Chapter 8 Comparing Quantities (Additional Questions)

To convert a fraction to a percentage, we multiply the fraction by $100\%$.

The given fraction is $\frac{3}{8}$.

Percentage $= \frac{3}{8} \times 100\%$.

$= \frac{3 \times 100}{8} \%$

Cancel out the common factors between $100$ and $8$. Both are divisible by $4$.

$100 \div 4 = 25$

$8 \div 4 = 2$

So, the expression becomes:

$= \frac{3 \times 25}{2} \%$

$= \frac{75}{2} \%$

Now, divide $75$ by $2$:

$75 \div 2 = 37.5$

Thus, the percentage is $37.5\%$.

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The correct option is (B) $37.5\%$.

To find $15\%$ of $\textsf{₹} 1200$, we can convert the percentage to a fraction or a decimal and then multiply by the amount.

$15\%$ as a fraction is $\frac{15}{100}$.

So, $15\%$ of $\textsf{₹} 1200$ is calculated as:

$\frac{15}{100} \times 1200$

We can simplify by cancelling out the zeros in the denominator and the number:

$\frac{15}{\cancel{100}} \times \cancel{1200}^{12}$

$= 15 \times 12$

Now, we perform the multiplication:

$\begin{array}{cc}& & 1 & 5 \\ \times & & 1 & 2 \\ \hline && 3 & 0 \\ & 1 & 5 & \times \\ \hline & 1 & 8 & 0 \\ \hline \end{array}$

So, $15 \times 12 = 180$.

Therefore, $15\%$ of $\textsf{₹} 1200$ is $\textsf{₹} 180$.

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The correct option is (A) $\textsf{₹} 180$.

To find the percentage of marks scored, we use the formula:

Percentage of marks $= \frac{\text{Marks Scored}}{\text{Total Marks}} \times 100\%$

Given:

Marks Scored $= 450$

Total Marks $= 600$

Substitute the values into the formula:

Percentage of marks $= \frac{450}{600} \times 100\%$

We can simplify the fraction $\frac{450}{600}$ by cancelling out common factors.

Cancel out the trailing zeros:

$\frac{450}{600} = \frac{45}{60}$

Both $45$ and $60$ are divisible by $15$ ($45 = 3 \times 15$, $60 = 4 \times 15$).

$\frac{\cancel{45}^{3}}{\cancel{60}_{4}} = \frac{3}{4}$

So, the percentage calculation becomes:

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

Now, multiply $\frac{3}{4}$ by $100$:

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

$\frac{100}{4} = 25$

So, $3 \times 25 = 75$.

Therefore, the percentage of marks scored by the student is $75\%$.

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The correct option is (C) $75\%$.

To find the percentage increase in the price, we first calculate the increase in price.

Increase in price $=$ New Price $-$ Original Price

Increase in price $=$ $\textsf{₹} 900 - \textsf{₹} 800 = \textsf{₹} 100$

Now, we calculate the percentage increase using the formula:

Percentage Increase $= \frac{\text{Increase}}{\text{Original Price}} \times 100\%$

Substitute the values:

Percentage Increase $= \frac{\textsf{₹} 100}{\textsf{₹} 800} \times 100\%$

Percentage Increase $= \frac{100}{800} \times 100\%$

Simplify the fraction:

$\frac{100}{800} = \frac{1}{8}$

So, Percentage Increase $= \frac{1}{8} \times 100\%$

Now, calculate $\frac{1}{8} \times 100$:

$\frac{100}{8} = 12.5$

Therefore, the percentage increase in the price is $12.5\%$.

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The correct option is (B) $12.5\%$.

Given:

Cost Price (CP) = $\textsf{₹} 1500$

Selling Price (SP) = $\textsf{₹} 1800$

Since the Selling Price is greater than the Cost Price, there is a profit.

Profit $=$ Selling Price $-$ Cost Price

Profit $=$ $\textsf{₹} 1800 - \textsf{₹} 1500$

Profit $=$ $\textsf{₹} 300$

To find the profit percentage, we use the formula:

Profit Percentage $= \frac{\text{Profit}}{\text{Cost Price}} \times 100\%$

Substitute the values:

Profit Percentage $= \frac{\textsf{₹} 300}{\textsf{₹} 1500} \times 100\%$

Profit Percentage $= \frac{300}{1500} \times 100\%$

Simplify the fraction by cancelling out common factors.

Cancel out the two trailing zeros in the numerator and the denominator:

$\frac{300}{1500} = \frac{3}{15}$

Simplify the fraction $\frac{3}{15}$ by dividing both numerator and denominator by $3$:

$\frac{\cancel{3}^{1}}{\cancel{15}_{5}} = \frac{1}{5}$

So, the profit percentage is:

Profit Percentage $= \frac{1}{5} \times 100\%$

Calculate $\frac{1}{5} \times 100$:

$\frac{100}{5} = 20$

Therefore, the profit percentage is $20\%$.

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The correct option is (C) $20\%$.

Given:

Selling Price (SP) = $\textsf{₹} 750$

Loss Percentage = $25\%$

We need to find the Cost Price (CP).

When there is a loss, the Selling Price is calculated as a percentage of the Cost Price after deducting the loss percentage.

SP = CP $\times (100\% - \text{Loss Percentage})$

In terms of fraction or decimal:

SP = CP $\times (1 - \frac{\text{Loss Percentage}}{100})$

Substitute the given values into the formula:

$750 = \text{CP} \times (1 - \frac{25}{100})$

$750 = \text{CP} \times (1 - 0.25)$

$750 = \text{CP} \times 0.75$

To find CP, divide the Selling Price by the decimal equivalent of $(100\% - \text{Loss Percentage})$:

$\text{CP} = \frac{750}{0.75}$

We can write $0.75$ as the fraction $\frac{75}{100}$, which simplifies to $\frac{3}{4}$.

$\text{CP} = \frac{750}{\frac{3}{4}}$

When dividing by a fraction, we multiply by its reciprocal:

$\text{CP} = 750 \times \frac{4}{3}$

Now, we perform the multiplication. We can simplify by dividing $750$ by $3$ first:

$\text{CP} = \cancel{750}^{250} \times \frac{4}{\cancel{3}_{1}}$

$\text{CP} = 250 \times 4$

$\text{CP} = 1000$

So, the cost price of the article was $\textsf{₹} 1000$.

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The correct option is (B) $\textsf{₹} 1000$.

Given:

Marked Price (MP) = $\textsf{₹} 2500$

Discount Percentage = $10\%$

We need to find the Selling Price (SP).

The discount is calculated on the Marked Price.

Discount Amount $=$ Discount Percentage of Marked Price

Discount Amount $= 10\%$ of $\textsf{₹} 2500$

Discount Amount $= \frac{10}{100} \times 2500$

Simplify by cancelling out the zeros:

Discount Amount $= \frac{10}{\cancel{100}} \times \cancel{2500}^{25}$

Discount Amount $= 10 \times 25$

Discount Amount $= 250$

So, the discount amount is $\textsf{₹} 250$.

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

Selling Price $=$ Marked Price $-$ Discount Amount

Selling Price $=$ $\textsf{₹} 2500 - \textsf{₹} 250$

Selling Price $=$ $\textsf{₹} 2250$

Alternatively, we can calculate the Selling Price directly. If a $10\%$ discount is offered, the Selling Price is $(100\% - 10\%) = 90\%$ of the Marked Price.

Selling Price $= 90\%$ of $\textsf{₹} 2500$

Selling Price $= \frac{90}{100} \times 2500$

Simplify by cancelling out the zeros:

Selling Price $= \frac{90}{\cancel{100}} \times \cancel{2500}^{25}$

Selling Price $= 90 \times 25$

Selling Price $= 2250$

So, the selling price of the jacket is $\textsf{₹} 2250$.

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The correct option is (A) $\textsf{₹} 2250$.

Given:

Marked Price (MP) = $\textsf{₹} 500$

Discount Amount = $\textsf{₹} 50$

We need to find the Discount Percentage.

The formula for Discount Percentage is:

Discount Percentage $= \frac{\text{Discount Amount}}{\text{Marked Price}} \times 100\%$

Substitute the given values into the formula:

Discount Percentage $= \frac{\textsf{₹} 50}{\textsf{₹} 500} \times 100\%$

Discount Percentage $= \frac{50}{500} \times 100\%$

Simplify the fraction by cancelling out common factors.

Cancel out the trailing zeros:

$\frac{50}{500} = \frac{5}{50}$

Simplify the fraction $\frac{5}{50}$ by dividing both numerator and denominator by $5$:

$\frac{\cancel{5}^{1}}{\cancel{50}_{10}} = \frac{1}{10}$

So, the discount percentage is:

Discount Percentage $= \frac{1}{10} \times 100\%$

Calculate $\frac{1}{10} \times 100$:

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

Therefore, the discount percentage is $10\%$.

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The correct option is (B) $10\%$.

Given:

Price including GST = $\textsf{₹} 1180$

GST Percentage = $18\%$

We need to find the original price of the article before GST.

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

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

GST Amount $=$ GST Percentage of Original Price

GST Amount $= 18\%$ of $P$

GST Amount $= \frac{18}{100} \times P = 0.18 P$

Price including GST $=$ Original Price $+$ GST Amount

$1180 = P + 0.18 P$

$1180 = (1 + 0.18) P$

$1180 = 1.18 P$

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

$P = \frac{1180}{1.18}$

To perform the division, we can remove the decimal from the denominator by multiplying both the numerator and denominator by $100$:

$P = \frac{1180 \times 100}{1.18 \times 100}$

$P = \frac{118000}{118}$

Now, perform the division:

$\frac{118000}{118} = \frac{118 \times 1000}{118}$

Cancel out the $118$ in the numerator and denominator:

$P = \frac{\cancel{118} \times 1000}{\cancel{118}}$

$P = 1000$

So, the original price of the article before GST was $\textsf{₹} 1000$.

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The correct option is (A) $\textsf{₹} 1000$.

Given:

Principal (P) = $\textsf{₹} 5000$

Rate of Interest (R) = $8\%$ per annum

Time (T) = 3 years

To Find:

Simple Interest (SI)

Solution:

The formula for Simple Interest is:

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

Substitute the given values into the formula:

$SI = \frac{5000 \times 8 \times 3}{100}$

Calculate the numerator:

$5000 \times 8 \times 3 = 5000 \times 24 = 120000$

Now divide by $100$:

$SI = \frac{120000}{100}$

Cancel out the two zeros from the numerator and the denominator:

$SI = \frac{1200\cancel{00}}{\cancel{100}}$

$SI = 1200$

The simple interest is $\textsf{₹} 1200$.

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The correct option is (B) $\textsf{₹} 1200$.

Given:

Principal (P) = $\textsf{₹} 10000$

Rate of Interest (R) = $10\%$ per annum

Time (n) = 2 years

Compounding frequency = Annually

To Find:

Amount (A)

Solution:

The formula for the amount when interest is compounded annually is:

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

Substitute the given values into the formula:

$A = 10000(1 + \frac{10}{100})^2$

Simplify the term inside the parenthesis:

$1 + \frac{10}{100} = 1 + 0.1 = 1.1$

So, the formula becomes:

$A = 10000(1.1)^2$

Calculate $(1.1)^2$:

$(1.1)^2 = 1.1 \times 1.1 = 1.21$

Now, substitute this back into the equation:

$A = 10000 \times 1.21$

$A = 12100$

The amount after 2 years is $\textsf{₹} 12100$.

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The correct option is (B) $\textsf{₹} 12100$.

Given:

Current Population (P) = 40000

Annual Increase Rate (R) = $5\%$

Time (n) = 2 years

To Find:

Population after 2 years

Solution:

This problem can be solved using the formula for compound growth, similar to compound interest.

Population after $n$ years $=$ Current Population $\times (1 + \frac{\text{Annual Increase Rate}}{100})^n$

Let the population after 2 years be $P_n$.

$P_n = 40000 \times (1 + \frac{5}{100})^2$

Simplify the term inside the parenthesis:

$1 + \frac{5}{100} = 1 + 0.05 = 1.05$

So, the formula becomes:

$P_n = 40000 \times (1.05)^2$

Calculate $(1.05)^2$:

$(1.05)^2 = 1.05 \times 1.05$

$\begin{array}{cc}& & 1 & . & 0 & 5 \\ \times & & 1 & . & 0 & 5 \\ \hline && & & 2 & 5 \\ & & 0 & 0 & 0 & \times \\ 1 & 0 & 5 & \times & \times \\ \hline 1 & . & 1 & 0 & 2 & 5 \\ \hline \end{array}$

So, $(1.05)^2 = 1.1025$.

Now, substitute this back into the equation:

$P_n = 40000 \times 1.1025$

To multiply $40000$ by $1.1025$, we can multiply $4$ by $11025$ and then adjust the decimal.

$4 \times 11025 = 44100$

Since $1.1025$ has four decimal places, and $40000$ has four zeros, the multiplication is straightforward:

$40000 \times 1.1025 = 40000 \times \frac{11025}{10000} = 4 \times 11025 = 44100$

The population after 2 years will be 44100.

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The correct option is (B) 44100.

Given:

Current Value (Principal) = $\textsf{₹} 500000$

Depreciation Rate (R) = $10\%$ per annum

Time (n) = 1 year

To Find:

Value after 1 year

Solution:

When the value depreciates, we use a formula similar to the compound amount formula, but with subtraction instead of addition in the parenthesis.

Value after $n$ years $=$ Current Value $\times (1 - \frac{\text{Depreciation Rate}}{100})^n$

Let the value after 1 year be $V_1$.

$V_1 = 500000 \times (1 - \frac{10}{100})^1$

Simplify the term inside the parenthesis:

$1 - \frac{10}{100} = 1 - 0.1 = 0.9$

So, the formula becomes:

$V_1 = 500000 \times (0.9)^1$

$V_1 = 500000 \times 0.9$

$V_1 = 500000 \times \frac{9}{10}$

Cancel out one zero from $500000$ and $10$:

$V_1 = 50000 \times 9$

$V_1 = 450000$

The value of the car after 1 year will be $\textsf{₹} 450000$.

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The correct option is (A) $\textsf{₹} 450000$.

Let's analyze each statement:

(A) Profit is calculated as Selling Price - Cost Price.

