Chapter 7 Comparing Quantities

Solution:

1. Ratio of girls to boys:

Given, the number of girls is 18 and they are 60% of the total number of students.

Let the total number of students be $S$.

$60\%$ of $S = 18$

$\frac{60}{100} \times S = 18$

$S = 18 \times \frac{100}{60}$

$S = 18 \times \frac{10}{6}$

$S = \cancel{18}^3 \times \frac{10}{\cancel{6}_1}$

$S = 3 \times 10$

$S = 30$

Total number of students in the class is 30.

Number of boys = Total students - Number of girls

Number of boys = $30 - 18 = 12$

The ratio of the number of girls to the number of boys is:

Ratio = Number of girls : Number of boys

Ratio = $18 : 12$

To simplify the ratio, divide both numbers by their greatest common divisor, which is 6.

Ratio = $\frac{18}{6} : \frac{12}{6}$

Ratio = $3 : 2$

2. Cost per head including teachers:

Distance to the picnic site = 55 km.

Transport charge rate = $\textsf{₹}$ 12 per km.

Total transport cost = Distance $\times$ Rate per km

Total transport cost = $55 \times 12$

Total transport cost = $\textsf{₹}$ 660.

Total cost of refreshments = $\textsf{₹}$ 4280.

Total cost of the picnic = Total transport cost + Total cost of refreshments

Total cost = $\textsf{₹} 660 + \textsf{₹} 4280$

Total cost = $\textsf{₹}$ 4940.

Total number of students = 30.

Number of teachers going with the class = 2.

Total number of people going = Total students + Number of teachers

Total number of people = $30 + 2 = 32$.

Cost per head = Total cost / Total number of people

Cost per head = $\frac{\textsf{₹} 4940}{32}$

Cost per head = $\textsf{₹}$ 154.375.

3. Percentage of distance covered and left:

Total distance to the picnic site = 55 km.

Distance covered to the first stop = 22 km.

Percentage of total distance covered = $\left(\frac{\text{Distance covered}}{\text{Total distance}}\right) \times 100\%$

Percentage covered = $\left(\frac{22}{55}\right) \times 100\%$

Percentage covered = $\left(\frac{\cancel{22}^2}{\cancel{55}^5}\right) \times 100\%$

Percentage covered = $\frac{2}{5} \times 100\%$

Percentage covered = $2 \times 20\%$

Percentage covered = $40\%$.

Distance left to be covered = Total distance - Distance covered

Distance left = $55 - 22 = 33$ km.

Percentage of distance left = $\left(\frac{\text{Distance left}}{\text{Total distance}}\right) \times 100\%$

Percentage left = $\left(\frac{33}{55}\right) \times 100\%$

Percentage left = $\left(\frac{\cancel{33}^3}{\cancel{55}^5}\right) \times 100\%$

Percentage left = $\frac{3}{5} \times 100\%$

Percentage left = $3 \times 20\%$

Percentage left = $60\%$.

Alternatively, Percentage left = $100\%$ - Percentage covered = $100\% - 40\% = 60\%$.

Solution:

To find the ratio of two quantities, the units must be the same.

(a) Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.

Both speeds are given in km per hour, so the units are the same.

Ratio = Speed of cycle : Speed of scooter

Ratio = 15 km/h : 30 km/h

Ratio = $\frac{15}{30}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 15.

Ratio = $\frac{\cancel{15}^1}{\cancel{30}_2}$

Ratio = $\frac{1}{2}$

The ratio is $1:2$.

(b) 5 m to 10 km

The units are meters (m) and kilometers (km). We need to convert one to match the other. Let's convert kilometers to meters.

1 km = 1000 m

10 km = $10 \times 1000$ m = 10000 m

Now find the ratio of 5 m to 10000 m.

Ratio = 5 m : 10000 m

Ratio = $\frac{5}{10000}$

Simplify the fraction by dividing the numerator and denominator by 5.

Ratio = $\frac{\cancel{5}^1}{\cancel{10000}_{2000}}$

Ratio = $\frac{1}{2000}$

The ratio is $1:2000$.

(c) 50 paise to ₹ 5

The units are paise and rupees ($\textsf{₹}$). We need to convert one to match the other. Let's convert rupees to paise.

$\textsf{₹}$ 1 = 100 paise

$\textsf{₹}$ 5 = $5 \times 100$ paise = 500 paise

Now find the ratio of 50 paise to 500 paise.

Ratio = 50 paise : 500 paise

Ratio = $\frac{50}{500}$

Simplify the fraction by dividing the numerator and denominator by 50.

