CBSE Class 8 Mathematics A Square and A Cube Long Answer and Case Study Questions with Answers

Long Answer Questions

Question 1. Queen Ratnamanjuri’s will sends 100 people to toggle 100 lockers, and at the end only the square-numbered lockers are open.
(i) Write all the factor pairs (“partner factors”) of 24 and of 36, and count the factors of each.
(ii) Using the idea of partner factors, explain why most numbers have an even number of factors but a square number has an odd number of factors.
(iii) Hence explain why exactly the square-numbered lockers stay open, and list the lockers that remain open out of the first 50.
Answer:
(i) 24: 1 × 24, 2 × 12, 3 × 8, 4 × 6. Factors: 1, 2, 3, 4, 6, 8, 12, 24 — 8 factors.
36: 1 × 36, 2 × 18, 3 × 12, 4 × 9, 6 × 6. Factors: 1, 2, 3, 4, 6, 9, 12, 18, 36 — 9 factors.
(ii) Every factor has a partner factor, and the two multiply to give the number. So factors come in pairs and the count is normally even; 24 has four pairs, giving 8 factors.
In a square, one pair has equal members, like 6 × 6 for 36. That factor is its own partner and is counted only once, so the total count becomes odd.
(iii) A locker is toggled once by each person whose number is a factor of the locker number. An odd number of toggles leaves it open and an even number leaves it closed.
So exactly the square-numbered lockers stay open. Out of the first 50 they are 1, 4, 9, 16, 25, 36 and 49 — seven lockers.
In simple words: Factors pair up, except the repeated factor of a square, so only squares have an odd number of factors and their lockers stay open.

Teacher's Note:
1. Write factor pairs to count factors without missing any.
2. Odd toggles from closed means open.

Question 2. The units digit of a number carries useful information about squares.
(i) Write the squares of 1 to 10 and list all the digits that appear in the units place of these squares.
(ii) Which digits can never appear in the units place of a perfect square? Use this to decide which of 1372, 2025, 4683 and 5476 are definitely not perfect squares.
(iii) Explain, with the example of 26 and 36, why the units digit alone cannot prove that a number is a perfect square.
(iv) Write the largest perfect square that is less than 1000.
Answer:
(i) 1, 4, 9, 16, 25, 36, 49, 64, 81, 100. Their units digits are 1, 4, 9, 6, 5, 6, 9, 4, 1, 0, so a square always ends in 0, 1, 4, 5, 6 or 9.
(ii) The digits 2, 3, 7 and 8 never appear. 1372 ends in 2 and 4683 ends in 3, so 1372 and 4683 are definitely not perfect squares.
2025 ends in 5 and 5476 ends in 6, so the test does not rule them out; in fact 2025 = 45² and 5476 = 74².
(iii) Both 26 and 36 end in 6. 36 = 6² is a square, but 26 is not, since 5² = 25 and 6² = 36. A suitable units digit only allows a square; it cannot prove one.
(iv) 31² = 961 and 32² = 1024 > 1000, so the largest square below 1000 is 961.
In simple words: Squares end only in 0, 1, 4, 5, 6 or 9; a wrong digit rules a number out, but a right digit proves nothing.

Teacher's Note:
1. The test is useful only in the negative form.
2. 31² = 961 is a handy fact.

Question 3. The picture shows how a square grows into the next square by adding an inverted L-shaped layer of unit squares.
(i) Write the number of unit squares added at each new layer for the first five squares, and say what kind of numbers these are.
(ii) State the rule connecting the sum of the first n odd numbers with n², and write the formula for the nth odd number.
(iii) Using the rule, find the sum of all odd numbers from 1 to 39.
(iv) How many consecutive odd numbers starting from 1 add up to 625?
Answer:
(i) The layers add 1, 3, 5, 7, 9 unit squares. These are the consecutive odd numbers.
Each inverted L adds one square along the new row, one along the new column and one at the corner, so it has 2 more squares than the previous L.
(ii) Adding the odd numbers in order builds the squares one after another: 1 + 3 + 5 + ⋯ + (2n − 1) = n².
So the sum of the first n odd numbers is n², and the nth odd number is 2n − 1.
(iii) 2n − 1 = 39 gives n = 20, so the sum is 20² = 400.
(iv) n² = 625 and 25 × 25 = 625, so n = 25. The first 25 odd numbers (1 to 49) add up to 625.
In simple words: Each L-layer is the next odd number, so adding n odd numbers gives n².

