Practice · Sum fractional parts of a whole · Sensim Math

Variant 1 easy

At a fruit stand, $\frac{2}{3}$ of all the tangerines were sold and $\frac{1}{4}$ of all of them were eaten. The claim is that the tangerines left over are less than $\frac{1}{6}$ of the original amount of tangerines.

Decide whether this claim is correct.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Of all the tangerines, 2 of every 3 were sold and 1 of every 4 were eaten. Someone claims the leftovers are under 1 over 6 of the original. We must judge that claim.

Givens

$\frac{2}{3}$ of the whole was sold.
$\frac{1}{4}$ of the whole was eaten.
The claim: the leftover is less than $\frac{1}{6}$ .

Unknowns

Whether the claim holds.

Constraints

Both fractions are of the ORIGINAL amount, not of what was left after the other.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram

Because both shares are taken from the same whole, they simply add. Find how much left the stand altogether, take that from 1, and compare what remains with the claimed bound.

3 · Execute3 carry out the plan

1Add the parts that are gone

#7 Identify Subproblems 5.NF.A.1

Sold and eaten are both fractions of the original, so put them over 12 and add.

\frac{2}{3} + \frac{1}{4} = \frac{11}{12}

\frac{11}{12}

of the tangerines are gone.

2Subtract to find the leftover

#7 Identify Subproblems 5.NF.A.2

The whole is 1, so what is left is 1 minus what went.

1 - \frac{11}{12} = \frac{1}{12}

\frac{1}{12}

of the original is left.

3Compare the leftover with $\frac{1}{6}$

#1 Draw a Diagram 5.NF.A.2

Put both over a common denominator to see which is bigger.

\frac{1}{12} \ < \ \frac{1}{6}

The leftover is less than the claimed bound, so the claim is correct.

Answer: The claim is correct: the leftover is

\frac{1}{12}

, which is less than

\frac{1}{6}

.

4 · Reviewdoes it hold up?

Sanity-check with a number: out of 12 tangerines, 2 are sold and 1 eaten, leaving 9 -- which is $\frac{1}{12}$ of the 12.

Another way: Draw one bar for the whole, shade $\frac{2}{3}$ and then $\frac{1}{4}$ of the same bar, and measure the unshaded strip against $\frac{1}{6}$ .

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two shares over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions — Subtracting from the whole and judging the claim.

💡Takeaway. When two shares come from the same whole they add. It is only when one is taken from what the other left that they multiply.

Variant 2 easy

At a fruit stand, $\frac{5}{8}$ of all the tangerines were sold and $\frac{1}{4}$ of all of them were eaten. The claim is that the tangerines left over are less than $\frac{1}{6}$ of the original amount of tangerines.

Decide whether this claim is correct.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Of all the tangerines, 5 of every 8 were sold and 1 of every 4 were eaten. Someone claims the leftovers are under 1 over 6 of the original. We must judge that claim.

Givens

$\frac{5}{8}$ of the whole was sold.
$\frac{1}{4}$ of the whole was eaten.
The claim: the leftover is less than $\frac{1}{6}$ .

Unknowns

Whether the claim holds.

Constraints

Both fractions are of the ORIGINAL amount, not of what was left after the other.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram

Because both shares are taken from the same whole, they simply add. Find how much left the stand altogether, take that from 1, and compare what remains with the claimed bound.

3 · Execute3 carry out the plan

1Add the parts that are gone

#7 Identify Subproblems 5.NF.A.1

Sold and eaten are both fractions of the original, so put them over 8 and add.

\frac{5}{8} + \frac{1}{4} = \frac{7}{8}

\frac{7}{8}

of the tangerines are gone.

2Subtract to find the leftover

#7 Identify Subproblems 5.NF.A.2

The whole is 1, so what is left is 1 minus what went.

1 - \frac{7}{8} = \frac{1}{8}

\frac{1}{8}

of the original is left.

3Compare the leftover with $\frac{1}{6}$

#1 Draw a Diagram 5.NF.A.2

Put both over a common denominator to see which is bigger.

\frac{1}{8} \ < \ \frac{1}{6}

The leftover is less than the claimed bound, so the claim is correct.

