Chapter 1 Fractions in Disguise (Class 8 - Latest Maths NCERT (Ganita Prakash II) Solutions)

To Find:

Conversion of given fractions into percentages.

Solution:

In the Indian school system, a common method to convert a fraction into a percentage is to multiply the fraction by $100$ and add the percent symbol ($\%$).

(i) $\frac{3}{5}$

$\text{Percentage} = 3 \times 20 = 60\%$

(ii) $\frac{7}{14}$

$\text{Percentage} = \frac{1}{\cancel{2}_1} \times \cancel{100}^{50} = 50\%$

(iii) $\frac{9}{20}$

$\text{Percentage} = \frac{9}{\cancel{20}_1} \times \cancel{100}^{5}$

$\text{Percentage} = 9 \times 5 = 45\%$

(iv) $\frac{72}{150}$

$\text{Percentage} = \frac{72}{\cancel{150}_3} \times \cancel{100}^{2}$

$\text{Percentage} = \frac{\cancel{72}^{24}}{3} \times 2$

$\text{Percentage} = 24 \times 2 = 48\%$

(v) $\frac{1}{3}$

$\text{Percentage} = \frac{1}{3} \times 100 = \frac{100}{3}\%$

On dividing $100$ by $3$, we get a recurring decimal:

$\text{Percentage} = 33.33\dots\% = 33\frac{1}{3}\%$

(vi) $\frac{5}{11}$

$\text{Percentage} = \frac{5}{11} \times 100 = \frac{500}{11}\%$

By long division:

$\text{Percentage} \approx 45.45\%$

Summary Table:

Given:

Total number of marbles with Nandini = $25$

Number of white marbles = $15$

To Find:

Percentage of white marbles.

Solution:

The percentage is calculated by the formula:

$\text{Percentage} = \frac{\text{Value}}{\text{Total Value}} \times 100$

Substituting the values:

We can simplify the fraction $\frac{15}{25}$ by dividing both the numerator and the denominator by their HCF, which is $5$.

$\text{Percentage} = \frac{\cancel{15}^3}{\cancel{25}_5} \times 100$

$\text{Percentage} = \frac{3}{\cancel{5}_1} \times \cancel{100}^{20}$

$\text{Percentage} = 3 \times 20 = 60\%$

Conclusion:

Nandini has $60\%$ white marbles. Therefore, the correct option is (iv).

Given:

Total number of students in the school = $80$

Number of students who come by walking = $15$

To Find:

The percentage of students who walk to school.

Solution:

To find the percentage, we express the number of walking students as a fraction of the total students and then multiply by $100$.

Substituting the values:

$\text{Percentage} = \frac{15}{80} \times 100$

Simplifying the fraction by cancelling common factors:

$\text{Percentage} = \frac{\cancel{15}^{3}}{\cancel{80}_{16}} \times 100$

$\text{Percentage} = \frac{3 \times \cancel{100}^{25}}{\cancel{16}_{4}}$

$\text{Percentage} = \frac{3 \times 25}{4} = \frac{75}{4}$

$\text{Percentage} = 18.75\%$

Solution:

Given:

1. A long-distance running track represented as a number line from "Start" ($0\%$) to "Finish" ($100\%$).

2. Positions of four runners $A, B, C,$ and $D$ on the track after $15$ minutes.

3. Percentage options provided in the boxes: $55\%$, $20\%$, $38\%$, $72\%$, $84\%$, and $93\%$.

To Find:

Match each runner ($A, B, C,$ and $D$) with their approximately completed percentage of the race based on their relative positions on the track.

Step-by-Step Analysis & Matching:

By observing the proportional spacing between the "Start" ($0\%$) and "Finish" ($100\%$) markers:

1. Runner A: Positioned earliest along the track among the labeled runners.

2. Runner B: Positioned ahead of A, approaching the first third of the track.

3. Runner C: Positioned past the halfway mark towards the final stretch.

4. Runner D: Positioned closest to the "Finish" line.

Summary Matching Table:

Solution:

We will compare these using logical reasoning and benchmarks ($0, \frac{1}{2}, 1$).

(i) $50\%$ ____ $5\%$

$50\%$ represents half of a whole, whereas $5\%$ is a very small fraction (one-twentieth).

Result: $50\% > 5\%$

(ii) $\frac{5}{10}$ ____ $50\%$

$\frac{5}{10}$ simplified is $\frac{1}{2}$. We know that $50\%$ also represents $\frac{1}{2}$ of the total.

Result: $\frac{5}{10} = 50\%$

(iii) $\frac{3}{11}$ _____ $61\%$

$\frac{3}{11}$ is less than $\frac{3}{6}$ (which would be $50\%$). Since $3$ is much less than half of $11$, this fraction is clearly less than $50\%$. On the other hand, $61\%$ is greater than $50\%$.

Result: $\frac{3}{11} < 61\%$

(iv) $30\%$ ____ $\frac{1}{3}$

We know that $\frac{1}{3}$ is approximately $33.33\%$. Comparing $30\%$ with $33.33\%$, it is evident that $30\%$ is smaller.

Result: $30\% < \frac{1}{3}$

Summary:

1. (i) $>$

2. (ii) $=$

3. (iii) $<$

4. (iv) $<$

Part (i):

From the given bar model, the total length representing $100\%$ is divided into $5$ equal parts.

Each part represents $\frac{100\%}{5} = 20\%$

For the right-side model, the total length corresponding to $100\%$ is represented by the total value $75$, which is divided into $5$ equal parts.

Thus, the value of $1$ part (or $20\%$) is given by $\frac{75}{5} = 15$.

The length marked with $?$ spans $4$ parts out of $5$, which represents $80\%$.