Ratio = $\frac{\cancel{50}^1}{\cancel{500}_{10}}$

Ratio = $\frac{1}{10}$

The ratio is $1:10$.

Solution:

To convert a ratio to a percentage, we write the ratio as a fraction and then multiply by 100%.

(a) 3 : 4

The ratio 3 : 4 can be written as the fraction $\frac{3}{4}$.

To convert this fraction to a percentage, multiply by 100%:

Percentage $= \frac{3}{4} \times 100\%$

Percentage $= 3 \times \frac{100}{4}\%$

Percentage $= 3 \times 25\%$

Percentage $= 75\%$

(b) 2 : 3

The ratio 2 : 3 can be written as the fraction $\frac{2}{3}$.

To convert this fraction to a percentage, multiply by 100%:

Percentage $= \frac{2}{3} \times 100\%$

Percentage $= \frac{200}{3}\%$

This can be expressed as a mixed number or a repeating decimal.

Percentage $= 66\frac{2}{3}\%$

or

Percentage $= 66.\overline{6}\%$

Solution:

Total number of students = 25.

Percentage of students interested in mathematics = 72%.

The percentage of students who are not interested in mathematics is found by subtracting the percentage of interested students from the total percentage (100%).

Percentage of students not interested in mathematics = $100\% - 72\% = 28\%$.

Now, we find the number of students who are not interested in mathematics by calculating 28% of the total number of students.

Number of students not interested = 28% of 25

Number of students not interested $= \frac{28}{100} \times 25$

We can simplify the calculation:

Number of students not interested $= \frac{28}{\cancel{100}_{4}} \times \cancel{25}^{1}$

Number of students not interested $= \frac{28}{4}$

Number of students not interested $= 7$

Thus, 7 students are not interested in mathematics.

Solution:

Let the total number of matches played by the football team be $M$.

The number of matches won is 10.

The win percentage is given as 40%.

The win percentage is calculated as:

Win Percentage $= \left(\frac{\text{Number of matches won}}{\text{Total number of matches played}}\right) \times 100\%$

So, we have:

$40\% = \left(\frac{10}{M}\right) \times 100\%$

We can write the percentage as a fraction:

$\frac{40}{100} = \frac{10}{M}$

To solve for $M$, we can cross-multiply:

$40 \times M = 10 \times 100$

$40M = 1000$

Divide both sides by 40:

$M = \frac{1000}{40}$

$M = \frac{100}{4}$

$M = 25$

Thus, the total number of matches played by the team was 25.

Solution:

Let the total amount of money Chameli had in the beginning be $x$ $\textsf{₹}$.

Percentage of money spent = 75%.

The percentage of money left is the total percentage (100%) minus the percentage spent.

Percentage of money left = $100\% - 75\% = 25\%$.

We are given that the amount of money left is $\textsf{₹}$ 600.

So, 25% of the original amount $x$ is equal to $\textsf{₹}$ 600.

25% of $x = \textsf{₹} 600$

We can write the percentage as a fraction:

$\frac{25}{100} \times x = 600$

Simplify the fraction $\frac{25}{100}$ to $\frac{1}{4}$.

$\frac{1}{4} \times x = 600$

To find $x$, multiply both sides of the equation by 4:

$x = 600 \times 4$

$x = 2400$

Thus, Chameli had $\textsf{₹}$ 2400 in the beginning.

Solution:

Given information:

Percentage of people who like cricket = 60%

Percentage of people who like football = 30%

The remaining percentage of people like other games.

Total percentage of people in the city = 100%.

First, we find the percentage of people who like other games.

Percentage of people who like other games = Total percentage - (Percentage who like cricket + Percentage who like football)

Percentage of people who like other games = $100\% - (60\% + 30\%)$

Percentage of people who like other games = $100\% - 90\%$

Percentage of people who like other games = $10\%$.

Now, we find the exact number of people for each category.

Total number of people in the city = 50 lakh.

In the Indian numbering system, 50 lakh is written as 50,00,000.

Total number of people = 50,00,000.

Number of people who like cricket:

This is 60% of the total number of people.

Number liking cricket = 60% of 50,00,000

Number liking cricket $= \frac{60}{100} \times 50,00,000$

Number liking cricket $= \frac{60}{\cancel{100}} \times 50000\cancel{00}$

Number liking cricket $= 60 \times 50000$

Number liking cricket $= 30,00,000$.

Number of people who like football:

This is 30% of the total number of people.

Number liking football = 30% of 50,00,000

Number liking football $= \frac{30}{100} \times 50,00,000$

Number liking football $= \frac{30}{\cancel{100}} \times 50000\cancel{00}$

Number liking football $= 30 \times 50000$

Number liking football $= 15,00,000$.

