Class 8 Ganita Prakash
Fractions in Disguise
Achievement Test - Part 7
Welcome to Part 7 of the Class 8 Fractions in Disguise Test Module!
In this section, we expose the mathematical illusions of Depreciation and Successive Percentages. We will see why increasing a price by 50% and then dropping it by 50% does not return you to your original price. Review your textbook solutions (Pages 24-27), test your logic with the 4-Pillar MCQs, and try the extra practice questions.[cite: 29, 30, 31]
In this section, we expose the mathematical illusions of Depreciation and Successive Percentages. We will see why increasing a price by 50% and then dropping it by 50% does not return you to your original price. Review your textbook solutions (Pages 24-27), test your logic with the 4-Pillar MCQs, and try the extra practice questions.[cite: 29, 30, 31]
1 TV Depreciation Pg 25
A TV is bought at a price of ₹21,000. After 1 year, the value of the TV depreciates by 5%. Find the value of the TV after one year.[cite: 29]
Concept:
1
Depreciation means a reduction in value. The new value will be $(100\% - 5\%) = 95\%$ of the original price.[cite: 29]
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Hint:
1
Calculate $5\%$ of $21000$ and subtract it from $21000$.
2
$5\%$ of $21000 = \frac{5}{100} \times 21000 = 5 \times 210$.
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Mistakes to Avoid:
!
Adding the value: Depreciation is a loss of value. Do not calculate $21000 \times 1.05$.
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Solution:
Initial Value = ₹21,000
Depreciation Amount = $5\%$ of $21,000$
$= \frac{5}{100} \times 21000 = 1050$
New Value = $21000 - 1050 = 19,950$
Ans: The value of the TV is ₹19,950.
🎯 Marking Scheme:
• 1 Mark for calculating depreciation (1050).
• 1 Mark for finding the final value.[cite: 29]
• 1 Mark for calculating depreciation (1050).
• 1 Mark for finding the final value.[cite: 29]
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2 Population Decay Pg 25
The population of a village was observed to be reducing by about 10% every decade. If the current population is 1250, what is the expected population after 3 decades?[cite: 29]
Concept:
1
This is exponential decay. Every decade, the new population is $90\%$ (or $0.9$) of the previous decade's population. Formula: $A = P(1-r)^t$.[cite: 29]
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Hint:
1
Calculate $1250 \times (0.9)^3$.
2
$(0.9 \times 0.9 \times 0.9) = 0.729$. Multiply $1250 \times 0.729$.
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Mistakes to Avoid:
!
Simple reduction: Reducing by $10\%$ three times is NOT the same as reducing by $30\%$. $1250 \times 0.70 = 875$, which is mathematically incorrect because the base drops each decade.
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Solution:
Initial Population = $1250$
Decay Rate = $10\%$ per decade ($0.9$ multiplier)
Population after 3 decades = $1250 \times (0.9)^3$
$= 1250 \times 0.729 = 911.25$
Rounding to the nearest whole person, the expected population is around 911 (or 910 based on approximation).
Ans: The expected population is ~911.
🎯 Marking Scheme:
• 1 Mark for exponential setup $(0.9)^3$.
• 1 Mark for final calculation.[cite: 29]
• 1 Mark for exponential setup $(0.9)^3$.
• 1 Mark for final calculation.[cite: 29]
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3 The Bakery Discount Pg 26
A bakery called Cakely is offering a "30% + 20% discount" on all cakes. Another bakery called Cakify is offering a 50% discount on all cakes. Which option is cheaper?[cite: 30]
Concept:
1
A successive discount of $30\% + 20\%$ means you first take $30\%$ off, and then take $20\%$ off the *new, reduced* price. It does not equal a flat $50\%$ discount.[cite: 30]
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Hint:
1
Assume a cake costs ₹100.
2
For Cakely: $100 \times 0.70 = 70$. Then $70 \times 0.80 = 56$.
3
For Cakify: $100 \times 0.50 = 50$. Compare 56 and 50.
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Mistakes to Avoid:
!
Adding the percentages: $30\% + 20\% = 50\%$. This is mathematically false in commerce because the base changes after the first discount.
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Solution:
Assume the cake price is ₹100.
1. Cakely (30% + 20%):
First discount (30%): $100 - 30 = \text{₹}70$.
Second discount (20% of 70): $70 - (0.2 \times 70) = 70 - 14 = \text{₹}56$.