Value for $?$ = $4 \times 15 = 60$

Part (ii):

From the given bar model, the total length representing $100\%$ is divided into $10$ equal parts.

Each part represents $\frac{100\%}{10} = 10\%$

For the right-side model, the total value corresponding to $100\%$ is $90$, which is divided into $10$ equal parts.

Thus, the value of $1$ part (or $10\%$) is given by $\frac{90}{10} = 9$.

The length marked with $?$ spans $6$ parts out of $10$, which represents $60\%$.

Value for $?$ = $6 \times 9 = 54$

Part (iii):

From the given bar model, the total length representing $100\%$ is divided into $4$ equal parts.

Each part represents $\frac{100\%}{4} = 25\%$

For the right-side model, the total value corresponding to $100\%$ is $140$, which is divided into $4$ equal parts.

Thus, the value of $1$ part (or $25\%$) is given by $\frac{140}{4} = 35$.

The length marked with $?$ spans $3$ parts out of $4$, which represents $75\%$.

Value for $?$ = $3 \times 35 = 105$

(i) $25\%$ of $160$

We know that $25\% = \frac{25}{100} = \frac{1}{4}$.

Therefore, $25\%$ of $160 = \frac{1}{4} \times 160 = 40$.

(ii) $16\%$ of $250$

$16\%$ of $250 = \frac{16}{100} \times 250 = \frac{4}{25} \times 250 = 4 \times 10 = 40$.

(iii) $62\%$ of $360$

Calculating the percentage value directly:

$62\%$ of $360 = \frac{62}{100} \times 360 = \frac{62 \times 36}{10} = \frac{2232}{10} = 223.2$.

(iv) $140\%$ of $40$

Calculating the percentage value directly:

$140\%$ of $40 = \frac{140}{100} \times 40 = \frac{14}{10} \times 40 = 14 \times 4 = 56$.

(v) $1\%$ of $1$ hour

Converting $1$ hour into minutes, we get $1 \text{ hour} = 60 \text{ minutes}$.

Now, finding $1\%$ of $60 \text{ minutes}$:

$1\%$ of $60 = \frac{1}{100} \times 60 = \frac{60}{100} = 0.6 \text{ minutes}$, which is equal to $0.6 \times 60 = 36 \text{ seconds}$.

(vi) $7\%$ of $10$ kg

Converting $10 \text{ kg}$ into grams, we get $10 \text{ kg} = 10000 \text{ g}$, or we can keep it in kilograms.

Finding $7\%$ of $10 \text{ kg}$:

$7\%$ of $10 = \frac{7}{100} \times 10 = \frac{70}{100} = 0.7 \text{ kg}$ (or $700 \text{ g}$).

Given:

Total volume of deep orange paint = $60$ ml.

Fraction of red paint used = $\frac{3}{4}$.

To Find:

The volume of red paint in ml.

Solution:

The volume of red paint can be found by multiplying the total volume by the given fraction.

Simplifying the calculation:

Solution:

(i) $50\%$ of $510$ _____ $50\%$ of $515$

Since the percentage is the same ($50\%$) and $510 < 515$, taking half of a smaller number will always result in a smaller value than taking half of a larger number.

Therefore, $50\%$ of $510$ $\mathbf{<}$ $50\%$ of $515$.

(ii) $37\%$ of $148$ _____ $73\%$ of $148$

Since the base quantity is the same ($148$) and $37\% < 73\%$, taking a smaller percentage of the same number yields a smaller result.

Therefore, $37\%$ of $148$ $\mathbf{<}$ $73\%$ of $148$.

(iii) $29\%$ of $43$ _____ $92\%$ of $110$

Both the percentage and the base quantity on the left-hand side ($29\%$ of $43$) are significantly smaller than those on the right-hand side ($92\%$ of $110$).

Left-hand side is approximately $\approx 0.3 \times 43 \approx 12.9$.

Right-hand side is approximately $\approx 0.9 \times 110 \approx 99$.

Therefore, $29\%$ of $43$ $\mathbf{<}$ $92\%$ of $110$.

(iv) $30\%$ of $40$ _____ $40\%$ of $50$

Let us compute both values:

$30\%$ of $40 = \frac{30}{100} \times 40 = 12$

$40\%$ of $50 = \frac{40}{100} \times 50 = 20$

Comparing the results, $12 < 20$.

Therefore, $30\%$ of $40$ $\mathbf{<}$ $40\%$ of $50$.

(v) $45\%$ of $200$ _____ $10\%$ of $490$

Let us compute both values:

$45\%$ of $200 = \frac{45}{100} \times 200 = 90$

$10\%$ of $490 = \frac{10}{100} \times 490 = 49$

Comparing the results, $90 > 49$.

Therefore, $45\%$ of $200$ $\mathbf{>}$ $10\%$ of $490$.

(vi) $30\%$ of $80$ _____ $24\%$ of $64$

Let us compute both values:

$30\%$ of $80 = \frac{30}{100} \times 80 = 24$

$24\%$ of $64 = \frac{24}{100} \times 64 = 15.36$

Comparing the results, $24 > 15.36$.

Therefore, $30\%$ of $80$ $\mathbf{>}$ $24\%$ of $64$.

Solution:

(i) Given $30\%$ of $k = 70$:

To find $60\%$, we double the value: $70 \times 2 = 140$.

To find $90\%$, we triple the value: $70 \times 3 = 210$.

To find $120\%$, we quadruple the value: $70 \times 4 = 280$.

Ans: 140, 210, 280

(ii) Given $100\%$ of $m = 215$:

To find $10\%$, divide by $10$: $215 \div 10 = 21.5$.

To find $1\%$, divide by $100$: $215 \div 100 = 2.15$.