Teacher's Note:
1. Count each new shade in the picture to see 1, 3, 5, 7.
2. Find n from the last odd number using 2n − 1.

Question 4. Study the cubes of the numbers from 1 to 10.
(i) Write the cubes of 1 to 10 and list the digits that appear in the units place of these cubes.
(ii) Compare this list with the units digits possible for perfect squares, and state the important difference.
(iii) The chapter states that each prime factor of a number appears three times over in the prime factorisation of its cube. Verify this statement for the number 12 and its cube 1728.
(iv) How many perfect cubes have one digit, how many have two digits, and how many have three digits?
Answer:
(i) 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000. Units digits: 1, 8, 7, 4, 5, 6, 3, 2, 9, 0 — all ten digits.
(ii) Cubes can end in any digit 0 to 9, but squares only in 0, 1, 4, 5, 6 or 9. So the units digit can never show that a number is not a cube.
Also each digit occurs exactly once among these cubes, so the units digit of a cube tells us the units digit of its cube root.
(iii) 12 = 2 × 2 × 3 = 2² × 3.
1728 = 12 × 12 × 12 = (2² × 3) × (2² × 3) × (2² × 3) = 2⁶ × 3³.
In 12, the prime 2 occurs twice and 3 once; in 1728, 2 occurs 2 × 3 = 6 times and 3 occurs 1 × 3 = 3 times. So each prime appears three times over — verified.
(iv) One-digit cubes: 1, 8 — 2. Two-digit: 27, 64 — 2. Three-digit: 125, 216, 343, 512, 729 — 5.
In simple words: Cubes can end in any digit, each prime is tripled in a cube, and there are 2, 2 and 5 cubes with one, two and three digits.

Teacher's Note:
1. The units digit of a cube matches its root in a one-to-one way.
2. Powers multiply by 3 when we cube.

Question 5. Prime factorisation is a reliable way of testing a perfect square.
(i) Find the prime factorisation of 2916 and show that it is a perfect square. Write its square root.
(ii) Find the prime factorisation of 4500 and explain, using the grouping of prime factors, why it is not a perfect square.
(iii) Find the smallest number by which 4500 must be multiplied so that the product is a perfect square, and write the square root of that product.
Answer:
(i) 2916 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 = 2² × 3⁶.
In pairs: (2 × 2) × (3 × 3) × (3 × 3) × (3 × 3). So 2916 = (2 × 3 × 3 × 3) × (2 × 3 × 3 × 3) = 54².
Hence 2916 is a perfect square and √2916 = 54.
(ii) 4500 = 2 × 2 × 3 × 3 × 5 × 5 × 5 = 2² × 3² × 5³.
The 2s and 3s pair up, but one 5 is left without a partner, so the factors cannot be split into two identical groups. 4500 is not a perfect square.
(iii) One more 5 is needed, so the smallest multiplier is 5: 4500 × 5 = 22500 = 2² × 3² × 5⁴.
√22500 = 2 × 3 × 5² = 150. Check: 150 × 150 = 22500.
In simple words: 2916 pairs up completely so it is 54², while 4500 needs one more 5 to become 22500 = 150².

Teacher's Note:
1. Every prime must have an even power in a square.
2. Halve the powers to get the root.

