NCERT Solutions Ganita Prakash (Part 1) Chapter 6 .4 This Way or That Way, All Ways Lead to the Bay — Figure it Out

Book page 1546 Updated on2026-09-05

Q1.

Compute these products using the suggested identity. (i) 46² using Identity 1A for (a + b)² (ii) 397 × 403 using Identity 1C for (a + b) (a – b) (iii) 91² using Identity 1B for (a – b)² (iv) 43 × 45 using Identity 1C for (a + b) (a – b)

Answer

(i) 46² = (40 + 6)²
= 40² + 2 × 40 × 6 + 6²
= 1600 + 480 + 36 = 2116

(ii) 397 × 403 = (400 – 3)(400 + 3)
= 400² – 3²
= 160000 – 9 = 159991

(iii) 91² = (100 – 9)²
= 100² + 9² – 2 × 100 × 9
= 10000 + 81 – 1800 = 8281

(iv) 43 × 45 = (44 – 1)(44 + 1)
= 44² – 1²
= 1936 – 1 = 1935

Tip: for (iv) the midpoint of 43 and 45 is 44. Identity 1C works for any two numbers by taking a as their midpoint and b as half their difference — but it only saves effort when a² is easy.

Q2.

Use either a suitable identity or the distributive property to find each of the following products. (i) (p – 1) (p + 11) (ii) (3a – 9b) (3a + 9b) (iii) –(2y + 5) (3y + 4) (iv) (6x + 5y)² (v) (2x – 1⁄2)² (vi) (7p) × (3r) × (p + 2)

Answer

(i) (p – 1)(p + 11) = p² + 11p – p – 11
= p² + 10p – 11 (distributive property)

(ii) (3a – 9b)(3a + 9b) = (3a)² – (9b)²
= 9a² – 81b² (Identity 1C)

(iii) (2y + 5)(3y + 4) = 6y² + 8y + 15y + 20 = 6y² + 23y + 20
So –(2y + 5)(3y + 4) = –6y² – 23y – 20

(iv) (6x + 5y)² = (6x)² + 2(6x)(5y) + (5y)²
= 36x² + 60xy + 25y² (Identity 1A)

(v) (2x – 1⁄2)² = (2x)² + (1⁄2)² – 2 × 2x × 1⁄2
= 4x² – 2x + 1⁄4 (Identity 1B)

(vi) (7p) × (3r) × (p + 2) = 21pr (p + 2)
= 21p²r + 42pr

Spot checks at p = x = y = a = b = r = 1: (i) 0 × 12 = 0 and 1 + 10 – 11 = 0 ✓; (v) at x = 1: (1.5)² = 2.25 and 4 – 2 + 0.25 = 2.25 ✓

Tip: in (vi) multiply the two single terms first. 7p × 3r = 21pr is one term, and only then does the bracket need distributing.

Q3.

For each statement identify the appropriate algebraic expression(s). (i) Two more than a square number. 2 + s, (s + 2)², s² + 2, s² + 4, 2s², 2²s (ii) The sum of the squares of two consecutive numbers. m² + n², (m + n)², m² + 1, m² + (m + 1)², m² + (m – 1)², (m + (m + 1))², (2m)² + (2m + 1)²

Answer

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

Expression	What it really says
2 + s	two more than s — but s need not be a square
(s + 2)²	the square of “two more than s”
s² + 2	two more than the square number s² ✓
s² + 4	four more than a square number
2s²	twice a square number
2²s	4s, four times s

(ii) The sum of the squares of two consecutive numbers: m² + (m + 1)²

Two other listed expressions also describe such a sum, and it is worth saying why:

m² + (m – 1)² — the consecutive pair is (m – 1) and m. As m runs over all numbers this gives exactly the same family of values, so it is also acceptable.
(2m)² + (2m + 1)² — 2m and 2m + 1 are consecutive, so this is a sum of squares of two consecutive numbers. But it only reaches the pairs that start at an even number, so it is less general.
m² + n² is any two squares; (m + n)² and (m + (m + 1))² are squares of sums, not sums of squares; m² + 1 is a square plus 1.

Check it yourself: at m = 4, m² + (m + 1)² = 16 + 25 = 41 = 4² + 5² ✓, while (m + (m + 1))² = 81, which is 9² — the square of the sum, a different thing entirely.

Q4.

Consider any 2 by 2 square of numbers in a calendar, as shown in the figure. Find products of numbers lying along each diagonal — 4 × 12 = 48, 5 × 11 = 55. Do this for the other 2 by 2 squares. What do you observe about the diagonal products? Explain why this happens. Hint: Label the numbers in each 2 by 2 square as a, (a + 1), a + 7, (a + 8).

Answer

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.

