Chapter 1 exercise answers: A Square and A Cube

Open NCERT p. 10Matches NCERT’s answer

1. A perfect square's last digit can only be 0, 1, 4, 5, 6 or 9.
2. 2032 ends in 2, 2048 ends in 8, and 1027 ends in 7 — none of these last digits is possible for a perfect square, so all three are ruled out.
3. 1089 ends in 9, which is allowed, so check it directly: 33² = 1089.
4. So 1089 is a perfect square, but 2032, 2048 and 1027 are not.

Answer(i) 2032, (ii) 2048 and (iii) 1027 are not perfect squares; (iv) 1089 is a perfect square (33²).

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Open NCERT p. 10Matches NCERT’s answer

1. The last digit of a square depends only on the last digit of the original number.
2. 64² = 4096 (ends in 6); 108² = 11664 (ends in 4); 292² = 85264 (ends in 4); 36² = 1296 (ends in 6).
3. So 108² and 292² are the ones ending in 4.

Answer108² = 11664 and 292² = 85264 both end in 4.

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Open NCERT p. 10Matches NCERT’s answer

1. Use (n + 1)² = n² + 2n + 1 with n = 125.
2. 2 × 125 + 1 = 251.
3. So 126² = 15625 + 251, which is option (iv).

Answer(iv) 15625 + 251, since 126² = 125² + 2×125 + 1.

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1. For a square, area = side², so side = √area.
2. √441 = 21, because 21² = 441.

Answer21 m

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Open NCERT p. 10Matches NCERT’s answer

1. Any such number must be a multiple of the LCM of 4, 9 and 10.
2. LCM(4, 9, 10) = 180 = 2² × 3² × 5.
3. For a perfect square every prime's power must be even. 5 has power 1 (odd), so multiply by one more 5.
4. 180 × 5 = 900 = 30², and no smaller multiple of 180 is a perfect square.

Answer900

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Open NCERT p. 10Matches NCERT’s answer

1. Prime factorise: 9408 = 2⁶ × 3 × 7².
2. For a perfect square every power must be even. 2⁶ and 7² are already fine, but 3 is alone.
3. Multiply by 3 to pair it up: 9408 × 3 = 28224 = 2⁶ × 3² × 7².
4. √28224 = 2³ × 3 × 7 = 168.

AnswerMultiply by 3; the product 28224 has square root 168.

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(i) 16 and 17

1. Between n² and (n + 1)² there are always exactly 2n whole numbers.
2. Here n = 16, so 2 × 16 = 32.

Answer32

(ii) 99 and 100

1. Here n = 99, so 2 × 99 = 198.

Answer198

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Open NCERT p. 10Matches NCERT’s answer

1. In each line the third term is the product of the first two numbers being squared: 1×2=2, 2×3=6, 3×4=12.
2. And the number on the right is one more than that product: 1×2+1=3, 2×3+1=7, 3×4+1=13.
3. So the pattern is n² + (n+1)² + [n(n+1)]² = [n(n+1)+1]².
4. For n=4: third term stays 4×5=20 (already given), and the answer is 4×5+1 = 21.
5. For n=9: third term is 9×10=90, and the answer is 9×10+1 = 91.

Answer4² + 5² + 20² = 21²; 9² + 10² + 90² = 91².

Open NCERT p. 11Checked by computer

1. The picture is a 9×9 arrangement of tiles (81 tiles), alternately a straight 5×5 grid and the same grid rotated to sit on its corner.
2. Every tile, straight or rotated, is itself divided into 5×5 = 25 tiny squares.
3. Total tiny squares = 81 × 25 = 2025.
4. 81 = 3⁴ and 25 = 5², so 2025 = 3⁴ × 5².

Answer2025 tiny squares; 2025 = 3⁴ × 5².

Open NCERT p. 16Matches NCERT’s answer

1. 27000 = 2³ × 3³ × 5³ = (2×3×5)³ = 30³.
2. 10648 = 2³ × 11³ = (2×11)³ = 22³.

Answer∛27000 = 30, ∛10648 = 22.

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Open NCERT p. 16Matches NCERT’s answer

1. Prime factorise: 1323 = 3³ × 7².
2. For a perfect cube every power must be a multiple of 3; 7² needs one more 7 to reach 7³.
3. Multiply by 7: 1323 × 7 = 9261 = 21³.

Answer7

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Open NCERT p. 16One way to think about it

(i) The cube of any odd number is even.

1. odd × odd × odd is always odd, so the cube of an odd number is odd, not even.
2. Example: 3³ = 27, which is odd.

In shortFalse

(ii) There is no perfect cube that ends with 8.

1. A number ending in 2 always has a cube ending in 8.
2. For example 2³ = 8 and 12³ = 1728, both perfect cubes ending in 8.

In shortFalse

(iii) The cube of a 2-digit number may be a 3-digit number.

1. The smallest 2-digit number is 10, and 10³ = 1000, which already has 4 digits.
2. So no 2-digit number's cube has only 3 digits.

In shortFalse

(iv) The cube of a 2-digit number may have seven or more digits.

1. The largest 2-digit number is 99, and 99³ = 970299, which has only 6 digits.
2. So no 2-digit number's cube reaches 7 digits.

In shortFalse

(v) Cube numbers have an odd number of factors.

1. A perfect square always has an odd number of factors, but a cube is not always a square.
2. 8 = 2³ has factors 1, 2, 4, 8 — 4 factors, an even count.

In shortFalse

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Open NCERT p. 16Matches NCERT’s answer

1. Split each number after the thousands place: this brackets it between two multiples of ten, since 10³=1000, 20³=8000, 30³=27000, 40³=64000.
2. The last digit of the cube fixes the last digit of the root: 1→1, 3→7, 7→3, 8→2 (and 0→0, 2→8, 4→4, 5→5, 6→6, 9→9).
3. 1331 is between 1000 and 8000, so the tens digit is 1; it ends in 1, so the root ends in 1 → 11.
4. 4913 is between 1000 and 8000, so the tens digit is 1; it ends in 3, so the root ends in 7 → 17.
5. 12167 is between 8000 and 27000, so the tens digit is 2; it ends in 7, so the root ends in 3 → 23.
6. 32768 is between 27000 and 64000, so the tens digit is 3; it ends in 8, so the root ends in 2 → 32.

Answer11, 17, 23, 32

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Open NCERT p. 17Matches NCERT’s answer

1. Compute each difference: 67³ − 66³ = 13267.
2. 43³ − 42³ = 5419.
3. 67² − 66² = 133.
4. 43² − 42² = 85.
5. 13267 is the largest of the four, so (i) is the greatest.

Answer(i) 67³ − 66³ = 13267 is the greatest.

Watch this explained “A sum against a product”, 9:13 into Cube roots, and what successive differences expose · हिंदी में देखें