Question 6. Cube roots can be found by prime factorisation and also by intelligent guessing.
(i) Using prime factorisation, show that 9261 is a perfect cube and find ∛9261.
(ii) Show that 2000 is not a perfect cube by grouping its prime factors into triplets.
(iii) Guess ∛13824 without full factorisation: use the units digit of 13824 to decide the units digit of the cube root, and the part of the number before the last three digits to decide the tens digit. Verify your guess.
Answer:
(i) 9261 = 3 × 3087 = 3 × 3 × 1029 = 3 × 3 × 3 × 343 = 3³ × 7³.
The factors form three identical groups (3 × 7), so 9261 = (3 × 7)³ = 21³ and ∛9261 = 21.
(ii) 2000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 = 2⁴ × 5³. We can form the triplets (2 × 2 × 2) and (5 × 5 × 5), but one 2 is left over. So 2000 is not a perfect cube.
(iii) 13824 ends in 4. Among the cubes of 0 to 9, only 4³ = 64 ends in 4, so the cube root ends in 4.
Leaving out the last three digits gives 13. Since 2³ = 8 < 13 < 27 = 3³, the tens digit is 2.
So ∛13824 = 24. Verification: 24 × 24 × 24 = 576 × 24 = 13824.
In simple words: 9261 = 21³, 2000 has a spare 2 so it is not a cube, and 13824 = 24³ by guessing digits.

Teacher's Note:
1. Each cube root digit can be guessed from a separate part of the number.
2. Always multiply back to verify a guess.

Question 7. A square tray is completely covered by small square tiles as shown. The tray is divided into 3 rows and 3 columns of equal square blocks, and each block is made of a 4 × 4 arrangement of tiny square tiles.
(i) Find the number of tiny tiles along one edge of the tray and the total number of tiny tiles.
(ii) Write the prime factorisation of this total and show from it that the total is a perfect square.
(iii) How many tiny tiles would the tray hold if each block were made of a 5 × 5 arrangement instead?
(iv) How many whole numbers lie between the total found in (i) and the total found in (iii)?
Answer:
(i) One edge has 3 blocks with 4 tiny tiles each, so it carries 3 × 4 = 12 tiny tiles. Total = 12 × 12 = 144.
(ii) 144 = 2 × 2 × 2 × 2 × 3 × 3 = 2⁴ × 3².
The factors split into two identical groups: 144 = (2 × 2 × 3) × (2 × 2 × 3) = 12². So 144 is a perfect square with √144 = 12.
(iii) Each edge would carry 3 × 5 = 15 tiles, so the total would be 15 × 15 = 225.
(iv) Between 12² = 144 and 15² = 225 there are 225 − 144 − 1 = 80 whole numbers.
In simple words: The tray has 12 × 12 = 144 tiny tiles; with 5 × 5 blocks it would have 225, and 80 numbers lie between them.

Teacher's Note:
1. Find tiles along one edge first, then square it.
2. Count between two numbers by subtracting and taking 1 away.

Question 8. A number has to be a multiple of 9, 12 and 16 at the same time.
(i) Find the LCM of 9, 12 and 16 and write its prime factorisation.
(ii) Find the smallest perfect square that is divisible by 9, 12 and 16, and write its square root. Explain how the prime factorisation shows this at once.
(iii) Find the smallest perfect cube that is divisible by 9, 12 and 16, and write its cube root. Explain which prime factors had to be added and why.
Answer:
(i) 9 = 3², 12 = 2² × 3, 16 = 2⁴. Taking the highest power of each prime, LCM = 2⁴ × 3² = 16 × 9 = 144.
(ii) Any number divisible by 9, 12 and 16 is a multiple of 144, so it contains at least 2⁴ and 3².
In a perfect square every prime has an even power, and in 144 = 2⁴ × 3² both powers 4 and 2 are already even. So no extra factor is needed.
144 = (2² × 3) × (2² × 3) = 12², so the smallest square is 144 and √144 = 12. Check: 144 ÷ 9 = 16, 144 ÷ 12 = 12, 144 ÷ 16 = 9.
(iii) In a perfect cube every power must be a multiple of 3. The power of 2 must rise from 4 to 6 and that of 3 from 2 to 3, so we multiply by 2² × 3 = 12.
2⁶ × 3³ = 64 × 27 = 1728, and ∛1728 = 12. Check: 1728 ÷ 9 = 192, 1728 ÷ 12 = 144, 1728 ÷ 16 = 108.
In simple words: The LCM 144 is already a square (12²), and multiplying it by 12 gives the smallest cube 1728 = 12³.