This statement is true. Profit occurs when the selling price (SP) is greater than the cost price (CP). The profit amount is given by $SP - CP$.

(B) Loss is calculated as Cost Price - Selling Price.

This statement is true. Loss occurs when the cost price (CP) is greater than the selling price (SP). The loss amount is given by $CP - SP$.

(C) Discount is calculated on the Cost Price.

This statement is false. Discounts are usually given on the Marked Price (or List Price) of an item, not on its cost price. The selling price is obtained after deducting the discount from the marked price.

(D) Simple Interest is calculated only on the Principal amount.

This statement is true. Simple interest is calculated based solely on the initial principal amount for the entire duration of the loan or investment. Unlike compound interest, the interest earned in previous periods is not added to the principal for calculating interest in subsequent periods.

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Based on the analysis, the true statements are (A), (B), and (D).

The correct options are (A) Profit is calculated as Selling Price - Cost Price., (B) Loss is calculated as Cost Price - Selling Price., and (D) Simple Interest is calculated only on the Principal amount.

Analysis of Assertion (A):

Assertion (A) states that compound interest is always greater than or equal to simple interest for the same principal, period, and rate.

For a period of 1 year, the simple interest and compound interest are equal.

For a period greater than 1 year and a positive interest rate, compound interest earns interest on the principal amount as well as on the interest accumulated in previous periods. Simple interest only earns interest on the initial principal.

Therefore, for periods greater than 1 year and a positive interest rate, compound interest is greater than simple interest. If the interest rate is zero or the period is 1 year, they are equal.

Thus, the statement "Compound interest on a certain sum is always greater than or equal to simple interest on the same sum for the same period and rate" is TRUE.

Analysis of Reason (R):

Reason (R) states that compound interest includes interest on the accumulated interest from previous periods, whereas simple interest is calculated only on the initial principal.

This is the fundamental difference between compound interest and simple interest. Compound interest is calculated on the principal amount plus the accumulated interest from past periods, leading to exponential growth. Simple interest is calculated solely on the initial principal amount, leading to linear growth.

Thus, the statement is TRUE.

Relationship between A and R:

Reason (R) explains the mechanism by which compound interest is calculated (interest on interest) and how it differs from simple interest (interest only on principal). This difference is precisely why compound interest grows faster than simple interest over longer periods (for positive rates) and is equal for the first period or zero rate, which directly supports and explains Assertion (A).

Therefore, Reason (R) is the correct explanation for Assertion (A).

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Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

The correct option is (A) Both A and R are true, and R is the correct explanation of A.

Let's match each term with its correct formula:

(i) Simple Interest (SI)

The formula for Simple Interest is the product of Principal, Rate, and Time, divided by 100.

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

This matches formula (b) $P \times \frac{R}{100} \times T$.

(i) $\leftrightarrow$ (b)

(ii) Amount (A) in Simple Interest

The Amount in Simple Interest is the sum of the Principal and the Simple Interest earned.

$A = P + SI$

This matches formula (d) $P + SI$.

(ii) $\leftrightarrow$ (d)

(iii) Amount (A) in Compound Interest

The formula for the Amount when interest is compounded annually is given by $P$ multiplied by $(1 + \frac{R}{100})$ raised to the power of the time period $T$.

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

This matches formula (a) $P \left(1 + \frac{R}{100}\right)^T$.

(iii) $\leftrightarrow$ (a)

(iv) Compound Interest (CI)

The Compound Interest is the difference between the final Amount and the initial Principal.

$CI = A - P$

This matches formula (c) $A - P$.

(iv) $\leftrightarrow$ (c)

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Combining the matches:

(i) $\leftrightarrow$ (b)

(ii) $\leftrightarrow$ (d)

(iii) $\leftrightarrow$ (a)

(iv) $\leftrightarrow$ (c)

This sequence matches option (A).

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The correct option is (A) (i)-(b), (ii)-(d), (iii)-(a), (iv)-(c).

Given:

Principal (P) = $\textsf{₹} 20000$

Rate of Interest (R) = $10\%$ per annum

Time (n) = 2 years

Compounding frequency = Annually

To Find:

The formula to calculate the amount (A) Mr. Sharma needs to repay when interest is compounded annually.

Solution:

The formula for the amount (A) when interest is compounded annually is:

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

Substitute the given values of P, R, and n into this formula:

$P = 20000$

$R = 10$

$n = 2$

So, the formula becomes:

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

Let's compare this formula with the given options:

(A) $A = 20000 \times \left(1 + \frac{10}{100}\right)^2$ - This matches the derived formula for the amount with compound interest.

(B) $A = 20000 \times \frac{10}{100} \times 2$ - This is the formula for Simple Interest (SI).

(C) $A = 20000 + \left(20000 \times \frac{10}{100} \times 2\right)$ - This is the formula for the Amount in Simple Interest ($A = P + SI$).

(D) $A = 20000 (1 - \frac{10}{100})^2$ - This is the formula for the value after depreciation.

Therefore, the correct formula for calculating the amount Mr. Sharma needs to repay with compound interest is given by option (A).

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The correct option is (A) $A = 20000 \times \left(1 + \frac{10}{100}\right)^2$.

Given from Question 17 Case Study:

Principal (P) = $\textsf{₹} 20000$

Rate of Interest (R) = $10\%$ per annum

Time (n) = 2 years

Compounding frequency = Annually

To Find:

Total Amount (A) Mr. Sharma needs to repay after 2 years.

Solution:

The formula for the amount (A) when interest is compounded annually is:

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

Substitute the given values into the formula:

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

Simplify the term inside the parenthesis:

$1 + \frac{10}{100} = 1 + 0.1 = 1.1$

So, the formula becomes:

$A = 20000 (1.1)^2$

Calculate $(1.1)^2$:

$(1.1)^2 = 1.1 \times 1.1 = 1.21$

Now, substitute this back into the equation:

$A = 20000 \times 1.21$

Perform the multiplication:

$A = 20000 \times 1.21 = 20000 \times \frac{121}{100}$

Cancel out the two zeros from $20000$ and $100$:

$A = 200 \times 121$

$200 \times 121 = 2 \times 100 \times 121 = 2 \times 12100 = 24200$

The total amount Mr. Sharma needs to repay after 2 years is $\textsf{₹} 24200$.

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The correct option is (B) $\textsf{₹} 24200$.

Given:

Selling Price (SP) = $\textsf{₹} 600$

Discount Percentage = $20\%$

To Find:

Marked Price (MP)

Solution:

When a discount of $20\%$ is given on the Marked Price, the Selling Price is the remaining percentage of the Marked Price.

Percentage paid = $100\% - \text{Discount Percentage}$

Percentage paid = $100\% - 20\% = 80\%$

So, the Selling Price is $80\%$ of the Marked Price.

$SP = 80\%$ of MP

$600 = \frac{80}{100} \times \text{MP}$

$600 = 0.80 \times \text{MP}$

To find the Marked Price, divide the Selling Price by $0.80$:

$\text{MP} = \frac{600}{0.80}$

To perform the division without decimals, multiply both numerator and denominator by $100$:

$\text{MP} = \frac{600 \times 100}{0.80 \times 100}$

$\text{MP} = \frac{60000}{80}$

Cancel out a zero from the numerator and denominator:

$\text{MP} = \frac{6000}{8}$

Now, perform the division:

$\text{MP} = 750$

The marked price of the item was $\textsf{₹} 750$.

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The correct option is (B) $\textsf{₹} 750$.

To convert a decimal into a percentage, we multiply the decimal by $100\%$.

The given decimal is $0.045$.

Percentage $= 0.045 \times 100\%$

Multiplying a decimal by $100$ means moving the decimal point two places to the right.

$0.045 \times 100 = 4.5$

So, the percentage is $4.5\%$.

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The correct option is (C) $4.5\%$.

The question asks on which value the profit or loss percentage is always calculated.

Profit is the excess of Selling Price over Cost Price ($Profit = SP - CP$).

Loss is the excess of Cost Price over Selling Price ($Loss = CP - SP$).

Profit percentage is defined as $\frac{\text{Profit}}{\text{Cost Price}} \times 100\%$.

Loss percentage is defined as $\frac{\text{Loss}}{\text{Cost Price}} \times 100\%$.

In both formulas, the denominator is the Cost Price (CP).

Therefore, profit or loss percentage is always calculated on the Cost Price.

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Completing the sentence: Profit or loss percentage is always calculated on the Cost Price (CP).

The correct option is (C) Cost Price (CP).

Let's consider each option:

(A) Cost Price (CP): This is the price at which an article is bought. Profit and loss are calculated based on the relationship between Selling Price and Cost Price. So, CP is related to profit and loss.

(B) Selling Price (SP): This is the price at which an article is sold. Profit and loss are determined by comparing the Selling Price and Cost Price. So, SP is related to profit and loss.

(C) Marked Price (MP): This is the price listed on an article, often before any discount. While the marked price is used in calculating discounts which affect the selling price, the primary calculation of profit or loss is based on Cost Price and Selling Price. Discount is related to Marked Price and Selling Price ($Selling\;Price = Marked\;Price - Discount$). However, the question asks for a term NOT related to profit and loss. Profit and Loss themselves are derived directly from CP and SP.

(D) Profit: This is the gain made when Selling Price is greater than Cost Price. Profit is directly a result of a transaction involving buying and selling, and is a core concept in profit and loss calculations.

While Marked Price is indirectly related through discounts affecting the selling price, it is primarily a term used in the context of discounts. Cost Price, Selling Price, and Profit (or Loss) are the fundamental elements directly involved in the profit and loss calculation itself.

In the context of standard profit and loss calculations (Profit/Loss on Cost Price), Marked Price is often introduced when discounts are involved. However, CP, SP, Profit, and Loss are the most direct terms in the core definition and calculation of profit and loss.

Considering the options, Marked Price is the least directly involved in the fundamental definition of profit ($SP - CP$) and loss ($CP - SP$) compared to CP, SP, and Profit itself.

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The term that is NOT a fundamental element in the definition and calculation of profit and loss is the Marked Price (MP), although it is used in problems involving discounts that lead to a Selling Price.

The correct option is (C) Marked Price (MP).

Given:

Original Value (P) = $\textsf{₹} 100000$

Depreciation Rate (R) = $10\%$ per annum

Time (n) = 2 years

To Find:

Value after 2 years ($V_n$)

Solution:

When the value of an asset depreciates, we use the formula similar to the compound amount formula, but the rate is subtracted from 100%.

The formula for the value of an asset after $n$ years due to depreciation is:

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

Substitute the given values into the formula:

$P = 100000$

$R = 10$

$n = 2$

$V_2 = 100000 \left(1 - \frac{10}{100}\right)^2$

Simplify the term inside the parenthesis:

$1 - \frac{10}{100} = 1 - 0.10 = 0.90$

So, the formula becomes:

$V_2 = 100000 (0.9)^2$

Calculate $(0.9)^2$:

$(0.9)^2 = 0.9 \times 0.9 = 0.81$

Now, substitute this back into the equation:

$V_2 = 100000 \times 0.81$

Perform the multiplication:

$V_2 = 100000 \times \frac{81}{100}$

Cancel out the two zeros from $100000$ and $100$:

$V_2 = 1000 \times 81$

$V_2 = 81000$

The value of the machine after 2 years will be $\textsf{₹} 81000$.

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The correct option is (B) $\textsf{₹} 81000$.

Given:

Marked Price of Dining Table = $\textsf{₹} 18000$

Marked Price of Set of Chairs = $\textsf{₹} 12000$

To Find:

Total marked price of the items Mr. Gupta bought.

Solution:

The total marked price is the sum of the marked prices of all the items bought.

Total Marked Price $=$ Marked Price of Dining Table $+$ Marked Price of Set of Chairs

Total Marked Price $=$ $\textsf{₹} 18000 + \textsf{₹} 12000$

Total Marked Price $=$ $\textsf{₹} 30000$

The total marked price of the items Mr. Gupta bought is $\textsf{₹} 30000$.

---

The correct option is (C) $\textsf{₹} 30000$.

Given from Question 24 Case Study:

Marked Price of Dining Table = $\textsf{₹} 18000$

Marked Price of Set of Chairs = $\textsf{₹} 12000$

Total Marked Price = $\textsf{₹} 18000 + \textsf{₹} 12000 = \textsf{₹} 30000$

Discount Percentage = $15\%$

To Find:

Total Discount Amount Mr. Gupta received.

Solution:

The discount is calculated on the total marked price.

Discount Amount $=$ Discount Percentage of Total Marked Price

Discount Amount $= 15\%$ of $\textsf{₹} 30000$

Discount Amount $= \frac{15}{100} \times 30000$

Simplify the expression by cancelling out the zeros:

Discount Amount $= \frac{15}{\cancel{100}} \times \cancel{30000}^{300}$

Discount Amount $= 15 \times 300$

Perform the multiplication:

$15 \times 300 = 4500$

The total discount amount Mr. Gupta received is $\textsf{₹} 4500$.

---

The correct option is (A) $\textsf{₹} 4500$.

Given from Question 24 Case Study:

Total Marked Price (MP) = $\textsf{₹} 30000$

Discount Percentage = $15\%$

GST Percentage = $5\%$ (applicable on the discounted price)

To Find:

Final amount Mr. Gupta has to pay.

Solution:

First, calculate the discount amount.

Discount Amount $=$ Discount Percentage of Total Marked Price

Discount Amount $= 15\%$ of $\textsf{₹} 30000$

Discount Amount $= \frac{15}{100} \times 30000$

Discount Amount $= 15 \times \frac{30000}{100}$

Discount Amount $= 15 \times 300 = 4500$

So, the total discount amount is $\textsf{₹} 4500$.

Next, calculate the price after discount, which is the Selling Price (SP).

Selling Price $=$ Total Marked Price $-$ Discount Amount

Selling Price $=$ $\textsf{₹} 30000 - \textsf{₹} 4500 = \textsf{₹} 25500$

Now, calculate the GST amount applicable on the Selling Price.

GST Amount $=$ GST Percentage of Selling Price

GST Amount $= 5\%$ of $\textsf{₹} 25500$

GST Amount $= \frac{5}{100} \times 25500$

GST Amount $= 5 \times \frac{25500}{100}$

GST Amount $= 5 \times 255$

$5 \times 255 = 1275$

So, the GST amount is $\textsf{₹} 1275$.

