← Sum fractional parts of a whole · Part-Whole Fraction Reasoning

Sum fractional parts of a whole · 12 practice problems

5.NF.A.15.NF.A.2

Generated variants — 12

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 easy answer: The claim is correct: the leftover is 112\frac{1}{12}, which is less than 16\frac{1}{6}.

At a fruit stand, 23\frac{2}{3} of all the tangerines were sold and 14\frac{1}{4} of all of them were eaten. The claim is that the tangerines left over are less than 16\frac{1}{6} of the original amount of tangerines.

Decide whether this claim is correct.

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
  • 23\frac{2}{3} of the whole was sold.
  • 14\frac{1}{4} of the whole was eaten.
  • The claim: the leftover is less than 16\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.
23+14=1112\frac{2}{3} + \frac{1}{4} = \frac{11}{12}
1112\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.
1−1112=1121 - \frac{11}{12} = \frac{1}{12}
112\frac{1}{12} of the original is left.

3Compare the leftover with 16\frac{1}{6}

#1 Draw a Diagram 5.NF.A.2
Put both over a common denominator to see which is bigger.
112 < 16\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 112\frac{1}{12}, which is less than 16\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 112\frac{1}{12} of the 12.

Another way: Draw one bar for the whole, shade 23\frac{2}{3} and then 14\frac{1}{4} of the same bar, and measure the unshaded strip against 16\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 answer: The claim is correct: the leftover is 18\frac{1}{8}, which is less than 16\frac{1}{6}.

At a fruit stand, 58\frac{5}{8} of all the tangerines were sold and 14\frac{1}{4} of all of them were eaten. The claim is that the tangerines left over are less than 16\frac{1}{6} of the original amount of tangerines.

Decide whether this claim is correct.

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
  • 58\frac{5}{8} of the whole was sold.
  • 14\frac{1}{4} of the whole was eaten.
  • The claim: the leftover is less than 16\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.
58+14=78\frac{5}{8} + \frac{1}{4} = \frac{7}{8}
78\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.
1−78=181 - \frac{7}{8} = \frac{1}{8}
18\frac{1}{8} of the original is left.

3Compare the leftover with 16\frac{1}{6}

#1 Draw a Diagram 5.NF.A.2
Put both over a common denominator to see which is bigger.
18 < 16\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 18\frac{1}{8}, which is less than 16\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 18\frac{1}{8} of the 8.

Another way: Draw one bar for the whole, shade 58\frac{5}{8} and then 14\frac{1}{4} of the same bar, and measure the unshaded strip against 16\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 answer: The claim is correct: the leftover is 110\frac{1}{10}, which is less than 18\frac{1}{8}.

At a fruit stand, 12\frac{1}{2} of all the tangerines were sold and 25\frac{2}{5} of all of them were eaten. The claim is that the tangerines left over are less than 18\frac{1}{8} of the original amount of tangerines.

Decide whether this claim is correct.

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
  • 12\frac{1}{2} of the whole was sold.
  • 25\frac{2}{5} of the whole was eaten.
  • The claim: the leftover is less than 18\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.
12+25=910\frac{1}{2} + \frac{2}{5} = \frac{9}{10}
910\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.
1−910=1101 - \frac{9}{10} = \frac{1}{10}
110\frac{1}{10} of the original is left.

3Compare the leftover with 18\frac{1}{8}

#1 Draw a Diagram 5.NF.A.2
Put both over a common denominator to see which is bigger.
110 < 18\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 110\frac{1}{10}, which is less than 18\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 110\frac{1}{10} of the 10.

Another way: Draw one bar for the whole, shade 12\frac{1}{2} and then 25\frac{2}{5} of the same bar, and measure the unshaded strip against 18\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 answer: The claim is correct: the leftover is 29\frac{2}{9}, which is less than 14\frac{1}{4}.

At a fruit stand, 49\frac{4}{9} of all the tangerines were sold and 13\frac{1}{3} of all of them were eaten. The claim is that the tangerines left over are less than 14\frac{1}{4} of the original amount of tangerines.

Decide whether this claim is correct.

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
  • 49\frac{4}{9} of the whole was sold.
  • 13\frac{1}{3} of the whole was eaten.
  • The claim: the leftover is less than 14\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.
49+13=79\frac{4}{9} + \frac{1}{3} = \frac{7}{9}
79\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.
1−79=291 - \frac{7}{9} = \frac{2}{9}
29\frac{2}{9} of the original is left.

3Compare the leftover with 14\frac{1}{4}

#1 Draw a Diagram 5.NF.A.2
Put both over a common denominator to see which is bigger.
29 < 14\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 29\frac{2}{9}, which is less than 14\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 29\frac{2}{9} of the 9.

