3 The student incorrectly The student correctly The student answers The student correctly answers both parts and answers one part and one part correctly but answers using any 4.OA.3 shows no reasoning. shows little reasoning. shows solid reasoning place value strategy: 4.NBT.5 in both problems, or 400 the student shows 4.NBT.6 Each teacher some reasoning with received 72 correct answers for pencils. both parts. 4 The student incorrectly The student incorrectly The student The student correctly represents division solves the numeric decomposes incorrectly decomposes and 4.NBT.6 using place value disks equation but shows in one place value or divides using the place and incorrectly solves some understanding of does not include the value disks and numerically. the place value chart remainder. provides a numerical and use of the answer of 142 with a algorithm. remainder of 1. 5 The student answers The student correctly The student correctly The student solves all fewer than two parts solves two parts, solves three parts with parts correctly using 4.NBT.6 correctly, showing little showing little evidence understanding of place any place value to no evidence of place of place value value strategies, or the strategy: value strategies. strategies. student correctly solves 1,772 all four parts but does 761 with a not show solid remainder of 4 evidence of place value 1,624 understanding. 731 6 The student incorrectly The student correctly The student answers all The student correctly answers two or more of answers two of four four parts correctly but answers all four parts 4.MD.3 the four parts, showing parts, showing some shows little reasoning showing solid evidence 4.OA.1 little to no reasoning. reasoning. in Part (a), or the of place value student answers three understanding: 4.OA.2 of four parts correctly 4.OA.3 1,638 square showing solid reasoning meters of flooring 4.NBT.5 and understanding (estimate 40 × 40 4.NBT.6 mathematically. = 1,600 square m). It is a reasonable because the answer and estimate have a difference of only 38 square meters. 528 more small posters than large posters.
3 4.OA.3 4.NBT.5 4.NBT.6 The student incorrectly answers both parts and shows no reasoning. The student correctly answers one part and shows little reasoning. The student answers one part correctly but shows solid reasoning in both problems, or the student shows some reasoning with correct answers for both parts. The student correctly answers using any place value strategy: 400 Each teacher received 72 pencils. 4 4.NBT.6 The student incorrectly represents division using place value disks and incorrectly solves numerically. The student incorrectly solves the numeric equation but shows some understanding of the place value chart and use of the algorithm. The student decomposes incorrectly in one place value or does not include the remainder. The student correctly decomposes and divides using the place value disks and provides a numerical answer of 142 with a remainder of 1. 5 4.NBT.6 The student answers fewer than two parts correctly, showing little to no evidence of place value strategies. The student correctly solves two parts, showing little evidence of place value strategies. The student correctly solves three parts with understanding of place value strategies, or the student correctly solves all four parts but does not show solid evidence of place value understanding. The student solves all parts correctly using any place value strategy: 1,772 761 with a remainder of 4 1,624 731 6 4.MD.3 4.OA.1 4.OA.2 4.OA.3 4.NBT.5 4.NBT.6 The student incorrectly answers two or more of the four parts, showing little to no reasoning. The student correctly answers two of four parts, showing some reasoning. The student answers all four parts correctly but shows little reasoning in Part (a), or the student answers three of four parts correctly showing solid reasoning and understanding mathematically. The student correctly answers all four parts showing solid evidence of place value understanding: 1,638 square meters of flooring (estimate 40 × 40 = 1,600 square m). It is a reasonable because the answer and estimate have a difference of only 38 square meters. 528 more small posters than large posters.