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NCERT Solutions Ganita Prakash (Part 1) Chapter 2 .2 Exponential Notation and Operations · Magical Pond — In-text Questions

Book page 252 Updated on2026-09-05

Q1.
In the middle of a beautiful, magical pond lies a bright pink lotus. The number of lotuses doubles every day in this pond. After 30 days, the pond is completely covered with lotuses. On which day was the pond half full?
Answer

On day 29.

Number of lotuses on day 30 = full pond
Each day the number doubles, so day 30 = 2 × (day 29)
day 29 = half of day 30 = half the pond
Why it happens: this is the most surprising fact about doubling, and the reason it is worth a whole page. With 29 days gone the pond looks only half covered — it seems there is plenty of time left. But everything that took 29 days to build is matched in the single day that follows. Exponential growth stays invisible for a long time and then finishes in one step.
Did you know? Turn the question round: on which day was the pond one-quarter covered? Day 28. One-eighth? Day 27. Going back is halving, exactly as going forward is doubling.
Q2.
If the pond is completely covered by lotuses on the 30th day, how much of it is covered by lotuses on the 29th day?
Answer

Half the pond is covered on day 29.

lotuses on day 30 = 2 × lotuses on day 29
so lotuses on day 29 = (1/2) × lotuses on day 30
= half the full pond
Why it happens: you do not need to know how many lotuses "full" means. The doubling rule fixes the ratio between consecutive days, and the ratio alone answers the question. This is the same reasoning as "the thickness after 10 folds is 1024 times the thickness before, wherever you start".
Q3.
Write the number of lotuses (in exponential form) when the pond was — (i) fully covered (ii) half covered
Answer

The pond starts with one lotus on day 0, and doubles each day.

Day0123…2930
Lotuses1 = 202 = 214 = 228 = 23…229230
(i) fully covered = 230 lotuses ( = 1,07,37,41,824)
(ii) half covered = 229 lotuses ( = 53,68,70,912)
Why it happens: the exponent is a count of days. Writing 230 instead of 1,07,37,41,824 keeps the structure in view — you can see at once that the answer to (ii) is one power lower, because 230 ÷ 2 = 229.
Q4.
There is another pond in which the number of lotuses triples every day. When both the ponds had no flowers, Damayanti placed a lotus in the doubling pond. After 4 days, she took all the lotuses from there and put them in the tripling pond. How many lotuses will be in the tripling pond after 4 more days?
Answer

24 × 34 = 16 × 81 = 1296 lotuses.

After the first 4 days (doubling):
1 × 2 × 2 × 2 × 2 = 24 = 16 lotuses

These 16 go into the tripling pond. After 4 more days:
24 × 3 × 3 × 3 × 3 = 24 × 34
= 16 × 81 = 1296
Why it happens: each pond simply multiplies whatever is in it by its own factor, once per day. Four days of doubling contribute a factor 24, four days of tripling contribute 34, and the two factors multiply. The starting single lotus contributes a factor of 1 and disappears from the working.
Q5.
What if Damayanti had changed the order in which she placed the flowers in the lakes? How many lotuses would be there?
Answer

Exactly the same — 1296 lotuses.

Tripling first, then doubling:
1 × 34 × 24 = 81 × 16 = 1296
Doubling first, then tripling:
1 × 24 × 34 = 16 × 81 = 1296
Why it happens: the total is a product of eight factors — four 2s and four 3s — and multiplication does not care about order. So the order of the ponds is irrelevant. This is worth noticing: for additive processes the order can matter a great deal, but for repeated multiplication it never does.
Q6.
Can this product be expressed as an exponent mⁿ, where m and n are some counting numbers?
Answer

Yes — 64.

1 × 34 × 24 = (3 × 3 × 3 × 3) × (2 × 2 × 2 × 2)
Regroup one 3 with one 2 at a time:
= (3 × 2) × (3 × 2) × (3 × 2) × (3 × 2)
= (3 × 2)4 = 64 = 1296 ✓

So m = 6 and n = 4. The general rule this demonstrates is

ma × na = (mn)a, where a is a counting number.
Why it happens: the rule works only because the two powers have the same exponent. Four 3s and four 2s can be paired off exactly, with nothing left over. With 35 × 24 the pairing would leave a spare 3, giving 64 × 3 — still simplified, but not a single power.
Check it yourself: use the rule on 25 × 55. It equals (2 × 5)5 = 105 = 1,00,000 — far easier than 32 × 3125.
Q7.
Use this observation to compute the value of 2⁵ × 5⁵.
Answer
25 × 55 = (2 × 5)5
= 105
= 1,00,000 (one lakh)

Check the long way: 25 = 32, 55 = 3125, and 32 × 3125 = 1,00,000. ✓

Why it happens: pairing each 2 with a 5 makes five 10s, and powers of 10 are the numbers we read most easily. This is exactly why any number ending in zeros can be split off as a power of 10 — the trailing zeros count the 2–5 pairs in its prime factorisation.
Q8.
Simplify 10⁴/5⁴ and write it in exponential form.
Answer
104 ÷ 54 = (10 × 10 × 10 × 10) ÷ (5 × 5 × 5 × 5)
Pair each 10 with a 5:
= (10 ÷ 5) × (10 ÷ 5) × (10 ÷ 5) × (10 ÷ 5)
= (10 ÷ 5)4 = 24 = 16

Check: 104 = 10000, 54 = 625, and 10000 ÷ 625 = 16. ✓

In general: ma ÷ na = (m ÷ n)a (n ≠ 0)
Why it happens: this is the division twin of the previous rule, and it works for the same reason — the equal exponents let every factor on top be paired with exactly one factor below. Read from right to left it is just as useful: (m/n)a can be split as ma/na.
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