Class 8 - Ganita Prakash Chapter 1 - Part 8
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Class 8 Ganita Prakash Fractions in Disguise
Achievement Test - Part 8
ADVANCED APPLICATIONS I Chapter Achievement Test 1.1 x% of y = y% of x SP = CP(1 + Margin) Part 8: Complex Algebra & Multi-Step Logic
Welcome to Part 8 of the Class 8 Fractions in Disguise Test Module!

In this advanced section, we combine all previous rules to solve multi-step problems, prove algebraic properties (like $x\%$ of $y = y\%$ of $x$), and handle combined populations, taxes, and complex cost equations. Review your textbook solutions (Pages 28-29), tackle the 4-Pillar MCQs, and try the extra practice questions.[cite: 32, 33]
1 Bengaluru Population Pg 28
The population of Bengaluru in 2025 is about 250% of its population in 2000. If the population in 2000 was 50 lakhs, what is the population in 2025?[cite: 32]
Concept:
1
"Is 250% of" means you directly multiply the base by $\frac{250}{100}$ (or 2.5).[cite: 32]
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Hint:
1
Calculate $2.5 \times 50$.
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Mistakes to Avoid:
!
Adding to the base: It says "is 250% *of*", not "increased *by* 250%". Do not do $50 + (2.5 \times 50)$.
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Solution:
Base Population (2000) = $50\text{ Lakhs}$
Population (2025) = $250\%$ of 50 Lakhs
$= \frac{250}{100} \times 50$
$= 2.5 \times 50 = 125\text{ Lakhs}$

Ans: The population in 2025 is 125 Lakhs.
🎯 Marking Scheme:
• 1 Mark for setting up $2.5 \times 50$.
• 1 Mark for final answer (125 Lakhs).[cite: 32]
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2 GST Phone Expression Pg 28
The price of a mobile phone is ₹8,250. A GST of 18% is added to the price. Which of the following gives the final price of the phone including the GST?
(i) $8250 + 18$
(iii) $8250 + \frac{18}{100}$
(v) $8250 \times 1.18$[cite: 32]
Concept:
1
A tax is an addition to the base price. Final Price = Base + ($18\%$ of Base) = $118\%$ of Base. $118\% = 1.18$.[cite: 32]
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Hint:
1
Look for the expression that multiplies the base by $1.18$.
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Mistakes to Avoid:
!
Adding flat numbers: Options (i) and (iii) simply add the number 18 or 0.18 to the price. Tax is a proportion of the price, not a flat scalar addition.
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Solution:
Base Price = ₹8250
Final Price = $100\% \text{ Base} + 18\% \text{ GST} = 118\%$
$118\% = \frac{118}{100} = 1.18$
Therefore, Final Price = $8250 \times 1.18$

Ans: Option (v) is correct.
🎯 Marking Scheme:
• 1 Mark for selecting Option (v).[cite: 32]
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3 Mice Population Multipliers Pg 28
The monthly percentage change in population of mice is given: Month 1 change +5%, Month 2 change -2%, and Month 3 change -3%. If initial is $p$, which expression gives the final population after three months?[cite: 32]
Concept:
1
Successive percentage changes act as chained multipliers on the base. A $+5\%$ change is a multiplier of $1.05$. A $-2\%$ change is $0.98$. A $-3\%$ change is $0.97$.[cite: 32]
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Hint:
1
Multiply $p \times (1 + 0.05) \times (1 - 0.02) \times (1 - 0.03)$.
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Mistakes to Avoid:
!
Adding the percentages: Do not calculate $+5\% - 2\% - 3\% = 0\%$ net change. The base changes every month, so it's not a 0% net change.
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Solution:
Month 1 Multiplier = $100\% + 5\% = 105\% = 1.05$
Month 2 Multiplier = $100\% - 2\% = 98\% = 0.98$
Month 3 Multiplier = $100\% - 3\% = 97\% = 0.97$