2 × 2 square	a × (a + 8)	(a + 1) × (a + 7)	Difference
4, 5 / 11, 12	4 × 12 = 48	5 × 11 = 55	7
3, 4 / 10, 11	3 × 11 = 33	4 × 10 = 40	7
16, 17 / 23, 24	16 × 24 = 384	17 × 23 = 391	7
9, 10 / 16, 17	9 × 17 = 153	10 × 16 = 160	7

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).

a (a + 8) = a² + 8a
(a + 1)(a + 7) = a² + 7a + a + 7 = a² + 8a + 7
Difference = (a² + 8a + 7) – (a² + 8a) = 7

Why it happens: both products contain the same a² + 8a, since 1 + 7 = 8 = 0 + 8. The only leftover is the corner product 1 × 7 = 7, which the anti-diagonal has and the main diagonal does not. The 7 comes straight from the length of the week.

Try This: take a 2 by 2 square in a multiplication grid instead. There the labels are pq, p(q + 1), (p + 1)q, (p + 1)(q + 1), and the two diagonal products come out equal. The difference is that a calendar adds, while a grid multiplies.

Q5.

Verify which of the following statements are true. (i) (k + 1) (k + 2) – (k + 3) is always 2. (ii) (2q + 1) (2q – 3) is a multiple of 4. (iii) Squares of even numbers are multiples of 4, and squares of odd numbers are 1 more than multiples of 8. (iv) (6n + 2)² – (4n + 3)² is 5 less than a square number.

Answer

(i) FALSE.

(k + 1)(k + 2) – (k + 3) = k² + 3k + 2 – k – 3
= k² + 2k – 1

It equals 2 only when k² + 2k – 3 = 0, i.e. k = 1 (or k = –3). At k = 2 it is 7, at k = 3 it is 14.

(ii) FALSE.

(2q + 1)(2q – 3) = 4q² – 4q – 3
= 4(q² – q) – 3

This is 3 less than a multiple of 4, so it is never a multiple of 4. In fact 2q + 1 and 2q – 3 are both odd, and an odd × odd is odd — it cannot even be a multiple of 2. Check: q = 1 gives –3, q = 2 gives 5, q = 3 gives 21.

(iii) TRUE.

Even number = 2n: (2n)² = 4n² — a multiple of 4 ✓
Odd number = 2n + 1: (2n + 1)² = 4n² + 4n + 1
= 4n(n + 1) + 1

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.

(6n + 2)² – (4n + 3)² = (36n² + 24n + 4) – (16n² + 24n + 9)
= 20n² – 5

For the statement to hold, 20n² would have to be a square number. But 20n² = 4 × 5n², and in 5n² the prime 5 appears an odd number of times, so 20n² is never a perfect square unless n = 0. At n = 1 the value is 15, and 15 + 5 = 20 is not a square. So the statement is true only in the trivial case n = 0.

Tip: one example that works never proves a claim — (i) is true at k = 1 and false everywhere else. Expanding is what settles it.

Q6.

A number leaves a remainder of 3 when divided by 7, and another number leaves a remainder of 5 when divided by 7. What is the remainder when their sum, difference, and product are divided by 7?

Answer

Write the two numbers in the form the condition gives:

n₁ = 7a + 3, n₂ = 7b + 5

Sum

n₁ + n₂ = 7a + 7b + 8 = 7(a + b + 1) + 1
Remainder = 1

Difference

n₁ – n₂ = 7(a – b) + 3 – 5 = 7(a – b) – 2
= 7(a – b – 1) + 5
Remainder = 5

(Taken the other way round, n₂ – n₁ = 7(b – a) + 2 leaves remainder 2.)

Product

n₁ n₂ = (7a + 3)(7b + 5)
= 49ab + 35a + 21b + 15
= 7(7ab + 5a + 3b + 2) + 1
Remainder = 1

Check with n₁ = 10, n₂ = 12: sum 22 = 7 × 3 + 1 ✓; difference –2 ≡ 5 (mod 7), and 12 – 10 = 2 ✓; product 120 = 7 × 17 + 1 ✓

Why it happens: every term with a factor 7 disappears on division. What is left is decided entirely by the remainders 3 and 5 — 3 + 5 = 8 gives remainder 1, and 3 × 5 = 15 gives remainder 1. This is why remainders can be added and multiplied on their own.

Q7.

Choose three consecutive numbers, square the middle one, and subtract the product of the other two. Repeat the same with other sets of numbers. What pattern do you notice? How do we write this as an algebraic equation? Expand both sides of the equation to check that it is a true identity.

Answer

Three numbers	Middle squared	Product of the others	Difference
4, 5, 6	25	24	1
9, 10, 11	100	99	1
19, 20, 21	400	399	1
–3, –2, –1	4	3	1

Pattern: the answer is always 1.

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

n² – (n – 1)(n + 1) = 1

Expanding both sides:

(n – 1)(n + 1) = n² – 1 (Identity 1C)
n² – (n² – 1) = n² – n² + 1 = 1 ✓

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

Why it happens: the outer two numbers are the same distance 1 from the middle, so their product is n² – 1² by Identity 1C. It always falls exactly 1 short of n². Move the two numbers to distance d instead, and the gap becomes d².

Q8.

What is the algebraic expression describing the following steps — add any two numbers. Multiply this by half of the sum of the two numbers? Prove that this result will be half of the square of the sum of the two numbers.

