NCERT Solutions for Class 8 Maths Chapter 6 We Distribute Yet Things Multiply (PDF) – 2027

The product increases by 27, the second number.

Then the product increases by 23, the first number.

The product increases by 27 + 23 + 1 = 51.

Yes. Write the two numbers as a and b. Then

Putting 23 for a and 27 for b gives exactly the three increases above: 23, 27 and 51.

Treat the whole bracket (a + 1) as a single term and distribute the second bracket over it.

So the increase in the product ab is a + b + 1.

The same answer — which is the point of trying it.

There is a change, and it is b – a – 1.

With a = 23, b = 27:

No. The change is b – a – 1, so the product decreases whenever b ≤ a.

Nothing changes — the expressions stay correct, because integers also obey distributivity: x (y + z) = xy + xz for any integers x, y, z.

And for (a + 1)(b – 1) = ab + b – a – 1 with a = –5, b = 8: –4 × 7 = –28, while –40 + 8 + 5 – 1 = –28 ✓

The two numbers become a + m and b + n, and the product changes by an + bm + mn.

Subtracting the old product ab leaves the change: an + bm + mn.

Yes — a decrease is an increase by a negative number, so no new identity is needed.

This is exactly the expression found earlier by direct expansion.

(i) Take m = –2, n = 3.

Direct check, keeping the subtraction: (a – 2)(b + 3) = a(b + 3) – 2(b + 3) = ab + 3a – 2b – 6 ✓

(ii) Take m = –3, n = –4.

Direct check: (a – 3)(b – 4) = a(b – 4) – 3(b – 4) = ab – 4a – 3b + 12 ✓

Multiply each term of the first bracket by each term of the second, letting the sign rules do the work.

Distributivity is not limited to two terms inside a bracket — every term gets multiplied.

Now simplify each term using exponent notation:

No. Two terms can be combined into one only when they are like terms — that is, when they have exactly the same letter-numbers.

Since ba = ab, the middle two are like terms:

Simplify each term:

Now collect the like terms:

In the frame shown, the middle entry is 4 × 6 = 24, so p is the row number and q is the column number of the centre. The row above is p – 1 and the row below is p + 1; the column to the left is q – 1 and to the right is q + 1.

Each term of the first bracket multiplies each term of the second; the integer sign rules fix the signs.

Set the change to zero and solve.

So any pair in which the second number is 4 more than twice the first will work.

(i) Break up the second bracket:

(4ab and ab are like terms: 4ab + ab = 5ab.)

(ii)

(–12y and 7y are like terms: –12y + 7y = –5y.)

The pattern: the second bracket gains one term each time (powers of a going down, powers of b going up), and the products are a² – b², a³ – b³, a⁴ – b⁴. The next identity is therefore

Check by expanding:

Split 11 as 10 + 1 and distribute.

Rule. Write the units digit as it is. Then, moving left, write the sum of each pair of neighbouring digits. Finally write the leading digit. Carry as usual.

Worked line for 3874 × 11, right to left:

Yes — for exactly the same reason.

Multiplying by 100 shifts every digit two places, so in the addition each column pairs digits that are two places apart instead of adjacent.

Written as a column sum with dcba = 3874:

Read off the digits d = 3, c = 8, b = 7, a = 4 and apply the pattern d | c | (b + d) | (a + c) | b | a.

Check: 387400 + 3874 = 391274 ✓

By 101: copy the last two digits, then in each further place write the sum of the digit there and the digit two places to its right; finish with the leading two digits. In symbols, dcba × 101 = d | c | (b + d) | (a + c) | b | a.

By 1001: the shift is three places, so pair digits three apart:

By 10001: the shift is four places. For a four-digit dcba the two blocks do not overlap at all:

Yes, but 3600 + 25 is not enough — two rectangles are missing.

3600 + 300 + 300 + 25 = 4225 ✓

The cut is in a different place, but the four pieces still add up to 4225.

No, not always. The difference between the two is the cross term.

