In-text — Identities 1B and 1C — We Distribute Yet Things Multiply

1. 13 marksGanita Prakash Cl-8 Part 1, in-text, page 147

   Expand the following using both Identity 1B, (a − b)² = a² − 2ab + b², and by applying the distributive property: (i) (b − 6)² (ii) (−2a + 3)² (iii) (7y − ¾z)²

   Hint. Identity 1B needs you to name a and b first; the distributive route means writing the bracket twice and multiplying out.

   (i) (b − 6)² By Identity 1B with a = b and b = 6: = b² − 2(b)(6) + 6² = b² − 12b + 36 By distribution: (b − 6)(b − 6) = b² − 6b − 6b + 36 = b² − 12b + 36 ✓

   (ii) (−2a + 3)² By Identity 1B, reading it as (3 − 2a)² with first term 3 and second 2a: = 3² − 2(3)(2a) + (2a)² = 9 − 12a + 4a² = 4a² − 12a + 9 By distribution: (−2a + 3)(−2a + 3) = 4a² − 6a − 6a + 9 = 4a² − 12a + 9 ✓ Notice (−2a)(−2a) = +4a², since a negative times a negative is positive.

   (iii) (7y − ¾z)² By Identity 1B with a = 7y and b = ¾z: = (7y)² − 2(7y)(¾z) + (¾z)² = 49y² − (42/4)yz + (9/16)z² = 49y² − (21/2)yz + (9/16)z² By distribution: (7y − ¾z)(7y − ¾z) = 49y² − (21/4)yz − (21/4)yz + (9/16)z², and the two middle terms add to (21/2)yz ✓

   Why the book asks for both methods. The identity is faster, but it only helps if you correctly identify a and b — including any coefficient, so the second term in (iii) is ¾z and not just z. Expanding by distribution is slower but never depends on spotting the right form, which makes it the reliable check.

   The commonest error is dropping the middle term entirely and writing (b − 6)² = b² + 36. Always remember there are three terms, not two.

   ✦ (i) b² − 12b + 36 (ii) 4a² − 12a + 9 (iii) 49y² − (21/2)yz + (9/16)z²

2. 22 marksGanita Prakash Cl-8 Part 1, in-text, page 148

   Use Identity 1C, (a + b)(a − b) = a² − b², to calculate 98 × 102 and 45 × 55.

   Hint. Find the number exactly halfway between the two factors — that is your a.

   The method: find the midpoint of the two numbers. If they are equally spaced either side of some value a, then they can be written as a − b and a + b, and their product collapses to a² − b².

   98 × 102 The midpoint of 98 and 102 is 100, and each lies 2 away. 98 × 102 = (100 − 2)(100 + 2) = 100² − 2² = 10000 − 4 = 9996

   45 × 55 The midpoint of 45 and 55 is 50, and each lies 5 away. 45 × 55 = (50 − 5)(50 + 5) = 50² − 5² = 2500 − 25 = 2475

   Why this is worth doing. Both products are awkward by long multiplication but trivial once written this way, since squaring 100 or 50 is immediate. The identity turns a two-digit multiplication into a subtraction.

   When it applies: only when the two numbers are equidistant from a convenient round value — so their sum must be even, otherwise the midpoint is not a whole number. 98 + 102 = 200 ✓ and 45 + 55 = 100 ✓

   ✦ 98 × 102 = 100² − 2² = 9996, and 45 × 55 = 50² − 5² = 2475.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 1, Reprint 2026-27 (hegp106.pdf). The chapter develops the distributive property into the three standard identities — 1A (a+b)², 1B (a−b)², 1C (a+b)(a−b) — via a multiplication-grid model, then applies them to fast mental multiplication and to area/tile patterns. Every expansion here was independently re-expanded term by term and every numeric answer recomputed before comparison with the book's printed answer key. TWO NOTES: (1) the twelve 'Mind the Mistake, Mend the Mistake' items on page 150 have NO answers in the printed key — each has been worked out from first principles here, including identifying which four of the twelve are in fact already correct, and this is stated openly in the solution; (2) the circle-pattern activity in §6.4 ('This Way or That Way') is omitted because the circle counts cannot be recovered from the text without the printed figure.. Questions are referenced from the NCERT textbook for identification.