Finally, calculate the total amount Mr. Gupta has to pay by adding the GST amount to the Selling Price.

Final Amount $=$ Selling Price $+$ GST Amount

Final Amount $=$ $\textsf{₹} 25500 + \textsf{₹} 1275$

Final Amount $=$ $\textsf{₹} 26775$

---

The final amount Mr. Gupta has to pay is $\textsf{₹} 26775$.

The correct option is (B) $\textsf{₹} 26775$.

To convert a percentage to a fraction, we divide the percentage value by $100$.

$62.5\% = \frac{62.5}{100}$

To remove the decimal point in the numerator, we can multiply both the numerator and the denominator by $10$ (since there is one digit after the decimal point).

$\frac{62.5}{100} = \frac{62.5 \times 10}{100 \times 10} = \frac{625}{1000}$

Now, we need to simplify the fraction $\frac{625}{1000}$ to its lowest terms. We can do this by dividing both the numerator and the denominator by their greatest common divisor (GCD).

Both numbers are divisible by $5$, as they end in $5$ and $0$.

$\frac{625 \div 5}{1000 \div 5} = \frac{125}{200}$

Again, both numbers are divisible by $5$.

$\frac{125 \div 5}{200 \div 5} = \frac{25}{40}$

Again, both numbers are divisible by $5$.

$\frac{25 \div 5}{40 \div 5} = \frac{5}{8}$

The numbers $5$ and $8$ have no common factors other than $1$. So, the fraction $\frac{5}{8}$ is in its lowest terms.

Alternatively, we could have found the GCD of $625$ and $1000$.

Prime factorization of $625 = 5^4$

Prime factorization of $1000 = 10^3 = (2 \times 5)^3 = 2^3 \times 5^3$

The common factors are $5^3$.

GCD$(625, 1000) = 5^3 = 125$

Now, divide the numerator and denominator by $125$:

$\frac{625 \div 125}{1000 \div 125} = \frac{5}{8}$

---

The fraction $62.5\%$ in its lowest terms is $\frac{5}{8}$.

The correct option is (A) $\frac{5}{8}$.

Given:

The ratio of two quantities is $4:5$.

Let the first quantity be $Q_1$ and the second quantity be $Q_2$.

We are given the ratio $Q_1 : Q_2 = 4 : 5$.

This means $\frac{Q_1}{Q_2} = \frac{4}{5}$.

To Find:

The percentage of the first quantity compared to the second.

This can be expressed as $\frac{Q_1}{Q_2} \times 100\%$.

Solution:

We already have the ratio as a fraction: $\frac{Q_1}{Q_2} = \frac{4}{5}$.

Now, convert this fraction to a percentage by multiplying by $100\%$.

Percentage $= \frac{4}{5} \times 100\%$

Calculate $\frac{4}{5} \times 100$:

$\frac{4}{5} \times 100 = 4 \times \frac{100}{5}$

$\frac{100}{5} = 20$

So, $4 \times 20 = 80$.

Therefore, the percentage of the first quantity compared to the second is $80\%$.

---

The correct option is (C) $80\%$.

Given:

Total Marks = 500

Percentage Scored = $80\%$

To Find:

Marks scored by the student.

Solution:

To find the marks scored by the student, we need to calculate $80\%$ of the total marks.

Marks Scored $= 80\%$ of Total Marks

Marks Scored $= 80\%$ of 500

Marks Scored $= \frac{80}{100} \times 500$

Simplify the expression by cancelling out the zeros:

Marks Scored $= \frac{80}{\cancel{100}} \times \cancel{500}^{5}$

Marks Scored $= 80 \times 5$

Marks Scored $= 400$

The student scored 400 marks.

---

The correct option is (A) 400.

Given:

Original Population = 10000

New Population = 9500

To Find:

Percentage decrease in population.

Solution:

First, calculate the decrease in population.

Decrease in Population $=$ Original Population $-$ New Population

Decrease in Population $=$ $10000 - 9500 = 500$

Now, calculate the percentage decrease using the formula:

Percentage Decrease $= \frac{\text{Decrease}}{\text{Original Population}} \times 100\%$

Substitute the values:

Percentage Decrease $= \frac{500}{10000} \times 100\%$

Simplify the fraction by cancelling out common factors.

Cancel out the two zeros in the numerator and two zeros in the denominator:

$\frac{\cancel{500}}{\cancel{10000}_{100}} = \frac{5}{100}$

So, Percentage Decrease $= \frac{5}{100} \times 100\%$

Percentage Decrease $= 5\%$

The percentage decrease in population is $5\%$.

---

The correct option is (A) $5\%$.

Given:

False weight used = 960 gm

True weight that should be used = 1 kg = 1000 gm

The shopkeeper sells the goods at the cost price of the 1000 gm, but he only gives 960 gm of goods.

This means the shopkeeper's cost for 960 gm is the same as the selling price for 960 gm (which is equal to the cost price of 1000 gm).

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

Then, the CP of 1000 gm is $\textsf{₹} 1000$.

The shopkeeper sells 960 gm but charges the customer for 1000 gm at the cost price.

So, the Selling Price (SP) for 960 gm is equal to the CP of 1000 gm, which is $\textsf{₹} 1000$.

The shopkeeper's actual Cost Price for the goods he sells (960 gm) is $960 \times \textsf{₹} 1 = \textsf{₹} 960$.

So, for the 960 gm of goods sold:

Cost Price (CP) = $\textsf{₹} 960$

Selling Price (SP) = $\textsf{₹} 1000$

The shopkeeper makes a profit because SP > CP.

Profit $=$ SP $-$ CP

Profit $=$ $\textsf{₹} 1000 - \textsf{₹} 960 = \textsf{₹} 40$

The profit percentage is calculated on the actual cost price of the goods sold.

Profit Percentage $= \frac{\text{Profit}}{\text{Cost Price}} \times 100\%$

Profit Percentage $= \frac{40}{960} \times 100\%$

Simplify the fraction $\frac{40}{960}$:

$\frac{40}{960} = \frac{4}{96}$

Divide both numerator and denominator by 4:

$\frac{\cancel{4}^{1}}{\cancel{96}_{24}} = \frac{1}{24}$

So, Profit Percentage $= \frac{1}{24} \times 100\%$

Calculate $\frac{100}{24}$:

$\frac{100}{24} = \frac{50}{12} = \frac{25}{6}$

Convert $\frac{25}{6}$ to a decimal or mixed number:

$\frac{25}{6} = 4 \frac{1}{6}$

As a decimal, $\frac{1}{6} \approx 0.1666...$

So, $4 \frac{1}{6}\% \approx 4.1666...\%$

The profit percentage is $4.1\overline{6}\%$.

---

The profit percentage is approximately $4.16\%$.

The correct option is (B) $4.16\%$.

Explanation:

When interest is compounded half-yearly, it means the interest is calculated and added to the principal twice a year (every six months).

Let the annual rate of interest be $R\%$ and the time period be $T$ years.

For half-yearly compounding:

1. The rate of interest applied for each half-year period is half of the annual rate.

Rate per compounding period $= \frac{\text{Annual Rate}}{2} = \frac{R}{2}\%$.

So, the annual rate is halved.

2. The number of compounding periods in $T$ years is twice the number of years, as there are two half-years in each year.

Number of compounding periods $= \text{Time in Years} \times 2 = T \times 2 = 2T$.

So, the time period (in terms of number of compounding periods) is doubled.

The formula for the amount when compounded half-yearly becomes:

$A = P \left(1 + \frac{R/2}{100}\right)^{2T}$ or $A = P \left(1 + \frac{R}{200}\right)^{2T}$

---

Therefore, if the rate of interest is compounded half-yearly, the annual rate is halved and the time period (number of periods) is doubled.

The correct option is (A) halved, doubled.

Given:

Principal (P) = $\textsf{₹} 1000$

Annual Rate of Interest (R) = $20\%$ per annum

Time (T) = 1 year

Compounding frequency = Half-yearly

To Find:

Compound Interest (CI)

Solution:

When interest is compounded half-yearly, we adjust the rate and time:

Rate per half-year $= \frac{\text{Annual Rate}}{2} = \frac{20\%}{2} = 10\% = \frac{10}{100} = 0.1$

Number of half-year periods in 1 year $= \text{Time in Years} \times 2 = 1 \times 2 = 2$ periods.

The formula for the amount (A) when compounded half-yearly is:

$A = P \left(1 + \frac{\text{Rate per half-year}}{100}\right)^{\text{Number of half-year periods}}$

$A = P \left(1 + \frac{R/2}{100}\right)^{2T}$

Substitute the given values:

$A = 1000 \left(1 + \frac{10}{100}\right)^{2}$

Simplify the term inside the parenthesis:

$1 + \frac{10}{100} = 1 + 0.1 = 1.1$

So, the formula becomes:

$A = 1000 (1.1)^2$

Calculate $(1.1)^2$:

$(1.1)^2 = 1.1 \times 1.1 = 1.21$

Now, substitute this back into the equation:

$A = 1000 \times 1.21$

$A = 1210$

The amount after 1 year (compounded half-yearly) is $\textsf{₹} 1210$.

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

$CI = A - P$

$CI = \textsf{₹} 1210 - \textsf{₹} 1000$

$CI = \textsf{₹} 210$

---

The compound interest is $\textsf{₹} 210$.

The correct option is (B) $\textsf{₹} 210$.

Given:

Current Value (P) = $\textsf{₹} 500000$

Annual Increase Rate (R) = $10\%$ per annum

Time (n) = 3 years

To Find:

Value after 3 years ($V_n$)

Solution:

When a value increases at a certain rate annually, we use the compound growth formula:

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

Substitute the given values into the formula:

$P = 500000$

$R = 10$

$n = 3$

$V_3 = 500000 \left(1 + \frac{10}{100}\right)^3$

Simplify the term inside the parenthesis:

$1 + \frac{10}{100} = 1 + 0.1 = 1.1$

So, the formula becomes:

$V_3 = 500000 (1.1)^3$

Calculate $(1.1)^3$:

$(1.1)^3 = 1.1 \times 1.1 \times 1.1 = (1.21) \times 1.1$

$\begin{array}{cc}& & 1 & . & 2 & 1 \\ \times & & & . & 1 & 1 \\ \hline && & 1 & 2 & 1 \\ & 1 & 2 & 1 & \times & \\ \hline 1 & . & 3 & 3 & 1 \\ \hline \end{array}$

So, $(1.1)^3 = 1.331$.

Now, substitute this back into the equation:

$V_3 = 500000 \times 1.331$

Perform the multiplication:

$V_3 = 500000 \times 1.331 = 500000 \times \frac{1331}{1000}$

Cancel out three zeros from $500000$ and $1000$:

$V_3 = 500 \times 1331$

$V_3 = 5 \times 100 \times 1331 = 5 \times 133100$

$5 \times 133100 = 665500$

The value of the piece of land after 3 years will be $\textsf{₹} 665500$.

---

The correct option is (C) $\textsf{₹} 665500$.

Analysis of Assertion (A):

Assertion (A) states that Discount is always subtracted from the Marked Price to get the Selling Price.

The definition of discount is a reduction in price given on the marked price of an article. The selling price is the price at which the article is sold after applying the discount.

The relationship is given by the formula:

Selling Price $=$ Marked Price $-$ Discount Amount

So, Assertion (A) is TRUE.

---

Analysis of Reason (R):

Reason (R) states that Discount is a reduction offered on the list price or marked price.

The list price or marked price is the price at which the item is listed for sale. A discount is a reduction from this price to encourage sales.

This statement correctly defines discount in the context of marked price. So, Reason (R) is TRUE.

---

Relationship between A and R:

Reason (R) defines what a discount is – a reduction on the marked price. Assertion (A) describes how this reduction affects the final selling price – by being subtracted from the marked price. The definition provided in Reason (R) directly explains why the subtraction method in Assertion (A) is used to calculate the selling price. The discount amount, which is a reduction from the marked price (R), is taken away from the marked price to arrive at the selling price (A).

Therefore, Reason (R) is the correct explanation for Assertion (A).

---

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

The correct option is (A) Both A and R are true, and R is the correct explanation of A.

Given: Discount = $20\%$, Profit = $25\%$, Profit Amount = $\textsf{₹} 125$.

---

Profit is calculated on Cost Price (CP).

$25\%$ of CP $= \textsf{₹} 125$

$\frac{25}{100} \times \text{CP} = 125$

$\text{CP} = 125 \times \frac{100}{25} = 125 \times 4 = \textsf{₹} 500$

---

Selling Price (SP) $=$ CP $+$ Profit Amount

SP $=$ $\textsf{₹} 500 + \textsf{₹} 125 = \textsf{₹} 625$

---

SP is obtained after $20\%$ discount on Marked Price (MP).

SP $=$ MP $\times (100\% - \text{Discount}\%)$

$\textsf{₹} 625 = \text{MP} \times (100\% - 20\%)$

$\textsf{₹} 625 = \text{MP} \times 80\%$

$\textsf{₹} 625 = \text{MP} \times \frac{80}{100}$

$\text{MP} = 625 \times \frac{100}{80} = 625 \times \frac{10}{8} = 625 \times \frac{5}{4}$

$\text{MP} = \frac{3125}{4} = 781.25$

The calculated Marked Price is $\textsf{₹} 781.25$. This value is not listed in the options.

Comparing the calculated value with the options, $\textsf{₹} 800$ (Option D) is numerically closest to $\textsf{₹} 781.25$.

---

Given the discrepancy and the need to select from the options, the closest value is chosen.

The correct option is (D) $\textsf{₹} 800$.

Given:

Annual Rate of Interest (R) = $8\%$ per annum

Compounding frequency = Quarterly

To Find:

Rate of interest per compounding period.

Solution:

When interest is compounded quarterly, the annual interest rate is divided by the number of quarters in a year. There are 4 quarters in a year.

Rate of interest per compounding period $= \frac{\text{Annual Rate}}{\text{Number of compounding periods per year}}$

Rate of interest per compounding period $= \frac{8\%}{4}$

Rate of interest per compounding period $= 2\%$

The rate of interest for each quarterly period is $2\%$.

