The distributive property, and using it to compute faster

What to assume they know

• Every expression can be rewritten as a sum of terms — an expression re-set as a sum of terms
• Commutative and associative: why order and grouping are free — the terms of a sum may be reordered and regrouped
• Removing a bracket after a minus sign flips every term inside — removing a bracket that stands behind a minus
• Multiplication as repeated addition, and as rows-by-columns
• Two-digit by two-digit multiplication, well enough to feel which ones are hard

What they should be able to do

• State that multiplying a total is the same as multiplying each part and adding, and that the same holds for a difference
• Justify that statement by counting one arrangement in two ways, rather than by citing a rule
• Write the expression a situation forces when a repeated cost has two components
• Distinguish an expression where the property applies from one where it does not, such as a product with a lone number added on afterwards
• Split a factor into a sum or a difference so that a product becomes easier, and choose the split that helps
• Use one known product to obtain a neighbouring product without multiplying again
• Name the property, and say where in the chapter that name is actually printed

Where it usually goes wrong

• "Distributive means you may always multiply into a bracket, whatever is around it." The factor has to be multiplying the whole bracket. 5 × 4 + 3 has a lone 3 added on afterwards, and the chapter prints the inequality for exactly this reason.
• "It only works when the bracket contains a plus." The chapter runs the difference case straight after the addition case, with its own picture.
• "When the bracket holds a difference, both parts get added." A student who writes the second product with a plus sign will be caught by exercise 2(c) on p.42, which sets a bracketed difference against a sum of two products.
• "The distributive property is a rule about brackets." It is a statement about counting one collection two ways. The parade picture makes that visible; the rule follows from it.
• "Splitting a factor is always the fast route." It is fast when one piece is round. Splitting 47 as 23 and 24 helps nobody. The chapter's closing question asks the student to decide which products qualify.
• "Because the two forms are equal, they cost the same to compute." The whole of the tinkering section rests on their not costing the same — that is the reason to rewrite.
• "(4 + 3) × 5 must be worked bracket-first, so it is a different calculation." Same value, different route. Both are legitimate; one may be quicker.

Questions to check understanding

• Fill blanks and operation boxes so that a bracketed product and a sum of products agree, across sixteen variants including subtraction cases (Part I, p.41 and p.42)
• Put <, > or = between a bracketed expression and an expanded one by reasoning rather than by evaluating (Part I, p.42, and again p.43)
• Produce several different ways of reaching a stated value using a factor and a bracketed sum (Part I, p.42)
• Total the numbers in a pictured arrangement in at least two ways and describe each way as an expression (Part I, p.42)
• Choose, from a list of candidate expressions, the ones that describe a stated situation (Part I, p.43)
• Identify which of several expressions equal a given one without computing, using terms and bracket removal (Part I, p.44)

Examples worth working on the board

Two of them carry a wrong candidate that the chapter prints on purpose, and one carries a printed error — see the last item, which is not optional reading.