Teacher's Note:
1. Start from the LCM and adjust the powers.
2. Squares need even powers; cubes need multiples of 3.

Question 9. When G. H. Hardy visited the ailing Srinivasa Ramanujan, he mentioned that his taxi carried the dull number 1729, and Ramanujan at once said it was the smallest number expressible as the sum of two cubes in two different ways.
(i) Verify by calculation that 1³ + 12³ and 9³ + 10³ are both equal to 1729, and explain what the word “taxicab number” means.
(ii) Express 4104 as a sum of two positive cubes in two different ways.
(iii) Express 13832 as a sum of two positive cubes in two different ways.
(iv) Write the greatest perfect cube that is less than 1729 and the smallest perfect cube that is greater than 1729, and state how far 1729 is from each of them.
Answer:
(i) 1³ + 12³ = 1 + 1728 = 1729 and 9³ + 10³ = 729 + 1000 = 1729. Both give 1729.
A taxicab number is a number that can be written as a sum of two cubes in two different ways; the name comes from the taxi in which Hardy travelled.
(ii) 4104 = 2³ + 16³ = 8 + 4096 and 4104 = 9³ + 15³ = 729 + 3375.
(iii) 13832 = 2³ + 24³ = 8 + 13824 and 13832 = 18³ + 20³ = 5832 + 8000.
So 1729, 4104 and 13832 are the first three taxicab numbers.
(iv) 12³ = 1728 and 13³ = 2197, and 1728 < 1729 < 2197.
The greatest cube below 1729 is 1728 (1729 is just 1 more), and the smallest cube above it is 2197 (2197 − 1729 = 468 more).
In simple words: 1729, 4104 and 13832 can each be split into two cubes in two ways, and 1729 sits just 1 above 12³.

Teacher's Note:
1. Keep a list of cubes up to 24³ handy for such checks.
2. 1729 is not a cube itself.

Question 10. Write the first six perfect squares and the first six perfect cubes in two rows and take successive differences down the levels.
(i) Show that for the squares the differences become equal after two levels, and for the cubes they become equal after three levels. Write the constant value in each case.
(ii) Using algebra, show that (n + 1)² − n² = 2n + 1, and explain how this result matches the odd-number pattern of squares.
(iii) Using algebra, show that (n + 1)³ − n³ = 3n² + 3n + 1.
(iv) Use the two results to find 50² − 49² and 50³ − 49³, and state which of the two is greater and why this is expected.
Answer:
(i) Squares: 1, 4, 9, 16, 25, 36. Level 1: 3, 5, 7, 9, 11. Level 2: 2, 2, 2, 2 — constant after two levels, value 2.
Cubes: 1, 8, 27, 64, 125, 216. Level 1: 7, 19, 37, 61, 91. Level 2: 12, 18, 24, 30. Level 3: 6, 6, 6 — constant after three levels, value 6.
(ii) (n + 1)² − n² = n² + 2n + 1 − n² = 2n + 1.
2n + 1 is always odd, and for n = 1, 2, 3, … it gives 3, 5, 7, … So each new square is the previous one plus the next odd number — the odd-number pattern.
(iii) (n + 1)³ − n³ = n³ + 3n² + 3n + 1 − n³ = 3n² + 3n + 1.
(iv) With n = 49: 50² − 49² = 2 × 49 + 1 = 99.
50³ − 49³ = 3 × 49² + 3 × 49 + 1 = 7203 + 147 + 1 = 7351.
The cube difference is far greater. This is expected because 2n + 1 grows like a first-degree expression, while 3n² + 3n + 1 grows like a second-degree one.
In simple words: Square differences settle after two levels at 2 and cube differences after three levels at 6; 50³ − 49³ = 7351 is far bigger than 99.

Teacher's Note:
1. Expanding (n + 1)³ = n³ + 3n² + 3n + 1 is the key step.
2. Check (iv) directly: 2500 − 2401 = 99.

