CBSE Class 8 Mathematics A Square and A Cube MCQs Set 11

Answer: (b) 196

Explanation:
1. 14 × 14 = 196, so 196 = 14² is a perfect square.
2. 162 and 208 lie between consecutive squares (12² = 144, 13² = 169, 14² = 196, 15² = 225), so they are not squares.
3. 250 ends in a single zero, and a square always ends in an even number of zeros, so 250 is not a square.

Teacher's Note:
1. Locate the number between two consecutive squares first.
2. One zero at the end is enough to rule out a perfect square.

Answer: (d) 7

Explanation:
1. The units digit of a perfect square is always one of 0, 1, 4, 5, 6 or 9.
2. The digits 2, 3, 7 and 8 never appear there, so the answer is 7.

Teacher's Note:
1. Remember the 'forbidden' units digits 2, 3, 7 and 8.
2. Squaring 0 to 9 once is enough to see this pattern.

Answer: (a) 6

Explanation:
1. 6 × 6 × 6 = 36 × 6 = 216, so 6³ = 216.
2. Hence ∛216 = 6.

Teacher's Note:
1. The cube root asks which number multiplied by itself three times gives 216.
2. 36 is the square of 6, a common wrong choice.

Answer: (c) 150

Explanation:
1. 64 = 4³, 125 = 5³ and 216 = 6³.
2. 150 lies between 5³ = 125 and 6³ = 216, and no whole number lies between 5 and 6, so 150 is not a perfect cube.

Teacher's Note:
1. Learn the cubes from 1³ to 10³.
2. A number between two consecutive cubes cannot be a cube.

Answer: (b) ghana

Explanation:
1. Ghana meant the solid cube and also the product of a number with itself three times.
2. Varga meant the square, pada was used for the root, and varga-varga meant the fourth power.

Teacher's Note:
1. Ghana means 'solid', which matches the solid cube.
2. Varga-varga is 'square of a square', the fourth power.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A is true: the only factor of 1 is 1 itself, so it has exactly one factor, which is an odd count.
2. R is true: factors come in partner pairs, and a count is odd only when one pair has equal partners, which happens only for squares.
3. 1 = 1 × 1 is a square, so R explains A.

Teacher's Note:
1. 1 is the smallest square number, 1².
2. Partner factors explain why squares have odd counts.

Answer: (c) A is true, but R is false.

Explanation:
1. A is true: 1729 = 1³ + 12³ = 9³ + 10³, the smallest number that is a sum of two cubes in two ways. It is called the Hardy–Ramanujan number.
2. R is false: 12³ = 1728 and 13³ = 2197, so 1729 lies between two consecutive cubes and is not a cube.

Teacher's Note:
1. 1729 is a sum of two cubes, not a cube itself.
2. It is just 1 more than 12³ = 1728.

Answer: (a) Only Statement I is true.

Explanation:
1. The sum of the first n odd numbers is n², so the first 7 odd numbers add up to 7² = 49. Statement I is true.
2. The nth odd number is 2n − 1, so the 7th odd number is 2 × 7 − 1 = 13, not 15. Statement II is false.
3. So only Statement I is true.

Teacher's Note:
1. List them to check: 1, 3, 5, 7, 9, 11, 13.
2. 15 is the 8th odd number.

Answer: (d) 24

Explanation:
1. 12² = 144 and 13² = 169.
2. The numbers strictly between them are 145 to 168, that is 169 − 144 − 1 = 24 numbers.
3. This agrees with the rule that 2n numbers lie between n² and (n + 1)²: 2 × 12 = 24.

Teacher's Note:
1. Use the shortcut 2n for the count between n² and (n + 1)².
2. Do not include the two squares themselves.

Answer: (b) 2116

Explanation:
1. Each square is the previous square plus the next odd number.
2. The 46th odd number is 2 × 46 − 1 = 91.
3. So 46² = 2025 + 91 = 2116.

Teacher's Note:
1. (n + 1)² = n² + (2n + 1), here 2025 + 91.
2. Check: 46 × 46 = 2116.

