In-text — Fast Multiplication Using the Distributive Property — We Distribute Yet Things Multiply

1. 13 marksGanita Prakash Cl-8 Part 1, in-text, page 144

   Evaluate (i) 94 × 11, (ii) 495 × 11, (iii) 3279 × 11, (iv) 4791256 × 11 in a single line.

   Hint. 11 = 10 + 1, so the number is added to itself shifted one place — which means adding neighbouring digits.

   Why the shortcut works. N × 11 = N × (10 + 1) = N0 + N, so the number is added to itself shifted one place left. Lining them up, each digit of the answer is the sum of a digit and its neighbour, with the outer digits unchanged. For a four-digit number dcba:

   d c b a 0

   •   d c b a = d (c+d) (b+c) (a+b) a

   The rule: write the last digit, then add each pair of neighbouring digits working right to left, then write the first digit — carrying wherever a sum reaches 10.

   (i) 94 × 11 Outer digits 9 and 4; middle 9 + 4 = 13, so write 3 and carry 1 into the leading digit: 9+1 = 10, giving 1034. (Check: 940 + 94 = 1034 ✓)

   (ii) 495 × 11 Last digit 5. Then 9 + 5 = 14 → write 4, carry 1. Then 4 + 9 + 1 = 14 → write 4, carry 1. Leading 4 + 1 = 5. = 5445 (Check: 4950 + 495 = 5445 ✓)

   (iii) 3279 × 11 Last digit 9. Then 7 + 9 = 16 → 6, carry 1. Then 2 + 7 + 1 = 10 → 0, carry 1. Then 3 + 2 + 1 = 6. Leading 3. = 36069 (Check: 32790 + 3279 = 36069 ✓)

   (iv) 4791256 × 11 Working right to left: last digit 6; 5+6 = 11 → 1 carry 1; 2+5+1 = 8; 1+2 = 3; 9+1 = 10 → 0 carry 1; 7+9+1 = 17 → 7 carry 1; 4+7+1 = 12 → 2 carry 1; leading 4+1 = 5. = 52703816 (Check: 47912560 + 4791256 = 52703816 ✓)

   ✦ (i) 1034 (ii) 5445 (iii) 36069 (iv) 52703816 — each obtained by adding neighbouring digits, since 11 = 10 + 1.

2. 24 marksGanita Prakash Cl-8 Part 1, in-text, page 144

   What could be a general rule to multiply a number by 101 and write the product in one line? Extend this rule for 1001, 10001, … Use it to find (i) 89 × 101, (ii) 949 × 101, (iii) 265831 × 1001, (iv) 1111 × 1001, (v) 9734 × 99 and (vi) 23478 × 999.

   Hint. 101 = 100 + 1 and 1001 = 1000 + 1; but 99 = 100 − 1 and 999 = 1000 − 1.

   The general rule. For a number N, N × 101 = N × (100 + 1) = N00 + N — add N to itself shifted two places. N × 1001 = N000 + N — shifted three places. N × 10001 = N0000 + N — shifted four places.

   In each case the shift equals the number of zeros in the multiplier. When the shift is at least as large as the number of digits in N, the two copies do not overlap at all, and the answer is simply N written twice.

   For 99 and 999 the trick flips to subtraction: N × 99 = N × (100 − 1) = N00 − N, and N × 999 = N000 − N.

   (i) 89 × 101 = 8900 + 89 = 8989 (The digits overlap nowhere, so 89 simply repeats.)

   (ii) 949 × 101 = 94900 + 949 = 95849 (Here the number has 3 digits but the shift is only 2, so the copies overlap by one place and a carry appears.)

   (iii) 265831 × 1001 = 265831000 + 265831 = 266096831

   (iv) 1111 × 1001 = 1111000 + 1111 = 1112111

   (v) 9734 × 99 = 973400 − 9734 = 963666

   (vi) 23478 × 999 = 23478000 − 23478 = 23454522

   The neat special case: when N has no more digits than the shift, N × 1001 is just N written twice — for example 265 × 1001 = 265265. This is why 1001 = 7 × 11 × 13 gives the well-known divisibility trick for 7, 11 and 13 on six-digit repeats.

   ✦ (i) 8989 (ii) 95849 (iii) 266096831 (iv) 1112111 (v) 963666 (vi) 23454522 — using N × 10…01 = shifted N + N, and N × 9…9 = shifted N − N.

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.