In-text Questions — Same-Base Powers and the Lotus Ponds — Power Play

1. 12 marksGanita Prakash Cl-8 Part 1, Math Talk in §2.2 (the 3⁷ = 3² × 3⁵ observation)

   3⁷ can also be written as 3² × 3⁵. Can you reason out why? Use this observation to compute (i) 2⁹ (ii) 5⁷ (iii) 4⁶.

   Hint. Split the repeated multiplication into two smaller groups and count how many 3's are in each.

   Step 1 — Explain the identity. 3⁷ means 3 multiplied by itself 7 times; splitting those seven 3's into a group of 2 and a group of 5 gives (3 × 3) × (3 × 3 × 3 × 3 × 3) = 3² × 3⁵ — the total count of 3's is unchanged.

   Step 2 — Compute 2⁹ by splitting as 2⁴ × 2⁵. 2⁴ × 2⁵ = 16 × 32 = 512.

   Step 3 — Compute 5⁷ by splitting as 5³ × 5⁴. 5³ × 5⁴ = 125 × 625 = 78125.

   Step 4 — Compute 4⁶ by splitting as 4³ × 4³. 4³ × 4³ = 64 × 64 = 4096.

   ✦ Answer: 3⁷ = 3² × 3⁵ because splitting the seven repeated factors into two groups doesn't change how many there are in total; 2⁹ = 512, 5⁷ = 78125, 4⁶ = 4096.

   Where students slip. Adding the exponents of the two groups incorrectly, or splitting into groups that don't add up to the original exponent — the two exponents in the split must always sum to the original power.

2. 22 marksGanita Prakash Cl-8 Part 1, Math Talk in §2.2 (power of a power)

   Write the following expressions as a power of a power, in at least two different ways: (i) 8⁶ (ii) 7¹⁵ (iii) 9¹⁴ (iv) 5⁸

   Hint. Look for a way to write the base itself as a power (like 8 = 2³), as well as a way to split the exponent into a product of two smaller numbers.

   Step 1 — (i) 8⁶. Splitting the exponent: 8⁶ = (8²)³. Rewriting the base: since 8 = 2³, 8⁶ = 2¹⁸ = (2³)⁶.

   Step 2 — (ii) 7¹⁵. Splitting the exponent: 7¹⁵ = (7³)⁵, and also 7¹⁵ = (7⁵)³.

   Step 3 — (iii) 9¹⁴. Splitting the exponent: 9¹⁴ = (9²)⁷. Rewriting the base: since 9 = 3², 9¹⁴ = 3²⁸ = (3²)¹⁴.

   Step 4 — (iv) 5⁸. Splitting the exponent: 5⁸ = (5²)⁴, and also 5⁸ = (5⁴)².

   ✦ Answer: (i) 8⁶ = (8²)³ = (2³)⁶ (ii) 7¹⁵ = (7³)⁵ = (7⁵)³ (iii) 9¹⁴ = (9²)⁷ = (3²)¹⁴ (iv) 5⁸ = (5²)⁴ = (5⁴)².

   Where students slip. Multiplying the two exponents incorrectly when checking a power-of-a-power split — (aᵐ)ⁿ = aᵐⁿ, so the two exponents used in any valid split must multiply back to the original exponent.

3. 32 marksGanita Prakash Cl-8 Part 1, Magical Pond activity in §2.2

   The number of lotuses in a pond doubles every day; after 30 days it is completely covered. On which day was it half covered? Write the number of lotuses (in exponential form) when the pond was (i) fully covered (ii) half covered.

   Hint. If the number of lotuses doubles overnight, what must yesterday's count have been to reach today's full count?

   Step 1 — Find the half-covered day. Since the number of lotuses doubles every day, the day before it's full must have had exactly half as many lotuses — day 29.

   Step 2 — Write the fully covered count. Starting from 1 lotus on day 1 and doubling each day after, the pond is fully covered with 2³⁰ lotuses on day 30.

   Step 3 — Write the half-covered count. Half of 2³⁰ is 2²⁹.

   ✦ Answer: The pond was half covered on day 29; fully covered = 2³⁰ lotuses, half covered = 2²⁹ lotuses.

   Where students slip. Guessing the half-covered day is day 15 (half of 30) — doubling growth means the halfway point in quantity is only one day before the end, not halfway through the number of days.