Final Population = $p \times 1.05 \times 0.98 \times 0.97$

Ans: Statement (ii) is the correct expression.
🎯 Marking Scheme:
• 1 Mark for converting % to decimal multipliers.
• 1 Mark for selecting the correct chained multiplication expression.[cite: 32]
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4 Margin vs Discount Trap Pg 28
A shopkeeper initially set the price of a product with a 35% profit margin. Due to poor sales, he decided to offer a 30% discount on the selling price. Will he make a profit or a loss? Give reasons for your answer.[cite: 32]
Concept:
1
Assume a base Cost Price (CP) of 100 to make it easy. First apply the $35\%$ increase to get the Marked Price. Then apply the $30\%$ discount (multiply by $0.70$) to that Marked Price. Compare final to 100.[cite: 32]
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Hint:
1
Let CP = 100. Marked Price = 135.
2
Discounted Price = $135 \times 0.70$.
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Mistakes to Avoid:
!
Assuming $+35\% - 30\% = +5\%$ Profit: The 30% discount is applied to the inflated 135, so 30% of 135 is much larger than 30% of 100. It wipes out the profit!
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Solution:
Let Cost Price (CP) = 100.
Marked Price (after 35% margin) = $100 + 35 = 135$

Final Selling Price (after 30% discount on MP):
$\text{SP} = 135 \times 0.70 = 94.5$

Since the final SP (94.5) is less than the CP (100), he makes a loss.
Loss = $100 - 94.5 = 5.5\%$

Ans: He will make a loss of 5.5%.
🎯 Marking Scheme:
• 1 Mark for calculating Marked Price correctly.
• 1 Mark for calculating the final discounted price.
• 1 Mark for stating it is a Loss.[cite: 32]
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5 Commutative Algebra Pg 29
What is 5% of 40? What is 40% of 5? What do you notice? Can you make a general statement and justify it using algebra, comparing $x\%$ of $y$ and $y\%$ of $x$?[cite: 33]
Concept:
1
Because multiplication is commutative ($A \times B = B \times A$), percentages are reversible! $x\%$ of $y$ will always yield the exact same numerical result as $y\%$ of $x$.[cite: 33]
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Hint:
1
Write out $x\%$ of $y$ algebraically: $\frac{x}{100} \times y$.
2
Write out $y\%$ of $x$ algebraically: $\frac{y}{100} \times x$. Are they the same?
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Mistakes to Avoid:
!
Just giving numerical answers: The question specifically asks you to *justify it using algebra*. You must provide the $x$ and $y$ proof to get full marks.
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Solution:
1. Numerical test:
$5\%$ of $40 = \frac{5}{100} \times 40 = 2$.
$40\%$ of $5 = \frac{40}{100} \times 5 = 2$.
We notice they are exactly the same.

2. Algebraic Proof:
$x\%$ of $y = \frac{x}{100} \times y = \frac{xy}{100}$
$y\%$ of $x = \frac{y}{100} \times x = \frac{yx}{100}$
Since $xy = yx$, we prove that $x\%$ of $y = y\%$ of $x$.
🎯 Marking Scheme:
• 1 Mark for calculating the numbers correctly (2).
• 1 Mark for the correct algebraic proof $\frac{xy}{100}$.[cite: 33]
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6 Excursion Ratios Pg 29
A school is organising an excursion. 40% of them are Grade 8 students and the rest are Grade 9 students. Among these Grade 8 students, 60% are girls. (i) What percentage of the total students are Grade 8 girls? (ii) If the total number of students is 160, how many are Grade 8 girls?[cite: 33]
Concept:
1
This is a nested percentage problem ("percentage of a percentage"). You need $60\%$ OF the $40\%$. Mathematically, multiply the two percentage fractions.[cite: 33]
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Hint:
1
(i) Calculate $0.60 \times 0.40$.
2
(ii) Multiply that total percentage by 160.
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Mistakes to Avoid:
!
Calculating 60% of total: The 60% refers *only* to the Grade 8 students, not the entire excursion pool.
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Solution:
(i) Percentage of total students:
$= 60\%$ of $40\%$ of Total
$= 0.60 \times 0.40 = 0.24 = 24\%$