Answer: (a) (i)-(s), (ii)-(q), (iii)-(p), (iv)-(r)

Explanation:
1. √121 = 11, so (i)-(s).
2. ∛64 = 4, since 4³ = 64, so (ii)-(q).
3. √144 = 12, so (iii)-(p).
4. ∛343 = 7, since 7³ = 343, so (iv)-(r).
5. The correct match is option (a).

Teacher's Note:
1. Notice which roots are square roots and which are cube roots.
2. √64 would be 8, but ∛64 is 4.

Answer: (c) 49

Explanation:
1. The array has 7 rows with 7 dots in each row.
2. Total = 7 × 7 = 7² = 49 dots.

Teacher's Note:
1. Count one row and one column; in a square array they are equal.
2. Square numbers get their name from such square arrays.

Answer: (d) 81

Explanation:
1. Only square numbers have an odd number of factors.
2. 81 = 9² is the only square in the list; its factors are 1, 3, 9, 27 and 81, which is 5 factors.
3. 48, 56 and 72 are not squares, so each has an even number of factors. Answer: 81.

Teacher's Note:
1. Look for the perfect square instead of listing all factors.
2. In a square, the root is its own partner factor.

Answer: (c) split into three identical groups

Explanation:
1. A perfect cube is a number of the form n × n × n, so its prime factors split into three identical groups.
2. For example, 3375 = (3 × 5) × (3 × 5) × (3 × 5) = 15³.
3. Splitting into two identical groups is the test for a perfect square.

Teacher's Note:
1. Square: two equal groups; cube: three equal groups.
2. Equivalently, every prime appears a multiple of 3 times.

Answer: (a) 25

Explanation:
1. 10 + 15 = 25 = 5².
2. This fits the pattern that the sum of two consecutive triangular numbers is always a perfect square.

Teacher's Note:
1. Triangular numbers: 1, 3, 6, 10, 15, 21, …
2. Two consecutive triangles fit together to make a square.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. A is true: 2048 ends in 8, and no perfect square ends in 2, 3, 7 or 8.
2. R is true: an even number times itself is even.
3. But being even does not stop a number from being a square (4, 16 and 36 are even squares), so R does not explain A.

Teacher's Note:
1. The reason for A is the units digit, not evenness.
2. Always ask whether R actually leads to A.

Answer: (d) A is false, but R is true.

Explanation:
1. A is false: cubes can end in any digit from 0 to 9, for example 2³ = 8 and 3³ = 27.
2. R is true: a number ending in 2 has a cube ending in 8 (2³ = 8, 12³ = 1728).

Teacher's Note:
1. Cubes of 0 to 9 end in ten different digits.
2. R itself gives a counter-example to A.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. 1156 = 2 × 2 × 17 × 17 = (2 × 17) × (2 × 17) = 34², so A is true.
2. R states exactly this grouping into two identical groups, which is the test for a perfect square.
3. So R is true and correctly explains A.

Teacher's Note:
1. Pairing prime factors is the main test for squares.
2. Check: 34 × 34 = 1156.

Answer: (c) Both statements are true.

Explanation:
1. 9² = 81, 11² = 121 and 19² = 361 all end in 1, so Statement I is true.
2. 14² = 196 and 26² = 676 end in 6 (4 × 4 = 16, 6 × 6 = 36), so Statement II is true.
3. So both statements are true.

Teacher's Note:
1. Only the units digits decide the units digit of a square.
2. 1 × 1 = 1 and 9 × 9 = 81 both end in 1.

Answer: (b) Only Statement II is true.

Explanation:
1. 15 and 35 end in 5 but are not squares, so Statement I is false.
2. A square ending in 5 comes from a number ending in 5, and such squares are 25, 225, 625, 1225, 2025, … — all with 2 in the tens place. Statement II is true.
3. So only Statement II is true.

Teacher's Note:
1. A units digit can rule a square out but cannot prove one.
2. Squares of numbers ending in 5 always end in 25.

Answer: (b) 7

Explanation:
1. 1323 = 3 × 3 × 3 × 7 × 7.
2. The 3s already form a triplet, but 7 appears only twice, so one more 7 is needed.
3. 1323 × 7 = 9261 = 21³. The smallest multiplier is 7.

