Class 8 - Ganita Prakash Chapter 1 - Part 6
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Class 8 Ganita Prakash Fractions in Disguise
Achievement Test - Part 6
SIMPLE & COMPOUND GROWTH Chapter Achievement Test 1.1 LINEAR: p(1 + rt) EXPONENTIAL: p(1 + r)α΅— Part 6: Interest & Population Dynamics
Welcome to Part 6 of the Class 8 Fractions in Disguise Test Module!

In this crucial section, we transition into Module III: Growth & Decay. We will explore the mathematical difference between Simple Growth (Linear) and Compound Growth (Exponential). You will compare bank deposits, deduce growth formulas, and project population sizes over time. Review your textbook solutions (Pages 20-24) and test your mastery![cite: 25, 26, 27, 28]
1 Bank Deposit Comparison Pg 21
If one deposits ₹6000 in the bank at 10% p.a., what is the amount after 3 years and the percentage gain with and without compounding?[cite: 25, 26]
Concept:
1
Without compounding (Simple Growth), interest is calculated only on the original principal: $A = P + (P \times r \times t)$.[cite: 25]
2
With compounding (Exponential Growth), interest is calculated on the principal *plus* previous interest: $A = P(1+r)^t$.[cite: 26]
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Hint:
1
Without: Interest per year = $6000 \times 0.1 = 600$. For 3 years: $600 \times 3 = 1800$.
2
With: Multiply by $1.1$ three times: $6000 \times 1.1 \times 1.1 \times 1.1$.
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Mistakes to Avoid:
!
Confusing 1.1 with 1.3: $(1.1)^3$ is $1.331$, not $1.3$. Compounding always yields a higher multiplier than simple addition.
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Solution:
1. Without Compounding:
Amount = $6000 + (6000 \times 0.10 \times 3)$
$= 6000 + 1800 = \text{₹}7800$
Percentage Gain = $\frac{7800}{6000} \times 100 = 130\%$

2. With Compounding:
Amount = $6000 \times (1.1)^3 = 6000 \times 1.331 = \text{₹}7986$
Percentage Gain = $\frac{7986}{6000} \times 100 = 133.1\%$

Ans: No Comp: ₹7800 (130%). Comp: ₹7986 (133.1%).
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2 Yahapur & Wahapur Banks Pg 22
Compare how much one gets if they deposit ₹20,000 for 2 years at 10% p.a. (Bank of Yahapur) and for 4 years at 5% p.a. (Bank of Wahapur) with and without compounding.[cite: 26, 27]
Concept:
1
Apply the formulas. Simple: $P + (P \times r \times t)$. Compound: $P(1+r)^t$.[cite: 26, 27]
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Hint:
1
Yahapur Simple: $20000 + (20000 \times 0.1 \times 2)$. Wahapur Simple: $20000 + (20000 \times 0.05 \times 4)$.
2
Yahapur Comp: $20000 \times (1.1)^2$. Wahapur Comp: $20000 \times (1.05)^4$.
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Mistakes to Avoid:
!
Assuming equal returns: While both simple interest scenarios yield 20% total ($10 \times 2 = 20$, and $5 \times 4 = 20$), the compound interest scenarios diverge significantly because of compounding frequency.
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Solution:
Bank of Yahapur (10%, 2 years):
Without Comp: $20000 + (20000 \times 0.1 \times 2) = \text{₹}24,000$
With Comp: $20000 \times (1.1)^2 = 20000 \times 1.21 = \text{₹}24,200$

Bank of Wahapur (5%, 4 years):
Without Comp: $20000 + (20000 \times 0.05 \times 4) = \text{₹}24,000$
With Comp: $20000 \times (1.05)^4 = 20000 \times 1.215506 = \text{₹}24,310.125$

Ans: Wahapur compound yields the highest return.
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3 Formula Selection Pg 24
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) $p \times 6 \times 4$ (iv) $p \times \frac{0.06}{100} \times 4$ (vii) $p + (p \times 0.06 \times 4)$[cite: 28]
Concept:
1
When compounding is NOT done, it's Simple Interest. Total Amount = Principal + Simple Interest. Formula: $A = p + (p \times r \times t)$.[cite: 28]
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Hint:
1
Convert 6% to a decimal. $6\% = 0.06$.
2
Find the option that adds $p$ to the product of $p$, $0.06$, and $4$.
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Mistakes to Avoid:
!
Forgetting the principal: Option (i) only calculates the interest (and does it wrong by using 6 instead of 0.06). You must add the base principal back to get the *total amount*.
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Solution:
Rate ($r$) = $6\% = 0.06$
Time ($t$) = $4$