Number of people who like other games:

This is 10% of the total number of people.

Number liking other games = 10% of 50,00,000

Number liking other games $= \frac{10}{100} \times 50,00,000$

Number liking other games $= \frac{10}{\cancel{100}} \times 50000\cancel{00}$

Number liking other games $= 10 \times 50000$

Number liking other games $= 5,00,000$.

Let's verify the total number of people:

Total people = People liking cricket + People liking football + People liking other games

Total people = $30,00,000 + 15,00,000 + 5,00,000$

Total people = $45,00,000 + 5,00,000$

Total people = $50,00,000$. This matches the given total number of people.

Answers:

Percentage of people who like other games is 10%.

The number of people who like cricket is 30,00,000.

The number of people who like football is 15,00,000.

The number of people who like other games is 5,00,000.

Solution:

Given:

Marked Price (MP) = $\textsf{₹} 840$

Selling Price (SP) = $\textsf{₹} 714$

The Discount is the difference between the Marked Price and the Selling Price.

Discount = Marked Price - Selling Price

Discount = $\textsf{₹} 840 - \textsf{₹} 714$

Discount = $\textsf{₹} 126$.

The Discount Percentage is calculated on the Marked Price.

Discount % = $\left(\frac{\text{Discount}}{\text{Marked Price}}\right) \times 100\%$

Discount % = $\left(\frac{126}{840}\right) \times 100\%$

We can simplify the fraction $\frac{126}{840}$.

$\frac{126}{840} = \frac{\cancel{126}^{\cancel{63}^{\cancel{9}^3}}}{\cancel{840}_{\cancel{420}_{\cancel{60}_{20}}}} = \frac{3}{20}$.

Alternatively, $\frac{126}{840} = \frac{126 \div 42}{840 \div 42} = \frac{3}{20}$. (Since $42 \times 3 = 126$ and $42 \times 20 = 840$).

Now, calculate the percentage:

Discount % = $\frac{3}{20} \times 100\%$

Discount % = $3 \times \frac{\cancel{100}^{5}}{\cancel{20}_{1}}\%$

Discount % = $3 \times 5\%$

Discount % = $15\%$

The discount is $\textsf{₹} 126$ and the discount percentage is $15\%$.

Solution:

Given:

List Price (Marked Price) of the frock = $\textsf{₹} 220$

Discount percentage = 20%

First, we calculate the amount of discount.

Discount Amount = Discount % of List Price

Discount Amount $= 20\% \text{ of } \textsf{₹} 220$

Discount Amount $= \frac{20}{100} \times 220$

Discount Amount $= \frac{1}{5} \times 220$

Discount Amount $= \frac{\cancel{220}^{44}}{\cancel{5}_{1}}$

Discount Amount $= 44$

So, the amount of discount is $\textsf{₹} 44$.

Next, we calculate the Sale Price (Selling Price).

Sale Price = List Price - Discount Amount

Sale Price $= \textsf{₹} 220 - \textsf{₹} 44$

Sale Price $= \textsf{₹} 176$.

The amount of discount is $\textsf{₹} 44$ and the sale price is $\textsf{₹} 176$.

Solution:

Given:

Cost of the pair of roller skates = $\textsf{₹} 450$

Sales Tax rate = 5%

The Sales Tax is calculated on the original cost of the item.

Sales Tax Amount = Sales Tax % of Cost

Sales Tax Amount $= 5\% \text{ of } \textsf{₹} 450$

Sales Tax Amount $= \frac{5}{100} \times 450$

Sales Tax Amount $= \frac{1}{20} \times 450$

Sales Tax Amount $= \frac{\cancel{450}^{45}}{\cancel{20}_{2}} = \frac{45}{2}$

Sales Tax Amount $= 22.50$

So, the Sales Tax amount is $\textsf{₹} 22.50$.

The Bill Amount is the original cost plus the Sales Tax.

Bill Amount = Cost + Sales Tax Amount

Bill Amount $= \textsf{₹} 450 + \textsf{₹} 22.50$

Bill Amount $= \textsf{₹} 472.50$.

The bill amount is $\textsf{₹} 472.50$.

Solution:

Given:

Price of the air cooler including VAT = $\textsf{₹} 3300$

VAT rate = 10%

Let the price of the air cooler before VAT was added be $P$ $\textsf{₹}$.

The VAT amount is 10% of the original price $P$.

VAT Amount = $10\% \text{ of } P = \frac{10}{100} \times P = 0.10 P$.

The price including VAT is the original price plus the VAT amount.

