Jin needs $$4\frac{2}{5}$$ cups total and has $$2\frac{4}{5}$$ cups. The amount to buy is $$4\frac{2}{5} - 2\frac{4}{5} = 1\frac{3}{5}$$ cups. Choice A gives the correct amount but incorrectly states it as total flour. Choice B makes an arithmetic error. Choice D incorrectly adds the amounts instead of subtracting.

When you see word problems involving adding and subtracting mixed numbers, break down each step carefully and track what's happening to the rope.

Start with the original length: $$7\frac{2}{9}$$ feet. First, you cut off $$2\frac{8}{9}$$ feet, so subtract: $$7\frac{2}{9} - 2\frac{8}{9}$$. Since $$\frac{2}{9} < \frac{8}{9}$$, you need to borrow from the whole number. Convert $$7\frac{2}{9}$$ to $$6\frac{11}{9}$$ (since $$7\frac{2}{9} = 6 + 1\frac{2}{9} = 6\frac{11}{9}$$). Now subtract: $$6\frac{11}{9} - 2\frac{8}{9} = 4\frac{3}{9}$$.

Next, add the new piece: $$4\frac{3}{9} + 1\frac{4}{9} = 5\frac{7}{9}$$ feet.

Choice A ($$4\frac{3}{9}$$ feet) represents only the length after cutting, forgetting to add the new piece. Choice C ($$11\frac{5}{9}$$ feet) likely comes from adding all three measurements instead of subtracting the cut piece: $$7\frac{2}{9} + 2\frac{8}{9} + 1\frac{4}{9}$$. Choice D ($$6\frac{1}{9}$$ feet) could result from calculation errors when borrowing during subtraction or adding fractions incorrectly.

Choice B correctly shows $$5\frac{7}{9}$$ feet as the final length.

Study tip: In multi-step fraction problems, work through each operation separately and double-check your borrowing when subtracting mixed numbers. Make sure you're following the story—subtract what's removed, add what's attached.

When you see a word problem involving mixed numbers, you need to identify what operation to use and work systematically through the steps. This problem asks how much sugar remains after using some, so you'll subtract the amounts used from the starting amount.

Start with $$9\frac{2}{8}$$ cups and subtract both amounts used. First, add the sugar used: $$3\frac{1}{8} + 2\frac{7}{8}$$. Add the whole numbers: $$3 + 2 = 5$$. Add the fractions: $$\frac{1}{8} + \frac{7}{8} = \frac{8}{8} = 1$$. So the total used is $$5 + 1 = 6$$ cups.

Now subtract: $$9\frac{2}{8} - 6 = 3\frac{2}{8}$$ cups remaining. This matches answer D.

Looking at the wrong answers: Answer A ($$3\frac{4}{8}$$) likely comes from incorrectly adding the fractions $$\frac{2}{8} + \frac{1}{8} + \frac{7}{8} = \frac{10}{8}$$, but forgetting to convert $$\frac{10}{8}$$ to $$1\frac{2}{8}$$. Answer B ($$15\frac{2}{8}$$) results from adding all three amounts instead of subtracting: $$9\frac{2}{8} + 3\frac{1}{8} + 2\frac{7}{8}$$. Answer C ($$6\frac{0}{8}$$) comes from only calculating how much sugar was used total, not how much remains.

When solving mixed number word problems, always read carefully to determine the correct operation, then work step-by-step: handle whole numbers and fractions separately, and remember to convert improper fractions back to mixed numbers when needed.

Monday and Tuesday total: $$3\frac{1}{6} + 2\frac{5}{6} = 6\frac{0}{6} = 6$$ miles. Wednesday distance: $$6 - 1\frac{2}{6} = 4\frac{4}{6}$$ miles. Choice B makes an error in the subtraction. Choice C incorrectly adds $$1\frac{2}{6}$$ instead of subtracting. Choice D represents a calculation error in the final subtraction step.

Start with the remaining water: $$5\frac{1}{4}$$ gallons. Add the new water: $$5\frac{1}{4} + 2\frac{2}{4} = 7\frac{3}{4}$$ gallons. Choice B incorrectly adds all three amounts mentioned. Choice C gives only the amount that was used initially. Choice D represents the difference between what was added and what remained.

First, subtract the ribbon used: $$2\frac{3}{8} - 1\frac{5}{8} = \frac{6}{8}$$ yards remaining. Then add the new ribbon: $$\frac{6}{8} + 3\frac{2}{8} = 4\frac{0}{8} = 4$$ yards. Choice B incorrectly adds all three amounts. Choice C represents an improper fraction that wasn't simplified. Choice D adds without subtracting the used ribbon first.

When you see a multi-step word problem involving mixed numbers, break it down into clear steps: find what Luis ate total, then calculate what his sister ate based on that.

First, add Luis's lunch and dinner: $$1\frac{2}{3} + 2\frac{1}{3}$$. Since the denominators are the same, add the whole numbers (1 + 2 = 3) and the fractions ($$\frac{2}{3} + \frac{1}{3} = \frac{3}{3} = 1$$). Luis ate $$3 + 1 = 4$$ slices total.

Next, his sister ate $$1\frac{1}{3}$$ slices less than Luis's total. This means subtraction: $$4 - 1\frac{1}{3}$$. Convert 4 to a mixed number with thirds: $$4 = 3\frac{3}{3}$$. Now subtract: $$3\frac{3}{3} - 1\frac{1}{3} = 2\frac{2}{3}$$ slices.