---

The correct option is (C) $2\%$.

Given from Question 37 Case Study:

Principal (P) = $\textsf{₹} 15000$

Annual Rate (R) = $8\%$ per annum

Time (T) = 9 months

Compounding frequency = Quarterly

Derived from Question 37:

Rate per compounding period ($r$) $= \frac{8\%}{4} = 2\% = 0.02$

Number of compounding periods ($n$) in 9 months $= \frac{9 \text{ months}}{3 \text{ months/quarter}} = 3$ periods

To Find:

Amount (A) after 9 months.

Solution:

The formula for the amount when compounded is:

$A = P (1 + r)^n$

Substitute the values of P, r, and n:

$A = 15000 (1 + 0.02)^3$

$A = 15000 (1.02)^3$

Calculate $(1.02)^3$:

$(1.02)^2 = 1.02 \times 1.02 = 1.0404$

$(1.02)^3 = (1.02)^2 \times 1.02 = 1.0404 \times 1.02$

$\begin{array}{cc}& & 1 & . & 0 & 4 & 0 & 4 \\ \times & & & . & & 1 & 0 & 2 \\ \hline && & & 0 & 0 & 0 & 8 \\ && & 2 & 0 & 8 & 0 & 8 & \times \\ & 1 & 0 & 4 & 0 & 4 & \times & \times \\ \hline 1 & . & 0 & 6 & 1 & 2 & 0 & 8 \\ \hline \end{array}$

So, $(1.02)^3 = 1.061208$.

Now, substitute this back into the equation for A:

$A = 15000 \times 1.061208$

$A = 15 \times 1000 \times 1.061208$

$A = 15 \times 1061.208$

$\begin{array}{cc}& & 1 & 0 & 6 & 1 & . & 2 & 0 & 8 \\ \times & & & & & 1 & 5 & & & \\ \hline && 5 & 3 & 0 & 6 & . & 0 & 4 & 0 \\ 1 & 0 & 6 & 1 & 2 & . & 0 & 8 & \times \\ \hline 1 & 5 & 9 & 1 8 & . & 1 & 2 & 0 & 0 \\ \hline \end{array}$

Rounding to two decimal places for currency, the amount is $\textsf{₹} 15918.12$.

---

The amount Mrs. Rao will get after 9 months is $\textsf{₹} 15918.12$.

The correct option is (A) $\textsf{₹} 15918.12$.

Given:

$20\%$ of a number is 400.

To Find:

$50\%$ of the same number.

Solution:

Let the number be $x$.

We are given that $20\%$ of $x$ is 400.

$20\% \times x = 400$

Convert the percentage to a decimal or fraction:

$\frac{20}{100} \times x = 400$

$0.20 \times x = 400$

To find the number $x$, divide 400 by 0.20:

$x = \frac{400}{0.20} = \frac{400}{1/5} = 400 \times 5 = 2000$

The number is 2000.

---

Now we need to find $50\%$ of this number (2000).

$50\%$ of $2000 = \frac{50}{100} \times 2000$

$50\%$ of $2000 = 0.50 \times 2000$

$0.50 \times 2000 = \frac{1}{2} \times 2000 = 1000$

Alternatively, since we know $20\%$ of the number is 400, we can find $50\%$ using ratios.

If $20\%$ corresponds to 400,

Then $1\%$ corresponds to $\frac{400}{20} = 20$.

So, $50\%$ corresponds to $50 \times (\text{value of } 1\%) = 50 \times 20 = 1000$.

$50\%$ of the number is 1000.

---

The correct option is (B) 1000.

Given:

Principal (P) = $\textsf{₹} 10000$

Rate of Interest (R) = $5\%$ per annum

Time (T or n) = 2 years

To Find:

Difference between Compound Interest (CI) and Simple Interest (SI) for 2 years.

Solution:

First, calculate the Simple Interest (SI).

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

$SI = \frac{10000 \times 5 \times 2}{100}$

$SI = \frac{100000}{100}$

$SI = \textsf{₹} 1000$

---

Next, calculate the Compound Interest (CI).

First, find the Amount (A) when compounded annually.

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

$A = 10000 \left(1 + \frac{5}{100}\right)^2$

$A = 10000 (1 + 0.05)^2$

$A = 10000 (1.05)^2$

$A = 10000 \times 1.1025$

$A = \textsf{₹} 11025$

Now, calculate the Compound Interest (CI).

$CI = A - P$

$CI = \textsf{₹} 11025 - \textsf{₹} 10000 = \textsf{₹} 1025$

---

Finally, find the difference between Compound Interest and Simple Interest.

Difference $=$ CI $-$ SI

Difference $=$ $\textsf{₹} 1025 - \textsf{₹} 1000 = \textsf{₹} 25$

---

For 2 years, the difference between CI and SI can also be calculated using the formula: Difference $= P \left(\frac{R}{100}\right)^2$

Difference $= 10000 \left(\frac{5}{100}\right)^2$

Difference $= 10000 \left(\frac{1}{20}\right)^2$

Difference $= 10000 \times \frac{1}{400}$

Difference $= \frac{10000}{400} = \frac{100}{4} = 25$

The difference is $\textsf{₹} 25$.

---

The correct option is (A) $\textsf{₹} 25$.

To convert a number (fraction or decimal) into a percentage, we multiply it by $100\%$.

---

(a) Convert $\frac{4}{5}$ into a percentage:

Percentage $= \frac{4}{5} \times 100\%$

Percentage $= \frac{400}{5}\%$

Percentage $= 80\%$

Thus, $\frac{4}{5} = \mathbf{80\%}$.

---

(b) Convert $0.075$ into a percentage:

Percentage $= 0.075 \times 100\%$

Percentage $= 7.5\%$

Thus, $0.075 = \mathbf{7.5\%}$.

---

(c) Convert $2\frac{1}{4}$ into a percentage:

First, convert the mixed number into an improper fraction.

$2\frac{1}{4} = \frac{(2 \times 4) + 1}{4} = \frac{8 + 1}{4} = \frac{9}{4}$

Now, convert the fraction into a percentage:

Percentage $= \frac{9}{4} \times 100\%$

Percentage $= \frac{900}{4}\%$

Percentage $= 225\%$

Thus, $2\frac{1}{4} = \mathbf{225\%}$.

To convert a percentage into a fraction, we divide the percentage by $100$ and remove the $\%$ symbol. Then we simplify the fraction to its lowest terms. To convert a percentage into a decimal, we divide the percentage by $100$ and remove the $\%$ symbol.

---

(a) Convert $60\%$ into a fraction and a decimal:

Fraction form: $60\% = \frac{60}{100}$

To simplify the fraction, divide the numerator and the denominator by their greatest common divisor, which is $20$.

$\frac{\cancel{60}^{3}}{\cancel{100}_{5}} = \frac{3}{5}$

So, the fraction in simplest form is $\mathbf{\frac{3}{5}}$.

Decimal form: $60\% = \frac{60}{100} = 0.60$

So, the decimal form is $\mathbf{0.6}$.

---

(b) Convert $12.5\%$ into a fraction and a decimal:

Fraction form: $12.5\% = \frac{12.5}{100}$

To remove the decimal from the numerator, multiply both the numerator and the denominator by $10$.

$\frac{12.5 \times 10}{100 \times 10} = \frac{125}{1000}$

To simplify the fraction, divide the numerator and the denominator by their greatest common divisor, which is $125$.

$\frac{\cancel{125}^{1}}{\cancel{1000}_{8}} = \frac{1}{8}$

So, the fraction in simplest form is $\mathbf{\frac{1}{8}}$.

Decimal form: $12.5\% = \frac{12.5}{100} = 0.125$

So, the decimal form is $\mathbf{0.125}$.

To find a percentage of a quantity, we convert the percentage into a fraction (by dividing by $100$) or a decimal and then multiply it by the given quantity.

---

(a) Find $30\%$ of $\textsf{₹}1500$:

$30\%$ of $\textsf{₹}1500 = \frac{30}{100} \times \textsf{₹}1500$

$= \frac{30}{\cancel{100}_{1}} \times \textsf{₹}\cancel{1500}^{15}$

$= 30 \times \textsf{₹}15$

$= \textsf{₹}450$

Thus, $30\%$ of $\textsf{₹}1500$ is $\mathbf{\textsf{₹}450}$.

---

(b) Find $15\%$ of $250$ kg:

$15\%$ of $250$ kg $= \frac{15}{100} \times 250$ kg

$= \frac{15}{\cancel{100}_{2}} \times \cancel{250}^{5}$ kg

$= \frac{15 \times 5}{2}$ kg

$= \frac{75}{2}$ kg

$= 37.5$ kg

Thus, $15\%$ of $250$ kg is $\mathbf{37.5 \text{ kg}}$.

Given:

Percentage of boys in the school $= 45\%$

Total number of students in the school $= 1200$

---

To Find:

The number of girls in the school.

---

Solution:

The total percentage of students in the school is $100\%$.

The percentage of girls in the school is the total percentage of students minus the percentage of boys.

Percentage of girls $= (100 - 45)\%$

Percentage of girls $= 55\%$

Now, we need to find $55\%$ of the total number of students.

Number of girls $= 55\%$ of $1200$

Number of girls $= \frac{55}{100} \times 1200$

Number of girls $= \frac{55}{\cancel{100}_{1}} \times \cancel{1200}^{12}$

Number of girls $= 55 \times 12$

Number of girls $= 660$

Alternatively, we could find the number of boys first and subtract it from the total number of students.

Number of boys $= 45\%$ of $1200$

Number of boys $= \frac{45}{100} \times 1200$

Number of boys $= \frac{45}{\cancel{100}_{1}} \times \cancel{1200}^{12}$

Number of boys $= 45 \times 12$

Number of boys $= 540$

Number of girls $=$ Total number of students $-$ Number of boys

Number of girls $= 1200 - 540$

Number of girls $= 660$

Thus, the number of girls in the school is $\mathbf{660}$.

Given:

Original price of petrol $= \textsf{₹}90$ per litre

New price of petrol $= \textsf{₹}99$ per litre

---

To Find:

The percentage increase in the price of petrol.

---

Solution:

First, calculate the increase in the price of petrol.

Increase in price $=$ New price $-$ Original price

Increase in price $= \textsf{₹}99 - \textsf{₹}90$

Increase in price $= \textsf{₹}9$

Now, calculate the percentage increase using the formula:

Percentage increase $= \frac{\text{Increase in price}}{\text{Original price}} \times 100\%$

Percentage increase $= \frac{\textsf{₹}9}{\textsf{₹}90} \times 100\%$

Percentage increase $= \frac{9}{90} \times 100\%$

Percentage increase $= \frac{\cancel{9}^{1}}{\cancel{90}_{10}} \times 100\%$

Percentage increase $= \frac{1}{10} \times 100\%$

Percentage increase $= \frac{\cancel{100}^{10}}{\cancel{10}_{1}}\%$

Percentage increase $= 10\%$

Thus, the percentage increase in the price of petrol is $\mathbf{10\%}$.

Given:

Original price of the scooter (Cost Price) $= \textsf{₹}40,000$

Selling price of the scooter $= \textsf{₹}36,000$

---

To Find:

The percentage decrease in the price.

---

Solution:

Since the selling price ($\textsf{₹}36,000$) is less than the original price ($\textsf{₹}40,000$), there is a decrease in the price.

First, calculate the decrease in the price of the scooter.

Decrease in price $=$ Original price $-$ Selling price

Decrease in price $= \textsf{₹}40,000 - \textsf{₹}36,000$

Decrease in price $= \textsf{₹}4,000$

Now, calculate the percentage decrease using the formula:

Percentage decrease $= \frac{\text{Decrease in price}}{\text{Original price}} \times 100\%$

Percentage decrease $= \frac{\textsf{₹}4,000}{\textsf{₹}40,000} \times 100\%$

Percentage decrease $= \frac{4000}{40000} \times 100\%$

Percentage decrease $= \frac{\cancel{4000}^{1}}{\cancel{40000}_{10}} \times 100\%$

Percentage decrease $= \frac{1}{10} \times 100\%$

Percentage decrease $= \frac{\cancel{100}^{10}}{\cancel{10}_{1}}\%$

Percentage decrease $= 10\%$

Thus, the percentage decrease in the price is $\mathbf{10\%}$.

Given:

Cost Price (CP) of the fan $= \textsf{₹}800$

Profit percentage $= 15\%$

---

To Find:

The selling price (SP) of the fan.

---

Solution:

The selling price (SP) is calculated by adding the profit amount to the cost price (CP).

First, we need to calculate the amount of profit.

Profit amount $= 15\%$ of the Cost Price

Profit amount $= 15\%$ of $\textsf{₹}800$

Profit amount $= \frac{15}{100} \times \textsf{₹}800$

Profit amount $= \frac{15}{\cancel{100}_{1}} \times \textsf{₹}\cancel{800}^{8}$

Profit amount $= 15 \times \textsf{₹}8$

Profit amount $= \textsf{₹}120$

Now, we can find the selling price.

Selling Price (SP) $=$ Cost Price (CP) $+$ Profit amount

SP $= \textsf{₹}800 + \textsf{₹}120$

SP $= \textsf{₹}920$

Alternatively, the selling price can be calculated directly as $(100 + \text{Profit \%})\%$ of the Cost Price.

SP $= (100 + 15)\%$ of $\textsf{₹}800$

SP $= 115\%$ of $\textsf{₹}800$

SP $= \frac{115}{100} \times \textsf{₹}800$

SP $= \frac{115}{\cancel{100}_{1}} \times \textsf{₹}\cancel{800}^{8}$

SP $= 115 \times \textsf{₹}8$

SP $= \textsf{₹}920$

Thus, the selling price of the fan is $\mathbf{\textsf{₹}920}$.

Given:

Selling Price (SP) of the table $= \textsf{₹}1500$

Loss percentage $= 10\%$

---

To Find:

The cost price (CP) of the table.