• Example 15 — the hotel bill (Part I, §2.2, pp.38–39). Lhamo and Norbu each order a vegetable cutlet at ₹43 and a rasgulla at ₹24. The chapter first offers 2 × 43 + 24, splits it into its two terms, and rejects it: that expression adds one rasgulla to two cutlets. The bracketed form is 2 × (43 + 24), and paying for two of each gives 2 × 43 + 2 × 24. A follow-up printed on p.39 adds a third friend, Sangmu, ordering the same.
• Example 16 — the Republic Day parade (p.39). Scouts march in 4 rows of 5; guides in 3 rows of 5. The total is written as 4 × 5 + 3 × 5, and also as (4 + 3) × 5 by counting the rows first. The chapter prints the term split 4 × 5 and 3 × 5 with the values 20 and 15 under them.
• The near miss (p.40). The chapter asserts that 5 × 4 + 3 and 5 × (4 + 3) are not equal and asks the reader to explain why. It then asks whether 5 × (4 + 3), 5 × (3 + 4) and (3 + 4) × 5 all agree. The first question is the load-bearing one — it draws the boundary of the property.
• The general argument, addition case (p.40). Ten copies of 98 followed by three more copies, written out in full with braces naming the two runs, so that the total is visibly thirteen copies. The chapter concludes with the two bracketed forms.
• The general argument, subtraction case (p.40). Fourteen copies of 10 with the last six struck through, leaving eight, and the two bracketed forms that follow.
• The property in words (p.40). A single bold line: taking a multiple of a total, or of a difference, gives the same answer as taking the multiples and then totalling or differencing them. Paraphrase it; do not lift it.
• Example 17 — one product from its neighbour (p.41). Given that 53 × 18 is 954, find 63 × 18. The chapter splits 63 as 53 and 10, and adds the two products.
• Example 18 — splitting downward (p.41). Evaluate 97 × 25 by writing 97 as 100 less 3, giving 100 × 25 – 3 × 25. The chapter does not print the value — it says to find it.
• Three products to practise on (p.41): 95 × 8, 104 × 15, 49 × 50. Each invites a different split. The chapter then asks whether this beats the ordinary multiplication procedure, and which other products would suit it — a genuinely open question and a good closing beat.
• Two counting pictures (p.42, exercise 4). A three-by-three arrangement mixing squares marked 4 with circles marked 8 — five 4s and four 8s; and a four-by-four arrangement of circles marked 5 and 6, eight of each, laid out so that the colours alternate in a symmetric pattern. Each asks for the total by two separate routes, which is this topic's whole point applied backwards. Counts read from p.42.
• A printed error. On p.40, in the line that rewrites the property with the factors swapped, the middle product is set as 9 × 83 where the argument requires 98 × 3. As printed the equality is false: the intended statement is that ten 98s and three 98s make thirteen 98s. This was confirmed against the printed page p.40 and reading the printed line, so it is the book's own typesetting slip and not an artefact of text extraction. Use 98 × 10 + 98 × 3 in the explanation and, if the chapter is shown, do not reproduce the printed line.

Figures to have open

• The two thali plates (Part I, p.39). Two identical plates, each with a cutlet compartment labelled ₹43 and a rasgulla compartment labelled ₹24, with one brace grouping the plates and another grouping the like items across plates. Redraw. The double bracing is the figure's argument, and it was checked against p.39.
• The parade array (Part I, p.39). Thirty-five marching figures in five columns: three rows of one colour above four rows of another, each row-group braced and labelled with its product, and a single outer brace labelled with the bracketed form. Note for the artist: the printed figure puts the three-row group on top, although the text names the four-row group first. Counts and layout confirmed against the printed p.39.
• The repeated-addition strips (Part I, p.40). One line of copies with two braces beneath naming the two runs; and a second line with the surplus copies struck through. Redraw both — they carry the generalisation and are the only reason it is not an assertion.
• The two counting pictures (Part I, p.42). A three-by-three mix of squares and circles, and a four-by-four colour-alternating grid. Redraw with the numerals legible; the arrangement is what suggests the two different groupings.
• No figure accompanies Examples 17 and 18 in the book; a split-factor schematic for sections 9 and 10 is added here. Confirmed against the printed p.41.

Where this sits in the book

• NCERT Ganita Prakash, Class 7, Part I, printed Chapter 2 "Arithmetic Expressions", §2.2 "Reading and Evaluating Complex Expressions". Two unnumbered subheadings carry this topic: "Removing Brackets — II", pp.38–40, and "Tinker the Terms II", pp.40–44. Example 15 begins on p.38 and is resolved on p.39; Example 16 is on p.39; the general argument and the bold statement of the property are on p.40; Examples 17 and 18 are on p.41; the exercise sets run pp.41–44.
• Chapter SUMMARY, Part I, p.44 — the sixth bullet, and the only place in the chapter where the property is called by name.
• Forward pointer: Part II, printed Chapter 2 "Operations with Integers", §2.2, where the property is restated over the integers; and Part I, printed Chapter 4 §4.2, where it is carried into letter-numbers.