CBSE Class 8 Mathematics A Square and A Cube Short Answer Questions with Answers

Short Answer Questions

Question 1. In Queen Ratnamanjuri’s locker puzzle, locker number 49 is left open at the end while locker number 50 is found closed. List the factors of each number and explain, using the number of times a locker is toggled, why this happens.
Answer:
A locker is toggled once by each person whose number is a factor of the locker number, so the number of toggles equals the number of factors.
Factors of 49: 1, 7, 49 — 3 factors (odd), so locker 49 ends up open.
Factors of 50: 1, 2, 5, 10, 25, 50 — 6 factors (even), so locker 50 ends up closed.
49 = 7 × 7 is a square; 7 is its own partner factor, which makes its factor count odd.
In simple words: 49 has an odd number of factors because it is a square, so its locker ends open; 50 has an even number, so it ends closed.

Teacher's Note:
1. Closed lockers start closed, so an odd number of toggles means open.
2. Pair the factors to see why squares are special.

Question 2. (i) Write both integer square roots of 81 and of 144.
(ii) Explain why every perfect square has two integer square roots, and state which one is used in this chapter.
Answer:
(i) The square roots of 81 are 9 and −9; those of 144 are 12 and −12.
(ii) A negative number multiplied by itself also gives a positive product, so (−x) × (−x) = x × x = y.
Hence every perfect square y has two integer square roots, one positive and one negative.
In this chapter only the positive square root is used, so we write √81 = 9 and √144 = 12.
In simple words: Both x and −x square to the same number, but √ always means the positive root.

Teacher's Note:
1. Minus times minus gives plus.
2. The √ symbol always stands for the positive root.

Question 3. Four numbers are written on the board: 2057, 3364, 4928 and 7803. Using only the units digit, say which of them are definitely not perfect squares and give the reason. What can you say about the remaining number?
Answer:
A perfect square can end only in 0, 1, 4, 5, 6 or 9.
2057 ends in 7, 4928 ends in 8 and 7803 ends in 3, so 2057, 4928 and 7803 are definitely not perfect squares.
3364 ends in 4, so the test cannot rule it out; the units digit can only prove that a number is not a square, never that it is one.
On checking, 58² = 3364, so 3364 is actually a perfect square.
In simple words: Numbers ending in 3, 7 or 8 cannot be squares; 3364 passes the test and turns out to be 58².

Teacher's Note:
1. The units-digit test only works in one direction.
2. Confirm a possible square by finding its root.

Question 4. (i) Using the odd-number pattern, find the value of 1 + 3 + 5 + ⋯ + 23 without adding the terms one by one.
(ii) Express 81 as a sum of consecutive odd numbers starting from 1.
Answer:
(i) The nth odd number is 2n − 1. From 2n − 1 = 23 we get n = 12, so 23 is the 12th odd number.
The sum of the first n odd numbers is n², so 1 + 3 + 5 + ⋯ + 23 = 12² = 144.
(ii) 81 = 9², so it is the sum of the first 9 odd numbers:
81 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17.
In simple words: Twelve odd numbers add to 12² = 144, and nine odd numbers add to 9² = 81.

Teacher's Note:
1. First find how many odd numbers there are.
2. The last odd number in (ii) is 2 × 9 − 1 = 17.

Question 5. What is a perfect cube? List all the perfect cubes that lie between 1 and 250, and explain why 9 is not a perfect cube.
Answer:
A perfect cube is a number got by multiplying a natural number by itself three times, that is, a number of the form n³.
Perfect cubes up to 250: 1, 8, 27, 64, 125 and 216 (1³ to 6³).
9 is not a cube because 2³ = 8 and 3³ = 27, and there is no natural number between 2 and 3. So no number from 9 to 26 is a cube.
In simple words: Cubes are numbers like n × n × n; 9 lies between 2³ and 3³, so it is not a cube.

Teacher's Note:
1. 7³ = 343 is already more than 250.
2. Cubes are much more spread out than squares.