Price including VAT = Original Price + VAT Amount

$\textsf{₹} 3300 = P + 0.10 P$

$\textsf{₹} 3300 = (1 + 0.10) P$

$\textsf{₹} 3300 = 1.10 P$

To find $P$, divide the price including VAT by 1.10.

$P = \frac{3300}{1.10}$

To remove the decimal from the denominator, multiply both the numerator and the denominator by 100.

$P = \frac{3300 \times 100}{1.10 \times 100} = \frac{330000}{110}$

$P = \frac{\cancel{330000}^{3000}}{\cancel{110}_1}$

So, $P = 3000$.

Alternatively:

If the VAT is 10%, it means the price including VAT is 100% (original price) + 10% (VAT) = 110% of the original price.

So, $110\%$ of Original Price $= \textsf{₹} 3300$.

$\frac{110}{100} \times \text{Original Price} = 3300$

Original Price $= 3300 \times \frac{100}{110}$

Original Price $= 3300 \times \frac{10}{11}$

Original Price $= \frac{\cancel{3300}^{300} \times 10}{\cancel{11}_1}$

Original Price $= 300 \times 10 = 3000$

The price of the air cooler before VAT was added was $\textsf{₹} 3000$.

Solution:

Given:

Price of the article including GST = $\textsf{₹} 784$

GST rate = 12%

Let the price of the article before GST was added be $P$ $\textsf{₹}$.

The GST amount is 12% of the original price $P$.

GST Amount = $12\% \text{ of } P = \frac{12}{100} \times P = 0.12 P$.

The price including GST is the original price plus the GST amount.

Price including GST = Original Price + GST Amount

$\textsf{₹} 784 = P + 0.12 P$

$\textsf{₹} 784 = (1 + 0.12) P$

$\textsf{₹} 784 = 1.12 P$

To find $P$, divide the price including GST by 1.12.

$P = \frac{784}{1.12}$

To remove the decimal from the denominator, multiply both the numerator and the denominator by 100.

$P = \frac{784 \times 100}{1.12 \times 100} = \frac{78400}{112}$

Now, perform the division.

$P = 700$. (Since $112 \times 7 = 784$, so $112 \times 700 = 78400$).

Alternatively:

If the GST is 12%, it means the price including GST is 100% (original price) + 12% (GST) = 112% of the original price.

So, $112\%$ of Original Price $= \textsf{₹} 784$.

$\frac{112}{100} \times \text{Original Price} = 784$

Original Price $= 784 \times \frac{100}{112}$

Original Price $= \frac{78400}{112}$

Original Price $= 700$.

The price of the article before GST was added was $\textsf{₹} 700$.

Solution:

Given:

Discount rate = 10%

Marked Price (MP) of a pair of jeans = $\textsf{₹} 1450$

Marked Price (MP) of one shirt = $\textsf{₹} 850$

First, calculate the total marked price of the items the customer wants to buy.

Marked Price of two shirts $= 2 \times \textsf{₹} 850 = \textsf{₹} 1700$.

Total Marked Price = Marked Price of jeans + Marked Price of two shirts

Total Marked Price $= \textsf{₹} 1450 + \textsf{₹} 1700 = \textsf{₹} 3150$.

Next, calculate the amount of discount offered on the total marked price.

Discount Amount = 10% of Total Marked Price

Discount Amount $= 10\% \text{ of } \textsf{₹} 3150$

Discount Amount $= \frac{10}{100} \times 3150$

Discount Amount $= \frac{1}{\cancel{10}} \times 315\cancel{0}$

Discount Amount $= 315$.

The discount amount is $\textsf{₹} 315$.

Finally, calculate the amount the customer has to pay (Sale Price).

Amount to pay = Total Marked Price - Discount Amount

Amount to pay $= \textsf{₹} 3150 - \textsf{₹} 315$

Amount to pay $= \textsf{₹} 2835$.

The customer would have to pay $\textsf{₹} 2835$.

Alternate Solution:

When a discount of 10% is offered on the marked price, the customer pays the remaining percentage of the marked price, which is $100\% - 10\% = 90\%$.

Total Marked Price = $\textsf{₹} 1450 \text{ (Jeans)} + (2 \times \textsf{₹} 850) \text{ (Shirts)}$

Total Marked Price = $\textsf{₹} 1450 + \textsf{₹} 1700 = \textsf{₹} 3150$.

Amount to pay = 90% of Total Marked Price

Amount to pay $= 90\% \text{ of } \textsf{₹} 3150$

Amount to pay $= \frac{90}{100} \times 3150$

Amount to pay $= \frac{9}{\cancel{10}} \times 315\cancel{0}$

Amount to pay $= 9 \times 315$

Amount to pay $= 2835$.