---

Solution:

When there is a loss, the Selling Price (SP) is related to the Cost Price (CP) by the formula:

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

Substitute the given values into the formula:

$1500 = CP \times \frac{(100 - 10)}{100}$

$1500 = CP \times \frac{90}{100}$

$1500 = CP \times \frac{9}{10}$

To find CP, rearrange the equation:

$CP = 1500 \times \frac{10}{9}$

$CP = \frac{15000}{9}$

Simplify the fraction:

$CP = \frac{\cancel{15000}^{5000}}{\cancel{9}_{3}}$

$CP = \frac{5000}{3}$

Thus, the cost price of the table is $\mathbf{\textsf{₹}\frac{5000}{3}}$.

---

Alternate Solution:

Let the Cost Price (CP) be $x$.

The loss is $10\%$ of the Cost Price.

Loss amount $= 10\%$ of $x = \frac{10}{100} \times x = \frac{1}{10} x$

The Selling Price (SP) is the Cost Price minus the Loss amount.

$SP = CP - \text{Loss amount}$

$1500 = x - \frac{1}{10} x$

Combine the terms on the right side:

$1500 = x \left(1 - \frac{1}{10}\right)$

$1500 = x \left(\frac{10 - 1}{10}\right)$

$1500 = x \times \frac{9}{10}$

Solve for $x$:

$x = 1500 \times \frac{10}{9}$

$x = \frac{15000}{9}$

$x = \frac{5000}{3}$

So, the cost price is $\textsf{₹}\frac{5000}{3}$.

Given:

Marked Price (MP) of the dress $= \textsf{₹}2000$

Discount percentage $= 10\%$

---

To Find:

The discount amount and the selling price (SP) of the dress.

---

Solution:

The discount amount is calculated as the discount percentage of the Marked Price.

Discount amount $= 10\%$ of Marked Price

Discount amount $= 10\%$ of $\textsf{₹}2000$

Discount amount $= \frac{10}{100} \times \textsf{₹}2000$

Discount amount $= \frac{10}{\cancel{100}_{1}} \times \textsf{₹}\cancel{2000}^{20}$

Discount amount $= 10 \times \textsf{₹}20$

Discount amount $= \textsf{₹}200$

The selling price is the Marked Price minus the Discount amount.

Selling Price (SP) $=$ Marked Price (MP) $-$ Discount amount

SP $= \textsf{₹}2000 - \textsf{₹}200$

SP $= \textsf{₹}1800$

Thus, the discount amount is $\mathbf{\textsf{₹}200}$ and the selling price of the dress is $\mathbf{\textsf{₹}1800}$.

Given:

Selling Price (SP) of the article $= \textsf{₹}900$

Discount percentage $= 10\%$

---

To Find:

The marked price (MP) of the article.

---

Solution:

When a discount is offered on the marked price, the selling price is obtained by subtracting the discount amount from the marked price.

The discount amount is calculated as the discount percentage of the Marked Price.

Discount amount $= 10\%$ of MP

Discount amount $= \frac{10}{100} \times MP$

Discount amount $= 0.1 \times MP$

The formula relating Selling Price, Marked Price, and Discount is:

Selling Price (SP) $=$ Marked Price (MP) $-$ Discount amount

Substitute the given SP and the expression for the discount amount:

$900 = MP - 0.1 \times MP$

Factor out MP from the right side:

$900 = MP \times (1 - 0.1)$

$900 = MP \times 0.9$

To find the Marked Price, divide the Selling Price by $0.9$:

$MP = \frac{900}{0.9}$

Convert the decimal $0.9$ to a fraction $\frac{9}{10}$:

$MP = \frac{900}{\frac{9}{10}}$

$MP = 900 \times \frac{10}{9}$

Perform the multiplication and simplification:

$MP = \frac{\cancel{900}^{100}}{\cancel{9}_{1}} \times 10$

$MP = 100 \times 10$

$MP = 1000$

Thus, the marked price of the article is $\mathbf{\textsf{₹}1000}$.

---

Alternate Solution:

A discount of $10\%$ on the marked price means that the selling price is $(100 - 10)\% = 90\%$ of the marked price.

So, SP $= 90\%$ of MP

Substitute the given Selling Price:

$900 = \frac{90}{100} \times MP$

$900 = \frac{9}{10} \times MP$

To find MP, multiply both sides of the equation by the reciprocal of $\frac{9}{10}$, which is $\frac{10}{9}$:

$MP = 900 \times \frac{10}{9}$

$MP = \frac{900 \times 10}{9}$

$MP = \frac{9000}{9}$

$MP = 1000$

Thus, the marked price of the article is $\mathbf{\textsf{₹}1000}$.

Given:

Principal (P) $= \textsf{₹}8000$

Rate of Interest (R) $= 5\%$ per annum

Time (T) $= 3$ years

---

To Find:

Simple Interest (SI)

---

Solution:

The formula for calculating Simple Interest is:

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

Substitute the given values into the formula:

$SI = \frac{8000 \times 5 \times 3}{100}$

First, multiply the values in the numerator:

$8000 \times 5 \times 3 = 8000 \times 15 = 120000$

Now, divide by $100$:

$SI = \frac{120000}{100}$

$SI = 1200$

Thus, the Simple Interest on a principal of $\textsf{₹}8000$ at $5\%$ per annum for $3$ years is $\mathbf{\textsf{₹}1200}$.

Given:

Principal (P) $= \textsf{₹}5000$

Rate of Interest (R) $= 10\%$ per annum

Time (n) $= 2$ years

Compounding frequency is annually.

---

To Find:

The amount (A) obtained after 2 years.

---

Solution:

When interest is compounded annually, the formula for the amount (A) is:

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

Substitute the given values into the formula:

$A = 5000 \left(1 + \frac{10}{100}\right)^2$

$A = 5000 \left(1 + 0.1\right)^2$

$A = 5000 \left(1.1\right)^2$

Calculate $(1.1)^2$:

$(1.1)^2 = 1.1 \times 1.1 = 1.21$

Now, substitute this value back into the equation for A:

$A = 5000 \times 1.21$

Perform the multiplication:

$A = 5000 \times \frac{121}{100}$

$A = 50 \times 121$

$A = 6050$

Thus, the amount obtained after 2 years is $\mathbf{\textsf{₹}6050}$.

Given:

The word 'EXAMINATION'.

The word 'Mathematics'.

---

To Find:

The percentage of 'Mathematics' in the word 'EXAMINATION'.

---

Solution:

First, let's count the total number of letters in the word 'EXAMINATION'.

EXAMINATION has 12 letters.

The letters in 'EXAMINATION' are: E, X, A, M, I, N, A, T, I, O, N.

The distinct letters are: E, X, A, M, I, N, T, O.

The count of each letter in 'EXAMINATION' is:

A: 2 times

I: 2 times

N: 2 times

E: 1 time

X: 1 time

M: 1 time

T: 1 time

O: 1 time

Total letters = $2+2+2+1+1+1+1+1 = 12$

---

Now, let's look at the word 'Mathematics'.

The letters in 'Mathematics' are: M, A, T, H, E, M, A, T, I, C, S.

The distinct letters are: M, A, T, H, E, I, C, S.

Comparing the letters required for 'Mathematics' with the letters available in 'EXAMINATION', we notice that the word 'Mathematics' contains the letters 'H', 'C', and 'S'.

These letters ('H', 'C', 'S') are not present in the word 'EXAMINATION'.

Also, 'Mathematics' requires 2 'M's and 2 'T's, while 'EXAMINATION' only has 1 'M' and 1 'T'.

Since it is impossible to form the word 'Mathematics' using only the letters available in 'EXAMINATION' (because some required letters are missing and some needed letters are not available in sufficient quantity), the word 'Mathematics' is not present in 'EXAMINATION'.

Therefore, the percentage of 'Mathematics' in the word 'EXAMINATION' is $\mathbf{0\%}$.

To find the net percentage change, we can assume an original price for the article. Let's assume the original price (Cost Price or initial price) of the article is $\textsf{₹}100$.

---

Given:

Markup percentage $= 20\%$

Discount percentage $= 10\%$

---

To Find:

The net percentage change in the price.

---

Solution:

Assume the original price $= \textsf{₹}100$.

The shopkeeper marks up the price by $20\%$.

Markup amount $= 20\%$ of original price

Markup amount $= \frac{20}{100} \times \textsf{₹}100 = \textsf{₹}20$

The Marked Price (MP) is the original price plus the markup amount.

Marked Price $= \textsf{₹}100 + \textsf{₹}20 = \textsf{₹}120$

Now, a discount of $10\%$ is offered on the Marked Price.

Discount amount $= 10\%$ of Marked Price

Discount amount $= 10\%$ of $\textsf{₹}120$

Discount amount $= \frac{10}{100} \times \textsf{₹}120 = \frac{1}{10} \times \textsf{₹}120 = \textsf{₹}12$

The Selling Price (SP) is the Marked Price minus the discount amount.

Selling Price $= \textsf{₹}120 - \textsf{₹}12 = \textsf{₹}108$

The net change in price is the difference between the Selling Price and the original price.

Net change $= \text{Selling Price} - \text{Original Price}$

Net change $= \textsf{₹}108 - \textsf{₹}100 = \textsf{₹}8$

Since the Selling Price ($\textsf{₹}108$) is greater than the original price ($\textsf{₹}100$), there is a net increase in the price.

Percentage change $= \frac{\text{Net change}}{\text{Original Price}} \times 100\%$

Percentage change $= \frac{\textsf{₹}8}{\textsf{₹}100} \times 100\%$

Percentage change $= \frac{8}{100} \times 100\%$

Percentage change $= 8\%$

Since the change is positive, it is a percentage increase.

Thus, the net percentage change in the price is a $\mathbf{8\%}$ increase.

Given:

Cost price of 10 articles $=$ Selling price of 8 articles.

---

To Find:

The profit or loss percentage.

---

Solution:

Let the cost price of one article be CP and the selling price of one article be SP.

According to the problem statement:

Cost price of 10 articles $= 10 \times CP$

Selling price of 8 articles $= 8 \times SP$

We are given that:

From equation (i), we can find the ratio of SP to CP:

$\frac{SP}{CP} = \frac{10}{8} = \frac{5}{4}$

Since $\frac{SP}{CP} = \frac{5}{4}$ is greater than $1$ (specifically, $SP = \frac{5}{4} \times CP$, which means SP is greater than CP), there is a profit.

The profit is calculated as Profit $=$ SP $-$ CP.

To find the profit percentage, we use the formula:

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

Profit \% $= \frac{SP - CP}{CP} \times 100\%$

Profit \% $= \left(\frac{SP}{CP} - \frac{CP}{CP}\right) \times 100\%$

Profit \% $= \left(\frac{SP}{CP} - 1\right) \times 100\%$

Substitute the value of $\frac{SP}{CP}$ from the ratio:

Profit \% $= \left(\frac{5}{4} - 1\right) \times 100\%$

Profit \% $= \left(\frac{5 - 4}{4}\right) \times 100\%$

Profit \% $= \frac{1}{4} \times 100\%$

Profit \% $= 25\%$

Thus, there is a profit of $\mathbf{25\%}$.

---

Alternate Solution:

Let the cost price of 10 articles and the selling price of 8 articles both be $\textsf{₹}x$.

Cost price of 10 articles $= \textsf{₹}x$

Cost price of 1 article $= \textsf{₹}\frac{x}{10}$

Selling price of 8 articles $= \textsf{₹}x$

Selling price of 1 article $= \textsf{₹}\frac{x}{8}$

Since the selling price of 1 article ($\textsf{₹}\frac{x}{8}$) is greater than the cost price of 1 article ($\textsf{₹}\frac{x}{10}$) because $\frac{1}{8} > \frac{1}{10}$, there is a profit.

Profit on 1 article $=$ Selling price of 1 article $-$ Cost price of 1 article

Profit $= \textsf{₹}\frac{x}{8} - \textsf{₹}\frac{x}{10}$

Profit $= \textsf{₹}\left(\frac{x}{8} - \frac{x}{10}\right)$

Find a common denominator for the fractions (LCM of 8 and 10 is 40):

Profit $= \textsf{₹}\left(\frac{5x}{40} - \frac{4x}{40}\right)$

Profit $= \textsf{₹}\frac{5x - 4x}{40} = \textsf{₹}\frac{x}{40}$

Now, calculate the profit percentage:

Profit \% $= \frac{\text{Profit}}{\text{Cost price of 1 article}} \times 100\%$

Profit \% $= \frac{\textsf{₹}\frac{x}{40}}{\textsf{₹}\frac{x}{10}} \times 100\%$

Profit \% $= \frac{\frac{x}{40}}{\frac{x}{10}} \times 100\%$

Profit \% $= \frac{x}{40} \times \frac{10}{x} \times 100\%$

Cancel out $x$ and simplify the fraction:

Profit \% $= \frac{\cancel{x}}{\cancel{40}_{4}} \times \frac{\cancel{10}^{1}}{\cancel{x}} \times 100\%$

Profit \% $= \frac{1}{4} \times 100\%$

Profit \% $= 25\%$

Thus, there is a profit of $\mathbf{25\%}$.

Given:

Selling Price (including GST) of the shirt $= \textsf{₹}750$

GST Rate $= 5\%$

---

To Find:

The original price of the shirt before GST was added.

---

Solution:

Let the original price of the shirt before GST be $\textsf{₹}x$.

The GST amount is calculated as a percentage of the original price.

GST Amount $= 5\%$ of Original Price

GST Amount $= \frac{5}{100} \times x = 0.05x$

The selling price including GST is the sum of the original price and the GST amount.

Selling Price = Original Price + GST Amount

$750 = x + 0.05x$

Combine the terms involving $x$:

$750 = (1 + 0.05)x$

$750 = 1.05x$

To find $x$, divide the selling price by $1.05$:

$x = \frac{750}{1.05}$

To perform the division, we can write $1.05$ as $\frac{105}{100}$.

$x = \frac{750}{\frac{105}{100}}$

Inverting the denominator fraction and multiplying:

$x = 750 \times \frac{100}{105}$

Simplify the fraction $\frac{100}{105}$ by dividing numerator and denominator by 5:

$\frac{100}{105} = \frac{20}{21}$

$x = 750 \times \frac{20}{21}$

Simplify by dividing 750 and 21 by 3:

$x = \cancel{750}^{250} \times \frac{20}{\cancel{21}_{7}}$

$x = \frac{250 \times 20}{7}$

$x = \frac{5000}{7}$

Thus, the original price of the shirt before GST was $\mathbf{\textsf{₹}\frac{5000}{7}}$.