To find $6\%$, multiply $1\%$ by $6$: $2.15 \times 6 = 12.9$.

Ans: 21.5, 2.15, 12.9

(iii) Given $90\%$ of $n = 270$:

To find $9\%$, divide by $10$: $270 \div 10 = 27$.

To find $18\%$, multiply $9\%$ value by $2$: $27 \times 2 = 54$.

To find $100\%$, we find $10\%$ first ($30$) and multiply by $10$: $300$.

Ans: 27, 54, 300

(iv) Self-made questions:

1. $20\%$ of $x$ is $50$, find $10\%$, $40\%$, and $100\%$.

2. $100\%$ of $y$ is $400$, find $5\%$, $25\%$, and $150\%$.

Solution:

(i) $3$ is $x\%$ of $300$

Ans: 1

(ii) $x$ is $40\%$ of $4$

Ans: 1.6

(iii) $40$ is $80\%$ of $x$

Ans: 50

Summary:

(i) 1; (ii) 1.6; (iii) 50.

Given:

Quantity A: $10\%$ of a day

Quantity B: $1\%$ of a week

To Find:

Which duration is longer.

Solution:

First, let us convert both quantities into the same unit (hours) to compare them accurately, as is common in Indian mathematics problems.

Step 1: Calculating $10\%$ of a day

Converting $0.4$ hours to minutes: $0.4 \times 60 = 24 \text{ minutes}$. So, $10\%$ of a day is $2 \text{ hours } 24 \text{ minutes}$.

Step 2: Calculating $1\%$ of a week

Converting $0.68$ hours to minutes: $0.68 \times 60 \approx 40.8 \text{ minutes}$. So, $1\%$ of a week is approx $1 \text{ hour } 41 \text{ minutes}$.

Comparison:

Since $2.4 \text{ hours} > 1.68 \text{ hours}$, $10\%$ of a day is longer than $1\%$ of a week.

Challenge Questions for Peers:

1. Is $25\%$ of a $\textsf{₹} 1000$ note more than $10\%$ of a $\textsf{₹} 2000$ note and a $\textsf{₹} 500$ note combined?

2. Is $50\%$ of an hour longer than $5\%$ of a full day?

To find the percentage of fodder eaten by the bull each day, we use the formula:

$\text{Percentage eaten} = \left(\frac{\text{Fodder eaten}}{\text{Fodder given}}\right) \times 100\%$

Day 1: Fodder given = $2$, Fodder eaten = $1$

$\text{Percentage} = \left(\frac{1}{2}\right) \times 100\% = 50\%$

Day 2: Fodder given = $3$, Fodder eaten = $2$

$\text{Percentage} = \left(\frac{2}{3}\right) \times 100\% = 66.67\%$

Day 3: Fodder given = $4$, Fodder eaten = $3$

$\text{Percentage} = \left(\frac{3}{4}\right) \times 100\% = 75\%$

General expression for day $n$ (where $n$ is the day number):

Fodder given = $n + 1$, Fodder eaten = $n$

$\text{Percentage} = \left(\frac{n}{n+1}\right) \times 100\%$

On the $99$th day: Fodder given = $100$, Fodder eaten = $99$

$\text{Percentage} = \left(\frac{99}{100}\right) \times 100\% = 99\%$

Observation: We observe that as the days progress, the percentage of fodder eaten by the bull increases steadily and gets closer and closer to $100\%$, though it never actually reaches $100\%$ because there is always $1$ unit left uneaten.

Given:

Portion of plantation completed = $20\%$

Time taken = $18$ days

To Find:

Time taken to complete $100\%$ (the entire plantation).

Solution:

We can use the Unitary Method, which is the standard technique taught in Indian arithmetic.

Time taken for $20\%$ work = $18$ days

Importance of the Assumption:

The assumption that the rate of work stays the same is necessary because:

1. Environmental Factors: In Indian coffee estates (like those in Coorg or Wayanad), heavy rains or heatwaves can slow down work. If it rains, workers cannot pick berries at the same speed.

2. Labor Consistency: It assumes the number of workers remains the same and they don't get tired or fall ill over the long period of $90$ days.

3. Terrain: Some parts of the plantation might be on steeper slopes than others, making the work harder in those sections.

Without this assumption, we cannot use simple Direct Proportion to solve the problem.

Given:

Total duration of training = $90$ minutes

Percentage for Warm up = $10\%$

Percentage for Play = $80\%$

Percentage for Cool down = $10\%$

To Find:

The time duration (in minutes) for each specific activity.

Solution:

We calculate the time for each activity by finding the respective percentage of the total $90$ minutes.

1. Duration for Warm up:

$\text{Time} = \frac{1}{\cancel{10}} \times 9\cancel{0} = 9 \text{ minutes}$

2. Duration for Play:

$\text{Time} = \frac{8}{\cancel{10}} \times 9\cancel{0} = 8 \times 9 = 72 \text{ minutes}$

3. Duration for Cool down:

$\text{Time} = 9 \text{ minutes}$

Verification:

Total time = $9 + 72 + 9 = 90$ minutes. This matches the given total duration.

Given:

Percentage of world population in Northern Hemisphere = $90\%$

Approximate current world population (2024-25) $\approx 8.2$ billion

To Find:

Approximate number of people living in the Northern Hemisphere.

Solution:

We need to find $90\%$ of $8.2$ billion.

$\text{Population} = 0.9 \times 8.2$

$\text{Population} = 7.38$ billion

Final Answer:

Approximately $7.38$ billion people live in the Northern Hemisphere.

Given:

Proportion for $4$ people: Rava ($40\%$), Sugar ($40\%$), Ghee ($20\%$).

