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NCERT Solutions Ganita Prakash (Part 2) Chapter 1 –24Section 1.3 — Figure it Out

Book page 22 Updated on2026-09-05

Q1.
Bank of Yahapur offers an interest of 10% p.a. Compare how much one gets if they deposit ₹20,000 for a period of 2 years with compounding and without compounding annually.
Answer
Without compounding — the principal stays ₹20,000 every year
Interest each year = 10% of 20,000 = ₹2000
Interest for 2 years = 2 × 2000 = ₹4000
Amount = 20,000 + 4000 = ₹24,000

With compounding — each year's interest joins the principal
After year 1: 20,000 × 1.1 = ₹22,000
After year 2: 22,000 × 1.1 = ₹24,200
(or 20,000 × 1.1² = 20,000 × 1.21 = 24,200)

Compounding gives ₹200 more.

Why it happens: The extra ₹200 is exactly 10% of the first year's interest of ₹2000. In the compounding account that ₹2000 stays in the deposit and earns interest of its own during the second year; in the other account it is paid out and earns nothing. That is the whole difference between the two options — interest on interest.
Q2.
Bank of Wahapur offers an interest of 5% p.a. Compare how much one gets if one deposits ₹20,000 for a period of 4 years with compounding and without compounding annually.
Answer
Without compounding
Interest each year = 5% of 20,000 = ₹1000
Interest for 4 years = 4 × 1000 = ₹4000
Amount = ₹24,000

With compounding
Amount = 20,000 × (1.05)⁴
1.05² = 1.1025, so 1.05⁴ = 1.1025 × 1.1025 = 1.21550625
= 20,000 × 1.21550625 = ₹24,310.125 = ₹24,310.13 to the nearest paisa
YearOpening amountInterest at 5%Closing amount
1₹20,000₹1000₹21,000
2₹21,000₹1050₹22,050
3₹22,050₹1102.50₹23,152.50
4₹23,152.50₹1157.63₹24,310.13

Compounding gives about ₹310 more.

Q3.
Do you observe anything interesting in the solutions of the two questions above? Share and discuss.
Answer

Without compounding both deposits give exactly the same amount, ₹24,000 — but with compounding they do not.

10% for 2 years5% for 4 years
Total simple interest₹4000₹4000
Amount, no compounding₹24,000₹24,000
Amount, compounded₹24,200₹24,310.13
Without compounding the amount is p(1 + rt), and rt is the same for both:
0.10 × 2 = 0.20  and  0.05 × 4 = 0.20

With compounding the amount is p(1 + r)ᵗ, and these differ:
(1.10)² = 1.2100  but  (1.05)⁴ = 1.2155
Why it happens: Without compounding, only the product rt matters — interest is added on the same principal every time, so a high rate for a short time and a low rate for a long time balance out. With compounding, the number of times the interest is folded back matters too. The 5% account folds interest back four times instead of two, and each fold earns interest on all the earlier interest. More frequent compounding beats a higher rate applied fewer times, even when the plain totals agree.
Q4.
Jasmine invests amount ‘p’ for 4 years at an interest of 6% p.a. Which of the following expression(s) describe the total amount she will get after 4 years when compounding is not done? (i) p × 6 × 4 (ii) p × 0.6 × 4 (iii) p × (0.6/100) × 4 (iv) p × (0.06/100) × 4 (v) p × 1.6 × 4 (vi) p × 1.06 × 4 (vii) p + (p × 0.06 × 4)
Answer

Only (vii).

Without compounding, amount = p + interest
Interest = p × r × t = p × 0.06 × 4 = 0.24p
Amount = p + 0.24p = 1.24p — which is what (vii) says
OptionWhat it actually computes
(i) p × 6 × 424p — treats 6% as the number 6
(ii) p × 0.6 × 42.4p — uses 60%, not 6%
(iii) p × (0.6/100) × 40.024p — a hundred times too small
(iv) p × (0.06/100) × 40.0024p — divides by 100 twice over
(v) p × 1.6 × 46.4p
(vi) p × 1.06 × 44.24p — multiplies the whole amount by 4 instead of the interest
(vii) p + (p × 0.06 × 4)1.24p ✓
Why it happens: Options (iii) and (iv) are the same slip made twice — 6% is already 0.06, so writing 0.06/100 divides by 100 a second time. Option (vi) is the subtler error: p × 1.06 is the amount after one year, and multiplying that by 4 quadruples the whole deposit rather than repeating the interest. Without compounding the interest is added four times, but the principal is counted only once.
Q5.
The post office offers an interest of 7% p.a. How much interest would one get if one invests ₹50,000 for 3 years without compounding? How much more would one get if it was compounded?
Answer
Without compounding
Interest = p × r × t = 50,000 × 0.07 × 3
= ₹10,500

