In-text Questions — Perfect Squares — A Square and A Cube

1. 11 markGanita Prakash Cl-8 Part 1, Math Talk after the locker puzzle intro

   Does every number have an even number of factors?

   Hint. Think about what happens to a factor pair like 6 = 2 × 3 when the two factors are equal instead, like 4 = 2 × 2.

   Step 1 — Recall how factors pair up. Most numbers have factors that pair off, each pair multiplying to give the number, so factors normally come in twos.

   Step 2 — Find the exception. When a number is the product of a number with itself (like 4 = 2 × 2), that repeated factor only counts once, leaving one factor unpaired — so the total count is odd.

   ✦ Answer: No — numbers that are perfect squares have an odd number of factors, since one factor pairs with itself instead of a different partner.

   Where students slip. Assuming every factor pair is made of two different numbers — for a square number, one pair repeats the same factor twice, which is exactly what breaks the usual even-count pattern.

2. 21 markGanita Prakash Cl-8 Part 1, Math Talk after the locker puzzle intro

   Can you use this insight to find more numbers with an odd number of factors?

   Hint. Which locker numbers have exactly one factor pair where both factors are the same?

   Step 1 — Recall the source of an odd factor count. A number has an odd number of factors exactly when it has a repeated factor pair, i.e. when it is a perfect square.

   Step 2 — Generate more examples. 1×1, 2×2, 3×3, 4×4, and so on all give numbers (1, 4, 9, 16, ...) with this repeated-pair property.

   ✦ Answer: Yes — every perfect square (1, 4, 9, 16, 25, ...) has an odd number of factors, for the same reason as 36.

   Where students slip. Trying to find odd-factor numbers by trial and error instead of using the square-number pattern just established — every perfect square works, and nothing else does.

3. 32 marksGanita Prakash Cl-8 Part 1, prompt after the locker-toggling explanation

   Write the locker numbers (from 1 to 100) that remain open.

   Hint. A locker stays open only if it's toggled an odd number of times — tie this back to the factor-count rule just discussed.

   Step 1 — Connect toggling to factor count. A locker is toggled once per factor of its number, so it ends open exactly when its number has an odd count of factors.

   Step 2 — Identify which numbers those are. Only perfect squares have an odd number of factors.

   Step 3 — List the perfect squares from 1 to 100. 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 (that is, 1² through 10²).

   ✦ Answer: Lockers 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 remain open.

   Where students slip. Trying to simulate all 100 people toggling all 100 lockers by hand — the factor-count shortcut avoids this entirely.

4. 42 marksGanita Prakash Cl-8 Part 1, Math Talk on units digits of squares

   Which of the following numbers have the digit 6 in the units place? (i) 38² (ii) 34² (iii) 46² (iv) 56² (v) 74² (vi) 82²

   Hint. A square's units digit depends only on the units digit of the number being squared.

   Step 1 — Recall the rule. Only numbers ending in 4 or 6 produce a square ending in 6 (since 4² = 16 and 6² = 36, both end in 6).

   Step 2 — Check each option's units digit. (i) 38 ends in 8 → square ends in 4. (ii) 34 ends in 4 → square ends in 6. (iii) 46 ends in 6 → square ends in 6. (iv) 56 ends in 6 → square ends in 6. (v) 74 ends in 4 → square ends in 6. (vi) 82 ends in 2 → square ends in 4.

   ✦ Answer: (ii) 34², (iii) 46², (iv) 56², and (v) 74² all end in 6.

   Where students slip. Trying to compute every square in full — checking just the units digit of the original number is enough to answer this.

5. 51 markGanita Prakash Cl-8 Part 1, Math Talk on trailing zeros

   If a number contains 3 zeros at the end, how many zeros will its square have at the end?

   Hint. Check a small example first, like 10² or 100², and see how the zero-count changes.

   Step 1 — Test the pattern on a simple case. 10 has 1 trailing zero, and 10² = 100 has 2 trailing zeros — double the original count.