The customer would have to pay $\textsf{₹} 2835$.

Solution:

Given:

Price of the TV (Original Price) = $\textsf{₹} 13,000$

Sales Tax rate = 12%

The Sales Tax is calculated on the original price of the item.

Sales Tax Amount = Sales Tax % of Original Price

Sales Tax Amount $= 12\% \text{ of } \textsf{₹} 13000$

Sales Tax Amount $= \frac{12}{100} \times 13000$

Sales Tax Amount $= \frac{12}{\cancel{100}} \times 130\cancel{00}$

Sales Tax Amount $= 12 \times 130$

Sales Tax Amount $= 1560$.

So, the Sales Tax amount is $\textsf{₹} 1560$.

The amount Vinod will have to pay is the original price plus the Sales Tax.

Amount to pay = Original Price + Sales Tax Amount

Amount to pay $= \textsf{₹} 13000 + \textsf{₹} 1560$

Amount to pay $= \textsf{₹} 14560$.

Alternatively:

If a sales tax of 12% is charged on the original price, the final price is the original price (100%) plus the sales tax (12%), which is $100\% + 12\% = 112\%$ of the original price.

Amount to pay = 112% of Original Price

Amount to pay $= 112\% \text{ of } \textsf{₹} 13000$

Amount to pay $= \frac{112}{100} \times 13000$

Amount to pay $= \frac{112}{\cancel{100}} \times 130\cancel{00}$

Amount to pay $= 112 \times 130$

Amount to pay $= 14560$.

Vinod will have to pay $\textsf{₹} 14560$.

Solution:

Given:

Selling Price (Amount paid) = $\textsf{₹} 1600$

Discount percentage = 20%

We need to find the Marked Price (MP) of the skates.

Let the Marked Price be MP.

The discount amount is calculated on the Marked Price.

Discount Amount = 20% of MP

Discount Amount $= \frac{20}{100} \times \text{MP} = 0.20 \times \text{MP}$.

The Selling Price is the Marked Price minus the Discount Amount.

Selling Price = Marked Price - Discount Amount

$\textsf{₹} 1600 = \text{MP} - 0.20 \times \text{MP}$

$\textsf{₹} 1600 = (1 - 0.20) \times \text{MP}$

$\textsf{₹} 1600 = 0.80 \times \text{MP}$

To find MP, divide the Selling Price by 0.80.

MP $= \frac{1600}{0.80}$

To remove the decimal from the denominator, multiply both the numerator and the denominator by 100.

MP $= \frac{1600 \times 100}{0.80 \times 100} = \frac{160000}{80}$

MP $= \frac{\cancel{160000}^{16000}}{\cancel{80}_{8}} = \frac{16000}{8} = 2000$.

So, the marked price is $\textsf{₹} 2000$.

Alternatively:

If a discount of 20% is given on the marked price, it means the customer pays $100\% - 20\% = 80\%$ of the marked price.

So, 80% of Marked Price = Selling Price

$80\% \text{ of MP} = \textsf{₹} 1600$

$\frac{80}{100} \times \text{MP} = 1600$

$\frac{4}{5} \times \text{MP} = 1600$

To find MP, multiply both sides by the reciprocal of $\frac{4}{5}$, which is $\frac{5}{4}$.

MP $= 1600 \times \frac{5}{4}$

MP $= \frac{\cancel{1600}^{400} \times 5}{\cancel{4}_1}$

MP $= 400 \times 5 = 2000$.

The marked price is $\textsf{₹} 2000$.

The marked price of the pair of skates was $\textsf{₹} 2000$.

Solution:

Given:

Price of the hair-dryer including VAT = $\textsf{₹} 5,400$

VAT rate = 8%

We need to find the price of the hair-dryer before VAT was added.

Let the price of the hair-dryer before VAT was added be $P$ $\textsf{₹}$.

The VAT amount is 8% of the original price $P$.

VAT Amount = $8\% \text{ of } P = \frac{8}{100} \times P = 0.08 P$.

The price including VAT is the original price plus the VAT amount.

Price including VAT = Original Price + VAT Amount

$\textsf{₹} 5400 = P + 0.08 P$

$\textsf{₹} 5400 = (1 + 0.08) P$

$\textsf{₹} 5400 = 1.08 P$

To find $P$, divide the price including VAT by 1.08.

$P = \frac{5400}{1.08}$

To remove the decimal from the denominator, multiply both the numerator and the denominator by 100.

$P = \frac{5400 \times 100}{1.08 \times 100} = \frac{540000}{108}$

Now, perform the division: $540000 \div 108$. Since $108 \times 5 = 540$, $108 \times 5000 = 540000$.