Question 6. The chapter says that squares can also be taken of fractions and of decimals.
(i) Find the area of a square whose side is \( \frac{3}{5} \) units.
(ii) Find the area of a square whose side is 2.5 units, and write each of the two areas in the form (number)².
Answer:
The area of a square is side × side, and this works for fractions and decimals too.
(i) Area = \( \frac{3}{5} \times \frac{3}{5} = \frac{9}{25} \) = \( \frac{9}{25} \) square units.
(ii) Area = 2.5 × 2.5 = 6.25 square units.
In the form (number)²: \( \left(\frac{3}{5}\right)^2 = \frac{9}{25} \) and (2.5)² = 6.25.
In simple words: Square the side even when it is a fraction or a decimal: 9/25 and 6.25.

Teacher's Note:
1. Square both the numerator and the denominator.
2. These are squares, but not called perfect squares.

Question 7. Unit cubes of side 1 cm are stacked to build a solid cube of side 5 cm. Find the number of unit cubes in one layer, the number of layers, and the total number of unit cubes used.
Answer:
Each layer is a 5 × 5 square, so one layer has 25 unit cubes.
There are 5 layers, one above another.
Total = 25 × 5 = 5 × 5 × 5 = 5³ = 125 unit cubes.
In simple words: A 5 cm cube has 5 layers of 25 unit cubes, which is 125 in all.

Teacher's Note:
1. Layer count × cubes per layer gives the volume.
2. This is why n³ is called 'n cubed'.

Question 8. Section 1.3 of the chapter uses three old mathematical terms: varga, ghana and varga-varga. Write what each of them means, and state what Aryabhata (499 CE) wrote about varga.
Answer:
Varga — the square figure, its area, and also the square power (product of two equal quantities).
Ghana — the solid cube, and also the product of a number with itself three times.
Varga-varga — the fourth power of a number.
Aryabhata (499 CE) wrote that a square figure of four equal sides and the number representing its area are both called varga, and the product of two equal quantities is also varga. So the word for the square power came from the square figure.
In simple words: Varga is the square, ghana the cube, varga-varga the fourth power, and Aryabhata linked varga to both the square shape and its area.

Teacher's Note:
1. Link each word to a shape: varga to the square, ghana to the solid cube.
2. Varga-varga is 'the square of a square'.

Question 9. It is given that 45² = 2025. Using the fact that every square is the sum of consecutive odd numbers starting from 1, find 46². Show clearly how you obtain the odd number that has to be added.
Answer:
45² is the sum of the first 45 odd numbers. To reach 46² we add the 46th odd number.
The nth odd number is 2n − 1, so the 46th odd number is 2 × 46 − 1 = 91.
So 46² = 2025 + 91 = 2116.
Check: 46 × 46 = 2116.
In simple words: Add the 46th odd number, 91, to 2025 to get 46² = 2116.

Teacher's Note:
1. The odd number to add is 45 + 46 = 91.
2. Always verify with direct multiplication.

Question 10. By successively subtracting consecutive odd numbers starting from 1, show that 64 is a perfect square and write its square root. Repeat the process for 45 and state what it tells you.
Answer:
64 − 1 = 63, 63 − 3 = 60, 60 − 5 = 55, 55 − 7 = 48, 48 − 9 = 39, 39 − 11 = 28, 28 − 13 = 15, 15 − 15 = 0.
We reach 0 at the 8th step, so 64 = 8² and √64 = 8.
For 45: 45 − 1 = 44, 44 − 3 = 41, 41 − 5 = 36, 36 − 7 = 29, 29 − 9 = 20, 20 − 11 = 9, 9 − 13 = −4.
We go past 0 without landing on it, so 45 is not a perfect square.
In simple words: 64 reaches zero after 8 subtractions so √64 = 8, but 45 skips over zero so it is not a square.

Teacher's Note:
1. The number of steps to reach 0 is the square root.
2. Crossing below 0 means the number is not a square.

Question 11. (i) How many whole numbers lie strictly between 19² and 20²? Write the general rule for the count of numbers between the squares of n and n + 1.
(ii) How many perfect squares lie between 101 and 200?
Answer:
(i) 19² = 361 and 20² = 400, so the count is 400 − 361 − 1 = 38.
In general, 2n numbers lie between n² and (n + 1)²; here 2 × 19 = 38.
(ii) The squares between 101 and 200 are 11² = 121, 12² = 144, 13² = 169 and 14² = 196.
So there are 4 perfect squares.
In simple words: 38 numbers lie between 19² and 20², and only 121, 144, 169 and 196 are squares between 101 and 200.