Identity 1A: (a + b)² = a² + 2ab + b².

Test (i) at m = 4: 7² = 49 and 16 + 24 + 9 = 49 ✓

Using the distributive property

Using the identity

Using Identity 1A with a = 3j and b = 2k:

Using the distributive property

Yes. Draw a square of side 55 sitting inside a square of side 60.

Start with 60² and take away the two 60 × 5 rectangles along two sides. That removes the little 5 × 5 corner twice, so add it back once.

Multiply the bracket by itself, using the distributive property.

A subtraction is an addition of the opposite, so Identity 1A already covers it.

The only two things used are (–b)² = b² and a × (–b) = –ab.

Draw a square of side a. Mark off a square of side (a – b) inside it, in one corner.

(i) (b – 6)²

(ii) (–2a + 3)² — read it as (3 – 2a)²

(iii) (7y – 3⁄4 z)²

Yes — and the two new squares are built from the sum and the difference of the pair.

Pattern: 2(a² + b²) = (a + b)² + (a – b)².

Yes — adding them proves the pattern in one line.

Collect like terms: a² + a² = 2a², b² + b² = 2b², and 2ab – 2ab = 0.

Read each line as a difference of two squares and look at the numbers on the right.

So the guess is a² – b² = (a + b)(a – b). Expanding settles it:

It is a true identity, so the pattern always holds. You had already met it in Figure it Out 5 (i).

Look for the number midway between the two factors.

Start with a square of side a and cut out a square of side b from one corner. The area left is a² – b², and it is an L-shape.

Cut the L into two rectangles and slide one against the other:

• Rectangle 1: a wide, (a – b) tall
• Rectangle 2: b wide, (a – b) tall

Placing Rectangle 2 beside Rectangle 1 makes a single rectangle of width a + b and height a – b.

It is Identity 1C rearranged. Start from

and add b² to both sides:

Neither — they are always equal.

Or expand both and compare:

The two expansions are the same expression, so the two are equal for every a and b.

Use Identity 1C backwards: a² – b² = (a + b)(a – b), so we need two factors of 100 whose product is 100.

They hold for all of them. Both patterns were proved by expanding brackets, and expanding uses only distributivity and commutativity — rules that integers and fractions obey just as counting numbers do.

Pattern 1: 2(a² + b²) = (a + b)² + (a – b)²

Pattern 2: a² – b² = (a + b)(a – b)

Mistake. The factor –3p was not multiplied into the bracket at all — its sign was simply spread over the two terms.

Mistake. The 2 and the 3 were multiplied into the first term of each bracket but not the second.

Mistake. y + 2(y + 2) is a sum, not a product; “y” and “2” were wrongly gathered into a factor (y + 2).

Mistake. The cross term is missing — squaring a sum is not squaring each part.

No mistake — this is correct.

Mistake. The distributive property is multiplication over addition. Inside the bracket here there is a product, not a sum, so nothing may be distributed.

No mistake — this is correct.

Mistake. w² and w are unlike terms, so their coefficients cannot be added.

Mistake. 2a³ and 3a³ were combined correctly, but 6a²b and 6ab² are unlike terms and were wrongly merged into 12a²b².

No mistake — this is correct, and it is a model of how to set the work out.

Mistake. Only the first-with-first and last-with-last products were taken; the two cross products are missing.

No mistake — this is correct. Check by expanding again:

Counting the printed figures: Step 1 has 3 circles, Step 2 has 8, Step 3 has 15.

The next figure (Step 4) is a 5 by 5 block of circles with one corner circle missing — 24 circles.

Expression for Step k. Each figure is a (k + 1) by (k + 1) square with one circle removed:

Each figure is a square frame of tiles — a large square with a smaller square hole cut out of the middle.

The counts go up by 4 each time: 8, 12, 16, …

Several ways of seeing the frame, all landing on the same expression.

Two students see it two ways, and both are right.

Tadang’s method. The whole figure is a square of side (m + n), and the four rectangles each measure m by n.