(ii) Number of girls out of 160:
Number = $24\%$ of $160$
$= \frac{24}{100} \times 160 = 24 \times 1.6 = 38.4$

Ans: (i) 24%. (ii) 38.4 (Math theoretical answer).
🎯 Marking Scheme:
• 1 Mark for finding 24% correctly.
• 1 Mark for calculating 24% of 160.[cite: 33]
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7 Equating SP & CP Pg 29
A shopkeeper sells pencils at a price such that the selling price of 3 pencils is equal to the cost of 5 pencils. Does he make a profit or a loss? What is his profit or loss percentage?[cite: 33]
Concept:
1
Translate the sentence directly into algebra: $3 \times \text{SP} = 5 \times \text{CP}$. Rearrange to find the ratio of SP to CP.[cite: 33]
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Hint:
1
If $3 \text{SP} = 5 \text{CP}$, then $\text{SP} = \frac{5}{3} \text{CP}$.
2
Profit = $\text{SP} - \text{CP} = \frac{5}{3}\text{CP} - 1\text{CP} = \frac{2}{3}\text{CP}$. Calculate $\frac{2}{3}$ as a percentage.
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Mistakes to Avoid:
!
Assuming loss because 3 < 5: He only has to sell 3 pencils to recover the cost of 5! The 2 extra pencils are pure profit. It is a profit, not a loss.
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Solution:
$3 \times \text{SP} = 5 \times \text{CP}$
$\text{SP} = \frac{5}{3} \text{CP}$

Since SP > CP ($\frac{5}{3} > 1$), it is a Profit.
Profit = $\text{SP} - \text{CP} = \frac{5}{3}\text{CP} - \text{CP} = \frac{2}{3}\text{CP}$

Profit \% = $\frac{\text{Profit}}{\text{CP}} \times 100 = \frac{2}{3} \times 100 = 66.66\%$