$P = 5000$.

So, the price before VAT was added is $\textsf{₹} 5000$.

Alternatively:

If the VAT is 8%, then the price including VAT is the original price (100%) plus the VAT (8%), which is $100\% + 8\% = 108\%$ of the original price.

So, 108% of Original Price $= \textsf{₹} 5400$.

$\frac{108}{100} \times \text{Original Price} = 5400$

Original Price $= 5400 \times \frac{100}{108}$

Original Price $= \frac{\cancel{5400}^{50} \times 100}{\cancel{108}_1}$

Original Price $= 50 \times 100 = 5000$.

The price before VAT was added is $\textsf{₹} 5000$.

The price of the hair-dryer before VAT was added was $\textsf{₹} 5000$.

Solution:

Given:

Price of the article including GST = $\textsf{₹} 1239$

GST rate = 18%

We need to find the price of the article before GST was added.

Let the price of the article before GST was added be $P$ $\textsf{₹}$.

The GST amount is 18% of the original price $P$.

GST Amount = $18\% \text{ of } P = \frac{18}{100} \times P = 0.18 P$.

The price including GST is the original price plus the GST amount.

Price including GST = Original Price + GST Amount

$\textsf{₹} 1239 = P + 0.18 P$

$\textsf{₹} 1239 = (1 + 0.18) P$

$\textsf{₹} 1239 = 1.18 P$

To find $P$, divide the price including GST by 1.18.

$P = \frac{1239}{1.18}$

To remove the decimal from the denominator, multiply both the numerator and the denominator by 100.

$P = \frac{1239 \times 100}{1.18 \times 100} = \frac{123900}{118}$

Now, perform the division: $123900 \div 118$. Let's simplify the fraction.

$P = \frac{\cancel{123900}^{61950}}{\cancel{118}_{59}}$. Now divide 61950 by 59. $61950 \div 59 = 1050$.

So, $P = 1050$.

The price before GST was added is $\textsf{₹} 1050$.

Alternatively:

If the GST is 18%, then the price including GST is the original price (100%) plus the GST (18%), which is $100\% + 18\% = 118\%$ of the original price.

So, 118% of Original Price $= \textsf{₹} 1239$.

$\frac{118}{100} \times \text{Original Price} = 1239$

Original Price $= 1239 \times \frac{100}{118}$

Original Price $= \frac{123900}{118}$

Original Price $= \frac{\cancel{123900}^{61950}}{\cancel{118}_{59}} = \frac{61950}{59} = 1050$.

The price before GST was added is $\textsf{₹} 1050$.

The price of the article before GST was added was $\textsf{₹} 1050$.

Solution:

Given:

Principal (P) = $\textsf{₹} 10,000$

Rate of Interest (R) = 15% per annum

Time (T) = 2 years

We need to find the Simple Interest (SI) and the total Amount (A) to be paid.

The formula for Simple Interest is:

$SI = \frac{P \times R \times T}{100}$

Substitute the given values into the formula:

$SI = \frac{10000 \times 15 \times 2}{100}$

Simplify the expression:

$SI = \frac{10000 \times 30}{100}$

$SI = \frac{\cancel{10000}^{100} \times 30}{\cancel{100}_1}$

$SI = 100 \times 30$

$SI = 3000$

The simple interest on the sum is $\textsf{₹} 3000$.

The amount to be paid at the end of 2 years is the sum of the Principal and the Simple Interest.

Amount (A) = Principal (P) + Simple Interest (SI)

Amount = $\textsf{₹} 10000 + \textsf{₹} 3000$

Amount = $\textsf{₹} 13000$.

The simple interest is $\textsf{₹} 3000$ and the amount to be paid at the end of 2 years is $\textsf{₹} 13000$.

Solution:

Given:

Principal (P) = $\textsf{₹} 12600$

Time period (n) = 2 years

Rate of Interest (R) = 10% per annum

Compounding: Annually

We need to find the Compound Interest (CI).

We use the formula for the Amount (A) when interest is compounded annually:

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

Substitute the given values:

$A = 12600 \left(1 + \frac{10}{100}\right)^2$

$A = 12600 \left(1 + 0.10\right)^2$

$A = 12600 \left(1.10\right)^2$

$A = 12600 \times (1.10 \times 1.10)$

$A = 12600 \times 1.21$

Calculating the product $12600 \times 1.21$:

$12600 \times 1.21 = 126 \times 100 \times \frac{121}{100} = 126 \times 121$.