Total Amount = $P + \text{Interest}$
$= p + (p \times r \times t)$
$= p + (p \times 0.06 \times 4)$

Ans: Expression (vii) is correct.
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4 Post Office Difference Pg 24
The post office offers 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?[cite: 28]
Concept:
1
Calculate Simple Interest: $I = P \times r \times t$. Then calculate Compound Interest: $CI = P(1+r)^t - P$. Finally, subtract the two to find the difference.[cite: 28]
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Hint:
1
Simple Interest: $50000 \times 0.07 \times 3$.
2
Compound Amount: $50000 \times (1.07)^3$. Subtract 50000 to get Compound Interest.
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Mistakes to Avoid:
!
Comparing Amount vs Interest: The question asks "how much *more* (interest) would one get". Ensure you extract the interest from the compound amount before comparing, or just compare the two final amounts.
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Solution:
1. Without Compounding (Simple Interest):
Interest = $50000 \times 0.07 \times 3 = 3500 \times 3 = \text{₹}10,500$

2. With Compounding:
Amount = $50000 \times (1.07)^3 = 50000 \times 1.225043 = 61252.15$
Comp Interest = $61252.15 - 50000 = \text{₹}11,252.15$

3. Difference:
$11252.15 - 10500 = \text{₹}752.15$

Ans: SI is ₹10,500. One gets ₹752.15 more with CI.
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5 Population & Bacteria Pg 24
1. The population of a city is rising by 3% every year. Current population is 1.5 crore, expected after 3 years?
2. Bacteria increases at 2.5% per hour. Find count at the end of 2 hours if initial is 5,06,000.[cite: 28]
Concept:
1
Population and bacterial growth are classic examples of *Compound Growth*. The base increases every cycle. Formula: $A = P(1+r)^t$.[cite: 28]
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Hint:
1
Population: $1.5 \times (1.03)^3$.
2
Bacteria: $506000 \times (1.025)^2$. Note: $2.5\% = 0.025$.
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Mistakes to Avoid:
!
Treating 2.5% as 0.25: $2.5\% = \frac{2.5}{100} = 0.025$. Using 0.25 would mean a massive 25% growth rate!
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Solution:
1. City Population:
$A = 1.5 \times (1 + 0.03)^3$
$= 1.5 \times (1.03)^3 = 1.5 \times 1.092727 \approx 1.639\text{ Crore}$

2. Bacteria Count:
$A = 506000 \times (1 + 0.025)^2$
$= 506000 \times (1.025)^2 = 506000 \times 1.050625$
$= 531,616.25$ (Round to 531,616)