Answer: The claim is correct: the leftover is

\frac{1}{8}

, which is less than

\frac{1}{6}

.

4 · Reviewdoes it hold up?

Sanity-check with a number: out of 8 tangerines, 5 are sold and 1 eaten, leaving 2 -- which is $\frac{1}{8}$ of the 8.

Another way: Draw one bar for the whole, shade $\frac{5}{8}$ and then $\frac{1}{4}$ of the same bar, and measure the unshaded strip against $\frac{1}{6}$ .

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two shares over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions — Subtracting from the whole and judging the claim.

💡Takeaway. When two shares come from the same whole they add. It is only when one is taken from what the other left that they multiply.

Variant 3 easy

At a fruit stand, $\frac{1}{2}$ of all the tangerines were sold and $\frac{2}{5}$ of all of them were eaten. The claim is that the tangerines left over are less than $\frac{1}{8}$ of the original amount of tangerines.

Decide whether this claim is correct.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Of all the tangerines, 1 of every 2 were sold and 2 of every 5 were eaten. Someone claims the leftovers are under 1 over 8 of the original. We must judge that claim.

Givens

$\frac{1}{2}$ of the whole was sold.
$\frac{2}{5}$ of the whole was eaten.
The claim: the leftover is less than $\frac{1}{8}$ .

Unknowns

Whether the claim holds.

Constraints

Both fractions are of the ORIGINAL amount, not of what was left after the other.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram

Because both shares are taken from the same whole, they simply add. Find how much left the stand altogether, take that from 1, and compare what remains with the claimed bound.

3 · Execute3 carry out the plan

1Add the parts that are gone

#7 Identify Subproblems 5.NF.A.1

Sold and eaten are both fractions of the original, so put them over 10 and add.

\frac{1}{2} + \frac{2}{5} = \frac{9}{10}

\frac{9}{10}

of the tangerines are gone.

2Subtract to find the leftover

#7 Identify Subproblems 5.NF.A.2

The whole is 1, so what is left is 1 minus what went.

1 - \frac{9}{10} = \frac{1}{10}

\frac{1}{10}

of the original is left.

3Compare the leftover with $\frac{1}{8}$

#1 Draw a Diagram 5.NF.A.2

Put both over a common denominator to see which is bigger.

\frac{1}{10} \ < \ \frac{1}{8}

The leftover is less than the claimed bound, so the claim is correct.

Answer: The claim is correct: the leftover is

\frac{1}{10}

, which is less than

\frac{1}{8}

.

4 · Reviewdoes it hold up?

Sanity-check with a number: out of 10 tangerines, 1 are sold and 2 eaten, leaving 7 -- which is $\frac{1}{10}$ of the 10.

Another way: Draw one bar for the whole, shade $\frac{1}{2}$ and then $\frac{2}{5}$ of the same bar, and measure the unshaded strip against $\frac{1}{8}$ .

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two shares over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions — Subtracting from the whole and judging the claim.

💡Takeaway. When two shares come from the same whole they add. It is only when one is taken from what the other left that they multiply.

Variant 4 easy

At a fruit stand, $\frac{4}{9}$ of all the tangerines were sold and $\frac{1}{3}$ of all of them were eaten. The claim is that the tangerines left over are less than $\frac{1}{4}$ of the original amount of tangerines.

Decide whether this claim is correct.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Of all the tangerines, 4 of every 9 were sold and 1 of every 3 were eaten. Someone claims the leftovers are under 1 over 4 of the original. We must judge that claim.

Givens

$\frac{4}{9}$ of the whole was sold.
$\frac{1}{3}$ of the whole was eaten.
The claim: the leftover is less than $\frac{1}{4}$ .

Unknowns

Whether the claim holds.

Constraints

Both fractions are of the ORIGINAL amount, not of what was left after the other.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram

Because both shares are taken from the same whole, they simply add. Find how much left the stand altogether, take that from 1, and compare what remains with the claimed bound.

3 · Execute3 carry out the plan

1Add the parts that are gone

#7 Identify Subproblems 5.NF.A.1

Sold and eaten are both fractions of the original, so put them over 9 and add.