(i) Solution for 8 people:

When we increase the quantity of a dish to serve more people, the proportion (percentage) of each ingredient remains exactly the same to maintain the taste. Only the actual weight/volume of the ingredients increases.

Proportions for 8 people:

Rava: $40\%$

Sugar: $40\%$

Ghee: $20\%$

(ii) Solution for total weight of $2$ kg:

Total weight = $2$ kg = $2000$ grams.

1. Quantity of Rava:

$\text{Rava} = 40\% \text{ of } 2000 \text{ g}$

$\text{Rava} = \frac{40}{100} \times 2000 = 800 \text{ g}$

2. Quantity of Sugar:

$\text{Sugar} = 40\% \text{ of } 2000 \text{ g}$

$\text{Sugar} = \frac{40}{100} \times 2000 = 800 \text{ g}$

3. Quantity of Ghee:

$\text{Ghee} = 20\% \text{ of } 2000 \text{ g}$

$\text{Ghee} = \frac{20}{100} \times 2000 = 400 \text{ g}$

Verification:

Total weight = $800 + 800 + 400 = 2000$ g = $2$ kg.

Given:

Cost Price (CP) of the geometry box = $\textsf{₹} 75$

Selling Price (SP) of the geometry box = $\textsf{₹} 110$

To Find:

Profit margin with respect to the cost (Profit Percentage).

Solution:

First, we need to calculate the actual profit earned by the shopkeeper.

Now, we calculate the profit margin with respect to the cost price:

$\text{Profit Margin} = \frac{35}{75} \times 100$

Simplifying the fraction:

$\text{Profit Margin} = \frac{\cancel{35}^7}{\cancel{75}_{15}} \times 100$

$\text{Profit Margin} = \frac{7 \times \cancel{100}^{20}}{\cancel{15}_3} = \frac{140}{3}$

$\text{Profit Margin} \approx 46.67\%$

Given:

Cost of materials (CP) = $\textsf{₹} 475$

Desired Profit Margin = $50\%$

To Find:

Selling Price (SP) of the chair.

Solution:

First, we calculate the amount of profit the carpenter wants to earn.

$\text{Profit} = \frac{50}{100} \times 475 = \frac{1}{2} \times 475 = \textsf{₹} 237.50$

Now, we find the Selling Price:

$\text{SP} = 475 + 237.50$

$\text{SP} = \textsf{₹} 712.50$

Conclusion:

The carpenter should sell each chair for $\textsf{₹} 712.50$ to achieve a $50\%$ profit margin.

Alternate Solution:

We can use the direct formula for Selling Price:

$\text{SP} = \text{CP} \times \left( \frac{100 + \text{Gain}\%}{100} \right)$

$\text{SP} = 475 \times \frac{150}{100} = 475 \times 1.5 = \textsf{₹} 712.50$

Given:

Total Sales (Revenue/SP) = $\textsf{₹} 2.5$ crore

Profit Margin = $25\%$ (on cost)

To Find:

Total Expenditure (Cost Price).

Solution:

Let the total expenditure be $x$.

According to the problem, the sales figure includes the cost plus a $25\%$ profit on that cost.

$2.5 = x + 0.25x$

$2.5 = 1.25x$

Solving for $x$:

$x = 2$

Conclusion:

The total expenditure of the company was $\textsf{₹} 2$ crore. The profit earned was $\textsf{₹} 0.5$ crore.

Given:

Original Price (Marked Price) = $\textsf{₹} 300$

Discount Percentage = $25\%$

To Find:

Final Price Anwar has to pay.

Solution:

First, we calculate the discount amount in rupees.

$\text{Discount} = \frac{25}{100} \times 300 = \frac{1}{4} \times 300 = \textsf{₹} 75$

Now, we find the final price after subtracting the discount from the original price.

$\text{Final Price} = \textsf{₹} 225$

Conclusion:

Anwar will have to pay $\textsf{₹} 225$ to buy the shirt.

Alternate Solution:

Since the discount is $25\%$, Anwar only pays $(100\% - 25\%) = 75\%$ of the original price.

$\text{Final Price} = 75\% \text{ of } 300 = \frac{75}{100} \times 300 = 75 \times 3 = \textsf{₹} 225$.

Given:

Price in 2015 (Original Price) = $\textsf{₹} 60$

Price in 2025 (New Price) = $\textsf{₹} 100$

To Find:

Percentage increase in the price of petrol.

Solution:

First, we find the absolute increase in price:

Now, we calculate the percentage increase using the formula:

Substituting the values from equation (i):

Rounding off, we get $66.66\%$.

Conclusion:

The correct option is (iv) $66.66\%$.

Given:

Selling Price (after discount) = $\textsf{₹} 4,40,000$

Discount Percentage = $15\%$

To Find:

Original Price (Marked Price) of the car.

Solution:

Let the original price of the car be $x$.

If the discount is $15\%$, then the selling price is $100\% - 15\% = 85\%$ of the original price.

Simplifying the fraction by dividing by $5$:

Conclusion:

The original price of the car was approximately $\textsf{₹} 5,17,647.06$.

Alternate Solution:

We can use the formula: $\text{Marked Price} = \frac{\text{Selling Price} \times 100}{100 - \text{Discount}\%}$.

$\text{MP} = \frac{440000 \times 100}{100 - 15} = \frac{440000 \times 100}{85} = \textsf{₹} 5,17,647.06$.

Part 1: Percentage of votes for the winner

Given:

Total votes = $1600$

Winner's votes = $500$

Solution:

Part 2: Guessing the minimum number of candidates

Solution:

The winner got $31.25\%$ of the votes. In a standard election, a winner is the person with the highest number of votes (First-Past-The-Post system).