Case Study Based Questions

Question 11. A square vegetable garden. The members of a village self-help group are given a square plot of land for a vegetable garden. The record book of the panchayat says that the area of the plot is 1024 m², but the length of its side is not written anywhere. The group wants to put a wire fence all around the plot before sowing seeds, and the fencing wire costs 75 rupees per metre. Next year the panchayat plans to give the group a second square plot whose area will be exactly double the area of the present plot. A neighbouring farmer has a square field of area 1000 m² and wants to mark out the largest square inside it whose side is a whole number of metres.
(i) Find the length of a side of the garden plot.
(ii) Find the total cost of fencing the plot once, all the way round.
(iii) Will the second plot of area 2048 m² have a whole-number side? Give a reason.
(iv) What is the largest whole-number side of a square that can be marked inside the field of area 1000 m²?
Answer:
(i) Area = side × side, and 32 × 32 = 1024, so the side is √1024 = 32 m.
(ii) Perimeter = 4 × 32 = 128 m, so the cost = 128 × 75 = 9600 rupees.
(iii) No. 2048 = 2¹¹, and eleven 2s cannot be split into two equal groups, so 2048 is not a perfect square. Also 45² = 2025 < 2048 < 2116 = 46².
(iv) 31² = 961 and 32² = 1024, and 961 < 1000 < 1024. So the largest whole-number side is 31 m.
In simple words: The plot has side 32 m and costs 9600 rupees to fence; 2048 is not a square, and 31 m is the largest side inside 1000 m².

Teacher's Note:
1. Doubling a square area does not give a square area.
2. Perimeter of a square = 4 × side.

Question 12. Tiling a square hall. The floor of a square hall in a school measures 6 m along each wall. The headmistress decides to cover the whole floor with square ceramic tiles of side 25 cm, placed in neat rows and columns with no tile cut and no gap left. The shopkeeper shows her the tiles, which are sold in sealed boxes, and tells her the price per tile. He also offers square tiles of side 30 cm as an alternative for the same hall, saying that fewer of them would be needed.
(i) How many tiles of side 25 cm are needed along one edge of the hall?
(ii) How many such tiles are needed for the whole floor?
(iii) Write the prime factorisation of the number found in (ii) and show from it that the number is a perfect square.
(iv) If the hall were covered with the 30 cm tiles instead, how many tiles would be needed?
Answer:
(i) 6 m = 600 cm, so tiles along one edge = 600 ÷ 25 = 24.
(ii) The floor has 24 rows of 24 tiles, so it needs 24 × 24 = 576 tiles.
(iii) 576 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 = 2⁶ × 3².
The factors split into two identical groups: (2 × 2 × 2 × 3) × (2 × 2 × 2 × 3) = 24 × 24. So 576 is a perfect square with √576 = 24.
(iv) One edge needs 600 ÷ 30 = 20 tiles, so the floor needs 20 × 20 = 400 tiles.
In simple words: The hall needs 24 × 24 = 576 small tiles or 20 × 20 = 400 larger tiles.

Teacher's Note:
1. Convert metres to centimetres before dividing.
2. A square floor tiled with square tiles always uses a square number of tiles.

Question 13. Sugar cubes in a box. Rehana buys a box of sugar cubes for her tea stall. Each sugar cube is a cube of side 1 cm. Her younger brother Imran empties the box on the table and counts exactly 500 cubes. He then starts stacking them, layer upon layer, to build the largest possible solid cube, keeping the leftover cubes aside. He first practises by building a solid cube of side 5 cm, and then tries a solid cube of side 6 cm to see how the layers grow.
(i) How many sugar cubes are needed for the solid cube of side 5 cm?
(ii) For the solid cube of side 6 cm, how many cubes are there in one layer, how many layers are there, and how many cubes are needed in all?
(iii) What is the side of the largest solid cube Imran can build from 500 cubes, and how many cubes are left over?
(iv) How many more sugar cubes must he buy to build a solid cube of side 8 cm?
Answer:
(i) 5 × 5 × 5 = 5³ = 125 sugar cubes.
(ii) One layer is a 6 × 6 square, so it has 36 cubes; there are 6 layers, so the total is 6 × 6 × 6 = 216 cubes.
(iii) 7³ = 343 and 8³ = 512, and 343 < 500 < 512. So the largest cube has side 7 cm, and 500 − 343 = 157 cubes are left over.
(iv) A cube of side 8 cm needs 8³ = 512 cubes, so he must buy 512 − 500 = 12 more.
In simple words: Imran can build a 7 cm cube with 157 cubes left, and needs 12 more cubes for an 8 cm cube.