With compounding
Amount = 50,000 × (1.07)³
1.07² = 1.1449, so 1.07³ = 1.1449 × 1.07 = 1.225043
= 50,000 × 1.225043 = ₹61,252.15
Interest = 61,252.15 – 50,000 = ₹11,252.15

Extra from compounding = 11,252.15 – 10,500 = ₹752.15
Check it yourself: Year by year — 50,000 → 53,500 → 57,245 → 61,252.15. The yearly interest rises from ₹3500 to ₹3745 to ₹4007.15, because the principal it is charged on keeps growing.
Q6.
Giridhar borrows a loan of ₹12,500 at 12% per annum for 3 years without compounding and Raghava borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?
Answer

Giridhar pays more — by ₹362.50.

Giridhar — 12%, no compounding
Interest = 12,500 × 0.12 × 3 = ₹4500

Raghava — 10%, compounded annually
Amount = 12,500 × (1.1)³ = 12,500 × 1.331 = ₹16,637.50
Interest = 16,637.50 – 12,500 = ₹4137.50

Difference = 4500 – 4137.50 = ₹362.50
Why it happens: Compounding is not automatically the costlier deal. Over 3 years, 10% compounded multiplies the debt by 1.331 — an effective 33.1% — while 12% simple multiplies it by 1.36, an effective 36%. The 2 percentage points of extra rate outweigh what compounding adds over such a short term. Over a longer period the balance tips the other way: at 10% compounded for 10 years the multiplier is 2.594, well past 12% simple, which reaches only 2.2.
Q7.
Consider an amount ₹1000. If this grows at 10% p.a., how long will it take to double when compounding is done vs. when compounding is not done? Is compounding an example of exponential growth and not-compounding an example of linear growth?
Answer

Without compounding: 10 years. With compounding: 8 years. And yes — compounding is exponential growth, and not-compounding is linear growth.

No compounding: interest is ₹100 every year
Amount = 1000 + 100t. Doubling needs 100t = 1000
t = 10 years

With compounding: Amount = 1000 × (1.1)ᵗ
(1.1)⁷ = 1.9487 → ₹1948.72 — not yet doubled
(1.1)⁸ = 2.1436 → ₹2143.59 — past ₹2000
So it doubles during the 8th year → 8 years on annual compounding
End of year12467810
No compounding (₹)1100120014001600170018002000
Compounded (₹)110012101464.101771.561948.722143.592593.74
Why it happens: Without compounding the amount is p(1 + rt) — t appears once, multiplied by a constant, so the growth is a fixed ₹100 every year and the graph is a straight line. That is linear growth. With compounding the amount is p(1 + r)ᵗ — t is now an exponent, so each year multiplies the previous amount by 1.1 rather than adding to it. The yearly increase itself grows: ₹100 in year 1, ₹110 in year 2, ₹121 in year 3. That is exponential growth, and it is why the compounded column pulls further and further ahead the longer you leave it.
Q8.
The population of a city is rising by about 3% every year. If the current population is 1.5 crore, what is the expected population after 3 years?
Answer

A yearly percentage rise compounds, because each year's growth is measured on the population at the start of that year.

Population after 3 years = 1.5 crore × (1.03)³
1.03² = 1.0609, so 1.03³ = 1.0609 × 1.03 = 1.092727
= 1.5 × 1.092727 crore = 1.6390905 crore
≈ 1.64 crore (about 1,63,90,905 people)
Why it happens: Simply adding 3% three times — 9% of 1.5 crore = 1.635 crore — is close but slightly low. The extra 0.0041 crore (about 41,000 people) comes from the growth of the people added in years 1 and 2, who go on having children themselves. Over 3 years the gap is small; over 30 years, 1.03³⁰ = 2.43 against a simple 1.90, and the difference is enormous.
Q9.
In a laboratory, the number of bacteria in a certain experiment increases at the rate of 2.5% per hour. Find the number of bacteria at the end of 2 hours if the initial count is 5,06,000.
Answer
Count after 2 hours = 5,06,000 × (1.025)²
(1.025)² = 1.050625
= 5,06,000 × 1.050625
= 5,31,616.25 ≈ 5,31,616 bacteria

Step by step:

After 1 hour: 5,06,000 × 1.025 = 5,18,650
After 2 hours: 5,18,650 × 1.025 = 5,31,616.25
Careful: The first hour adds 12,650 bacteria and the second adds 12,966.25 — more, because the second 2.5% is taken on the larger count of 5,18,650. Since bacteria come in whole numbers, reporting about 5,31,616 is right; the decimal is an artefact of using a smooth percentage on a discrete count.
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