---

Alternate Solution:

The selling price includes the original price (which is $100\%$) plus the GST amount ($5\%$ of the original price).

So, the selling price represents $(100\% + 5\%) = 105\%$ of the original price.

Let the original price be $\textsf{₹}x$.

According to the problem:

$105\%$ of $x = \textsf{₹}750$

Convert the percentage to a fraction:

$\frac{105}{100} \times x = 750$

To find $x$, multiply both sides by $\frac{100}{105}$:

$x = 750 \times \frac{100}{105}$

Simplify the expression:

$x = \frac{750 \times 100}{105}$

Simplify the fraction $\frac{750 \times 100}{105}$. Divide 750 by 105 (common factor 15: $750 \div 15 = 50$, $105 \div 15 = 7$):

$x = \frac{\cancel{750}^{50} \times 100}{\cancel{105}_{7}}$

$x = \frac{50 \times 100}{7}$

$x = \frac{5000}{7}$

Thus, the original price of the shirt before GST was $\mathbf{\textsf{₹}\frac{5000}{7}}$.

Given:

Original price of the car $= \textsf{₹}5,00,000$

Depreciation rate $= 10\%$ per annum

Time period $= 1$ year

---

To Find:

The price of the car after 1 year.

---

Solution:

Depreciation is a decrease in the value of an asset over time.

The depreciation amount for the first year is $10\%$ of the original price.

Depreciation amount $= 10\%$ of $\textsf{₹}5,00,000$

Depreciation amount $= \frac{10}{100} \times \textsf{₹}5,00,000$

Depreciation amount $= \frac{1}{\cancel{10}_{1}} \times \textsf{₹}\cancel{5,00,000}^{50,000}$

Depreciation amount $= \textsf{₹}50,000$

The price of the car after 1 year is the original price minus the depreciation amount.

Price after 1 year $=$ Original price $-$ Depreciation amount

Price after 1 year $= \textsf{₹}5,00,000 - \textsf{₹}50,000$

Price after 1 year $= \textsf{₹}4,50,000$

Thus, the price of the car after 1 year is $\mathbf{\textsf{₹}4,50,000}$.

---

Alternate Solution:

After a depreciation of $10\%$, the remaining value of the car is $(100\% - 10\%) = 90\%$ of the original price.

Price after 1 year $= 90\%$ of Original price

Price after 1 year $= 90\%$ of $\textsf{₹}5,00,000$

Price after 1 year $= \frac{90}{100} \times \textsf{₹}5,00,000$

Price after 1 year $= \frac{9}{\cancel{10}_{1}} \times \textsf{₹}\cancel{5,00,000}^{50,000}$

Price after 1 year $= 9 \times \textsf{₹}50,000$

Price after 1 year $= \textsf{₹}4,50,000$

Thus, the price of the car after 1 year is $\mathbf{\textsf{₹}4,50,000}$.

Given:

Principal (P) $= \textsf{₹}1000$

Annual Rate of Interest (R) $= 8\%$ per annum

Time (T) $= 1$ year

Compounding frequency is half-yearly.

---

To Find:

Compound Interest (CI)

---

Solution:

When the interest is compounded half-yearly, the annual rate is divided by $2$ and the number of years is multiplied by $2$ to get the number of compounding periods.

Rate per half-year ($r$) $= \frac{\text{Annual Rate}}{2} = \frac{8\%}{2} = 4\%$

Number of compounding periods ($n$) $= \text{Time in years} \times 2 = 1 \times 2 = 2$ half-years

The formula for the amount (A) when compounded half-yearly is:

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

Substitute the given values:

$A = 1000 \left(1 + \frac{4}{100}\right)^2$

$A = 1000 \left(1 + 0.04\right)^2$

$A = 1000 \left(1.04\right)^2$

Calculate $(1.04)^2$:

$(1.04)^2 = 1.04 \times 1.04 = 1.0816$

Now, substitute this value back into the equation for A:

$A = 1000 \times 1.0816$

$A = 1081.60$

The amount obtained after 1 year compounded half-yearly is $\textsf{₹}1081.60$.

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

$CI = A - P$

$CI = \textsf{₹}1081.60 - \textsf{₹}1000$

$CI = \textsf{₹}81.60$

Thus, the Compound Interest is $\mathbf{\textsf{₹}81.60}$.

Given:

Current Population (P) $= 20,000$

Annual Increase Rate (R) $= 5\%$ per year

Time period (n) $= 1$ year

---

To Find:

Population after 1 year.

---

Solution:

The increase in population in 1 year is calculated as the percentage increase of the current population.

Increase in population $= 5\%$ of Current Population

Increase in population $= 5\%$ of $20,000$

Increase in population $= \frac{5}{100} \times 20000$

Increase in population $= \frac{5}{\cancel{100}_{1}} \times \cancel{20000}^{200}$

Increase in population $= 5 \times 200$

Increase in population $= 1000$

The population after 1 year is the sum of the current population and the increase in population.

Population after 1 year $=$ Current Population $+$ Increase in population

Population after 1 year $= 20000 + 1000$

Population after 1 year $= 21000$

Thus, the population after 1 year will be $\mathbf{21,000}$.

---

Alternate Solution:

When the population increases by a certain percentage per year, the population after 1 year can be found by multiplying the current population by $(1 + \frac{\text{Rate}}{100})$.

Population after 1 year $=$ Current Population $\times \left(1 + \frac{\text{Rate}}{100}\right)$

Population after 1 year $= 20000 \times \left(1 + \frac{5}{100}\right)$

Population after 1 year $= 20000 \times \left(1 + 0.05\right)$

Population after 1 year $= 20000 \times 1.05$

Population after 1 year $= 20000 \times \frac{105}{100}$

Population after 1 year $= \frac{20000 \times 105}{100}$

Population after 1 year $= \frac{\cancel{20000}^{200} \times 105}{\cancel{100}_{1}}$

Population after 1 year $= 200 \times 105$

Population after 1 year $= 21000$

Thus, the population after 1 year will be $\mathbf{21,000}$.

Given:

A sum of money (Principal, P) doubles itself in 10 years at simple interest.

Time (T$_1$) $= 10$ years.

Amount (A$_1$) $= 2 \times P$

Interest is Simple Interest (SI).

---

To Find:

Time (T$_2$) required for the Principal to become three times (Amount, A$_2 = 3P$) at the same rate (R).

---

Solution:

The formula for Simple Interest is $SI = \frac{P \times R \times T}{100}$.

The Amount is given by $A = P + SI$.

In the first case, the amount is $A_1 = 2P$ after $T_1 = 10$ years.

The Simple Interest earned is $SI_1 = A_1 - P = 2P - P = P$.

Using the SI formula for the first case:

$SI_1 = \frac{P \times R \times T_1}{100}$

Substitute the values $SI_1 = P$ and $T_1 = 10$:

$P = \frac{P \times R \times 10}{100}$

Now, we can solve for the rate (R).

$P = \frac{P \times R}{10}$

Multiply both sides by 10:

$10P = P \times R$

If $P \neq 0$, we can divide both sides by P:

$R = \frac{10P}{P}$

$R = 10$

So, the rate of simple interest is $10\%$ per annum.

---

In the second case, we want the amount to be $A_2 = 3P$ at the same rate $R = 10\%$. Let the required time be $T_2$.

The Simple Interest required is $SI_2 = A_2 - P = 3P - P = 2P$.

Using the SI formula for the second case:

$SI_2 = \frac{P \times R \times T_2}{100}$

Substitute the values $SI_2 = 2P$ and $R = 10$:

$2P = \frac{P \times 10 \times T_2}{100}$

$2P = \frac{P \times T_2}{10}$

Multiply both sides by 10:

$20P = P \times T_2$

If $P \neq 0$, we can divide both sides by P:

$T_2 = \frac{20P}{P}$

$T_2 = 20$

Thus, the sum of money will become three times itself in $\mathbf{20}$ years at the same rate of simple interest.

Given:

Marked Price (MP) of the article $= \textsf{₹}800$

Selling Price (SP) of the article $= \textsf{₹}720$

---

To Find:

The discount percentage offered.

---

Solution:

The discount amount is the difference between the Marked Price and the Selling Price.

Discount amount $=$ Marked Price $-$ Selling Price

Discount amount $= \textsf{₹}800 - \textsf{₹}720$

Discount amount $= \textsf{₹}80$

The discount percentage is calculated on the Marked Price using the formula:

Discount \% $= \frac{\text{Discount amount}}{\text{Marked Price}} \times 100\%$

Substitute the calculated discount amount and the given Marked Price into the formula:

Discount \% $= \frac{\textsf{₹}80}{\textsf{₹}800} \times 100\%$

Discount \% $= \frac{80}{800} \times 100\%$

Simplify the fraction $\frac{80}{800}$:

Discount \% $= \frac{\cancel{80}^{1}}{\cancel{800}_{10}} \times 100\%$

Discount \% $= \frac{1}{10} \times 100\%$

Discount \% $= \frac{\cancel{100}^{10}}{\cancel{10}_{1}}\%$

Discount \% $= 10\%$

Thus, the discount percentage offered is $\mathbf{10\%}$.

Given:

Purchase price of the television $= \textsf{₹}12,000$

Repair expenses $= \textsf{₹}200$

Selling price of the television $= \textsf{₹}13,500$

---

To Find:

The profit or loss percent.

---

Solution:

The effective cost price (CP) of the television includes the purchase price and the repair expenses.

Effective Cost Price $=$ Purchase price $+$ Repair expenses

Effective CP $= \textsf{₹}12,000 + \textsf{₹}200$

Effective CP $= \textsf{₹}12,200$

The Selling Price (SP) is $\textsf{₹}13,500$.

Compare the Selling Price and the Effective Cost Price:

SP ($\textsf{₹}13,500$) is greater than Effective CP ($\textsf{₹}12,200$).

Since SP $>$ CP, there is a profit.

Profit amount $=$ Selling Price $-$ Effective Cost Price

Profit amount $= \textsf{₹}13,500 - \textsf{₹}12,200$

Profit amount $= \textsf{₹}1,300$

Now, calculate the profit percentage using the formula:

Profit \% $= \frac{\text{Profit amount}}{\text{Effective Cost Price}} \times 100\%$

Profit \% $= \frac{\textsf{₹}1,300}{\textsf{₹}12,200} \times 100\%$

Profit \% $= \frac{1300}{12200} \times 100\%$

Simplify the fraction:

Profit \% $= \frac{\cancel{1300}^{13}}{\cancel{12200}_{122}} \times 100\%$

Profit \% $= \frac{13}{122} \times 100\%$

Profit \% $= \frac{1300}{122}\%$

Simplify $\frac{1300}{122}$ by dividing both numerator and denominator by 2:

$\frac{1300}{122} = \frac{650}{61}$

Profit \% $= \frac{650}{61}\%$

Convert the improper fraction to a mixed number or a decimal (optional):

$\frac{650}{61} = 10 \frac{40}{61}$ or approximately $10.656\%$

Thus, the profit percentage is $\mathbf{\frac{650}{61}\%}$ or $\mathbf{10 \frac{40}{61}\%}$.

Given:

Total quantity of apples bought $= 50$ kg

Cost price per kg of apples $= \textsf{₹}60$

Percentage of apples sold in the first part $= 60\%$

Profit percentage on the first part $= 10\%$

Percentage of apples sold in the second part $= 100\% - 60\% = 40\%$

Loss percentage on the second part $= 5\%$

---

To Find:

Total profit or loss amount and the overall profit or loss percentage.

---

Solution:

First, calculate the total cost price (CP) of the apples.

Total CP $=$ Quantity $\times$ CP per kg

Total CP $= 50 \text{ kg} \times \textsf{₹}60/\text{kg} = \textsf{₹}3000$

---

Now, consider the first part of the apples.

Quantity of apples in the first part $= 60\%$ of $50$ kg

Quantity $= \frac{60}{100} \times 50 \text{ kg} = 0.60 \times 50 \text{ kg} = 30 \text{ kg}$

Cost price of the first part $=$ Quantity $\times$ CP per kg

CP$_1 = 30 \text{ kg} \times \textsf{₹}60/\text{kg} = \textsf{₹}1800$

These apples were sold at a profit of $10\%$.

Selling price of the first part (SP$_1$) $=$ CP$_1 \times (1 + \frac{\text{Profit}\%}{100})$

SP$_1 = \textsf{₹}1800 \times (1 + \frac{10}{100}) = \textsf{₹}1800 \times (1 + 0.10) = \textsf{₹}1800 \times 1.10$

SP$_1 = \textsf{₹}1980$

---

Next, consider the second part of the apples.

Quantity of apples in the second part $=$ Total quantity $-$ Quantity in first part

Quantity $= 50 \text{ kg} - 30 \text{ kg} = 20 \text{ kg}$

Cost price of the second part $=$ Quantity $\times$ CP per kg

CP$_2 = 20 \text{ kg} \times \textsf{₹}60/\text{kg} = \textsf{₹}1200$

These apples were sold at a loss of $5\%$.

Selling price of the second part (SP$_2$) $=$ CP$_2 \times (1 - \frac{\text{Loss}\%}{100})$

SP$_2 = \textsf{₹}1200 \times (1 - \frac{5}{100}) = \textsf{₹}1200 \times (1 - 0.05) = \textsf{₹}1200 \times 0.95$

SP$_2 = \textsf{₹}1140$

---

Now, calculate the total selling price (SP) of all the apples.

Total SP $=$ SP$_1$ $+$ SP$_2$

Total SP $= \textsf{₹}1980 + \textsf{₹}1140 = \textsf{₹}3120$

---

Compare the total selling price with the total cost price.

Total CP $= \textsf{₹}3000$

Total SP $= \textsf{₹}3120$

Since Total SP $>$ Total CP, there is an overall profit.

Total Profit Amount $=$ Total SP $-$ Total CP

Total Profit Amount $= \textsf{₹}3120 - \textsf{₹}3000 = \textsf{₹}120$

---

Finally, calculate the overall profit percentage.