Ans: Pop: ~1.639 Cr. Bacteria: 531,616.
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1 Growth Identification
Which of the following scenarios models linear (simple) growth rather than exponential (compound) growth?
A. Bacteria multiplying by 5% an hour.
B. City population rising 2% a year.
C. Earning exactly ₹500 interest every year.
D. A virus spreading 10% daily.
Concept:
1
Linear growth adds a constant *amount* every cycle. Exponential growth multiplies by a constant *percentage* (which means the added amount grows larger over time).
Hint:
1
Look for the option that describes a fixed, unchanging addition.
Mistakes to Avoid:
!
Assuming all biology is linear: Biological growth (populations, viruses, bacteria) is almost always exponential.
Solution:
Earning exactly ₹500 every year means the growth is $P + 500t$. This is the equation of a straight line (linear).
Ans: Option C
2 Simple Interest
What is the total amount received if ₹5000 is invested at 10% p.a. without compounding for 2 years?
A. ₹5500
B. ₹6000
C. ₹6050
D. ₹1000
Concept:
1
Without compounding = Simple Interest. $A = P + (P \times r \times t)$.
Hint:
1
$10\%$ of 5000 is 500.
2
For 2 years, interest is $500 \times 2 = 1000$. Add to principal.
Mistakes to Avoid:
!
Finding the interest only: Option D is 1000, which is just the interest. The question asks for the *total amount*.
Solution:
Interest = $5000 \times 0.10 \times 2 = 1000$
Total Amount = $5000 + 1000 = \text{₹}6000$
Ans: Option B (₹6000)
3 Compound Interest
What is the total amount received if ₹5000 is invested at 10% p.a. WITH compounding for 2 years?
A. ₹5500
B. ₹6000
C. ₹6050
D. ₹6500
Concept:
1
With compounding, apply the formula $A = P(1+r)^t$.
Hint:
1
Calculate $5000 \times (1.1)^2$.
2
$(1.1)^2 = 1.21$. Multiply $5000 \times 1.21$.
Mistakes to Avoid:
!
Using simple interest: Option B is the Simple Interest answer. Compounding yields slightly more because you earn interest on the Year 1 interest.
Solution:
Amount = $5000 \times (1.1)^2$
$= 5000 \times 1.21 = \text{₹}6050$
Ans: Option C (₹6050)
4 Population Multiplier
A village grows by exactly 10% every year. Which expression represents its population after 3 years if the initial population is $P$?
A. $P + (0.3 \times P)$
B. $P \times 1.3$
C. $P(1.1)^3$
D. $P \times 1.33$
Concept:
1
Population growth compounds upon the previous year's total. It is an exponential model.
Hint:
1
Use $P(1+r)^t$ where $r=0.10$ and $t=3$.
Mistakes to Avoid:
!
Choosing linear options: Options A and B represent simple linear growth. Option D evaluates out to 1.33 but the true value of $1.1^3$ is $1.331$. Thus, the exact expression is C.
Solution:
Formula: $P(1+0.10)^3 = P(1.1)^3$
Ans: Option C
5 CI vs SI Difference
What is the difference between the Compound Interest and Simple Interest on ₹1,000 at 10% p.a. for 2 years?
A. ₹0
B. ₹10
C. ₹20
D. ₹100
Concept:
1
Calculate both values. SI = $(P \times r \times t)$. CI = $P(1+r)^t - P$. Subtract to find the difference.
Hint:
1
SI = $1000 \times 0.10 \times 2 = 200$.
2
CI = $1000 \times 1.21 - 1000 = 1210 - 1000 = 210$.
Mistakes to Avoid:
!
Calculation error: A common fast trick for 2-year difference is $D = P \times (r)^2$. If you use this, $1000 \times (0.1)^2 = 1000 \times 0.01 = 10$.
Solution:
SI = 200
Amount (Compound) = $1000 \times (1.1)^2 = 1210$. CI = 210
Difference = $210 - 200 = 10$
Ans: Option B (₹10)
1 Compounded Bank Loan
A farmer takes a loan of ₹10,000 from a bank at an interest rate of 8% per annum, compounded annually. How much total amount must he pay back to clear the loan after 2 years?
Concept:
1
Use the exponential compound formula: $A = P(1+r)^t$.
Hint:
1
$r = 8\% = 0.08$.
2
Calculate $10000 \times (1.08)^2$. $(1.08 \times 1.08 = 1.1664)$.
Mistakes to Avoid:
!
Simple Interest Error: $10000 + (10000 \times 0.08 \times 2) = 11600$. This ignores the compounding interest on the first year's interest.
Solution:
$A = 10000 \times (1 + 0.08)^2$
$= 10000 \times (1.08)^2 = 10000 \times 1.1664$
$= \text{₹}11,664$

Ans: He must pay back ₹11,664.
2 Projecting Town Growth
A small town has a population of 25,000. It is growing steadily at a rate of 4% per year. What will the population be after 2 years?
Concept:
1
Population growth is exponential. $A = P(1+r)^t$, where $r = 0.04$ and $t = 2$.
Hint:
1
Calculate $(1.04)^2 = 1.0816$.
2
Multiply $25000 \times 1.0816$.
Mistakes to Avoid:
!
Adding exactly 1000 each year: Year 1 growth is $4\%$ of $25000 = 1000$. But Year 2 growth is $4\%$ of $26000 = 1040$. Do not use simple linear growth!
Solution:
Population = $25000 \times (1.04)^2$
$= 25000 \times 1.0816 = 27,040$

Ans: The population will be 27,040.
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