4. 43 marksGanita Prakash Cl-8 Part 1, Magical Pond activity in §2.2 (the tripling pond)

   A single lotus is placed in a doubling pond. After 4 days, all its lotuses are moved to a tripling pond, where they grow for 4 more days. How many lotuses are in the tripling pond at the end? Would the count change if the order were swapped (tripling first, then doubling)?

   Hint. Work out the count after each 4-day stage separately, then regroup the final product.

   Step 1 — Grow in the doubling pond for 4 days. Starting from 1 lotus, doubling for 4 days gives 1 × 2 × 2 × 2 × 2 = 2⁴.

   Step 2 — Grow in the tripling pond for 4 more days. 2⁴ lotuses tripling for 4 days gives 2⁴ × 3 × 3 × 3 × 3 = 2⁴ × 3⁴.

   Step 3 — Check what happens if the order is swapped. Swapping gives 1 × 3⁴ × 2⁴ — the same two groups of factors, just multiplied in a different order, so the total is unchanged: 2⁴ × 3⁴ = (2 × 2 × 2 × 2) × (3 × 3 × 3 × 3), which regroups into (2 × 3) × (2 × 3) × (2 × 3) × (2 × 3) = (2 × 3)⁴ = 6⁴ = 1296.

   ✦ Answer: There are 2⁴ × 3⁴ = 6⁴ = 1296 lotuses either way — swapping the order of the two growth stages doesn't change the final count.

   Where students slip. Assuming the order of the two 4-day growth stages must matter — multiplication doesn't care about order, so doubling-then-tripling and tripling-then-doubling give exactly the same final count.

5. 52 marksGanita Prakash Cl-8 Part 1, following the mᵃ × nᵃ = (mn)ᵃ identity

   Use the identity mᵃ × nᵃ = (mn)ᵃ to compute the value of 2⁵ × 5⁵.

   Hint. Combine the two bases first, using the identity, rather than computing each power separately.

   Step 1 — Apply the mᵃ × nᵃ identity. 2⁵ × 5⁵ = (2 × 5)⁵ = 10⁵.

   Step 2 — Evaluate. 10⁵ = 100000.

   ✦ Answer: 2⁵ × 5⁵ = 10⁵ = 100000.

   Where students slip. Computing 2⁵ = 32 and 5⁵ = 3125 separately and multiplying them the long way — combining the bases first (into 10⁵) makes the arithmetic immediate.

6. 62 marksGanita Prakash Cl-8 Part 1, following the mᵃ ÷ nᵃ = (m/n)ᵃ identity

   Simplify 10⁴ ÷ 5⁴ and write it in exponential form.

   Hint. Use the identity mᵃ ÷ nᵃ = (m ÷ n)ᵃ to combine the bases before evaluating.

   Step 1 — Apply the identity. 10⁴ ÷ 5⁴ = (10 ÷ 5)⁴ = 2⁴.

   Step 2 — Evaluate. 2⁴ = 16.

   ✦ Answer: 10⁴ ÷ 5⁴ = 2⁴ = 16.

   Where students slip. Computing 10⁴ = 10000 and 5⁴ = 625 separately and dividing the long way — combining the bases first (into 2⁴) is far quicker.

7. 72 marksGanita Prakash Cl-8 Part 1, How Many Combinations activity in §2.3

   Roxie has 7 dresses, 2 hats, and 3 pairs of shoes. How many different ways can Roxie dress up?

   Hint. Each choice (dress, hat, shoes) is made independently, so multiply the number of options together.

   Step 1 — Identify the independent choices. Roxie picks one of 7 dresses, one of 2 hats, and one of 3 pairs of shoes, and each choice doesn't affect what's available for the others.

   Step 2 — Multiply the choices. 7 × 2 × 3 = 42, since each extra clothing category multiplies the total rather than adding to it.

   ✦ Answer: 42 different ways.

   Where students slip. Adding the three counts (7 + 2 + 3) instead of multiplying — since every combination of one dress, one hat and one pair of shoes is a distinct outfit, the counts must be multiplied, not added.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 1, Reprint 2026-27 (hegp102.pdf). Questions are scattered as 'Math Talk'/'Try This' prompts through the running text, plus two formal 'Figure it Out' blocks. Every answer here is checked against the book's own printed answer key at the end of the chapter.. Questions are referenced from the NCERT textbook for identification.