles$ in length and occupies an area of $\frac{7}{8}\ miles^2$ . How wide is this particular site?

Possible Answers:

$4\tfrac{1}{8}\ miles$

$4\tfrac{5}{8}\ miles$

$3\tfrac{3}{8}\ miles$

$6\tfrac{3}{8}\ miles$

$4\tfrac{3}{8}\ miles$

Correct answer:

$4\tfrac{3}{8}\ miles$

Explanation:

In order to solve this question, we need to first recall how to find the area of a rectangle.

$\text{Area}=\text{Length} \times \text{Width}$

Substitute in the given values in the equation and solve for $\text{Width}$ .

$\frac{7}{8}\ miles^2=\frac{1}{5}\ miles\times Width$

Divide both sides by $\frac{1}{5}\ miles$

$\frac{\frac{7}{8}\ miles^2}{\frac{1}{5}\ miles}=\frac{\frac{1}{5}\ miles \times Width}{\frac{1}{5}\ miles}$

Dividing by a fraction is the same as multiplying by its inverse or reciprocal.

Find the reciprocal of $\frac{1}{5}\ miles$

$\frac{1}{5}\ miles\rightarrow \frac{5}{1}\ miles$

Simplify and rewrite.

$Width=\frac{5}{1} \times \frac{7}{8}$

Multiply and solve.

$Width=\frac{35}{8}$

Reduce.

$Width=4\tfrac{3}{8}\ miles$

The width of the fracking site is $4\tfrac{3}{8}\ miles$

Report an Error

Example Question #1 : The Number System

Hydraulic fracturing is a process used by gas companies to rupture and collects pockets of gas trapped within pockets of shale rock. A particular shale fracking site is $\frac{1}{4}\ miles$ in length and occupies an area of $\frac{7}{9}\ miles^2$ . How wide is this particular site?

Possible Answers:

$9\tfrac{1}{3}\ miles$

$5\tfrac{1}{3}\ miles$

$3\tfrac{5}{9}\ miles$

$9\tfrac{5}{9}\ miles$

$3\tfrac{1}{9}\ miles$

Correct answer:

$3\tfrac{1}{9}\ miles$

Explanation:

In order to solve this question, we need to first recall how to find the area of a rectangle.

$\text{Area}=\text{Length} \times \text{Width}$

Substitute in the given values in the equation and solve for $\text{Width}$ .

$\frac{7}{9}\ miles^2=\frac{1}{4}\ miles\times Width$

Divide both sides by $\frac{1}{4}\ miles$

$\frac{\frac{7}{9}\ miles^2}{\frac{1}{4}\ miles}=\frac{\frac{1}{4}\ miles \times Width}{\frac{1}{4}\ miles}$

Dividing by a fraction is the same as multiplying by its inverse or reciprocal.

Find the reciprocal of $\frac{1}{4}\ miles$

$\frac{1}{4}\ miles\rightarrow \frac{4}{1}\ miles$

Simplify and rewrite.

$Width=\frac{4}{1} \times \frac{7}{9}$

Multiply and solve.

$Width=\frac{28}{9}$

Reduce.

$Width=3\tfrac{1}{9}\ miles$

The width of the fracking site is $3\tfrac{1}{9}\ miles$

Report an Error

Example Question #87 : Grade 6

Hydraulic fracturing is a process used by gas companies to rupture and collects pockets of gas trapped within pockets of shale rock. A particular shale fracking site is $\frac{1}{4}\ miles$ in length and occupies an area of $\frac{7}{8}\ miles^2$ . How wide is this particular site?

Possible Answers:

$2\tfrac{1}{3}\ miles$

$3\tfrac{2}{3}\ miles$

$2\tfrac{2}{3}\ miles$

$3\tfrac{1}{2}\ miles$

$3\tfrac{1}{3}\ miles$

Correct answer:

$3\tfrac{1}{2}\ miles$

Explanation:

In order to solve this question, we need to first recall how to find the area of a rectangle.

$\text{Area}=\text{Length} \times \text{Width}$

Substitute in the given values in the equation and solve for $\text{Width}$ .

$\frac{7}{8}\ miles^2=\frac{1}{4}\ miles\times Width$

Divide both sides by $\frac{1}{4}\ miles$

$\frac{\frac{7}{8}\ miles^2}{\frac{1}{4}\ miles}=\frac{\frac{1}{4}\ miles \times Width}{\frac{1}{4}\ miles}$

Dividing by a fraction is the same as multiplying by its inverse or reciprocal.

Find the reciprocal of $\frac{1}{4}\ miles$

$\frac{1}{4}\ miles\rightarrow \frac{4}{1}\ miles$

Simplify and rewrite.

$Width=\frac{4}{1} \times \frac{7}{8}$

Multiply and solve.

$Width=\frac{28}{8}$

Reduce.

$Width=3\tfrac{1}{2}\ miles$

The width of the fracking site is $3\tfrac{1}{2}\ miles$

Report an Error

Example Question #88 : Grade 6

Hydraulic fracturing is a process used by gas companies to rupture and collects pockets of gas trapped within pockets of shale rock. A particular shale fracking site is $\frac{1}{5}\ miles$ in length and occupies an area of $\frac{35}{41}\ miles^2$ . How wide is this particular site?