Answer

Let the two numbers be a and b.

Step 1 — add them: a + b
Step 2 — multiply by half of that sum: (a + b) × 1⁄2 (a + b)
Expression: 1⁄2 (a + b)²

Proof.

(a + b) × 1⁄2 (a + b) = 1⁄2 × (a + b)(a + b)
= 1⁄2 (a + b)²
= 1⁄2 (a² + 2ab + b²)
= a²⁄2 + ab + b²⁄2

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

Check at a = 3, b = 5: sum = 8, half the sum = 4, product = 32; and 1⁄2 × 8² = 32 ✓

Why it happens: the two steps multiply the sum by itself and by 1⁄2. Because multiplication can be regrouped in any order, that is the same as squaring the sum first and halving afterwards.

Q9.

Which is larger? Find out without fully computing the product. (i) 14 × 26 or 16 × 24 (ii) 25 × 75 or 26 × 74

Answer

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

14 × 26 = (16 – 2)(24 + 2)
= 16 × 24 + 16 × 2 – 2 × 24 – 2 × 2
= 16 × 24 + 32 – 48 – 4
= 16 × 24 – 20

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

(ii) 26 × 74 is larger.

25 × 75 = (26 – 1)(74 + 1)
= 26 × 74 + 26 – 74 – 1
= 26 × 74 – 49

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

Why it happens: both pairs have the same total (14 + 26 = 16 + 24 = 40; 25 + 75 = 26 + 74 = 100). Among pairs with a fixed sum, the product is largest when the two numbers are closest together — Identity 1C makes this exact: if the pair is (c – d, c + d) then the product is c² – d², which shrinks as d grows.

Check it yourself: for sum 40, the products are 20 × 20 = 400, 19 × 21 = 399, 18 × 22 = 396, 16 × 24 = 384, 14 × 26 = 364 — falling away by 1, 4, 16, 36, the squares of the gap from the middle.

Q10.

A tiny park is coming up in Dhauli. The plan is shown in the figure. The two square plots, each of area g² sq. ft., will have a green cover. All the remaining area is a walking path w ft. wide that needs to be tiled. Write an expression for the area that needs to be tiled.

Answer

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.

Length = w + g + 2w + g + w = 2g + 4w ft
Breadth = w + g + w = g + 2w ft

The park: two g × g green plots set in a path w ft wide, with 2w between them.

Area of the whole park = (2g + 4w)(g + 2w)
= 2g² + 4gw + 4gw + 8w²
= 2g² + 8gw + 8w²

Green cover = 2g²
Area to be tiled = 2g² + 8gw + 8w² – 2g²
= 8gw + 8w² = 8w(g + w) sq. ft.

Check with numbers: g = 10, w = 2 → park is 28 × 14 = 392 sq. ft., green is 200 sq. ft., path is 192 sq. ft.; and 8 × 2 × 12 = 192 ✓

Why it happens: the park is exactly two squares of side (g + 2w) placed side by side, so its area is 2(g + 2w)². The tiled area is therefore 2[(g + 2w)² – g²], and Identity 1C turns that bracket into (2g + 2w)(2w) — giving 8w(g + w) at once.

Q11.

For each pattern shown below, (i) Draw the next figure in the sequence. (ii) How many basic units are there in Step 10? (iii) Write an expression to describe the number of basic units in Step y.

Answer

Pattern (a) — the yellow S-shaped figures. Each figure is a square of side (y + 2) rearranged: a block of y rows and (y + 2) columns, with the top row pulled up into a vertical arm on the right and the bottom row pulled down into a vertical arm on the left.

Step y	Block	Two arms	Total
1	1 × 3 = 3	3 + 3	9 = 3²
2	2 × 4 = 8	4 + 4	16 = 4²
3	3 × 5 = 15	5 + 5	25 = 5²
4 (next figure)	4 × 6 = 24	6 + 6	36 = 6²

Number of basic units at Step y = y(y + 2) + 2(y + 2)
= (y + 2)(y + 2) = (y + 2)²
Step 10: (10 + 2)² = 144 units

Step 4 of pattern (a) in outline: a 4 × 6 block with an arm of 6 above and an arm of 6 below, 36 units in all.

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.

Step y	Block	Extra row	Total
1	2² = 4	1	5
2	3² = 9	2	11
3	4² = 16	3	19
4 (next figure)	5² = 25	4	29

Number of basic units at Step y = (y + 1)² + y = y² + 3y + 1
Step 10: 11² + 10 = 121 + 10 = 131 units

Why it happens: in both patterns the figure is “a square, adjusted”. Naming the square first — (y + 2)² in (a), (y + 1)² in (b) — turns a picture-counting problem into a one-line expression, which is the whole point of Section 6.4.

Tip: as printed, the Step 3 figure of pattern (a) shows six squares in each arm rather than five, which would give 27 instead of 25. Steps 1 and 2 both fit (y + 2)² exactly, so read the arms as (y + 2) units long — the answer (y + 2)² is the one the pattern intends.

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