Teacher's Note:
1. (n + 1)² − n² − 1 = 2n.
2. Squares become rarer as numbers grow.

Question 12. Find the smallest number by which 2352 must be multiplied so that the product becomes a perfect square. Use prime factorisation and also write the square root of the product.
Answer:
2352 = 2 × 2 × 2 × 2 × 3 × 7 × 7 = 2⁴ × 3 × 7².
The 2s and 7s pair up, but the single 3 has no partner, so the smallest multiplier is 3.
2352 × 3 = 7056 = 2⁴ × 3² × 7².
√7056 = 2² × 3 × 7 = 84. Check: 84 × 84 = 7056.
In simple words: Multiply by the unpaired 3 to get 7056, whose square root is 84.

Teacher's Note:
1. Multiply by the unpaired prime, not by a bigger number.
2. Halve each power to get the square root.

Question 13. The triangular numbers are 1, 3, 6, 10, 15, 21, 28, 36, … Show that 21 + 28 is a perfect square, describe the pattern connecting two consecutive triangular numbers, and write the next pair of triangular numbers with the square they give.
Answer:
21 + 28 = 49 = 7², so the sum is a perfect square.
Pattern: the sum of any two consecutive triangular numbers is a perfect square: 1 + 3 = 2², 3 + 6 = 3², 6 + 10 = 4², …, 21 + 28 = 7².
Next pair: 28 + 36 = 64 = 8².
In a picture, two triangular dot patterns fit together to make a square.
In simple words: Two neighbouring triangular numbers always add up to a square, such as 21 + 28 = 7² and 28 + 36 = 8².

Teacher's Note:
1. Triangular numbers grow by 2, 3, 4, 5, … each time.
2. A dot picture makes the pattern easy to see.

Question 14. The chapter shows that every cube is the sum of a set of consecutive odd numbers, for example 7 + 9 + 11 = 3³. Write the seven consecutive odd numbers whose sum is 7³ and verify the sum.
Answer:
7³ is the sum of 7 consecutive odd numbers, beginning at 43:
43 + 45 + 47 + 49 + 51 + 53 + 55.
The numbers are evenly spaced about the middle number 49, so the sum is 7 × 49 = 343.
Since 7³ = 343, the sum is correct.
In simple words: The seven odd numbers from 43 to 55 add up to 343, which is 7³.

Teacher's Note:
1. The middle term is 7² = 49.
2. Before 43 come 1 + 2 + … + 6 = 21 odd numbers used for 1³ to 6³.

Question 15. Is 1944 a perfect cube? Use prime factorisation to decide. Then find the smallest number by which it must be multiplied to make it a perfect cube, and write the cube root of the product.
Answer:
1944 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 = 2³ × 3⁵.
Grouping gives the triplets (2 × 2 × 2) and (3 × 3 × 3), but two 3s are left over. So 1944 is not a perfect cube.
One more 3 completes the triplet, so multiply by 3: 1944 × 3 = 5832 = 2³ × 3⁶.
5832 = (2 × 3 × 3)³ = 18³, so ∛5832 = 18.
In simple words: 1944 is short of one 3, so multiplying by 3 gives 5832 = 18³.

Teacher's Note:
1. Every prime power must be a multiple of 3.
2. 3⁵ needs one more 3 to become 3⁶.

Question 16. Evaluate \( \left(\frac{3}{4}\right)^3 \), (−5)³ and (0.2)³. Also state, with a reason, whether (0.2)³ is greater or less than (0.2)².
Answer:
\( \left(\frac{3}{4}\right)^3 = \frac{3 \times 3 \times 3}{4 \times 4 \times 4} = \frac{27}{64} \), so the value is \( \frac{27}{64} \).
(−5)³ = (−5) × (−5) × (−5) = −125.
(0.2)³ = 0.2 × 0.2 × 0.2 = 0.008.
(0.2)² = 0.04, which is greater than 0.008. Multiplying a number between 0 and 1 by itself once more makes it smaller, so (0.2)³ is less than (0.2)².
In simple words: The cubes are 27/64, −125 and 0.008, and 0.008 is less than 0.04.