   Step 2 — Apply the doubling rule. A number with 3 trailing zeros should give a square with 3 × 2 = 6 trailing zeros, since each trailing zero comes from one factor of 10 in the number, and squaring doubles how many of those factors appear.

   ✦ Answer: Six zeros.

   Where students slip. Assuming the zero count simply carries over unchanged (3 zeros → 3 zeros) — squaring doubles the count of trailing zeros, it doesn't preserve it.

6. 62 marksGanita Prakash Cl-8 Part 1, Math Talk following the 10²/20²/40²/100²... trailing-zero table

   What do you notice about the number of zeros at the end of a number and the number of zeros at the end of its square? Will this always happen? Can we say that squares can only have an even number of zeros at the end?

   Hint. Look at 700² = 490000 and 900² = 810000 alongside the smaller examples — does the doubling rule still hold?

   Step 1 — State the observed pattern. In every case (10², 20², 40², 100², 200², 700², 900²), the square's trailing-zero count is exactly double the original number's trailing-zero count.

   Step 2 — Confirm it always happens. This follows from how zeros come from factors of 10 = 2 × 5, since squaring the number squares each of those factors too, doubling their count.

   Step 3 — Answer the final question. Since the square's zero-count is always double a whole number, it must always be even — an odd number of trailing zeros is impossible for a perfect square.

   ✦ Answer: The square always has exactly double the number of trailing zeros as the original number; yes, this always happens; and yes, squares can only ever have an even number of trailing zeros.

   Where students slip. Treating this as a coincidence limited to the numbers in the table — the doubling is guaranteed by how squaring affects the factors of 10 in any number, not just the examples shown.

7. 71 markGanita Prakash Cl-8 Part 1, Math Talk on parity

   What can you say about the parity of a number and its square?

   Hint. Test one even and one odd number and see whether the parity carries over.

   Step 1 — Test an even number. 4 is even, and 4² = 16 is also even.

   Step 2 — Test an odd number. 5 is odd, and 5² = 25 is also odd, since multiplying two odd numbers together always gives an odd result.

   ✦ Answer: The square of an even number is always even, and the square of an odd number is always odd — a number and its square always share the same parity.

   Where students slip. Assuming squaring might flip parity in some cases — multiplying two numbers of the same parity (a number by itself) always preserves that parity.

8. 82 marksGanita Prakash Cl-8 Part 1, prompt after the square-root estimation discussion

   Find how many numbers lie between two consecutive perfect squares. Do you notice a pattern?

   Hint. Try a small case first, like between 16 (= 4²) and 25 (= 5²), then see if the count relates to 4.

   Step 1 — Try a small case. Between 16 and 25 lie 17, 18, 19, 20, 21, 22, 23, 24 — that's 8 numbers.

   Step 2 — Relate this to the smaller root. 8 = 2 × 4, and 4 is the square root of the smaller square (16). So there are 2n numbers between n² and (n+1)².

   Step 3 — State the general count. Between any two consecutive perfect squares p = n² and q = (n+1)², there are q − p − 1 = 2n numbers.

   ✦ Answer: There are 2n numbers between n² and (n+1)² (equivalently, q − p − 1 numbers, where p and q are the two consecutive squares) — yes, this is a consistent pattern.

   Where students slip. Guessing that the same fixed number of integers always lies between consecutive squares — the count actually grows as the squares get larger (it equals twice the smaller root).

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 1, Reprint 2026-27 (hegp101.pdf). This chapter's questions are scattered as 'Math Talk'/'Try This' prompts through the running text, plus two formal 'Figure it Out' exercise blocks. The book supplies its own answer key at the end of the chapter for every question that has one definite answer — that answer key is the ground truth this file is checked against. One question (Figure it Out — Squares, Q9, the tiny-squares picture) has no answer in the book's own key; it was solved here by rendering the actual figure and counting directly (9×8 = 72 tiles of a 5×5 grid each = 1800 tiny squares).. Questions are referenced from the NCERT textbook for identification.