Teacher's Note:
1. Find the largest cube not more than 500.
2. Layers × cubes in a layer gives the total.

Question 14. The hostel locker game. In a school hostel corridor there are 50 lockers, numbered 1 to 50, all of them closed. The 50 students of the hostel are also numbered 1 to 50. Student 1 opens every locker. Student 2 toggles every 2nd locker, student 3 toggles every 3rd locker, and so on, until student 50 has had a turn. Toggling means opening a closed locker and closing an open one. The warden promises a prize to the student who can say, before the game begins, exactly which lockers will be open at the end.
(i) Which lockers are open at the end of the game, and how many are there?
(ii) How many lockers are toggled exactly twice, and what kind of numbers are they?
(iii) How many times is locker number 30 toggled?
(iv) Which locker is toggled only once during the whole game?
Answer:
(i) A locker is toggled once for each factor of its number and stays open only if that count is odd, which happens only for perfect squares. Open lockers: 1, 4, 9, 16, 25, 36 and 49 — 7 lockers.
(ii) A locker toggled exactly twice has only the factors 1 and itself, so its number is a prime number. The primes up to 50 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47 — 15 lockers.
(iii) The factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30, so locker 30 is toggled 8 times (and ends closed).
(iv) Locker 1, since 1 has only one factor. Student 1 opens it and nobody touches it again.
In simple words: Square lockers stay open, prime lockers are touched twice, locker 30 is touched 8 times and locker 1 only once.

Teacher's Note:
1. Number of toggles = number of factors.
2. 1 is neither prime nor composite but is a square.

Question 15. A growing Rangoli. Meera draws a square rangoli on her courtyard using small square paper tiles, all of the same size. She begins with a single tile. To make the next square she adds an inverted L-shaped layer of tiles along the top and the right-hand side, and she keeps going in the same way, each new layer turning the square into the next bigger square. Her mother has given her a packet of 100 tiles, and Meera wants to use as many of them as she can without leaving an incomplete layer. Her elder sister then asks how many extra tiles would be needed to grow a 12 × 12 rangoli into a 13 × 13 one.
(i) How many tiles are added in the 7th layer? Write the rule for the number of tiles in the nth layer.
(ii) How many tiles are used in all once the 7th layer is complete?
(iii) Using the packet of 100 tiles, how many complete layers can Meera make?
(iv) How many extra tiles are needed to change a 12 × 12 rangoli into a 13 × 13 rangoli?
Answer:
(i) The layers hold 1, 3, 5, 7, … tiles, the consecutive odd numbers. The nth layer needs 2n − 1 tiles, so the 7th layer needs 2 × 7 − 1 = 13 tiles.
(ii) The sum of the first n odd numbers is n², so after 7 layers the rangoli is a 7 × 7 square using 7² = 49 tiles.
(iii) She needs n² ≤ 100. Since 10² = 100 exactly, she can make 10 complete layers, using all 100 tiles in a 10 × 10 rangoli.
(iv) Extra tiles = 13² − 12² = 169 − 144 = 25. This is also the 13th odd number, 2 × 13 − 1 = 25.
In simple words: Each layer is the next odd number, so 7 layers use 49 tiles, 100 tiles make 10 layers, and going from 12 × 12 to 13 × 13 needs 25 tiles.

Teacher's Note:
1. The nth layer is the nth odd number.
2. Total after n layers is n².