Teacher's Note:
1. The cube of a negative number is negative.
2. Numbers between 0 and 1 shrink when multiplied by themselves.

Question 17. The locker game of the chapter opening is played in a corridor of 150 lockers with 150 students.
(i) How many lockers remain open at the end?
(ii) Which locker numbers are toggled exactly three times? Explain what is special about them.
Answer:
(i) A locker stays open when its number has an odd number of factors, which happens only for perfect squares.
The squares up to 150 are 1², 2², …, 12² (12² = 144, 13² = 169 is too big). So 12 lockers remain open.
(ii) A locker is toggled exactly 3 times when its number has exactly 3 factors. Such a number is the square of a prime p, with factors 1, p and p².
Up to 150 these are 4, 9, 25, 49 and 121 (squares of 2, 3, 5, 7 and 11).
In simple words: Twelve square-numbered lockers stay open, and the squares of primes are touched exactly three times.

Teacher's Note:
1. Number of toggles = number of factors.
2. Only squares of primes have exactly three factors.

Question 18. Without using prime factorisation, estimate and then find √1764. Use the squares of the two nearest multiples of 10 to fix an interval, the units digit to shortlist the possibilities, and the value of 42² to confirm your answer.
Answer:
Step 1: 40² = 1600 and 50² = 2500, and 1600 < 1764 < 2500, so 40 < √1764 < 50.
Step 2: 1764 ends in 4, and only numbers ending in 2 or 8 have squares ending in 4. So the root is 42 or 48.
Step 3: 1764 is much nearer to 1600 than to 2500 (in fact 45² = 2025 > 1764), so the root is less than 45 and 48 is ruled out.
Step 4: Only 42 is left. Confirm: 42² = 42 × 42 = 1764. Hence √1764 = 42.
In simple words: The root lies between 40 and 45 and ends in 2 or 8, so it is 42.

Teacher's Note:
1. 45² = 2025 splits the interval into two halves.
2. Always confirm by squaring.

Question 19. In the Square Pairs game the numbers 1, 2, 3, …, 8 are to be placed in a row, each used once, so that the sum of every pair of adjacent numbers is a perfect square. Write all the pairs from 1 to 8 whose sum is a perfect square, and use your list to explain why such a row is impossible.
Answer:
The possible sums are at most 7 + 8 = 15, so the only useful squares are 4 and 9.
Pairs: 1 + 3 = 4, 1 + 8 = 9, 2 + 7 = 9, 3 + 6 = 9, 4 + 5 = 9.
4 can stand only next to 5, and 5 only next to 4; likewise 2 and 7 can stand only next to each other.
So the numbers fall into three separate groups {1, 3, 6, 8}, {2, 7} and {4, 5}, with no allowed pair joining two groups.
A single row would have to pass from one group to another, which is impossible. Hence no such row exists for 1 to 8.
In simple words: The allowed pairs split 1 to 8 into three groups that cannot be joined, so one row is impossible.

Teacher's Note:
1. Find the largest possible sum to know which squares matter.
2. A number with only one possible neighbour must sit at an end.

Question 20. The sum of the first n odd natural numbers is 784.
(i) Find n and the perfect square obtained.
(ii) Find the nth odd number used in this sum, and check your answer by subtracting it from 784.
Answer:
(i) The sum of the first n odd numbers is n², so n² = 784.
20² = 400 and 30² = 900, so n lies between 20 and 30; as 784 ends in 4, n ends in 2 or 8, so n is 22 or 28. Since 25² = 625 < 784, n > 25, so n = 28.
Check: 28² = 784. Hence n = 28 and the perfect square is 784 = 28².
(ii) The 28th odd number is 2 × 28 − 1 = 55.
Check: 784 − 55 = 729 = 27², which is the sum of the first 27 odd numbers. This confirms the answer.
In simple words: 784 = 28², so 28 odd numbers are added and the last one is 55.

Teacher's Note:
1. n² = sum, so n = √sum.
2. Removing the last odd number gives the previous square.