\frac{4}{9} + \frac{1}{3} = \frac{7}{9}

\frac{7}{9}

of the tangerines are gone.

2Subtract to find the leftover

#7 Identify Subproblems 5.NF.A.2

The whole is 1, so what is left is 1 minus what went.

1 - \frac{7}{9} = \frac{2}{9}

\frac{2}{9}

of the original is left.

3Compare the leftover with $\frac{1}{4}$

#1 Draw a Diagram 5.NF.A.2

Put both over a common denominator to see which is bigger.

\frac{2}{9} \ < \ \frac{1}{4}

The leftover is less than the claimed bound, so the claim is correct.

Answer: The claim is correct: the leftover is

\frac{2}{9}

, which is less than

\frac{1}{4}

.

4 · Reviewdoes it hold up?

Sanity-check with a number: out of 9 tangerines, 4 are sold and 1 eaten, leaving 4 -- which is $\frac{2}{9}$ of the 9.

Another way: Draw one bar for the whole, shade $\frac{4}{9}$ and then $\frac{1}{3}$ of the same bar, and measure the unshaded strip against $\frac{1}{4}$ .

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two shares over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions — Subtracting from the whole and judging the claim.

💡Takeaway. When two shares come from the same whole they add. It is only when one is taken from what the other left that they multiply.

Variant 5 medium

At a fruit stand, $\frac{3}{8}$ of all the tangerines were sold and $\frac{1}{2}$ of all of them were eaten. The claim is that the tangerines left over are less than $\frac{1}{10}$ of the original amount of tangerines.

Decide whether this claim is correct.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Of all the tangerines, 3 of every 8 were sold and 1 of every 2 were eaten. Someone claims the leftovers are under 1 over 10 of the original. We must judge that claim.

Givens

$\frac{3}{8}$ of the whole was sold.
$\frac{1}{2}$ of the whole was eaten.
The claim: the leftover is less than $\frac{1}{10}$ .

Unknowns

Whether the claim holds.

Constraints

Both fractions are of the ORIGINAL amount, not of what was left after the other.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram

Because both shares are taken from the same whole, they simply add. Find how much left the stand altogether, take that from 1, and compare what remains with the claimed bound.

3 · Execute3 carry out the plan

1Add the parts that are gone

#7 Identify Subproblems 5.NF.A.1

Sold and eaten are both fractions of the original, so put them over 8 and add.

\frac{3}{8} + \frac{1}{2} = \frac{7}{8}

\frac{7}{8}

of the tangerines are gone.

2Subtract to find the leftover

#7 Identify Subproblems 5.NF.A.2

The whole is 1, so what is left is 1 minus what went.

1 - \frac{7}{8} = \frac{1}{8}

\frac{1}{8}

of the original is left.

3Compare the leftover with $\frac{1}{10}$

#1 Draw a Diagram 5.NF.A.2

Put both over a common denominator to see which is bigger.

\frac{1}{8} \ \geq \ \frac{1}{10}

The leftover is not less than the claimed bound, so the claim is not correct.

Answer: The claim is not correct: the leftover is

\frac{1}{8}

, which is not less than

\frac{1}{10}

.

4 · Reviewdoes it hold up?

Sanity-check with a number: out of 8 tangerines, 3 are sold and 1 eaten, leaving 4 -- which is $\frac{1}{8}$ of the 8.

Another way: Draw one bar for the whole, shade $\frac{3}{8}$ and then $\frac{1}{2}$ of the same bar, and measure the unshaded strip against $\frac{1}{10}$ .

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two shares over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions — Subtracting from the whole and judging the claim.

💡Takeaway. When two shares come from the same whole they add. It is only when one is taken from what the other left that they multiply.

Variant 6 medium

At a fruit stand, $\frac{2}{5}$ of all the tangerines were sold and $\frac{3}{10}$ of all of them were eaten. The claim is that the tangerines left over are less than $\frac{1}{3}$ of the original amount of tangerines.

Decide whether this claim is correct.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Of all the tangerines, 2 of every 5 were sold and 3 of every 10 were eaten. Someone claims the leftovers are under 1 over 3 of the original. We must judge that claim.