Possible Answers:

$11\tfrac{4}{41}\ miles$

$4\tfrac{21}{41}\ miles$

$4\tfrac{11}{41}\ miles$

$2\tfrac{11}{41}\ miles$

$5\tfrac{11}{41}\ miles$

Correct answer:

$4\tfrac{11}{41}\ miles$

Explanation:

In order to solve this question, we need to first recall how to find the area of a rectangle.

$\text{Area}=\text{Length} \times \text{Width}$

Substitute in the given values in the equation and solve for $\text{Width}$ .

$\frac{35}{41}\ miles^2=\frac{1}{5}\ miles\times Width$

Divide both sides by $\frac{1}{5}\ miles$

$\frac{\frac{35}{41}\ miles^2}{\frac{1}{5}\ miles}=\frac{\frac{1}{5}\ miles \times Width}{\frac{1}{5}\ miles}$

Dividing by a fraction is the same as multiplying by its inverse or reciprocal.

Find the reciprocal of $\frac{1}{5}\ miles$

$\frac{1}{5}\ miles\rightarrow \frac{5}{1}\ miles$

Simplify and rewrite.

$Width=\frac{5}{1} \times \frac{35}{41}$

Multiply and solve.

$Width=\frac{175}{41}$

Reduce.

$Width=4\tfrac{11}{41}\ miles$

The width of the fracking site is $4\tfrac{11}{41}\ miles$

Report an Error

Example Question #89 : Grade 6

Hydraulic fracturing is a process used by gas companies to rupture and collect pockets of gas trapped within pockets of shale rock. A particular shale fracking site is $\frac{3}{4}\textup{ miles}$ in length and occupies an area of $\frac{1}{2}\textup{ miles}^2$ . How wide is this particular site?

Possible Answers:

$\frac{2}{3}\textup{ miles}$

$\frac{1}{3}\textup{ miles}$

$\frac{6}{7}\textup{ miles}$

$\frac{3}{5}\textup{ miles}$

$\frac{2}{5}\textup{ miles}$

Correct answer:

$\frac{2}{3}\textup{ miles}$

Explanation:

In order to solve this question, we need to first recall how to find the area of a rectangle.

$\text{Area}=\text{Length} \times \text{Width}$

Substitute in the given values in the equation and solve for $\text{Width}$ .

$\frac{1}{2}\ miles^2=\frac{3}{4}\ miles\times Width$

Divide both sides by $\frac{3}{4}\ miles$

$\frac{\frac{1}{2}\ miles^2}{\frac{3}{4}\ miles}=\frac{\frac{3}{4}\ miles \times Width}{\frac{3}{4}\ miles}$

Dividing by a fraction is the same as multiplying by its inverse or reciprocal.

Find the reciprocal of $\frac{3}{4}\ miles$

$\frac{3}{4}\ miles\rightarrow \frac{4}{3}\ miles$

Simplify and rewrite.

$Width=\frac{4}{3} \times \frac{1}{2}$

Multiply and solve.

$Width=\frac{4}{6}$

Reduce.

$Width=\frac{2}{3}\ miles$

The width of the fracking site is $\frac{2}{3}\ miles$

Report an Error

Example Question #90 : Grade 6

Hydraulic fracturing is a process used by gas companies to rupture and collects pockets of gas trapped within pockets of shale rock. A particular shale fracking site is $\frac{1}{7}\textup{ miles}$ in length and occupies an area of $\frac{3}{5}\textup{ miles}^2$ . How wide is this particular site?

Possible Answers:

$4\tfrac{2}{5}\textup{ miles}$

$4\tfrac{3}{5}\textup{ miles}$

$4\tfrac{1}{5}\textup{ miles}$

$4\tfrac{4}{5}\textup{ miles}$

$4\textup{ miles}$

Correct answer:

$4\tfrac{1}{5}\textup{ miles}$

Explanation:

In order to solve this question, we need to first recall how to find the area of a rectangle.

$\text{Area}=\text{Length} \times \text{Width}$

Substitute in the given values in the equation and solve for $\text{Width}$ .

$\frac{3}{5}\ miles^2=\frac{1}{7}\ miles\times Width$

Divide both sides by $\frac{1}{7}\ miles$

$\frac{\frac{3}{5}\ miles^2}{\frac{1}{7}\ miles}=\frac{\frac{1}{7}\ miles \times Width}{\frac{1}{7}\ miles}$

Dividing by a fraction is the same as multiplying by its inverse or reciprocal.

Find the reciprocal of $\frac{1}{7}\ miles$

$\frac{1}{7}\ miles\rightarrow \frac{7}{1}\ miles$

Simplify and rewrite.

$Width=\frac{7}{1} \times \frac{3}{5}$

Multiply and solve.

$Width=\frac{21}{5}$

Reduce.

$Width=4\tfrac{1}{5}\ miles$

The width of the fracking site is $4\tfrac{1}{5}\ miles$

Report an Error

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Common Core: 6th Grade Math : Grade 6

Study concepts, example questions & explanations for Common Core: 6th Grade Math

All Common Core: 6th Grade Math Resources

Example Questions

Example Question #2 : Solving Word Problems With One Unit Conversions

Example Question #82 : Grade 6

Example Question #83 : Grade 6

Example Question #481 : New Sat

Example Question #61 : How To Divide Fractions

Example Question #1 : The Number System

Example Question #87 : Grade 6

Example Question #88 : Grade 6

Example Question #89 : Grade 6

Example Question #90 : Grade 6

All Common Core: 6th Grade Math Resources

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