Another way: Draw one bar for the whole, shade 49\frac{4}{9} and then 13\frac{1}{3} of the same bar, and measure the unshaded strip against 14\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 answer: The claim is not correct: the leftover is 18\frac{1}{8}, which is not less than 110\frac{1}{10}.

At a fruit stand, 38\frac{3}{8} of all the tangerines were sold and 12\frac{1}{2} of all of them were eaten. The claim is that the tangerines left over are less than 110\frac{1}{10} of the original amount of tangerines.

Decide whether this claim is correct.

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
  • 38\frac{3}{8} of the whole was sold.
  • 12\frac{1}{2} of the whole was eaten.
  • The claim: the leftover is less than 110\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.
38+12=78\frac{3}{8} + \frac{1}{2} = \frac{7}{8}
78\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.
1−78=181 - \frac{7}{8} = \frac{1}{8}
18\frac{1}{8} of the original is left.

3Compare the leftover with 110\frac{1}{10}

#1 Draw a Diagram 5.NF.A.2
Put both over a common denominator to see which is bigger.
18 ≥ 110\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 18\frac{1}{8}, which is not less than 110\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 18\frac{1}{8} of the 8.

Another way: Draw one bar for the whole, shade 38\frac{3}{8} and then 12\frac{1}{2} of the same bar, and measure the unshaded strip against 110\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 answer: The claim is correct: the leftover is 310\frac{3}{10}, which is less than 13\frac{1}{3}.

At a fruit stand, 25\frac{2}{5} of all the tangerines were sold and 310\frac{3}{10} of all of them were eaten. The claim is that the tangerines left over are less than 13\frac{1}{3} of the original amount of tangerines.

Decide whether this claim is correct.

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
  • 25\frac{2}{5} of the whole was sold.
  • 310\frac{3}{10} of the whole was eaten.
  • The claim: the leftover is less than 13\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.
25+310=710\frac{2}{5} + \frac{3}{10} = \frac{7}{10}
710\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.
1−710=3101 - \frac{7}{10} = \frac{3}{10}
310\frac{3}{10} of the original is left.

3Compare the leftover with 13\frac{1}{3}

#1 Draw a Diagram 5.NF.A.2
Put both over a common denominator to see which is bigger.
310 < 13\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 310\frac{3}{10}, which is less than 13\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 310\frac{3}{10} of the 10.

Another way: Draw one bar for the whole, shade 25\frac{2}{5} and then 310\frac{3}{10} of the same bar, and measure the unshaded strip against 13\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 answer: The claim is correct: the leftover is 120\frac{1}{20}, which is less than 110\frac{1}{10}.

At a fruit stand, 34\frac{3}{4} of all the tangerines were sold and 15\frac{1}{5} of all of them were eaten. The claim is that the tangerines left over are less than 110\frac{1}{10} of the original amount of tangerines.

Decide whether this claim is correct.

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
  • 34\frac{3}{4} of the whole was sold.
  • 15\frac{1}{5} of the whole was eaten.
  • The claim: the leftover is less than 110\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.
34+15=1920\frac{3}{4} + \frac{1}{5} = \frac{19}{20}
1920\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.
1−1920=1201 - \frac{19}{20} = \frac{1}{20}
120\frac{1}{20} of the original is left.

3Compare the leftover with 110\frac{1}{10}

#1 Draw a Diagram 5.NF.A.2
Put both over a common denominator to see which is bigger.
120 < 110\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 120\frac{1}{20}, which is less than 110\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 120\frac{1}{20} of the 20.

Another way: Draw one bar for the whole, shade 34\frac{3}{4} and then 15\frac{1}{5} of the same bar, and measure the unshaded strip against 110\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 answer: The claim is correct: the leftover is 115\frac{1}{15}, which is less than 112\frac{1}{12}.

At a fruit stand, 35\frac{3}{5} of all the tangerines were sold and 13\frac{1}{3} of all of them were eaten. The claim is that the tangerines left over are less than 112\frac{1}{12} of the original amount of tangerines.

Decide whether this claim is correct.

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
  • 35\frac{3}{5} of the whole was sold.
  • 13\frac{1}{3} of the whole was eaten.
  • The claim: the leftover is less than 112\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.
35+13=1415\frac{3}{5} + \frac{1}{3} = \frac{14}{15}
1415\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.
1−1415=1151 - \frac{14}{15} = \frac{1}{15}
115\frac{1}{15} of the original is left.

3Compare the leftover with 112\frac{1}{12}

#1 Draw a Diagram 5.NF.A.2
Put both over a common denominator to see which is bigger.
115 < 112\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 115\frac{1}{15}, which is less than 112\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 115\frac{1}{15} of the 15.