Question 16. Lists on clay and words in Sanskrit. A museum display describes the history of squares and cubes. One panel shows clay tablets of the Babylonians, dated about 1700 BCE, carrying the first known lists of perfect squares and perfect cubes; the panel explains that these lists helped in land measurement and in architectural design. A second panel describes the Indian tradition: varga stood for the square figure, its area and the square power, ghana for the solid cube and for the product of a number with itself three times, and varga-varga for the fourth power. Aryabhata, in 499 CE, wrote that a square figure of four equal sides and the number representing its area are both called varga. A third panel explains the word mula.
(i) Who compiled the first known lists of perfect squares and perfect cubes, and about when?
(ii) What do the terms varga and ghana each stand for?
(iii) Find the varga of 3 and the varga-varga of 3.
(iv) The word mula means the root of a plant. Explain how this word gave us varga-mula and ghana-mula, and name the Arabic and Latin words that followed the same idea.
Answer:
(i) The Babylonians, about 1700 BCE, on clay tablets; the lists were used in land measurement and architectural design.
(ii) Varga stands for the square figure, its area and the square power (product of two equal quantities). Ghana stands for the solid cube and the product of a number with itself three times.
(iii) Varga of 3 = 3² = 9. Varga-varga of 3 = 3⁴ = 9² = 81.
(iv) Mula means the root of a plant, and so the basis or origin. It was used for taking roots: varga-mula is the square root and ghana-mula the cube root.
The same idea was followed in Arabic by jidhr and in Latin by radix, both meaning the root of a plant.
In simple words: Babylonians made the first lists; varga is square, ghana is cube, and mula (root) gave us square root and cube root.

Teacher's Note:
1. Varga-varga is the square of a square, so 3⁴.
2. Our word 'root' comes from radix.

Question 17. A cubical water tank. A housing society builds an underground water tank in the shape of a cube. The engineer’s report states that the tank holds exactly 15625 litres of water when it is full, but the edge of the tank is not written in the report. The secretary of the society wants the measurement of the edge in metres so that a cover can be made for the top face. He is reminded that 1 litre of water occupies 1000 cm³. A larger society nearby is planning a cubical tank with each edge double the edge of this tank.
(i) Find ∛15625.
(ii) Find the edge of the tank in centimetres and in metres.
(iii) What will be the capacity, in litres, of the larger tank whose edge is double this edge?
(iv) Show that 15625 is a perfect square as well as a perfect cube, and write its square root and its cube root.
Answer:
(i) 25 × 25 × 25 = 15625, so ∛15625 = 25.
(ii) Volume = 15625 × 1000 = 15625000 cm³ = 25³ × 10³ = 250³. So the edge is 250 cm = 2.5 m.
(iii) Doubling the edge multiplies the volume by 2³ = 8, so the capacity = 8 × 15625 = 125000 litres.
(iv) 15625 = 5 × 5 × 5 × 5 × 5 × 5 = 5⁶.
Two identical groups: (5³) × (5³) = 125 × 125, so √15625 = 125.
Three identical groups: (5²) × (5²) × (5²) = 25 × 25 × 25, so ∛15625 = 25. Hence 15625 is both a perfect square and a perfect cube.
In simple words: The tank holds 25³ litres, has an edge of 2.5 m, and 15625 = 5⁶ is both 125² and 25³.

Teacher's Note:
1. 1000 cm³ = 10³ cm³, which adds a factor of 10 to the edge.
2. A sixth power is both a square and a cube.

Question 18. A square formation at the sports meet. At a district sports meet, 1500 students are to stand in a square formation, that is, in a number of rows with exactly as many students in every row as there are rows. The physical training teacher first checks whether all 1500 students can form an exact square. She then decides to make the largest possible square formation and to seat the remaining students on the gallery steps. The organisers later suggest calling in a few more students from a nearby school, so that the next bigger square formation becomes possible.
(i) Find the largest square number that does not exceed 1500, and the number of students in each row of that formation.
(ii) How many students will be seated on the gallery steps?
(iii) How many more students must join so that a complete square formation of the next size becomes possible?
(iv) Give a clear reason, using factors, why 1500 is not a perfect square.
Answer:
(i) 38² = 1444 and 39² = 1521, and 1444 < 1500 < 1521. So the largest square is 1444, with 38 students in each of 38 rows.
(ii) 1500 − 1444 = 56 students sit on the gallery steps.
(iii) The next formation needs 39² = 1521 students, so 1521 − 1500 = 21 more must join.
(iv) 1500 = 2 × 2 × 3 × 5 × 5 × 5. The 2s pair up and two 5s pair up, but the 3 and the third 5 have no partners, so the factors cannot be split into two identical groups. Hence 1500 is not a perfect square (in short, 1500 = 100 × 15 and 15 is not a square).
In simple words: The largest formation is 38 × 38 = 1444, leaving 56 students; 21 more are needed for 39 × 39.