1. If there were only 2 candidates: The winner must have more than $50\%$ of the votes. Since $31.25\% < 50\%$, there must be more than 2 candidates.

2. If there were 3 candidates: The remaining $100\% - 31.25\% = 68.75\%$ would be shared by 2 others. The average for the others would be $68.75 \div 2 = 34.375\%$. If one of them got $34.375\%$, the person with $31.25\%$ would not be the winner.

3. If there were 4 candidates: The remaining $68.75\%$ is shared by 3 others. The average for others would be $68.75 \div 3 \approx 22.92\%$. Since $31.25\% > 22.92\%$, it is possible for the person with $31.25\%$ to be the winner.

Conclusion:

The winner got $31.25\%$ of the total votes. The minimum number of candidates who stood for the election must be 4.

Given:

Price of rice in $2024$ (Original Price) = $\textsf{₹} 38$

Price of rice in $2025$ (New Price) = $\textsf{₹} 42$

To Find:

Rate of inflation (Percentage increase in price).

Solution:

First, we calculate the increase in the price of rice:

Now, we find the percentage increase using the formula:

$\text{Inflation Rate} = \frac{\text{Increase}}{\text{Original Price}} \times 100$

Given:

Percentage increase = $20\%$

Resulting number after increase = $90$

To Find:

The original number.

Solution:

Let the original number be $x$.

According to the problem, the original number plus $20\%$ of itself equals $90$.

To simplify, multiply the numerator and denominator by $10$:

Conclusion:

The original number is 75.

Alternate Solution:

If a number is increased by $20\%$, it becomes $120\%$ of its original value.

Therefore, $120\% \text{ of } x = 90$.

$\frac{120}{100} \times x = 90 \Rightarrow x = \frac{90 \times 100}{120} = \frac{3 \times 100}{4} = 75$.

Given:

Selling Price (SP) of each buffalo = $\textsf{₹} 80,000$

Buffalo 1: Profit = $5\%$

Buffalo 2: Loss = $10\%$

To Find:

Overall profit or loss amount.

Solution:

First, we calculate the Cost Price (CP) for each buffalo separately.

For Buffalo 1 (Profit $5\%$):

$\text{CP}_1 = \frac{\text{SP} \times 100}{100 + \text{Profit}\%}$

$\text{CP}_1 = \frac{1600000}{21} \approx \textsf{₹} 76,190.48$

For Buffalo 2 (Loss $10\%$):

$\text{CP}_2 = \frac{\text{SP} \times 100}{100 - \text{Loss}\%}$

$\text{CP}_2 = \frac{800000}{9} \approx \textsf{₹} 88,888.89$

Calculation of Overall Result:

Total Selling Price (Total SP) = $80,000 + 80,000 = \textsf{₹} 1,60,000$

Total Cost Price (Total CP) = $76,190.48 + 88,888.89 = \textsf{₹} 1,65,079.37$

Since Total CP > Total SP, the milkman has incurred an overall loss.

Given:

Population of elephants in the last decade = $p$

Percentage increase in population = $5\%$

To Find:

The current population of elephants.

Solution:

An increase of $5\%$ means the current population is the sum of the original population and the additional $5\%$ growth.

Now, the current population is found by adding this increase to the original population:

Conclusion:

The correct option is (iv) $p \times 1.05$.

Given:

Percentage fall in demand = $85\%$

Solution:

When we say a quantity has "fallen by $85\%$", it means we subtract $85\%$ from the original $100\%$ value.

This means the current demand is $15\%$ of what it used to be a decade ago.

Evaluation of Statements:

1. Statement (i) says it is $85\%$, which is incorrect because $85\%$ is the part that was lost, not the part that remains.

2. Statement (iii) says "The demand now is $15\%$ of the demand a decade ago," which matches our calculation.

Conclusion:

The statement that means the same is (iii).

Given:

Principal ($P$) = $\textsf{₹} 20,000$

Rate of Interest ($R$) = $10\%$ p.a.

Time ($T$ or $n$) = $2$ years

To Find:

1. Amount without compounding (Simple Interest).

2. Amount with annual compounding (Compound Interest).

3. Comparison between the two.

Solution:

Case 1: Without Compounding (Simple Interest)

$SI = \frac{20000 \times 10 \times 2}{100}$

$SI = 200 \times 10 \times 2 = \textsf{₹} 4,000$

Total Amount ($A_{SI}$) = $P + SI$

Case 2: With Annual Compounding (Compound Interest)

$A_{CI} = 20000 \left(1 + \frac{10}{100}\right)^2$

$A_{CI} = 20000 \left(1.1\right)^2$

Comparison:

Difference = $A_{CI} - A_{SI}$

Difference = $24200 - 24000 = \textsf{₹} 200$

One gets $\textsf{₹} 200$ more if the interest is compounded annually.

Given:

Principal ($P$) = $\textsf{₹} 20,000$

Rate of Interest ($R$) = $5\%$ p.a.

Time ($T$ or $n$) = $4$ years

Solution:

Case 1: Without Compounding (Simple Interest)

$SI = 200 \times 20 = \textsf{₹} 4,000$

Total Amount ($A_{SI}$) = $20000 + 4000 = \textsf{₹} 24,000$

Case 2: With Annual Compounding (Compound Interest)

$A_{CI} = 20000 \times (1.05)^4$

$A_{CI} = 20000 \times 1.21550625$

Comparison:

Difference = $24310.13 - 24000 = \textsf{₹} 310.13$

In this case, one gets $\textsf{₹} 310.13$ more through compounding.