$126 \times 121 = 126 \times (100 + 20 + 1) = 126 \times 100 + 126 \times 20 + 126 \times 1$

$= 12600 + 2520 + 126 = 15120 + 126 = 15246$.

So, the total amount after 2 years is $A = \textsf{₹} 15246$.

The Compound Interest (CI) is the difference between the Amount and the Principal:

$CI = A - P$

$CI = \textsf{₹} 15246 - \textsf{₹} 12600$

$CI = \textsf{₹} 2646$.

The Compound Interest on ₹ 12600 for 2 years at 10% per annum compounded annually is $\textsf{₹} 2646$.

Solution:

Given:

Population in the year 1997 (Initial Population, P) = 20,000

Rate of increase (R) = 5% per annum

Time period (n) = From 1997 to 2000. This is $2000 - 1997 = 3$ years.

This problem involves population growth, which can be calculated similarly to compound interest where the increase is added to the population each year.

The formula for the population after $n$ years with a constant growth rate is:

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

Where:

A = Population after $n$ years

P = Initial population

R = Rate of increase per annum

n = Number of years

Substitute the given values into the formula:

$A = 20000 \left(1 + \frac{5}{100}\right)^3$

$A = 20000 \left(1 + 0.05\right)^3$

$A = 20000 \left(1.05\right)^3$

$A = 20000 \times (1.05 \times 1.05 \times 1.05)$

Calculate $(1.05)^3$:

$1.05 \times 1.05 = 1.1025$

$1.1025 \times 1.05 = 1.157625$

So, $A = 20000 \times 1.157625$

Multiply 20000 by 1.157625:

$A = 20000 \times 1.157625 = 2 \times 10000 \times 1.157625 = 2 \times 11576.25 = 23152.5$.

The population at the end of the year 2000 is calculated to be 23152.5.

Since population must be a whole number, in practical terms, this value would likely be rounded to the nearest whole number, which is 23153.

Alternate Calculation (Year by Year):

Population in 1997 = 20,000

Increase during 1998 = 5% of 20000 = $\frac{5}{100} \times 20000 = 5 \times 200 = 1000$.

Population at the end of 1998 = $20000 + 1000 = 21000$.

Increase during 1999 = 5% of 21000 = $\frac{5}{100} \times 21000 = 5 \times 210 = 1050$.

Population at the end of 1999 = $21000 + 1050 = 22050$.

Increase during 2000 = 5% of 22050 = $\frac{5}{100} \times 22050 = 0.05 \times 22050 = 1102.5$.

Population at the end of 2000 = $22050 + 1102.5 = 23152.5$.

This confirms the result from the formula.

The population at the end of the year 2000 was 23152.5.

Solution:

Given:

Original Price of the TV = $\textsf{₹} 21,000$

Rate of Depreciation = 5% per annum

Time period = 1 year

We need to find the value of the TV after one year due to depreciation.

Depreciation is a reduction in value. The depreciation amount for one year is 5% of the original price.

Depreciation Amount = 5% of $\textsf{₹} 21,000$

Depreciation Amount $= \frac{5}{100} \times 21000$

Depreciation Amount $= \frac{5}{\cancel{100}} \times 210\cancel{00}$

Depreciation Amount $= 5 \times 210$

Depreciation Amount $= 1050$.

The depreciation amount after one year is $\textsf{₹} 1050$.

The value of the TV after one year is the original price minus the depreciation amount.

Value after 1 year = Original Price - Depreciation Amount

Value after 1 year $= \textsf{₹} 21000 - \textsf{₹} 1050$

Value after 1 year $= \textsf{₹} 19950$.

Alternatively:

When the value of an item depreciates by 5%, the value after one year is the original value reduced by 5%. This means the value remaining is $100\% - 5\% = 95\%$ of the original price.

Value after 1 year = 95% of Original Price

Value after 1 year $= 95\% \text{ of } \textsf{₹} 21000$

Value after 1 year $= \frac{95}{100} \times 21000$

Value after 1 year $= \frac{95}{\cancel{100}} \times 210\cancel{00}$

Value after 1 year $= 95 \times 210$

Value after 1 year $= 19950$.

The value of the TV after one year is $\textsf{₹} 19950$.

Solution:

The population is increasing at a fixed rate per year, which means it follows a compound growth pattern. We can use a formula similar to the compound interest formula to solve this problem.

The formula for population after $n$ years is given by $A = P \left(1 + \frac{R}{100}\right)^n$, where $A$ is the population at the end of $n$ years, $P$ is the initial population, $R$ is the annual growth rate, and $n$ is the number of years.

(i) Find the population in 2001.

Let the population in 2001 be $P_{2001}$. This is our initial population for the period 2001 to 2003.