Overall Profit \% $= \frac{\text{Total Profit Amount}}{\text{Total Cost Price}} \times 100\%$

Overall Profit \% $= \frac{\textsf{₹}120}{\textsf{₹}3000} \times 100\%$

Overall Profit \% $= \frac{120}{3000} \times 100\%$

Simplify the fraction:

Overall Profit \% $= \frac{\cancel{120}^{12}}{\cancel{3000}_{300}} \times \cancel{100}^{1}\%$

Overall Profit \% $= \frac{12}{300} \times 100\% = \frac{12 \times 100}{300}\%$

Overall Profit \% $= \frac{1200}{300}\% = 4\%$

Thus, the shopkeeper has a total profit of $\mathbf{\textsf{₹}120}$ and an overall profit percentage of $\mathbf{4\%}$.

Given:

Marked Price (MP) of the refrigerator $= \textsf{₹}18,000$

Discount percentage offered $= 10\%$

Profit percentage made $= 20\%$

---

To Find:

The cost price (CP) of the refrigerator.

The profit percentage if sold without discount.

---

Solution (Part 1: Find Cost Price):

First, calculate the Selling Price (SP) after the $10\%$ discount on the Marked Price.

The selling price is the marked price minus the discount amount.

Discount amount $= 10\%$ of Marked Price

Discount amount $= 10\%$ of $\textsf{₹}18,000$

Discount amount $= \frac{10}{100} \times \textsf{₹}18,000 = 0.10 \times \textsf{₹}18,000 = \textsf{₹}1800$

Selling Price (SP) $=$ Marked Price $-$ Discount amount

SP $= \textsf{₹}18,000 - \textsf{₹}1800 = \textsf{₹}16,200$

Alternatively, the selling price is $(100 - \text{Discount}\%)\%$ of the Marked Price.

SP $= (100 - 10)\%$ of $\textsf{₹}18,000 = 90\%$ of $\textsf{₹}18,000$

SP $= \frac{90}{100} \times \textsf{₹}18,000 = 0.90 \times \textsf{₹}18,000 = \textsf{₹}16,200$

Now, we know the Selling Price ($\textsf{₹}16,200$) and the Profit Percentage ($20\%$). We can find the Cost Price (CP).

The formula relating SP, CP, and Profit% is: SP $= CP \times (1 + \frac{\text{Profit}\%}{100})$

Substitute the known values:

$\textsf{₹}16,200 = CP \times (1 + \frac{20}{100})$

$\textsf{₹}16,200 = CP \times (1 + 0.20)$

$\textsf{₹}16,200 = CP \times 1.20$

To find CP, divide the SP by $1.20$:

$CP = \frac{\textsf{₹}16,200}{1.20}$

$CP = \frac{16200}{1.2}$

$CP = \frac{162000}{12}$ (Multiplying numerator and denominator by 10)

Perform the division:

$CP = 13500$

Thus, the cost price of the refrigerator is $\mathbf{\textsf{₹}13,500}$.

---

Solution (Part 2: Find Profit Percentage without Discount):

If the shopkeeper sells the refrigerator without any discount, the Selling Price will be equal to the Marked Price.

New Selling Price (New SP) $=$ Marked Price (MP)

New SP $= \textsf{₹}18,000$

We found the Cost Price (CP) in the previous part:

CP $= \textsf{₹}13,500$

Since the New SP ($\textsf{₹}18,000$) is greater than the CP ($\textsf{₹}13,500$), there will be a profit.

New Profit amount $=$ New Selling Price $-$ Cost Price

New Profit amount $= \textsf{₹}18,000 - \textsf{₹}13,500 = \textsf{₹}4,500$

Now, calculate the new profit percentage based on the Cost Price:

New Profit \% $= \frac{\text{New Profit amount}}{\text{Cost Price}} \times 100\%$

New Profit \% $= \frac{\textsf{₹}4,500}{\textsf{₹}13,500} \times 100\%$

New Profit \% $= \frac{4500}{13500} \times 100\%$

Simplify the fraction $\frac{4500}{13500}$:

$\frac{\cancel{4500}^{45}}{\cancel{13500}_{135}} = \frac{\cancel{45}^{1}}{\cancel{135}_{3}} = \frac{1}{3}$

New Profit \% $= \frac{1}{3} \times 100\%$

New Profit \% $= \frac{100}{3}\%$

New Profit \% $= 33\frac{1}{3}\%$ or approximately $33.33\%$

Thus, if the shopkeeper sells it without any discount, his profit percentage will be $\mathbf{33\frac{1}{3}\%}$.

Given:

Principal (P) $= \textsf{₹}20,000$

Rate of Interest (R) $= 10\%$ per annum

Time (n) $= 3$ years

Compounding frequency is annually.

---

To Find:

Amount (A) and Compound Interest (CI).

Simple Interest (SI) for the same principal, rate, and time.

Comparison between CI and SI.

---

Solution (Part 1: Compound Interest):

The formula for the amount (A) when compounded annually is:

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

Substitute the given values:

$A = 20000 \left(1 + \frac{10}{100}\right)^3$

$A = 20000 \left(1 + 0.1\right)^3$

$A = 20000 \left(1.1\right)^3$

Calculate $(1.1)^3$:

$(1.1)^3 = 1.1 \times 1.1 \times 1.1 = 1.21 \times 1.1 = 1.331$

Now, substitute this value back into the equation for A:

$A = 20000 \times 1.331$

$A = 20000 \times \frac{1331}{1000}$

$A = 20 \times 1331$

$A = 26620$

The amount obtained after 3 years compounded annually is $\textsf{₹}26,620$.

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

$CI = A - P$

$CI = \textsf{₹}26,620 - \textsf{₹}20,000$

$CI = \textsf{₹}6,620$

---

Solution (Part 2: Simple Interest):

The formula for Simple Interest (SI) is:

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

Substitute the given values (P = $\textsf{₹}20,000$, R = $10\%$, T = $3$ years):

$SI = \frac{20000 \times 10 \times 3}{100}$

$SI = \frac{20000 \times 30}{100}$

$SI = \frac{600000}{100}$

$SI = 6000$

The Simple Interest for the same principal, rate, and time is $\textsf{₹}6,000$.

---

Comparison:

Compound Interest (CI) $= \textsf{₹}6,620$

Simple Interest (SI) $= \textsf{₹}6,000$

Compare CI and SI:

$CI > SI$

Difference $= CI - SI = \textsf{₹}6,620 - \textsf{₹}6,000 = \textsf{₹}620$

Thus, the Amount is $\mathbf{\textsf{₹}26,620}$ and the Compound Interest is $\mathbf{\textsf{₹}6,620}$. The Simple Interest for the same period is $\mathbf{\textsf{₹}6,000}$. The Compound Interest is $\mathbf{\textsf{₹}620}$ more than the Simple Interest.

Given:

Principal (P) $= \textsf{₹}12,000$

Annual Rate of Interest (R) $= 8\%$ per annum

Time (T) $= 1\frac{1}{2}$ years $= \frac{3}{2}$ years

Compounding frequency is half-yearly.

---

To Find:

Compound Interest (CI)

---

Solution:

When the interest is compounded half-yearly, the annual rate is divided by $2$ and the number of years is multiplied by $2$ to get the number of compounding periods.

Rate per half-year ($r$) $= \frac{\text{Annual Rate}}{2} = \frac{8\%}{2} = 4\%$

Number of compounding periods ($n$) $= \text{Time in years} \times 2 = \frac{3}{2} \times 2 = 3$ half-years

The formula for the amount (A) when compounded half-yearly is:

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

Substitute the given values:

$A = 12000 \left(1 + \frac{4}{100}\right)^3$

$A = 12000 \left(1 + 0.04\right)^3$

$A = 12000 \left(1.04\right)^3$

Calculate $(1.04)^3$:

$(1.04)^3 = 1.04 \times 1.04 \times 1.04 = 1.0816 \times 1.04 = 1.124864$

Now, substitute this value back into the equation for A:

$A = 12000 \times 1.124864$

Perform the multiplication:

$A = 13498.368$

The amount obtained after $1\frac{1}{2}$ years compounded half-yearly is $\textsf{₹}13498.368$. We can round this to $\textsf{₹}13498.37$ for currency.

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

$CI = A - P$

$CI = \textsf{₹}13498.37 - \textsf{₹}12000$

$CI = \textsf{₹}1498.37$

Thus, the Compound Interest is $\mathbf{\textsf{₹}1498.37}$ (rounded to two decimal places).

Given:

Population in the year 2020 (P$_0$) $= 1,25,000$

Annual increase rate (R) $= 4\%$ per annum

---

To Find:

Population in the year 2023.

Population in the year 2019.

---

Solution (Part 1: Population in 2023):

The population increases at a constant rate each year, similar to compound interest. The formula for population growth is $P_t = P_0 (1 + \frac{R}{100})^t$, where $P_t$ is the population after $t$ years, $P_0$ is the initial population, and R is the annual growth rate.

Here, the initial population is in 2020 (P$_0 = 1,25,000$), and we want to find the population in 2023. The time period ($t$) is $2023 - 2020 = 3$ years.

$t = 3$ years.

Substitute the values into the formula:

Population in 2023 $= 125000 \left(1 + \frac{4}{100}\right)^3$

Population in 2023 $= 125000 \left(1 + 0.04\right)^3$

Population in 2023 $= 125000 \left(1.04\right)^3$

Calculate $(1.04)^3$:

$(1.04)^3 = 1.04 \times 1.04 \times 1.04 = 1.0816 \times 1.04 = 1.124864$

Now, substitute this value back into the equation:

Population in 2023 $= 125000 \times 1.124864$

Perform the multiplication:

$125000 \times 1.124864 = 140608$

Thus, the population of the city in the year 2023 will be $\mathbf{1,40,608}$.

---

Solution (Part 2: Population in 2019):

Let the population in the year 2019 be P$_{2019}$. The population in 2020 ($1,25,000$) is the population in 2019 plus the increase from 2019 to 2020. The time period ($t$) from 2019 to 2020 is $1$ year.

Using the population growth formula, where P$_0$ is the population in 2019 and P$_t$ is the population in 2020 ($t=1$):

Population in 2020 $= \text{Population in 2019} \times (1 + \frac{R}{100})^1$

$125000 = P_{2019} \times (1 + \frac{4}{100})$

$125000 = P_{2019} \times (1 + 0.04)$

$125000 = P_{2019} \times 1.04$

To find P$_{2019}$, divide the population in 2020 by $1.04$:

$P_{2019} = \frac{125000}{1.04}$

$P_{2019} = \frac{125000}{\frac{104}{100}} = 125000 \times \frac{100}{104}$

$P_{2019} = \frac{12500000}{104}$

Simplify the fraction by dividing numerator and denominator by 8:

$12500000 \div 8 = 1562500$

$104 \div 8 = 13$

$P_{2019} = \frac{1562500}{13}$

Performing the division $\frac{1562500}{13} \approx 120192.307$

Since population must be a whole number, we usually round to the nearest whole number, but the problem implies a consistent annual rate, which might suggest leaving it as a fraction or understanding it's a model. However, in population problems, we generally expect a whole number answer if the initial population was a whole number. Let's recheck the calculation.

$P_{2019} = \frac{125000}{1.04} = \frac{125000}{104/100} = 125000 \times \frac{100}{104}$

$125000 \times \frac{100}{104} = \frac{12500000}{104}$.

Dividing $12500000$ by $104$ gives approximately $120192.3$. It's possible the rate is an average or the question expects an approximate answer, or there was a typo. Assuming the question intends a calculation based on the formula:

$P_{2019} = \frac{12500000}{104}$

Let's keep the fraction form as the exact answer.

Population in 2019 $= \frac{12500000}{104} = \frac{1562500}{13}$

However, a population must be an integer. Let's consider if there's another way to interpret the question or if it's standard to round. Given it's a short answer question, an exact answer might be expected if possible, but division by 13 won't result in a terminating decimal. Let's assume for the purpose of providing an answer formatted per rules that the division is carried out, acknowledging the real-world constraint of population being an integer.

Calculating the division: $\frac{1562500}{13} \approx 120192.30769...$

Rounding to the nearest whole number gives $120192$.

Let's state the exact answer as a fraction and the rounded integer answer.

Exact Population in 2019 $= \frac{1562500}{13}$

Approximate Population in 2019 $\approx 120192$

Given the context of typical school problems, rounding is usually expected for population figures when the result is not an integer. Let's provide the rounded answer.

Thus, the population of the city in the year 2023 will be $\mathbf{1,40,608}$, and the population in the year 2019 was approximately $\mathbf{1,20,192}$.

Given:

Original price of the machine (V$_0$) $= \textsf{₹}60,000$

Annual depreciation rate (R) $= 10\%$ per annum

Time period (t) $= 2$ years

---

To Find:

The value of the machine after 2 years (V$_2$).

The total depreciation in 2 years.

---

Solution:

The value of the machine depreciates at a constant rate each year. The formula for the value of an asset after depreciation is similar to the compound interest formula, but with a negative sign for the rate:

$V_t = V_0 \left(1 - \frac{R}{100}\right)^t$

Where:

$V_t$ is the value after $t$ years.

$V_0$ is the original value.

R is the annual depreciation rate.

$t$ is the time in years.

Substitute the given values into the formula to find the value after 2 years:

$V_2 = 60000 \left(1 - \frac{10}{100}\right)^2$

$V_2 = 60000 \left(1 - 0.10\right)^2$

$V_2 = 60000 \left(0.90\right)^2$

Calculate $(0.90)^2$:

$(0.90)^2 = 0.90 \times 0.90 = 0.81$

Now, substitute this value back into the equation for V$_2$:

$V_2 = 60000 \times 0.81$

Perform the multiplication:

$V_2 = 60000 \times \frac{81}{100}$

$V_2 = 600 \times 81$

$V_2 = 48600$

The value of the machine after 2 years is $\textsf{₹}48,600$.

---

Next, calculate the total depreciation in 2 years.

Total Depreciation $=$ Original Value $-$ Value after 2 years

Total Depreciation $= \textsf{₹}60,000 - \textsf{₹}48,600$

Total Depreciation $= \textsf{₹}11,400$

Thus, the value of the machine after 2 years is $\mathbf{\textsf{₹}48,600}$, and the value has depreciated by $\mathbf{\textsf{₹}11,400}$ in 2 years.