Observations and Discussion:

1. Equality of Simple Interest: Interestingly, in both Question 1 and Question 2, the Simple Interest earned is exactly the same ($\textsf{₹} 4,000$). This is because the product of Rate and Time is the same ($10 \times 2 = 20$ and $5 \times 4 = 20$).

2. Power of Time in Compounding: Even though the product of $R \times T$ is constant, the Compound Interest is higher in the second case ($\textsf{₹} 4,310.13$) compared to the first ($\textsf{₹} 4,200$). This shows that Time ($n$) has a more powerful effect in the compound interest formula because it sits in the exponent position.

Summary Table:

When compounding is not done, simple interest is applied. The total amount ($\text{A}$) is the sum of the principal ($p$) and the simple interest.

Simple interest for $4$ years at $6\%$ p.a. is calculated as:

$\text{Interest} = p \times 0.06 \times 4$

Adding this interest to the principal $p$ gives the total amount:

$\text{Total Amount} = p + (p \times 0.06 \times 4)$

This matches expression (vii).

Alternatively, the total amount can also be represented as:

$p \times 1.06 \times 4$

This matches expression (vi).

Therefore, both (vi) and (vii) are correct.

Given:

Principal Amount ($P$) = $\textsf{₹} 50,000$

Rate of Interest ($R$) = $7\%$ p.a.

Time period ($T$ or $n$) = $3$ years

To Find:

1. Simple Interest (without compounding).

2. The difference between Compound Interest and Simple Interest.

Solution:

Step 1: Calculating Interest without Compounding (Simple Interest)

Post offices in India are very popular for such safe investment schemes. The simple interest is:

Step 2: Calculating Interest with Annual Compounding

First, we find the total amount ($A$) using the compound interest formula:

Now, we find the Compound Interest ($CI$):

Step 3: Calculating the Difference

Conclusion:

One would get $\textsf{₹} 10,500$ interest without compounding. With compounding, one would get $\textsf{₹} 752.15$ more.

Given:

For both: Principal ($P$) = $\textsf{₹} 12,500$, Time ($n$ or $T$) = $3$ years.

Giridhar: Rate ($R_1$) = $12\%$ (Simple Interest).

Raghava: Rate ($R_2$) = $10\%$ (Compound Interest).

To Find:

Who pays more interest and the difference between their interests.

Solution:

Step 1: Interest paid by Giridhar (Simple Interest)

$I_G = \frac{12500 \times 12 \times 3}{100}$

$I_G = 125 \times 36 = \textsf{₹} 4,500$.

Step 2: Interest paid by Raghava (Compound Interest)

First, we calculate the Amount ($A$):

$A = 12500 \left(1 + \frac{10}{100}\right)^3$

$A = 12500 \times (1.1)^3$

$A = 12500 \times 1.331 = \textsf{₹} 16,637.50$.

Now, Compound Interest ($I_R$):

$I_R = 16637.50 - 12500 = \textsf{₹} 4,137.50$.

Step 3: Comparison

Comparing $I_G = \textsf{₹} 4,500$ and $I_R = \textsf{₹} 4,137.50$.

Difference = $4500 - 4137.50 = \textsf{₹} 362.50$.

Conclusion: Giridhar pays more interest by $\textsf{₹} 362.50$.

Solution:

Part 1: Time taken to double when compounding is NOT done (Simple Interest)

Principal ($P$) = $\textsf{₹}1000$.

To double the amount, the total amount ($A$) must be $2 \times 1000 = \textsf{₹}2000$.

Therefore, the Simple Interest ($\text{SI}$) earned must be $\textsf{₹}2000 - \textsf{₹}1000 = \textsf{₹}1000$.

Using the simple interest formula $\text{SI} = \frac{P \times r \times t}{100}$:

$1000 = \frac{1000 \times 10 \times t}{100}$

$1000 = 100 \times t$

$t = \frac{1000}{100} = 10 \text{ years}$

Thus, it takes $10$ years to double when compounding is not done.

Part 2: Time taken to double when compounding IS done (Compound Interest)

The formula for compound interest is:

$A = P \left(1 + \frac{r}{100}\right)^t$

Substituting $A = 2000$, $P = 1000$, and $r = 10\%$:

$2000 = 1000 \left(1 + \frac{10}{100}\right)^t$

$2 = (1.1)^t$

Using logarithms or trial and error:

For $t = 7$, $(1.1)^7 \approx 1.948$

For $t = 8$, $(1.1)^8 \approx 2.143$

Thus, it takes approximately $7$ to $8$ years (specifically about $7.27$ years) to double when compounding is done.

Part 3: Type of Growth

Yes, compounding is an example of exponential growth because the amount grows by a constant multiplication factor each year, causing the growth rate to accelerate over time.

Yes, not-compounding (Simple Interest) is an example of linear growth because the amount increases by a fixed absolute amount (constant addition) of interest every single year.

Given:

Current Population ($P$) = $1.5$ crore.

Growth Rate ($R$) = $3\%$ p.a., Time ($n$) = $3$ years.

Solution:

Population growth follows the Compound Interest formula as the increase happens on the updated population of each year.

$A = 1.5 \left(1 + \frac{3}{100}\right)^3$

$A = 1.5 \times (1.03)^3$

$A = 1.5 \times 1.092727 = 1.6390905 \text{ crore}$.

Rounding to two decimal places, the population is approximately $1.64$ crore.

Given:

Initial count ($P$) = $5,06,000$.

Growth rate ($R$) = $2.5\%$ per hour, Time ($n$) = $2$ hours.

Solution:

The increase in bacteria is a biological process that follows the compounding rule.

$A = 506000 \times (1.025)^2$

$A = 506000 \times 1.050625$

$A = 531616.25$.

Since the number of bacteria must be a whole number, we round it to $5,31,616$.