Givens

$\frac{2}{5}$ of the whole was sold.
$\frac{3}{10}$ of the whole was eaten.
The claim: the leftover is less than $\frac{1}{3}$ .

Unknowns

Whether the claim holds.

Constraints

Both fractions are of the ORIGINAL amount, not of what was left after the other.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram

Because both shares are taken from the same whole, they simply add. Find how much left the stand altogether, take that from 1, and compare what remains with the claimed bound.

3 · Execute3 carry out the plan

1Add the parts that are gone

#7 Identify Subproblems 5.NF.A.1

Sold and eaten are both fractions of the original, so put them over 10 and add.

\frac{2}{5} + \frac{3}{10} = \frac{7}{10}

\frac{7}{10}

of the tangerines are gone.

2Subtract to find the leftover

#7 Identify Subproblems 5.NF.A.2

The whole is 1, so what is left is 1 minus what went.

1 - \frac{7}{10} = \frac{3}{10}

\frac{3}{10}

of the original is left.

3Compare the leftover with $\frac{1}{3}$

#1 Draw a Diagram 5.NF.A.2

Put both over a common denominator to see which is bigger.

\frac{3}{10} \ < \ \frac{1}{3}

The leftover is less than the claimed bound, so the claim is correct.

Answer: The claim is correct: the leftover is

\frac{3}{10}

, which is less than

\frac{1}{3}

.

4 · Reviewdoes it hold up?

Sanity-check with a number: out of 10 tangerines, 2 are sold and 3 eaten, leaving 5 -- which is $\frac{3}{10}$ of the 10.

Another way: Draw one bar for the whole, shade $\frac{2}{5}$ and then $\frac{3}{10}$ of the same bar, and measure the unshaded strip against $\frac{1}{3}$ .

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two shares over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions — Subtracting from the whole and judging the claim.

💡Takeaway. When two shares come from the same whole they add. It is only when one is taken from what the other left that they multiply.

Variant 7 medium

At a fruit stand, $\frac{3}{4}$ of all the tangerines were sold and $\frac{1}{5}$ of all of them were eaten. The claim is that the tangerines left over are less than $\frac{1}{10}$ of the original amount of tangerines.

Decide whether this claim is correct.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Of all the tangerines, 3 of every 4 were sold and 1 of every 5 were eaten. Someone claims the leftovers are under 1 over 10 of the original. We must judge that claim.

Givens

$\frac{3}{4}$ of the whole was sold.
$\frac{1}{5}$ of the whole was eaten.
The claim: the leftover is less than $\frac{1}{10}$ .

Unknowns

Whether the claim holds.

Constraints

Both fractions are of the ORIGINAL amount, not of what was left after the other.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram

Because both shares are taken from the same whole, they simply add. Find how much left the stand altogether, take that from 1, and compare what remains with the claimed bound.

3 · Execute3 carry out the plan

1Add the parts that are gone

#7 Identify Subproblems 5.NF.A.1

Sold and eaten are both fractions of the original, so put them over 20 and add.

\frac{3}{4} + \frac{1}{5} = \frac{19}{20}

\frac{19}{20}

of the tangerines are gone.

2Subtract to find the leftover

#7 Identify Subproblems 5.NF.A.2

The whole is 1, so what is left is 1 minus what went.

1 - \frac{19}{20} = \frac{1}{20}

\frac{1}{20}

of the original is left.

3Compare the leftover with $\frac{1}{10}$

#1 Draw a Diagram 5.NF.A.2

Put both over a common denominator to see which is bigger.

\frac{1}{20} \ < \ \frac{1}{10}

The leftover is less than the claimed bound, so the claim is correct.

Answer: The claim is correct: the leftover is

\frac{1}{20}

, which is less than

\frac{1}{10}

.

4 · Reviewdoes it hold up?

Sanity-check with a number: out of 20 tangerines, 3 are sold and 1 eaten, leaving 16 -- which is $\frac{1}{20}$ of the 20.