2. Cakify (50% flat):
Discount = 50%. Final price = $\text{₹}50$.
Since $50 < 56$, the flat 50% discount is cheaper.
Ans: Cakify is cheaper.
🎯 Marking Scheme:
• 1 Mark for calculating Cakely's effective price.
• 1 Mark for the correct conclusion.[cite: 30]
• 1 Mark for calculating Cakely's effective price.
• 1 Mark for the correct conclusion.[cite: 30]
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4 Surbhi's Mishap Pg 27
After Surbhi bought cookware from the wholesaler, she kept a profit margin of 50%. To clear the stock, she offered a 50% discount, thinking she wouldn't make a loss. If she sold goods originally for ₹12,000 after discount, how much loss did she incur? What is the percentage loss?[cite: 31]
Concept:
1
A $+50\%$ markup followed by a $-50\%$ discount always results in a net loss. This happens because the $50\%$ discount is calculated on a much larger number (the marked-up price) than the original $50\%$ markup was calculated on.[cite: 31]
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Hint:
1
Let Cost Price = $x$. Marked Price = $1.5x$. Selling Price = $50\%$ of $1.5x = 0.75x$.
2
If $0.75x = 12000$, find $x$ (the original cost).
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Mistakes to Avoid:
!
Assuming Break-even: Surbhi assumed $+50\% - 50\% = 0$. This is the classic percentage trap. The bases are different.
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Solution:
Let original Cost Price = $x$.
Marked Price = $x + 0.5x = 1.5x$
Selling Price = $1.5x \times 0.5 = 0.75x$
We are given the Selling Price is ₹12,000.
$0.75x = 12000 \implies x = \frac{12000}{0.75} = 16,000$.
Original Cost (CP) = ₹16,000.
Loss = $\text{CP} - \text{SP} = 16000 - 12000 = \text{₹}4000$.
Loss \% = $\frac{4000}{16000} \times 100 = 25\%$.
Ans: She lost ₹4000, which is a 25% loss.
🎯 Marking Scheme:
• 1 Mark for defining SP as 0.75x.
• 1 Mark for calculating CP (16000).
• 1 Mark for final Loss amounts.[cite: 31]
• 1 Mark for defining SP as 0.75x.
• 1 Mark for calculating CP (16000).
• 1 Mark for final Loss amounts.[cite: 31]
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1 Successive Discounts
A ₹100 item has a 20% discount applied, followed by an additional 10% discount on the new price. What is the final price?
A. ₹70
B. ₹72
C. ₹80
D. ₹68
Concept:
1
Calculate successive discounts sequentially. First find the price after the 20% discount. Then apply the 10% discount to that *new* price.
Hint:
1
$100 - 20\% = 80$.
2
Now find $10\%$ of 80 and subtract it.
Mistakes to Avoid:
!
Adding the percentages: $20\% + 10\% = 30\%$. $100 - 30 = 70$. This is incorrect because the second discount is only applied to ₹80, not ₹100.
Solution:
After 20% discount: $100 \times 0.8 = 80$
After 10% discount on 80: $80 \times 0.9 = 72$
Ans: Option B (₹72)
2 Auto Depreciation
A car bought for ₹100,000 depreciates by 10% each year. What is its value after 2 years?
A. ₹80,000
B. ₹81,000
C. ₹90,000
D. ₹85,000
Concept:
1
Use the exponential decay formula: $A = P(1-r)^t$.
Hint:
1
Calculate $100000 \times (0.9)^2$.
Mistakes to Avoid:
!
Simple deduction: $10\% + 10\% = 20\%$. $100000 - 20000 = 80000$. This is incorrect because the second 10% is calculated on the already depreciated 90k value.
Solution:
Value = $100000 \times (0.9)^2 = 100000 \times 0.81$
$= \text{₹}81,000$
Ans: Option B (₹81,000)
3 The Markup/Markdown Trap
A merchant marks up his goods by 20%, and then offers a 20% discount on the marked price. What is his net result?
A. No change (0%)
B. 4% Profit
C. 4% Loss
D. 40% Loss
Concept:
1
Assume base is 100. Increase by 20% = 120. Then decrease 120 by 20%. Compare the final number to 100.
Hint:
1
$100 \times 1.2 = 120$.
2
$120 \times 0.8 = 96$.