Given:

Discount percentage offered $= 20\%$

Selling Price (SP) of the article $= \textsf{₹}2400$

Cost Price (CP) of the article $= \textsf{₹}1800$

---

To Find:

The marked price (MP) of the article.

The profit or loss percentage.

---

Solution (Part 1: Find Marked Price):

The selling price is obtained after deducting the discount from the marked price.

A $20\%$ discount means the selling price is $(100 - 20)\% = 80\%$ of the marked price.

So, SP $= 80\%$ of MP

Substitute the given Selling Price:

$\textsf{₹}2400 = \frac{80}{100} \times MP$

$\textsf{₹}2400 = 0.80 \times MP$

To find MP, divide the Selling Price by $0.80$:

$MP = \frac{\textsf{₹}2400}{0.80}$

$MP = \frac{2400}{0.8} = \frac{24000}{8}$ (Multiplying numerator and denominator by 10)

Perform the division:

$MP = 3000$

Thus, the marked price of the article is $\mathbf{\textsf{₹}3000}$.

---

Solution (Part 2: Find Profit or Loss Percentage):

We are given the Cost Price (CP) $= \textsf{₹}1800$ and the Selling Price (SP) $= \textsf{₹}2400$.

Compare the Selling Price and the Cost Price:

SP ($\textsf{₹}2400$) is greater than CP ($\textsf{₹}1800$).

Since SP $>$ CP, there is a profit.

Profit amount $=$ Selling Price $-$ Cost Price

Profit amount $= \textsf{₹}2400 - \textsf{₹}1800$

Profit amount $= \textsf{₹}600$

Now, calculate the profit percentage based on the Cost Price:

Profit \% $= \frac{\text{Profit amount}}{\text{Cost Price}} \times 100\%$

Profit \% $= \frac{\textsf{₹}600}{\textsf{₹}1800} \times 100\%$

Profit \% $= \frac{600}{1800} \times 100\%$

Simplify the fraction $\frac{600}{1800}$:

$\frac{\cancel{600}^{1}}{\cancel{1800}_{3}} = \frac{1}{3}$

Profit \% $= \frac{1}{3} \times 100\%$

Profit \% $= \frac{100}{3}\%$

Profit \% $= 33\frac{1}{3}\%$ or approximately $33.33\%$

Thus, the marked price is $\mathbf{\textsf{₹}3000}$ and the profit percentage is $\mathbf{33\frac{1}{3}\%}$.

Given:

Price of the shoes including GST $= \textsf{₹}3000$

GST Rate $= 12\%$

---

To Find:

The original price of the shoes before GST was added.

The amount of GST paid by Ramesh.

---

Solution:

Let the original price of the shoes before GST be $\textsf{₹}x$.

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

The GST amount is calculated as a percentage of the original price.

GST Amount $= 12\%$ of Original Price

GST Amount $= \frac{12}{100} \times x = 0.12x$

Price including GST $=$ Original Price $+$ GST Amount

$\textsf{₹}3000 = x + 0.12x$

Combine the terms involving $x$:

$3000 = (1 + 0.12)x$

$3000 = 1.12x$

To find $x$ (the original price), divide the price including GST by $1.12$:

$x = \frac{3000}{1.12}$

$x = \frac{3000}{\frac{112}{100}} = 3000 \times \frac{100}{112}$

$x = \frac{300000}{112}$

Simplify the fraction $\frac{300000}{112}$ by dividing numerator and denominator by their greatest common divisor. Both are divisible by 8.

$300000 \div 8 = 37500$

$112 \div 8 = 14$

$x = \frac{37500}{14}$

Simplify further by dividing numerator and denominator by 2.

$37500 \div 2 = 18750$

$14 \div 2 = 7$

$x = \frac{18750}{7}$

So, the original price of the shoes before GST was $\textsf{₹}\frac{18750}{7}$.

Now, find the amount of GST paid.

GST Amount $=$ Price including GST $-$ Original Price

GST Amount $= \textsf{₹}3000 - \textsf{₹}\frac{18750}{7}$

To subtract, find a common denominator (which is 7):

$3000 = \frac{3000 \times 7}{7} = \frac{21000}{7}$

GST Amount $= \textsf{₹}\frac{21000}{7} - \textsf{₹}\frac{18750}{7}$

GST Amount $= \textsf{₹}\frac{21000 - 18750}{7}$

GST Amount $= \textsf{₹}\frac{2250}{7}$

Alternatively, calculate GST amount as $12\%$ of the original price:

GST Amount $= 12\%$ of $\textsf{₹}\frac{18750}{7}$

GST Amount $= \frac{12}{100} \times \frac{18750}{7}$

GST Amount $= \frac{\cancel{12}^{3}}{\cancel{100}_{25}} \times \frac{18750}{7}$

GST Amount $= \frac{3}{25} \times \frac{18750}{7}$

Divide 18750 by 25: $18750 \div 25 = 750$

GST Amount $= 3 \times \frac{750}{7} = \frac{2250}{7}$

The original price is $\textsf{₹}\frac{18750}{7}$. As a decimal, $\frac{18750}{7} \approx \textsf{₹}2678.57$.

The GST amount is $\textsf{₹}\frac{2250}{7}$. As a decimal, $\frac{2250}{7} \approx \textsf{₹}321.43$.

Check: $\textsf{₹}2678.57 + \textsf{₹}321.43 = \textsf{₹}3000.00$ (approximately due to rounding).

Thus, the original price of the shoes before GST was $\mathbf{\textsf{₹}\frac{18750}{7}}$ and the amount of GST paid was $\mathbf{\textsf{₹}\frac{2250}{7}}$.

Given:

Simple Interest (SI) $= \textsf{₹}1600$

Time (T) $= 2$ years

Rate of Interest (R) $= 5\%$ per annum

---

To Find:

The Principal sum (P).

Compound Interest (CI) on the same sum for the same period at the same rate, compounded annually.

---

Solution (Part 1: Find the Principal Sum):

The formula for Simple Interest is:

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

Substitute the given values (SI = $\textsf{₹}1600$, R = $5\%$, T = $2$ years):

$1600 = \frac{P \times 5 \times 2}{100}$

$1600 = \frac{P \times 10}{100}$

$1600 = \frac{P}{10}$

To find P, multiply both sides by 10:

$P = 1600 \times 10$

$P = 16000$

Thus, the principal sum is $\mathbf{\textsf{₹}16,000}$.

---

Solution (Part 2: Find Compound Interest):

Now, we need to find the Compound Interest (CI) on the same sum (P = $\textsf{₹}16,000$) for the same period (n = 2 years) at the same rate (R = $5\%$) compounded annually.

The formula for the amount (A) when compounded annually is:

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

Substitute the values P = $\textsf{₹}16,000$, R = $5\%$, n = $2$ years:

$A = 16000 \left(1 + \frac{5}{100}\right)^2$

$A = 16000 \left(1 + 0.05\right)^2$

$A = 16000 \left(1.05\right)^2$

Calculate $(1.05)^2$:

$(1.05)^2 = 1.05 \times 1.05 = 1.1025$

Now, substitute this value back into the equation for A:

$A = 16000 \times 1.1025$

Perform the multiplication:

$A = 16000 \times \frac{11025}{10000}$

$A = 16 \times \frac{11025}{10}$

$A = \frac{176400}{10}$

$A = 17640$

The amount obtained after 2 years compounded annually is $\textsf{₹}17,640$.

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

$CI = A - P$

$CI = \textsf{₹}17,640 - \textsf{₹}16,000$

$CI = \textsf{₹}1,640$

Thus, the principal sum is $\mathbf{\textsf{₹}16,000}$ and the Compound Interest on the same sum for the same period at the same rate, compounded annually, is $\mathbf{\textsf{₹}1,640}$.

Given:

Principal (P) $= \textsf{₹}8000$

Time (T or n) $= 2$ years

Difference between CI and SI (CI - SI) $= \textsf{₹}128$

Compounding frequency is annually.

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To Find:

The Rate of Interest (R) per annum.

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Solution:

First, we calculate the Simple Interest (SI) for the given Principal, Time, and an unknown Rate R.

The formula for Simple Interest is:

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

Substitute the given values P $= \textsf{₹}8000$, T $= 2$ years:

$SI = \frac{8000 \times R \times 2}{100}$

$SI = \frac{16000 \times R}{100}$

$SI = 160R$

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Next, we calculate the Compound Interest (CI) for the same Principal, Time, and Rate, compounded annually.

The formula for the Amount (A) when compounded annually is:

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

Substitute the given values P $= \textsf{₹}8000$, n $= 2$ years:

$A = 8000 \left(1 + \frac{R}{100}\right)^2$

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

$CI = A - P$

$CI = 8000 \left(1 + \frac{R}{100}\right)^2 - 8000$

Expand the term $\left(1 + \frac{R}{100}\right)^2$:

$\left(1 + \frac{R}{100}\right)^2 = 1^2 + 2 \times 1 \times \frac{R}{100} + \left(\frac{R}{100}\right)^2 = 1 + \frac{2R}{100} + \frac{R^2}{10000}$

Substitute this back into the CI equation:

$CI = 8000 \left(1 + \frac{2R}{100} + \frac{R^2}{10000}\right) - 8000$

$CI = 8000 + 8000 \times \frac{2R}{100} + 8000 \times \frac{R^2}{10000} - 8000$

$CI = 8000 + 160R + 0.8R^2 - 8000$

$CI = 160R + 0.8R^2$

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We are given that the difference between CI and SI is $\textsf{₹}128$.

$CI - SI = 128$

Substitute the expressions for CI and SI:

$(160R + 0.8R^2) - (160R) = 128$

Simplify the equation:

$160R + 0.8R^2 - 160R = 128$

$0.8R^2 = 128$

Solve for $R^2$:

$R^2 = \frac{128}{0.8}$

$R^2 = \frac{1280}{8}$

$R^2 = 160$

To find R, take the square root of both sides:

$R = \sqrt{160}$

Simplify the square root:

$\sqrt{160} = \sqrt{16 \times 10} = \sqrt{16} \times \sqrt{10} = 4\sqrt{10}$

Since the rate must be positive, $R = 4\sqrt{10}$.

The rate of interest per annum is $\mathbf{4\sqrt{10}\%}$.

Given:

Selling Price (SP$_1$) when there is a loss $= \textsf{₹}2350$

Loss Percentage $= 6\%$

Target Profit Percentage $= 10\%$

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To Find:

The Selling Price (SP$_2$) required to gain $10\%$.

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Solution:

First, we need to find the Cost Price (CP) of the cycle using the information about the first sale (loss of $6\%$ at SP = $\textsf{₹}2350$).

When there is a loss, the Selling Price (SP) is related to the Cost Price (CP) by the formula:

$SP = CP \times \left(1 - \frac{\text{Loss}\%}{100}\right)$

Substitute the given values: SP$_1$ = $\textsf{₹}2350$, Loss\% = $6\%$.

$2350 = CP \times \left(1 - \frac{6}{100}\right)$

$2350 = CP \times \left(1 - 0.06\right)$

$2350 = CP \times 0.94$

To find CP, divide the Selling Price by $0.94$:

$CP = \frac{2350}{0.94}$

To perform the division, we can write $0.94$ as $\frac{94}{100}$ or multiply numerator and denominator by 100:

$CP = \frac{2350 \times 100}{0.94 \times 100}$

$CP = \frac{235000}{94}$

Perform the division:

$235000 \div 94 = 2500$

So, the cost price of the cycle is $\textsf{₹}2500$.

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Now, we need to find the Selling Price (SP$_2$) at which the shopkeeper can gain $10\%$. We use the Cost Price we just calculated (CP = $\textsf{₹}2500$) and the target Profit Percentage ($10\%$).

When there is a profit, the Selling Price (SP) is related to the Cost Price (CP) by the formula:

$SP = CP \times \left(1 + \frac{\text{Profit}\%}{100}\right)$

Substitute the values CP = $\textsf{₹}2500$, Profit\% = $10\%$.

$SP_2 = 2500 \times \left(1 + \frac{10}{100}\right)$

$SP_2 = 2500 \times \left(1 + 0.10\right)$

$SP_2 = 2500 \times 1.10$

$SP_2 = 2500 \times 1.1$

Perform the multiplication:

$SP_2 = 2750$

Thus, the shopkeeper should sell the cycle for $\mathbf{\textsf{₹}2750}$ to gain $10\%$.

Given:

Production in 2021 (P$_0$) $= 40,000$

Production in 2023 (P$_t$) $= 46,000$

Time period (t) $= 2023 - 2021 = 2$ years

The increase is compounded annually at a constant rate.

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To Find:

The annual rate of increase in production (R).

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Solution:

The problem involves a consistent percentage increase over time, which can be modeled using a formula similar to that for compound interest or population growth:

$P_t = P_0 \left(1 + \frac{R}{100}\right)^t$

Where:

$P_t$ is the production after $t$ years.

$P_0$ is the initial production.

R is the annual rate of increase (in percentage).

$t$ is the time period in years.

Substitute the given values into the formula:

$46000 = 40000 \left(1 + \frac{R}{100}\right)^2$

Divide both sides by $40000$ to isolate the term with R:

$\frac{46000}{40000} = \left(1 + \frac{R}{100}\right)^2$

Simplify the fraction on the left side:

$\frac{46}{40} = \left(1 + \frac{R}{100}\right)^2$

$\frac{23}{20} = \left(1 + \frac{R}{100}\right)^2$

To solve for the term inside the parenthesis, take the square root of both sides:

$\sqrt{\frac{23}{20}} = 1 + \frac{R}{100}$

Calculate the value of $\sqrt{\frac{23}{20}}$ or $\sqrt{1.15}$.

$\sqrt{1.15} \approx 1.0723805$

So, $1.0723805 \approx 1 + \frac{R}{100}$

Subtract 1 from both sides:

$1.0723805 - 1 \approx \frac{R}{100}$

$0.0723805 \approx \frac{R}{100}$

Multiply both sides by 100 to find R:

$R \approx 0.0723805 \times 100$

$R \approx 7.23805$

The annual rate of increase is approximately $7.238\%$. We can round this to one or two decimal places.

Rounding to two decimal places, $R \approx 7.24\%$.

Thus, the annual rate of increase in production is approximately $\mathbf{7.24\%}$.