Choice A ($$5\frac{1}{3}$$) happens if you accidentally add $$1\frac{1}{3}$$ to Luis's total instead of subtracting - watch for that key word "less." Choice B ($$4\frac{2}{3}$$) occurs if you only subtract the whole number 1 from Luis's total, ignoring the $$\frac{1}{3}$$ part. Choice D ($$3\frac{0}{3}$$ or simply 3) results from subtracting only the fraction $$\frac{1}{3}$$ and ignoring the whole number 1.

The correct answer is C.

Strategy tip: In word problems with "less than," "fewer than," or "decreased by," you're subtracting. Always double-check whether you need to add or subtract by rereading the problem's key phrases carefully.

When you see a multi-step word problem with mixed numbers, break it down into clear parts and add step by step.

First, find Emma's total walking distance to school and the library by adding the mixed numbers. Since both fractions have the same denominator (7), you can add them directly: $$2\frac{3}{7} + 1\frac{5}{7} = 3\frac{8}{7}$$. Since $$\frac{8}{7}$$ equals $$1\frac{1}{7}$$, this becomes $$4\frac{1}{7}$$ miles total.

Next, the problem states that her walk home was $$2\frac{1}{7}$$ miles farther than this total distance. So you need to add: $$4\frac{1}{7} + 2\frac{1}{7} = 6\frac{2}{7}$$ miles to walk home.

Looking at the wrong answers: Choice A ($$6\frac{3}{7}$$) likely comes from incorrectly adding $$\frac{1}{7} + \frac{1}{7} = \frac{3}{7}$$ instead of $$\frac{2}{7}$$. Choice B ($$2\frac{0}{7}$$) represents only the "farther" distance without adding it to the school-library total. Choice C ($$4\frac{1}{7}$$) gives you the school-library total but forgets to add the extra $$2\frac{1}{7}$$ miles for the home distance.

The correct answer is D: $$6\frac{2}{7}$$ miles.

Remember to read word problems carefully and identify each step: first find intermediate totals, then use those results in the final calculation. Don't rush—multi-step problems require you to use one answer to find the next.

This question tests 4th grade ability to add and subtract mixed numbers with like denominators, using strategies such as converting to improper fractions or using properties of operations (CCSS.4.NF.3.c). Mixed numbers combine a whole number and a fraction (like 2 1/4). To add or subtract mixed numbers with the same denominator, we can either: (1) convert both to improper fractions, add/subtract, then convert back to mixed number, or (2) add/subtract the whole number parts and fraction parts separately using properties of operations. For subtraction, if the fraction part being subtracted is larger than the fraction part starting with, we must regroup by converting 1 whole into fractions. To subtract 5 3/4 and 2 1/4, students can convert to improper fractions: 23/4 and 9/4, then subtract numerators: 14/4 = 3 2/4 = 3 1/2, or subtract separately: whole parts (5-2=3) and fraction parts (3/4 - 1/4=2/4=1/2), then combine: 3 1/2. Choice B is correct because subtracting separately: 5-2=3, 3/4-1/4=2/4=1/2, combining 3 1/2. This demonstrates understanding of mixed number operations with like denominators. Choice C represents subtracting the wholes correctly but subtracting fractions as 3/4 - 1/4 = 1/2 and then mistakenly reducing the whole by 1, which happens when students unnecessarily regroup. To help students: Practice both methods. For improper fraction method: convert mixed to improper (a×c+b)/c, add/subtract numerators keeping denominator same, convert result back to mixed (divide numerator by denominator for quotient=whole, remainder=numerator). For separate parts method: add/subtract wholes, add/subtract fractions, combine. For subtraction, watch for regrouping only when needed. Check answers: subtraction result should be < minuend. Watch for: arithmetic errors in fractions, or regrouping when not required.

This question tests 4th grade ability to add and subtract mixed numbers with like denominators, using strategies such as converting to improper fractions or using properties of operations (CCSS.4.NF.3.c). Mixed numbers combine a whole number and a fraction (like 1 2/8). To add or subtract mixed numbers with the same denominator, we can either: (1) convert both to improper fractions, add/subtract, then convert back to mixed number, or (2) add/subtract the whole number parts and fraction parts separately using properties of operations. For subtraction, if the fraction part being subtracted is larger than the fraction part starting with, we must regroup by converting 1 whole into fractions. To add 1 2/8 and 2 3/8, students can convert to improper: 10/8 and 19/8, add: 29/8 = 3 5/8, or add separately: wholes (1+2=3), fractions (2/8+3/8=5/8), combine: 3 5/8. Choice A is correct because adding separately: 1+2=3, 2/8+3/8=5/8, combining: 3 5/8, demonstrating understanding of mixed number operations with like denominators. Choice B represents adding denominators by mistake, which happens when students think denominators change in addition. To help students: Practice both methods. For improper fraction method: convert mixed to improper (a b/c = (a×c+b)/c), add/subtract numerators keeping denominator same, convert result back to mixed (divide numerator by denominator for quotient=whole, remainder=numerator). For separate parts method: add/subtract wholes, add/subtract fractions, combine. Check answers: addition result > either addend. Watch for: adding denominators, conversion errors, and not carrying over if needed.