Another way: Draw one bar for the whole, shade $\frac{3}{4}$ and then $\frac{1}{5}$ of the same bar, and measure the unshaded strip against $\frac{1}{10}$ .

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two shares over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions — Subtracting from the whole and judging the claim.

💡Takeaway. When two shares come from the same whole they add. It is only when one is taken from what the other left that they multiply.

Variant 8 medium

At a fruit stand, $\frac{3}{5}$ of all the tangerines were sold and $\frac{1}{3}$ of all of them were eaten. The claim is that the tangerines left over are less than $\frac{1}{12}$ of the original amount of tangerines.

Decide whether this claim is correct.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Of all the tangerines, 3 of every 5 were sold and 1 of every 3 were eaten. Someone claims the leftovers are under 1 over 12 of the original. We must judge that claim.

Givens

$\frac{3}{5}$ of the whole was sold.
$\frac{1}{3}$ of the whole was eaten.
The claim: the leftover is less than $\frac{1}{12}$ .

Unknowns

Whether the claim holds.

Constraints

Both fractions are of the ORIGINAL amount, not of what was left after the other.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram

Because both shares are taken from the same whole, they simply add. Find how much left the stand altogether, take that from 1, and compare what remains with the claimed bound.

3 · Execute3 carry out the plan

1Add the parts that are gone

#7 Identify Subproblems 5.NF.A.1

Sold and eaten are both fractions of the original, so put them over 15 and add.

\frac{3}{5} + \frac{1}{3} = \frac{14}{15}

\frac{14}{15}

of the tangerines are gone.

2Subtract to find the leftover

#7 Identify Subproblems 5.NF.A.2

The whole is 1, so what is left is 1 minus what went.

1 - \frac{14}{15} = \frac{1}{15}

\frac{1}{15}

of the original is left.

3Compare the leftover with $\frac{1}{12}$

#1 Draw a Diagram 5.NF.A.2

Put both over a common denominator to see which is bigger.

\frac{1}{15} \ < \ \frac{1}{12}

The leftover is less than the claimed bound, so the claim is correct.

Answer: The claim is correct: the leftover is

\frac{1}{15}

, which is less than

\frac{1}{12}

.

4 · Reviewdoes it hold up?

Sanity-check with a number: out of 15 tangerines, 3 are sold and 1 eaten, leaving 11 -- which is $\frac{1}{15}$ of the 15.

Another way: Draw one bar for the whole, shade $\frac{3}{5}$ and then $\frac{1}{3}$ of the same bar, and measure the unshaded strip against $\frac{1}{12}$ .

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two shares over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions — Subtracting from the whole and judging the claim.

💡Takeaway. When two shares come from the same whole they add. It is only when one is taken from what the other left that they multiply.

Variant 9 hard

At a fruit stand, $\frac{1}{3}$ of all the tangerines were sold and $\frac{5}{12}$ of all of them were eaten. The claim is that the tangerines left over are less than $\frac{1}{5}$ of the original amount of tangerines.

Decide whether this claim is correct.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Of all the tangerines, 1 of every 3 were sold and 5 of every 12 were eaten. Someone claims the leftovers are under 1 over 5 of the original. We must judge that claim.

Givens

$\frac{1}{3}$ of the whole was sold.
$\frac{5}{12}$ of the whole was eaten.
The claim: the leftover is less than $\frac{1}{5}$ .

Unknowns

Whether the claim holds.

Constraints

Both fractions are of the ORIGINAL amount, not of what was left after the other.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram

Because both shares are taken from the same whole, they simply add. Find how much left the stand altogether, take that from 1, and compare what remains with the claimed bound.

3 · Execute3 carry out the plan

1Add the parts that are gone

#7 Identify Subproblems 5.NF.A.1

Sold and eaten are both fractions of the original, so put them over 4 and add.

\frac{1}{4} + \frac{0}{1} = \frac{3}{4}

\frac{3}{4}

of the tangerines are gone.

2Subtract to find the leftover

#7 Identify Subproblems 5.NF.A.2

The whole is 1, so what is left is 1 minus what went.

1 - \frac{3}{4} = \frac{1}{4}

\frac{1}{4}

of the original is left.