Another way: Draw one bar for the whole, shade 35\frac{3}{5} and then 13\frac{1}{3} of the same bar, and measure the unshaded strip against 112\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 answer: The claim is not correct: the leftover is 14\frac{1}{4}, which is not less than 15\frac{1}{5}.

At a fruit stand, 13\frac{1}{3} of all the tangerines were sold and 512\frac{5}{12} of all of them were eaten. The claim is that the tangerines left over are less than 15\frac{1}{5} of the original amount of tangerines.

Decide whether this claim is correct.

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
  • 13\frac{1}{3} of the whole was sold.
  • 512\frac{5}{12} of the whole was eaten.
  • The claim: the leftover is less than 15\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.
14+01=34\frac{1}{4} + \frac{0}{1} = \frac{3}{4}
34\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.
1−34=141 - \frac{3}{4} = \frac{1}{4}
14\frac{1}{4} of the original is left.

3Compare the leftover with 15\frac{1}{5}

#1 Draw a Diagram 5.NF.A.2
Put both over a common denominator to see which is bigger.
14 ≥ 15\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 14\frac{1}{4}, which is not less than 15\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 14\frac{1}{4} of the 4.

Another way: Draw one bar for the whole, shade 13\frac{1}{3} and then 512\frac{5}{12} of the same bar, and measure the unshaded strip against 15\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 answer: The claim is not correct: the leftover is 110\frac{1}{10}, which is not less than 112\frac{1}{12}.

At a fruit stand, 710\frac{7}{10} of all the tangerines were sold and 15\frac{1}{5} of all of them were eaten. The claim is that the tangerines left over are less than 112\frac{1}{12} of the original amount of tangerines.

Decide whether this claim is correct.

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
  • 710\frac{7}{10} of the whole was sold.
  • 15\frac{1}{5} of the whole was eaten.
  • The claim: the leftover is less than 112\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.
710+15=910\frac{7}{10} + \frac{1}{5} = \frac{9}{10}
910\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.
1−910=1101 - \frac{9}{10} = \frac{1}{10}
110\frac{1}{10} of the original is left.

3Compare the leftover with 112\frac{1}{12}

#1 Draw a Diagram 5.NF.A.2
Put both over a common denominator to see which is bigger.
110 ≥ 112\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 110\frac{1}{10}, which is not less than 112\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 110\frac{1}{10} of the 10.

Another way: Draw one bar for the whole, shade 710\frac{7}{10} and then 15\frac{1}{5} of the same bar, and measure the unshaded strip against 112\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 answer: The claim is correct: the leftover is 112\frac{1}{12}, which is less than 18\frac{1}{8}.

At a fruit stand, 56\frac{5}{6} of all the tangerines were sold and 112\frac{1}{12} of all of them were eaten. The claim is that the tangerines left over are less than 18\frac{1}{8} of the original amount of tangerines.

Decide whether this claim is correct.

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
  • 56\frac{5}{6} of the whole was sold.
  • 112\frac{1}{12} of the whole was eaten.
  • The claim: the leftover is less than 18\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.
56+112=1112\frac{5}{6} + \frac{1}{12} = \frac{11}{12}
1112\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.
1−1112=1121 - \frac{11}{12} = \frac{1}{12}
112\frac{1}{12} of the original is left.

3Compare the leftover with 18\frac{1}{8}

#1 Draw a Diagram 5.NF.A.2
Put both over a common denominator to see which is bigger.
112 < 18\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 112\frac{1}{12}, which is less than 18\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 112\frac{1}{12} of the 12.

Another way: Draw one bar for the whole, shade 56\frac{5}{6} and then 112\frac{1}{12} of the same bar, and measure the unshaded strip against 18\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 answer: The claim is correct: the leftover is 16\frac{1}{6}, which is less than 15\frac{1}{5}.

At a fruit stand, 712\frac{7}{12} of all the tangerines were sold and 14\frac{1}{4} of all of them were eaten. The claim is that the tangerines left over are less than 15\frac{1}{5} of the original amount of tangerines.

Decide whether this claim is correct.

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
  • 712\frac{7}{12} of the whole was sold.
  • 14\frac{1}{4} of the whole was eaten.
  • The claim: the leftover is less than 15\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.
01+16=56\frac{0}{1} + \frac{1}{6} = \frac{5}{6}
56\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.
1−56=161 - \frac{5}{6} = \frac{1}{6}
16\frac{1}{6} of the original is left.

3Compare the leftover with 15\frac{1}{5}

#1 Draw a Diagram 5.NF.A.2
Put both over a common denominator to see which is bigger.
16 < 15\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 16\frac{1}{6}, which is less than 15\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 16\frac{1}{6} of the 6.

Another way: Draw one bar for the whole, shade 712\frac{7}{12} and then 14\frac{1}{4} of the same bar, and measure the unshaded strip against 15\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.