Teacher's Note:
1. Trap 1500 between two consecutive squares.
2. Unpaired prime factors show a number is not a square.

Question 19. Packing cubical cartons. A warehouse stores tea in cubical cartons of edge 60 cm. Inside each carton, smaller cubical boxes are packed so that they fill the carton completely, leaving no gap and with no box cut or tilted. The supplier makes cubical boxes of edge 15 cm and of edge 20 cm, and is thinking of introducing a box of edge 25 cm as well. The manager wants to know, for each size, whether a carton can be filled exactly and how many boxes would then be needed.
(i) How many boxes of edge 15 cm exactly fill one carton?
(ii) How many boxes of edge 20 cm exactly fill one carton?
(iii) Can boxes of edge 25 cm fill the carton exactly? Give a reason.
(iv) What is the smallest edge of a cubical carton that can be filled exactly by boxes of edge 15 cm and also, separately, by boxes of edge 20 cm? Justify your answer.
Answer:
(i) 60 ÷ 15 = 4 boxes fit along each edge, so the carton holds 4 × 4 × 4 = 4³ = 64 boxes.
(ii) 60 ÷ 20 = 3 boxes fit along each edge, so the carton holds 3³ = 27 boxes.
(iii) No. 60 ÷ 25 = 2.4 is not a whole number, so the boxes cannot be laid along an edge without cutting. A carton is filled exactly only when its edge is a whole-number multiple of the box edge.
(iv) The edge must be a common multiple of 15 and 20; the smallest is their LCM. 15 = 3 × 5 and 20 = 2² × 5, so LCM = 2² × 3 × 5 = 60 cm — exactly the carton the warehouse already uses (64 boxes of 15 cm or 27 boxes of 20 cm).
In simple words: A 60 cm carton holds 4³ = 64 boxes of 15 cm or 3³ = 27 boxes of 20 cm, but 25 cm boxes do not fit exactly.

Teacher's Note:
1. Count boxes along one edge first, then cube.
2. The LCM gives the smallest common edge.

Question 20. A photograph and its border. A square photograph of area 576 cm² is to be pasted at the centre of a square mount board, so that a border of uniform width is left all around the photograph. The framer first prepares a mount board that leaves a border of width 3 cm on all four sides. The customer, however, asks to see how the frame would look with a much thinner border, of width 1 cm. While working out the two designs, the framer notices something interesting: in both cases the area of the border alone turns out to be a perfect square.
(i) Find the side of the photograph.
(ii) Find the side and the area of the mount board with the 3 cm border, and check whether this area is a perfect square.
(iii) Find the area of the border alone in this design, and write its square root.
(iv) Work out the area of the border for a border of width 1 cm, and write its square root.
Answer:
(i) 24 × 24 = 576, so the side of the photograph is √576 = 24 cm.
(ii) The border is added on both sides, so the side of the board = 24 + 3 + 3 = 30 cm, and its area = 30 × 30 = 900 cm². Since 900 = 30², it is a perfect square.
(iii) Border area = 900 − 576 = 324 cm². 324 = 2 × 2 × 3 × 3 × 3 × 3 = (2 × 3 × 3)² = 18², so √324 = 18.
(iv) Side of board = 24 + 1 + 1 = 26 cm, area = 26 × 26 = 676 cm². Border area = 676 − 576 = 100 cm², and √100 = 10.
In simple words: The photo is 24 cm wide; the 3 cm border has area 324 = 18² and the 1 cm border has area 100 = 10².

Teacher's Note:
1. Add the border width on both sides of the photo.
2. Border area = outer area − inner area.