3Compare the leftover with $\frac{1}{5}$

#1 Draw a Diagram 5.NF.A.2

Put both over a common denominator to see which is bigger.

\frac{1}{4} \ \geq \ \frac{1}{5}

The leftover is not less than the claimed bound, so the claim is not correct.

Answer: The claim is not correct: the leftover is

\frac{1}{4}

, which is not less than

\frac{1}{5}

.

4 · Reviewdoes it hold up?

Sanity-check with a number: out of 4 tangerines, 1 are sold and 0 eaten, leaving 3 -- which is $\frac{1}{4}$ of the 4.

Another way: Draw one bar for the whole, shade $\frac{1}{3}$ and then $\frac{5}{12}$ of the same bar, and measure the unshaded strip against $\frac{1}{5}$ .

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two shares over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions — Subtracting from the whole and judging the claim.

💡Takeaway. When two shares come from the same whole they add. It is only when one is taken from what the other left that they multiply.

Variant 10 hard

At a fruit stand, $\frac{7}{10}$ of all the tangerines were sold and $\frac{1}{5}$ of all of them were eaten. The claim is that the tangerines left over are less than $\frac{1}{12}$ of the original amount of tangerines.

Decide whether this claim is correct.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Of all the tangerines, 7 of every 10 were sold and 1 of every 5 were eaten. Someone claims the leftovers are under 1 over 12 of the original. We must judge that claim.

Givens

$\frac{7}{10}$ of the whole was sold.
$\frac{1}{5}$ of the whole was eaten.
The claim: the leftover is less than $\frac{1}{12}$ .

Unknowns

Whether the claim holds.

Constraints

Both fractions are of the ORIGINAL amount, not of what was left after the other.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram

Because both shares are taken from the same whole, they simply add. Find how much left the stand altogether, take that from 1, and compare what remains with the claimed bound.

3 · Execute3 carry out the plan

1Add the parts that are gone

#7 Identify Subproblems 5.NF.A.1

Sold and eaten are both fractions of the original, so put them over 10 and add.

\frac{7}{10} + \frac{1}{5} = \frac{9}{10}

\frac{9}{10}

of the tangerines are gone.

2Subtract to find the leftover

#7 Identify Subproblems 5.NF.A.2

The whole is 1, so what is left is 1 minus what went.

1 - \frac{9}{10} = \frac{1}{10}

\frac{1}{10}

of the original is left.

3Compare the leftover with $\frac{1}{12}$

#1 Draw a Diagram 5.NF.A.2

Put both over a common denominator to see which is bigger.

\frac{1}{10} \ \geq \ \frac{1}{12}

The leftover is not less than the claimed bound, so the claim is not correct.

Answer: The claim is not correct: the leftover is

\frac{1}{10}

, which is not less than

\frac{1}{12}

.

4 · Reviewdoes it hold up?

Sanity-check with a number: out of 10 tangerines, 7 are sold and 1 eaten, leaving 2 -- which is $\frac{1}{10}$ of the 10.

Another way: Draw one bar for the whole, shade $\frac{7}{10}$ and then $\frac{1}{5}$ of the same bar, and measure the unshaded strip against $\frac{1}{12}$ .

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two shares over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions — Subtracting from the whole and judging the claim.

💡Takeaway. When two shares come from the same whole they add. It is only when one is taken from what the other left that they multiply.

Variant 11 hard

At a fruit stand, $\frac{5}{6}$ of all the tangerines were sold and $\frac{1}{12}$ of all of them were eaten. The claim is that the tangerines left over are less than $\frac{1}{8}$ of the original amount of tangerines.

Decide whether this claim is correct.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Of all the tangerines, 5 of every 6 were sold and 1 of every 12 were eaten. Someone claims the leftovers are under 1 over 8 of the original. We must judge that claim.

Givens

$\frac{5}{6}$ of the whole was sold.
$\frac{1}{12}$ of the whole was eaten.
The claim: the leftover is less than $\frac{1}{8}$ .

Unknowns

Whether the claim holds.