The population in 2003 is given as $A = 54,000$.

The time period from 2001 to 2003 is $n = 2003 - 2001 = 2$ years.

The rate of increase is $R = 5\%$ per annum.

Using the formula $A = P \left(1 + \frac{R}{100}\right)^n$, we have:

$54000 = P_{2001} \left(1 + \frac{5}{100}\right)^2$

$54000 = P_{2001} \left(1 + 0.05\right)^2$

$54000 = P_{2001} \left(1.05\right)^2$

$54000 = P_{2001} \times (1.05 \times 1.05)$

$54000 = P_{2001} \times 1.1025$

To find $P_{2001}$, divide 54000 by 1.1025:

$P_{2001} = \frac{54000}{1.1025}$

$P_{2001} = \frac{54000 \times 10000}{1.1025 \times 10000} = \frac{540000000}{11025}$

Performing the division $\frac{540000000}{11025}$ gives approximately 48979.59.

Since population must be a whole number, we round this value.

Rounding to the nearest whole number, the population in 2001 was approximately 48980.

(ii) What would be its population in 2005?

We take the population in 2003 as the initial population for this calculation.

Initial Population (P) = 54,000 (in 2003)

The time period from 2003 to 2005 is $n = 2005 - 2003 = 2$ years.

The rate of increase is $R = 5\%$ per annum.

We need to find the population in 2005 ($A_{2005}$).

Using the formula $A = P \left(1 + \frac{R}{100}\right)^n$, we have:

$A_{2005} = 54000 \left(1 + \frac{5}{100}\right)^2$

$A_{2005} = 54000 \left(1.05\right)^2$

$A_{2005} = 54000 \times (1.05 \times 1.05)$

$A_{2005} = 54000 \times 1.1025$

Calculating the product $54000 \times 1.1025$:

$54000 \times 1.1025 = 59535$.

The population in 2005 would be 59535.

Answers:

(i) The population in 2001 was approximately 48980.

(ii) The population in 2005 would be 59535.

Solution:

Given:

Initial count of bacteria (P) = $5,06,000$

Rate of increase (R) = $2.5\%$ per hour

Time period (n) = $2$ hours

We need to find the count of bacteria at the end of 2 hours. The count is increasing at a fixed rate per hour, so we use the compound growth formula:

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

Where:

A = Count of bacteria after $n$ hours

P = Initial count of bacteria

R = Rate of increase per hour

n = Number of hours

Substitute the given values into the formula:

$A = 506000 \left(1 + \frac{2.5}{100}\right)^2$

$A = 506000 \left(1 + 0.025\right)^2$

$A = 506000 \left(1.025\right)^2$

Calculate $(1.025)^2$:

$1.025 \times 1.025 = 1.050625$

Substitute this back into the equation for A:

$A = 506000 \times 1.050625$

Calculate the product $506000 \times 1.050625$:

$A = 531616.25 \ or\ 5,31,616 (approx)$

The count of bacteria at the end of 2 hours would be $5,31,616 (approx)$.

The count of bacteria at the end of 2 hours is 5,31,616 (approx).

Solution:

Given:

Original Price of the scooter = $\textsf{₹} 42,000$

Rate of Depreciation = 8% per annum

Time period = 1 year

We need to find the value of the scooter after one year.

Depreciation means a reduction in value. The depreciation amount for one year is 8% of the original price.

Depreciation Amount = 8% of $\textsf{₹} 42,000$

Depreciation Amount $= \frac{8}{100} \times 42000$

Depreciation Amount $= \frac{8}{\cancel{100}} \times 420\cancel{00}$

Depreciation Amount $= 8 \times 420$

Depreciation Amount $= 3360$.

The depreciation amount after one year is $\textsf{₹} 3360$.

The value of the scooter after one year is the original price minus the depreciation amount.

Value after 1 year = Original Price - Depreciation Amount

Value after 1 year $= \textsf{₹} 42000 - \textsf{₹} 3360$

Value after 1 year $= \textsf{₹} 38640$.

Alternatively:

When the value of an item depreciates by 8% per annum, the value remaining after one year is $100\% - 8\% = 92\%$ of the original price.

Value after 1 year = 92% of Original Price

Value after 1 year $= 92\% \text{ of } \textsf{₹} 42000$

Value after 1 year $= \frac{92}{100} \times 42000$

Value after 1 year $= \frac{92}{\cancel{100}} \times 420\cancel{00}$

Value after 1 year $= 92 \times 420$

Value after 1 year $= 38640$.

The value after one year is $\textsf{₹} 38640$.

The value of the scooter after one year is $\textsf{₹} 38640$.