Teacher's Note:
1. Complete each prime into a group of three.
2. Multiplying by 21 would spoil the triplet of 3s.

Answer: (d) 44 m

Explanation:
1. Side = √1936.
2. 40² = 1600 and 45² = 2025, so the root lies between 40 and 45.
3. 1936 ends in 6, so the root ends in 4 or 6; it must be 44. Check: 44 × 44 = 1936.
4. Each side is 44 m.

Teacher's Note:
1. Trap the number between two known squares.
2. The units digit narrows the choice to one number.

Answer: (d) (i)-(p), (ii)-(r), (iii)-(q), (iv)-(s)

Explanation:
1. 64 = 8² = 4³, so (i)-(p).
2. 196 = 14², but it lies between 5³ = 125 and 6³ = 216, so (ii)-(r).
3. 343 = 7³, but it lies between 18² = 324 and 19² = 361, so (iii)-(q).
4. 150 is neither a square nor a cube, so (iv)-(s). The answer is option (d).

Teacher's Note:
1. 64 is both 8² and 4³.
2. Check square and cube separately for each number.

Answer: (c) 19 unit squares

Explanation:
1. Each L-shaped strip adds the next odd number of unit squares, because (n + 1)² − n² = 2n + 1.
2. In the picture the strip has 6 + 5 = 11 = 2 × 5 + 1 squares.
3. From 9 × 9 to 10 × 10 the strip has 100 − 81 = 2 × 9 + 1 = 19 unit squares.

Teacher's Note:
1. Count the strip: one new row plus one new column, minus the shared corner.
2. 10 + 10 − 1 = 19 gives the same answer.

Answer: (d) 51³ − 50³

Explanation:
1. 29² − 28² = 57 and 51² − 50² = 101.
2. 29³ − 28³ = 24389 − 21952 = 2437.
3. 51³ − 50³ = 132651 − 125000 = 7651.
4. So 51³ − 50³ is the greatest.

Teacher's Note:
1. Differences of cubes grow much faster than differences of squares.
2. (n + 1)² − n² = 2n + 1 gives 57 and 101 quickly.

Answer: (a) Only Statement I is true.

Explanation:
1. No perfect square ends in 2, 3, 7 or 8, so Statement I is true.
2. Cubes can end in any digit; for example 7³ = 343 ends in 3. So Statement II is false.
3. So only Statement I is true.

Teacher's Note:
1. The units-digit test works for squares, not for cubes.
2. One counter-example is enough to reject a statement.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. A is true: the units digit of 37³ comes from 7³ = 343, which ends in 3. Indeed 37³ = 50653.
2. R is true: cubes can end with any digit.
3. But R only says the digit 3 is possible; it does not explain why the digit is 3. So R is not the correct explanation.

Teacher's Note:
1. Only the units digit of the number decides the units digit of its cube.
2. A general fact may be true without explaining a particular one.

Answer: (c) 22 cm

Explanation:
1. The square of the side must be at most 500.
2. 22² = 484 ≤ 500, but 23² = 529 > 500.
3. So the largest side is 22 cm.

Teacher's Note:
1. Find the largest square not more than the given area.
2. 25 cm would need 625 cm², which is too much.

Answer: (a) (i)-(q), (ii)-(p), (iii)-(r), (iv)-(s)

Explanation:
1. 36: 1, 2, 3, 4, 6, 9, 12, 18, 36 — 9 factors, so (i)-(q).
2. 81: 1, 3, 9, 27, 81 — 5 factors, so (ii)-(p).
3. 49: 1, 7, 49 — 3 factors, so (iii)-(r).
4. 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 — 10 factors, so (iv)-(s).
5. The answer is option (a).

Teacher's Note:
1. List factors in partner pairs to avoid missing any.
2. The three squares have odd counts; 48 has an even count.

Answer: (b) 189

Explanation:
1. The solid has 3 unit cubes along each edge, so it has 3³ = 27 unit cubes.
2. A cube with double the edge has 6 along each edge, so it needs 6³ = 216 unit cubes.
3. Extra cubes = 216 − 27 = 189.

Teacher's Note:
1. Doubling the edge makes the number of cubes 8 times.
2. 216 is the total, not the extra number.