Final Result:

The number of bacteria after 2 hours is $5,31,616$.

Given:

Population of Bengaluru in the year $2000 = 50$ lakhs

Percentage growth/comparison for $2025 = 250\%$ of the population in $2000$

To Find:

The population of Bengaluru in the year $2025$.

Solution:

To find the current population, we calculate $250\%$ of the base population from $2000$.

Given:

Total World Population $\approx 8.2$ billion (or $8,200$ million)

Germany $\approx 83$ million

India $\approx 1.46$ billion (or $1,460$ million)

Bangladesh $\approx 175$ million

USA $\approx 347$ million

Solution:

We calculate the percentage share of each country using the formula:

$\text{Percentage Share} = \frac{\text{Country Population}}{\text{World Population}} \times 100$

1. India:

2. USA:

3. Bangladesh:

4. Germany:

Matching Results:

Given:

Base price of mobile phone = $\textsf{₹} 8,250$

GST rate = $18\%$

Solution:

The final price is the base price plus the GST amount calculated on that base price.

$\text{Final Price} = \text{Base Price} + (18\% \text{ of Base Price})$

Alternatively, we can factor out $8250$:

Conclusion:

The correct expressions are (v) $8250 \times 1.18$ and (vi) $8250 + 8250 \times 0.18$.

Solution:

Let the initial population of mice be $p$.

Successive percentage changes applied to the population are:

• Month 1: An increase of $5\%$, which is represented by the multiplier $1 + \frac{5}{100} = 1.05$.

• Month 2: A decrease of $2\%$, which is represented by the multiplier $1 - \frac{2}{100} = 0.98$.

• Month 3: A decrease of $3\%$, which is represented by the multiplier $1 - \frac{3}{100} = 0.97$.

Therefore, the population after three months is calculated by multiplying the initial population by these successive multipliers:

$\text{Final Population} = p \times 1.05 \times 0.98 \times 0.97$

This shows that statement (ii) is true.

Now, let us evaluate the final population value:

$1.05 \times 0.98 \times 0.97 = 0.997785$

Since the overall multiplier ($0.997785$) is less than $1$, the final population is less than the initial population $p$ ($0.997785p < p$).

This shows that statement (vi) is true.

Thus, the correct statements are (ii) and (vi).

Given:

Initial profit margin = $35\%$ above cost price.

Discount offered = $30\%$ on the selling price.

To Find:

Whether the shopkeeper makes a profit or a loss.

Solution:

Let us assume the Cost Price (CP) of the product is $\textsf{₹} 100$.

Step 1: Calculate the initial Selling Price (SP).

$\text{Profit margin} = 35\%$.

Step 2: Calculate the discount and new Selling Price.

A $30\%$ discount is offered on the SP ($\textsf{₹} 135$).

$\text{Discount amount} = 30\% \text{ of } 135$

$\text{Discount amount} = \frac{30}{100} \times 135 = 0.3 \times 135 = \textsf{₹} 40.5$

Step 3: Compare CP and new SP.

CP = $\textsf{₹} 100$ and New SP = $\textsf{₹} 94.5$.

Since $\text{CP} > \text{SP}$, the shopkeeper makes a loss.

Conclusion:

The shopkeeper makes a loss of $5.5\%$. The reason is that the $30\%$ discount is calculated on a larger base value (the increased selling price), which results in a reduction that is greater than the original $35\%$ increase on the cost price.

Given:

A square grid divided into different regions A, B, C, D, and E.

To Find:

The percentage of the total area occupied by region ‘E’.

Solution:

Let us count the units based on the dot intervals in the grid.

1. The total grid size is $8$ units $\times$ $8$ units (based on $9$ dots horizontally and vertically).

2. Region ‘E’ is a right-angled triangle.

Its base is $4$ units (from the middle vertical line to the left edge).

Its height is $4$ units (from the bottom horizontal line to the middle horizontal divider).

3. Calculate the area of region ‘E’:

4. Calculate the percentage of area occupied by E:

Conclusion:

Region ‘E’ occupies $12.5\%$ of the total area.

Solution:

Let us calculate each pair to identify the pattern:

Case 1:

$5\%$ of $40 = \frac{5}{100} \times 40 = 2$

$40\%$ of $5 = \frac{40}{100} \times 5 = 2$

Case 2:

$25\%$ of $12 = \frac{25}{100} \times 12 = \frac{1}{4} \times 12 = 3$

$12\%$ of $25 = \frac{12}{100} \times 25 = \frac{12}{4} = 3$

Case 3:

$15\%$ of $60 = \frac{15}{100} \times 60 = 9$

$60\%$ of $15 = \frac{60}{100} \times 15 = 9$

Observation:

In all the cases above, the results of the two calculations are identical. This means $x\%$ of $y$ is always equal to $y\%$ of $x$.

Algebraic Justification:

Let us consider two numbers $x$ and $y$.

According to the Commutative Property of Multiplication, which is a fundamental concept in Indian Vedic Mathematics as well as modern algebra, we know that:

Therefore, from equations (i) and (ii), we can conclude that:

Given:

Grade 8 students = $40\%$ of total students.

Girls in Grade 8 = $60\%$ of Grade 8 students.

Solution (i):

To find the percentage of Grade 8 girls relative to the entire group, we multiply the two percentages.

So, $24\%$ of the total students are Grade 8 girls.

Solution (ii):

Total number of students = $160$.

Number of Grade 8 girls = $24\%$ of $160$.

Given:

Selling Price (SP) of $3$ pencils = Cost Price (CP) of $5$ pencils.

To Find:

Profit or Loss percentage.

Solution:

Let the Cost Price (CP) of $1$ pencil be $\textsf{₹} x$.

