Count groups several equivalent ways

2.OA.C.43.OA.A.1 · take · grade 3

Archetype: Decompose a Number into Parts and Factors · step in a 4-type progression

Find the choice that is not a correct way to count how many beads there are in all. Write its letter.

(A) Count with $7+7+7+7$ .
(B) Count with $7\times4$ .
(C) Count with $4\times7$ .
(D) Count by adding $7\times2$ four times.

The beads are arranged in $4$ rows with $7$ beads in each row. The round, colored beads form an array that is $7$ across and $4$ down, $28$ beads in all.

Show solution

Understand

There are 28 beads in an array of 4 rows with 7 beads in each row. Among four offered counting methods, we must find the one that does NOT give the correct total of 28.

Givens

The beads form an array 7 across and 4 down, 28 beads in all.
(A) 7 + 7 + 7 + 7; (B) 7 x 4; (C) 4 x 7; (D) add 7 x 2 four times.

Unknowns

Which lettered choice does not count to 28.

Constraints

A correct method must total exactly 28.

Plan

#1 Draw a Diagram · also uses: #2 Make a Systematic List

Picture the 7-by-4 array of beads, then evaluate each choice against the array and list the totals to spot the one that is not 28.

Execute

#1 Draw a Diagram 3.OA.A.1

The beads are 4 rows of 7, so the correct total is 7 x 4 = 28. Any correct method must match this.

7 \times 4 = 28

An array of equal rows is exactly the product of rows and columns, so 4 rows of 7 is 28.

#2 Make a Systematic List 2.OA.C.4

(A) 7 + 7 + 7 + 7 = 28, the four rows added. (B) 7 x 4 = 28 and (C) 4 x 7 = 28 are the same product in two orders. All three give 28.

7+7+7+7 = 28,\ 7\times4 = 28,\ 4\times7 = 28

Adding equal rows and multiplying rows by columns are just two views of the same array total.

#2 Make a Systematic List 3.OA.A.1

Choice (D) adds 7 x 2 four times: 7 x 2 = 14, and 14 + 14 + 14 + 14 = 56, which is not 28. So (D) is the incorrect method.

7 \times 2 = 14,\quad 14 \times 4 = 56 \neq 28

Each row holds 7 beads, not 7 x 2 = 14, so D double-counts every row and overshoots to 56.

Answer: D

Review

Choices A, B, C all equal 28, the bead total, while D equals 56, twice as many, so D is clearly the wrong way to count.

You could split the array into halves: two rows give 7 x 2 = 14, and the whole array is 14 + 14 = 28 (adding 7 x 2 only twice), showing that adding it four times is too many.

Standards · min grade 3

2.OA.C.4 Use addition to find the total number of objects in rectangular arrays — Adding equal rows of 7 to total the bead array.
3.OA.A.1 Interpret products of whole numbers as total number of objects in groups — Reading 4 rows of 7 as the product 28 and judging each counting method.

💡 Equal rows can be added or multiplied to the same total -- but counting each row as double overshoots, easy Grade 3 array sense!