Yusuf’s method. The shaded part is itself a square. Along the top edge the big side m + n is covered by one rectangle’s short side m followed by its long side n; the shaded square starts where that m ends and stops where the next rectangle’s m begins, so its side is (m + n) – m – m = n – m.

The two expansions are the same expression, so the two methods agree — as they must, since they measure one region.

Each of the three rectangles measures x by y. The slanting region lies between the top and bottom bars, on either side of the middle bar.

Anusha’s method. ABCD is the square of side x lying between the two bars; EFGH is the middle bar, x tall and y wide.

All three reduce to x² – xy, so they are equivalent.

Method 1 — measure the leftover rectangle directly.

Method 2 — whole rectangle minus the L. The L is a p × r arm plus an (s – r) × r foot.

(i) Two more than a square number: s² + 2

(ii) The sum of the squares of two consecutive numbers: m² + (m + 1)² (also m² + (m – 1)², where the pair is m – 1 and m).

Observation: in every 2 by 2 square of the calendar, the two diagonal products differ by exactly 7. The anti-diagonal product (top-right × bottom-left) is the larger one.

Why, in algebra. Label the square as the hint suggests: a, a + 1 on top; a + 7, a + 8 below (the number directly below a is a + 7, because a calendar week has 7 days).

(i) FALSE.

It is not always 2: at k = 2 it is 7, at k = 3 it is 14.

(ii) FALSE.

This is 3 less than a multiple of 4, so it is never a multiple of 4 (q = 1 gives –3, q = 2 gives 5, q = 3 gives 21).

(iii) TRUE.

Now n and n + 1 are consecutive, so one of them is even and n(n + 1) is even. Writing n(n + 1) = 2t gives 4 × 2t + 1 = 8t + 1 — exactly 1 more than a multiple of 8 ✓ (9 = 8 + 1, 25 = 24 + 1, 49 = 48 + 1, 81 = 80 + 1).

(iv) FALSE.

For the statement to hold, 20n² would have to be a square number. It is not: n = 1 gives 20, n = 2 gives 80, n = 3 gives 180. (It works only in the trivial case n = 0.)

Write the two numbers in the form the condition gives:

Sum

Difference

So the difference leaves remainder 5 when the first number (remainder 3) is the larger, e.g. 17 – 12 = 5; taken the other way round, n₂ – n₁ = 7(b – a) + 2 leaves remainder 2, e.g. 12 – 10 = 2.

Product

Pattern: the answer is always 1.

As an equation. Call the middle number n, so the three are n – 1, n, n + 1:

Expanding both sides:

Both sides are 1 for every n, so it is a true identity.

Let the two numbers be a and b.

Proof.

which is exactly half of (a + b)² — the square of the sum of the two numbers.

(i) 16 × 24 is larger. Compare them by moving from one to the other.

So 14 × 26 is 20 less than 16 × 24. (Indeed 364 and 384.)

(ii) 26 × 74 is larger.

So 25 × 75 is 49 less than 26 × 74. (Indeed 1875 and 1924.)

Read the plan across and down. The path runs all round the park and also between the two green squares, where two path widths meet.

Pattern (a): Step y is a y × (y + 2) block with an arm of (y + 2) units above it and another below it.

Pattern (b) — the blue squares. Each figure is a full (y + 1) by (y + 1) block with an extra part-row of y units underneath.

The 10-coin triangle has 4 rows: 1, 2, 3, 4 coins from the top. The upside-down triangle has 4, 3, 2, 1 from the top.

Lay the two triangles on top of each other: 7 coins already sit in the right places. Only 3 coins have to move.

• Move the single apex coin down to below the bottom row.
• Move the two end coins of the bottom row up beside the second row.

The 15-coin triangle has 5 rows. Superimposing the two triangles leaves 10 coins in place, so 5 coins must move — and 5 moves are enough.

Yes. Compare the “coins” column with the “moves” column above: the number of moves is always one third of the number of coins, rounded down.