Mistakes to Avoid:
!
Assuming break-even: $+20\% - 20\%$ does not equal 0. The discount is taken from a larger base, resulting in a net loss.
Solution:
Final = $100 \times 1.2 \times 0.8 = 96$.
Net change = $100 - 96 = 4$.
Ans: Option C (4% Loss)
4 Comparing Discounts
Which is a better deal for a buyer? A flat 40% discount, or successive discounts of 20% and 20%?
A. Flat 40%
B. Successive 20% and 20%
C. Both are equal
D. Depends on the price
Concept:
1
Calculate the final remaining price percentage for both. The lower the remaining percentage, the better the deal for the buyer.
Hint:
1
Flat 40% leaves $60\%$ to pay.
2
Succ $20+20$ leaves $0.8 \times 0.8 = 0.64$ ($64\%$) to pay.
Mistakes to Avoid:
!
Assuming equal: Successive discounts always yield a smaller total discount than their sum because the second discount is applied to a reduced amount.
Solution:
Flat 40%: Buyer pays 60.
Successive 20% & 20%: Buyer pays $100 \times 0.8 \times 0.8 = 64$.
60 is cheaper than 64.
Ans: Option A (Flat 40%)
5 Percentage Gain (Would You Rather)
If you deposit ₹200 and receive ₹600 back, what is your percentage gain?
A. 100%
B. 200%
C. 300%
D. 400%
Concept:
1
Gain = Final Amount - Initial Deposit. Gain % = $\frac{\text{Gain}}{\text{Initial}} \times 100$.
Hint:
1
Gain = $600 - 200 = 400$.
2
Calculate $\frac{400}{200} \times 100$.
Mistakes to Avoid:
!
Calculating Final Amount Percentage: $\frac{600}{200} \times 100 = 300\%$. The final amount is 300% of the initial deposit, but the actual *gain* is 200%.
Solution:
Gain = $600 - 200 = 400$
Gain \% = $\frac{400}{200} \times 100 = 2 \times 100 = 200\%$
Ans: Option B (200%)
1 Successive Salary Change
Rahul's salary of ₹50,000 was increased by 10% for good performance. However, due to company losses the next year, it was decreased by 10%. What is his final salary?
Concept:
1
Apply the changes sequentially. Final Salary = $P \times (1 + r_{increase}) \times (1 - r_{decrease})$.
Hint:
1
Calculate $50000 \times 1.1 \times 0.9$.
2
$1.1 \times 0.9 = 0.99$. Multiply $50000 \times 0.99$.
Mistakes to Avoid:
!
Assuming it returns to ₹50,000: $+10\%$ followed by $-10\%$ results in a net 1% loss, not a net 0% change.
Solution:
Final Salary = $50000 \times 1.1 \times 0.9$
$= 50000 \times 0.99$
$= \text{₹}49,500$
Ans: His final salary is ₹49,500.
2 Variable Depreciation
A laptop is bought for ₹40,000. Its value depreciates by 20% in the first year, and by 10% in the second year. What is its value after 2 years?
Concept:
1
When depreciation rates vary, multiply the initial value sequentially by each year's retained value factor. Final Value = $P \times (1 - r_1) \times (1 - r_2)$.
Hint:
1
Calculate $40000 \times 0.8 \times 0.9$.
2
$0.8 \times 0.9 = 0.72$. Multiply $40000 \times 0.72$.
Mistakes to Avoid:
!
Adding the depreciation rates: Do not calculate a flat 30% discount ($40000 \times 0.70 = 28000$). The rates are sequential.
Solution:
Value after Year 1 = $40000 \times 0.8 = 32000$
Value after Year 2 = $32000 \times 0.9 = 28800$
Ans: The value is ₹28,800.
×
Fractions in Disguise
Part 1: Fraction & % Equivalence
Pages 1 - 4
Part 2: % of a Quantity
Pages 6 - 9
Part 3: Thresholds & >100%
Pages 10 - 14
Part 4: Proportions, Profit & Loss
Pages 14 - 17
Part 5: Discounts, Taxes & Margins
Pages 18 - 20
Part 6: Simple & Compound Growth
Pages 20 - 24
Part 7: Depreciation & Successive %
Pages 24 - 27
Part 8: Advanced Applications I
Pages 28 - 29
Part 9: Advanced Apps II & Data
Pages 29 - 30