Constraints

Both fractions are of the ORIGINAL amount, not of what was left after the other.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram

Because both shares are taken from the same whole, they simply add. Find how much left the stand altogether, take that from 1, and compare what remains with the claimed bound.

3 · Execute3 carry out the plan

1Add the parts that are gone

#7 Identify Subproblems 5.NF.A.1

Sold and eaten are both fractions of the original, so put them over 12 and add.

\frac{5}{6} + \frac{1}{12} = \frac{11}{12}

\frac{11}{12}

of the tangerines are gone.

2Subtract to find the leftover

#7 Identify Subproblems 5.NF.A.2

The whole is 1, so what is left is 1 minus what went.

1 - \frac{11}{12} = \frac{1}{12}

\frac{1}{12}

of the original is left.

3Compare the leftover with $\frac{1}{8}$

#1 Draw a Diagram 5.NF.A.2

Put both over a common denominator to see which is bigger.

\frac{1}{12} \ < \ \frac{1}{8}

The leftover is less than the claimed bound, so the claim is correct.

Answer: The claim is correct: the leftover is

\frac{1}{12}

, which is less than

\frac{1}{8}

.

4 · Reviewdoes it hold up?

Sanity-check with a number: out of 12 tangerines, 5 are sold and 1 eaten, leaving 6 -- which is $\frac{1}{12}$ of the 12.

Another way: Draw one bar for the whole, shade $\frac{5}{6}$ and then $\frac{1}{12}$ of the same bar, and measure the unshaded strip against $\frac{1}{8}$ .

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two shares over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions — Subtracting from the whole and judging the claim.

💡Takeaway. When two shares come from the same whole they add. It is only when one is taken from what the other left that they multiply.

Variant 12 hard

At a fruit stand, $\frac{7}{12}$ of all the tangerines were sold and $\frac{1}{4}$ of all of them were eaten. The claim is that the tangerines left over are less than $\frac{1}{5}$ of the original amount of tangerines.

Decide whether this claim is correct.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Of all the tangerines, 7 of every 12 were sold and 1 of every 4 were eaten. Someone claims the leftovers are under 1 over 5 of the original. We must judge that claim.

Givens

$\frac{7}{12}$ of the whole was sold.
$\frac{1}{4}$ of the whole was eaten.
The claim: the leftover is less than $\frac{1}{5}$ .

Unknowns

Whether the claim holds.

Constraints

Both fractions are of the ORIGINAL amount, not of what was left after the other.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram

Because both shares are taken from the same whole, they simply add. Find how much left the stand altogether, take that from 1, and compare what remains with the claimed bound.

3 · Execute3 carry out the plan

1Add the parts that are gone

#7 Identify Subproblems 5.NF.A.1

Sold and eaten are both fractions of the original, so put them over 6 and add.

\frac{0}{1} + \frac{1}{6} = \frac{5}{6}

\frac{5}{6}

of the tangerines are gone.

2Subtract to find the leftover

#7 Identify Subproblems 5.NF.A.2

The whole is 1, so what is left is 1 minus what went.

1 - \frac{5}{6} = \frac{1}{6}

\frac{1}{6}

of the original is left.

3Compare the leftover with $\frac{1}{5}$

#1 Draw a Diagram 5.NF.A.2

Put both over a common denominator to see which is bigger.

\frac{1}{6} \ < \ \frac{1}{5}

The leftover is less than the claimed bound, so the claim is correct.

Answer: The claim is correct: the leftover is

\frac{1}{6}

, which is less than

\frac{1}{5}

.

4 · Reviewdoes it hold up?

Sanity-check with a number: out of 6 tangerines, 0 are sold and 1 eaten, leaving 5 -- which is $\frac{1}{6}$ of the 6.

Another way: Draw one bar for the whole, shade $\frac{7}{12}$ and then $\frac{1}{4}$ of the same bar, and measure the unshaded strip against $\frac{1}{5}$ .

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two shares over a common denominator.
5.NF.A.2 Solve word problems involving addition and subtraction of fractions — Subtracting from the whole and judging the claim.

💡Takeaway. When two shares come from the same whole they add. It is only when one is taken from what the other left that they multiply.

Sum fractional parts of a whole · 12 practice problems

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