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class 8 ganita prakash chapter 1
📖 Textbook Theory: Chaining Multipliers
When a quantity undergoes several percentage changes sequentially, you cannot just add or subtract the percentages. You must convert each percentage change into its respective decimal multiplier, and then multiply them all together in a chain.
The Multiplier Rule:
An increase of $X\%$ means the multiplier is $(1 + \frac{X}{100})$.
A decrease of $X\%$ means the multiplier is $(1 - \frac{X}{100})$.
4. The monthly percentage change in population (compared to the previous month) of mice in a lab is given: Month 1 change was +5%, Month 2 change was –2%, and Month 3 change was –3%. Which of the following statement(s) are true? The initial population is $p$.
1
Calculate the decimal multiplier for each month separately.
2
Multiply all three decimal multipliers together with the starting population '$p$'.
3
Evaluate the final multiplier. If the total multiplier is exactly $1.0$, the population is unchanged. If it is $> 1$, it grew. If it is $< 1$, it shrank.
1
$+5\% \implies (1 + 0.05) = 1.05$.
2
$-2\% \implies (1 - 0.02) = 0.98$.
3
$-3\% \implies (1 - 0.03) = 0.97$.
4
Calculate: $1.05 \times 0.98 \times 0.97$. Is the result greater or less than 1?
!
The Additive Trap: You might think "$+5\% - 2\% - 3\% = 0\%$, so the population is unchanged (Option iv)." This is the most common error in successive percentages! Because the base shrinks in months 2 and 3, taking 2% and 3% off a larger base removes more than was initially added.
Building the Equation
Base Population = $p$.
M1 (+5%): Multiplier = $1.05$.
M2 (-2%): Multiplier = $0.98$.
M3 (-3%): Multiplier = $0.97$.
Total Equation = $p \times 1.05 \times 0.98 \times 0.97$. This matches Option (ii).
Let's solve it:
$= p \times (1.05 \times 0.98) \times 0.97$
$= p \times (1.029) \times 0.97$
$= p \times 0.99813$.
Since the final multiplier ($0.99813$) is strictly less than $1.0$, the final population is less than the starting population $p$. This matches Option (vi).
Ans: Options (ii) and (vi) are true.
class 8 ganita prakash chapter 1
📖 Textbook Theory: The Markup/Discount Trap
When a shopkeeper marks up a price, the percentage is calculated on the Cost Price. When they offer a discount, the percentage is calculated on the Marked Price.
The Danger: Because the Marked Price is always larger than the Cost Price, the discount percentage is taking a bite out of a much larger number. A seemingly safe discount can easily plunge the transaction into a loss.
5. 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.
1
Whenever working with percentages without a stated currency amount, assume the Cost Price is exactly 100. This makes the math incredibly easy.
2
Apply the 35% markup to the 100 base. Then, apply the 30% discount to the newly established Marked Price. Compare the final price to 100.
1
Cost = 100. Markup = 35. Marked Price = 135.
2
Discount is 30% of 135. Calculate $\frac{30}{100} \times 135$. Subtract that number from 135.
3
Is the final number greater than or less than 100?
!
The "$+35 - 30$" Error: It is incredibly tempting to just say "$35\% \text{ profit} - 30\% \text{ discount} = 5\% \text{ profit}$". The human brain wants to add and subtract percentages like normal numbers. But you cannot do this, because the 30% discount is attacking a much larger number (135) than the 35% markup attacked (100)!
Proving the Loss
Let Cost Price (CP) = 100.
Step 1: Set the Marked Price
Markup = 35% of 100 = 35.
Marked Price = $100 + 35 = 135$.
Step 2: Apply the Discount
Discount = 30% of 135.
$= 0.30 \times 135 = 40.5$.
Final Selling Price (SP) = $135 - 40.5 = 94.5$.
Step 3: Conclusion
Since the SP ($94.5$) is less than the CP ($100$), he took a loss of $5.5\%$.
Ans: He will make a loss. Reason: The 30% discount was calculated on a larger base (135) compared to the base the markup was calculated on (100). Therefore, the discount removed more absolute value (40.5) than was originally added (35).
class 8 ganita prakash chapter 1
📖 Textbook Theory: Spatial Proportions
Percentages don't just apply to money or populations. They also apply to physical space and geometry. To find the percentage of an area, you must first define what the "100%" Whole is.
The Method: 1. Find the total area of the entire shape (the Base). 2. Find the area of the specific region. 3. Create the fraction $\left( \frac{\text{Region}}{\text{Total Base}} \right)$ and multiply by 100 to find the percentage.
6. What percentage of area is occupied by the region marked ‘E’ in the figure?
1
Count the dots to establish a coordinate system. The large outer square has 7 dots on each side, which means it is $6 \times 6$ square units.
2
Region A is a $3 \times 3$ square. Region E sits inside the $3 \times 3$ square below A.
3
Find the area of the large square, find the area of triangle E, and convert the fraction into a percentage.
1
Total area = $6 \times 6 = 36$ square units.
2
Region E is a triangle. The base is 3 units long. The height is 3 units tall.
3
Area of a triangle = $\frac{1}{2} \times \text{base} \times \text{height}$.
4
Divide the triangle's area by 36, then multiply by 100.
!
Visual Guessing: You might look at the picture and guess "it looks like 10%". But geometry requires precision. By counting the exact grid units, you remove all guesswork.
Geometric Calculation
Step 1: Total Area
The outer grid is 6 units wide and 6 units high.
Total Area = $6 \times 6 = 36$ units.
Step 2: Area of Region E
Region E is a triangle bounded by a $3 \times 3$ square.
Area of E = $\frac{1}{2} \times \text{base} \times \text{height}$.
Area of E = $\frac{1}{2} \times 3 \times 3 = 4.5$ units.
Step 3: Conversion to Percentage
Fraction = $\frac{4.5}{36}$. Let's multiply the top and bottom by 2 to clear the decimal.
Fraction = $\frac{9}{72} = \frac{1}{8}$.
Percentage = $\frac{1}{8} \times 100 = \frac{100}{8} = 12.5\%$.
Ans: Region E occupies exactly 12.5% of the area.