As per the question:

Now, we find the Selling Price of $1$ pencil:

Since $\frac{5}{3}x$ (approx $1.67x$) is greater than $x$, the shopkeeper makes a profit.

Now, calculate the Profit Percentage:

Given:

Increase in Year 1 = $3\%$.

Increase in Year 2 = $4\%$.

Solution:

This is a case of Successive Percentage Increase. Let the initial bus fare be $\textsf{₹} 100$.

Step 1: After the first year increase ($3\%$)

New Fare = $100 + (3\% \text{ of } 100) = \textsf{₹} 103$.

Step 2: After the second year increase ($4\%$)

The $4\%$ increase is applied to the new fare of $\textsf{₹} 103$.

Step 3: Calculate total percentage increase

Overall Percentage Increase = $7.12\%$.

Given:

Percentage increase in length = $10\%$

The area of the rectangle remains unchanged.

To Find:

The percentage decrease in the breadth of the rectangle.

Solution:

Let the initial length be $L$ and the initial breadth be $B$.

Initial Area ($A$) = $L \times B$

New Length ($L'$) = $L + 10\% \text{ of } L = 1.1L$

Let the new breadth be $B'$. Since the area remains the same:

Dividing both sides by $L$:

Now, we find the decrease in breadth:

$\text{Decrease} = B - B' = B - \frac{10}{11}B = \frac{1}{11}B$

Percentage decrease:

$\text{Percentage} \approx 9.09\%$

Given:

Total weight of the chips packet = $65$ g

Proportions: Potato ($70\%$), Vegetable oil ($24\%$), Salt ($3\%$), and Spices ($3\%$).

To Find:

The weight of each individual ingredient in grams.

Solution:

We find the weight of each ingredient by calculating its percentage of the total $65$ g weight.

1. Weight of Potato:

$\text{Weight} = 45.5 \text{ g}$

2. Weight of Vegetable Oil:

$\text{Weight} = 15.6 \text{ g}$

3. Weight of Salt:

$\text{Weight} = 1.95 \text{ g}$

4. Weight of Spices:

Summary of Ingredients:

(i) Solution: Effective Price per Item

Assume the price of one item is $\textsf{₹} 100$.

Shop A (Buy 1 get 1): You pay for $1$ item but get $2$ items. Total cost = $\textsf{₹} 100$.

$\text{Effective price} = \frac{100}{2} = \textsf{₹} 50$.

Shop B (Buy 2 get 1): You pay for $2$ items but get $3$ items. Total cost = $\textsf{₹} 200$.

$\text{Effective price} = \frac{200}{3} \approx \textsf{₹} 66.67$.

Shop C (Buy 3 get 1): You pay for $3$ items but get $4$ items. Total cost = $\textsf{₹} 300$.

$\text{Effective price} = \frac{300}{4} = \textsf{₹} 75$.

Order (Cheapest to Costliest): Shop A < Shop B < Shop C.

(ii) Solution: Percentage Discount

$\text{Discount } \% = \frac{\text{Free Items}}{\text{Total Items}} \times 100$

Shop A: $\frac{1}{2} \times 100 = 50\%$

Shop B: $\frac{1}{3} \times 100 = 33.33\%$

Shop C: $\frac{1}{4} \times 100 = 25\%$

(iii) Choice for 4 Items:

If we need $4$ items, let's calculate the total cost in each shop:

Shop A: We "Buy 2 and get 2 free". Total cost = $2 \times 100 = \textsf{₹} 200$.

Shop B: We "Buy 2 and get 1 free" (3 items) and then buy $1$ more. Total cost = $200 + 100 = \textsf{₹} 300$.

Shop C: We "Buy 3 and get 1 free". Total cost = $3 \times 100 = \textsf{₹} 300$.

Decision: I would choose Shop A because it offers the lowest total cost of $\textsf{₹} 200$ for 4 items.

Given:

To Find:

The number of left-handed people who must leave to make the new left-handed percentage $98\%$.

Solution:

First, we determine the number of left-handed and right-handed people initially.

Number of left-handed people = $99\%$ of $100 = 99$.

Let the number of left-handed people who leave be $x$.

New total number of people = $100 - x$.

New number of left-handed people = $99 - x$.

According to the question, the new percentage of left-handed people should be $98\%$.

Cross-multiplying the terms:

$100(99 - x) = 98(100 - x)$

$9900 - 100x = 9800 - 98x$

$9900 - 9800 = 100x - 98x$

Verification:

If $50$ left-handed people leave, the new total is $100 - 50 = 50$ people. The remaining left-handers are $99 - 50 = 49$.

Percentage = $\frac{49}{50} \times 100 = 98\%$. (Correct)

Conclusion:

$50$ left-handed people have to leave the room.

Let us analyze each statement based on the given bar graph:

• Statement (i): People in their twenties have the highest percentages ($37\%$ for males and $26\%$ for females) among all groups. Thus, this statement is valid.

• Statement (ii): Female percentages are consistently lower than male percentages across all age groups. Thus, this statement is valid.

• Statement (iii): The graph shows literacy percentages, not total population counts. Thus, this statement is invalid.

• Statement (iv): For people in their thirties, the male literacy rate is $25\%$ (exactly a quarter) and female is $14\%$. Neither group exceeds a quarter. Thus, this statement is invalid.

• Statement (v): For seniors, the rates are $4\%$ and $2\%$, which are both less than $10\%$ ($1$ in $10$). Thus, this statement is valid.

• Statement (vi): The percentages for people in their twenties are $37\%$ and $26\%$, which are well below $50\%$. Thus, this statement is invalid.

Therefore, the valid statements are (i), (ii), and (v).