Ans: Profit of 66.66% ($66 \frac{2}{3}\%$).
🎯 Marking Scheme:
• 1 Mark for setting up algebraic equality.
• 1 Mark for determining Profit.
• 1 Mark for correct percentage (66.66%).[cite: 33]
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1 Successive Identical Changes
If a population decreases by 10% in Year 1, and increases by 10% in Year 2, what is the net change after 2 years?
A. 0% Change
B. 1% Decrease
C. 1% Increase
D. 20% Decrease
Concept:
1
A decrease followed by an identical increase (or vice versa) always results in a net decrease. Multiply the factors: $0.90 \times 1.10$.
Hint:
1
$0.90 \times 1.10 = 0.99$.
2
$0.99$ means $99\%$, which is a $1\%$ drop from $100\%$.
Mistakes to Avoid:
!
Assuming 0%: As covered previously, $-10\% + 10\%$ is an illusion. The bases change.
Solution:
Final = $100 \times 0.9 \times 1.1 = 99$.
Change = $100 - 99 = 1\%$ decrease.
Ans: Option B
2 Percentage Reversal
Without using a calculator, what is 8% of 50?
A. 2
B. 4
C. 8
D. 40
Concept:
1
Use the commutative property: $x\%$ of $y = y\%$ of $x$.
Hint:
1
$8\%$ of $50$ is exactly the same as $50\%$ of $8$.
2
What is half of 8?
Mistakes to Avoid:
!
Doing the hard math: Trying to calculate $\frac{8 \times 50}{100}$ is prone to decimal placement errors. Flipping it to $50\%$ of 8 is instant.
Solution:
$8\%$ of $50 = 50\%$ of $8$.
$50\%$ is $\frac{1}{2}$. Half of 8 is 4.
Ans: Option B (4)
3 GST Math
A bill shows a base total of ₹500 and a final total of ₹560. What was the GST percentage applied?
A. 10%
B. 12%
C. 15%
D. 60%
Concept:
1
GST % = $\frac{\text{Tax Amount}}{\text{Base Total}} \times 100$. Find the tax amount first by subtracting.
Hint:
1
Tax Amount = $560 - 500 = 60$.
2
Calculate $\frac{60}{500} \times 100$.
Mistakes to Avoid:
!
Choosing 60%: The tax amount is 60, but the *percentage* is 60 out of 500, which is much smaller.
Solution:
Tax = $560 - 500 = 60$
GST \% = $\frac{60}{500} \times 100 = \frac{60}{5} = 12\%$
Ans: Option B (12%)
4 Nested Percentages
In a class, 50% of the students play sports. Of those who play sports, 20% play tennis. What percentage of the ENTIRE class plays tennis?
A. 10%
B. 20%
C. 30%
D. 70%
Concept:
1
Multiply the percentage fractions to find the intersection of the two groups. Total % = $20\%$ OF $50\%$.
Hint:
1
Calculate $0.20 \times 0.50$.
Mistakes to Avoid:
!
Answering 20%: 20% of the *sports players* play tennis. Since only half the class plays sports, the total percentage of tennis players in the whole class must be much lower.
Solution:
Total % = $20\%$ of $50\%$
$= 0.20 \times 0.50 = 0.10$
$0.10 = 10\%$
Ans: Option A (10%)
5 SP = CP Logic
The selling price of 4 items equals the cost price of 5 items. What is the profit percentage?
A. 10%
B. 20%
C. 25%
D. 125%
Concept:
1
$4 \text{SP} = 5 \text{CP} \implies \text{SP} = \frac{5}{4} \text{CP}$. Profit is $\text{SP} - \text{CP} = \frac{1}{4} \text{CP}$.
Hint:
1
What percentage is $\frac{1}{4}$?
Mistakes to Avoid:
!
Calculating 1/5: Using the 5 as the denominator will give 20%. Remember, profit is relative to 1 CP. If SP is $\frac{5}{4}$ of CP, the excess is $\frac{1}{4}$.
Solution:
$\text{SP} = 1.25 \text{CP}$. This means SP is $125\%$ of CP.
Profit = $125\% - 100\% = 25\%$
Ans: Option C (25%)
1 Successive Bank Deposits
Ravi's bank account pays 10% compound interest annually. He deposits ₹10,000 at the start of Year 1, and another ₹10,000 at the start of Year 2. What is the total balance at the end of Year 2?
Concept:
1
Calculate year by year. End of Year 1 = $(10000) \times 1.10$. Start of Year 2 = End of Year 1 + new deposit. End of Year 2 = (Start of Year 2) $\times 1.10$.
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Hint:
1
End of Year 1 = $11,000$.
2
Start of Year 2 = $11,000 + 10,000 = 21,000$. Calculate $10\%$ interest on 21,000.
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Mistakes to Avoid:
!
Compounding 20,000 for 2 years: The second 10,000 deposit only experiences 1 year of compounding, not two! Do not do $20000 \times (1.1)^2$.
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Solution:
End of Year 1: $10000 \times 1.1 = 11,000$
Start of Year 2 (new deposit): $11000 + 10000 = 21,000$
End of Year 2: $21000 \times 1.1 = 23,100$

Ans: The total balance is ₹23,100.
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2 Complex Profit Margins
A laptop is sold at a 20% discount on the Marked Price, but the shopkeeper still makes a 20% profit on his Cost Price. If the Cost Price is ₹25,000, what is the Marked Price?
Concept:
1
This problem links MP and CP through the Selling Price (SP). Step 1: Find SP using CP and Profit %. Step 2: Use SP to find MP using the Discount %.
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Hint:
1
SP = CP + $20\%$ Profit = $25000 \times 1.20$.
2
SP is also $80\%$ of the Marked Price. So, $0.8 \times \text{MP} = \text{SP}$.
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Mistakes to Avoid:
!
Adding the 20% discount to SP: Do not calculate the MP by doing $1.2 \times \text{SP}$. You must divide SP by $0.80$ to find the correct Marked Price base.
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Solution:
SP = $120\%$ of CP = $1.2 \times 25000 = 30,000$

We know: $\text{SP} = \text{MP} \times 0.80$
$30000 = \text{MP} \times 0.80$
$\text{MP} = \frac{30000}{0.8} = 37,500$

Ans: The Marked Price is ₹37,500.
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