Chapter 7 Comparing Quantities (Class 7 - Maths NCERT Exemplar Solutions)

Given:

Height of the first person = $1.50$ m

Height of the second person = $75$ cm

To Find:

The ratio of the heights of the two persons.

Solution:

To find the ratio, both quantities must be in the same unit. We know that:

$1$ m = $100$ cm

So, height of the first person in cm = $1.50 \times 100$ cm = $150$ cm

Now, the ratio of the heights is:

Ratio = $\frac{150 \text{ cm}}{75 \text{ cm}}$

Dividing both numerator and denominator by $75$:

Ratio = $\frac{\cancel{150}^2}{\cancel{75}_1}$ = $\frac{2}{1}$

Therefore, the ratio is $2 : 1$.

Correct Option: (c)

Given:

Total number of children = $50$

Number of boys = $20$

To Find:

The percentage of girls in the class.

Solution:

First, we find the number of girls in the class:

Number of girls = Total children - Number of boys

Number of girls = $50 - 20 = 30$

Now, we calculate the percentage of girls:

Percentage of girls = $\left( \frac{\text{Number of girls}}{\text{Total children}} \times 100 \right) \%$

Percentage of girls = $\left( \frac{30}{50} \times 100 \right) \%$

Percentage of girls = $(30 \times 2) \%$ = $60 \%$

Correct Option: (a)

Alternate Solution:

Percentage of boys = $\left( \frac{20}{50} \times 100 \right) \% = 40 \%$

Percentage of girls = $100 \% - \text{Percentage of boys}$

Percentage of girls = $100 \% - 40 \% = 60 \%$

Given:

Principal ($P$) = $\textsf{₹} 5000$

Rate of interest ($R$) = $15 \%$ per annum

Time ($T$) = $1$ month = $\frac{1}{12}$ year

To Find:

The simple interest ($I$).

Solution:

The formula for Simple Interest is:

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

Substituting the given values:

$I = \frac{5000 \times 15 \times \frac{1}{12}}{100}$

$I = \frac{50 \times 15}{12}$

$I = \frac{750}{12}$

$I = \textsf{₹} 62.50$

Correct Option: (d)

Solution:

When two ratios $a : b$ and $c : d$ are equivalent (i.e., $a : b = c : d$), then the four quantities $a, b, c, d$ are said to be in proportion.

Therefore, the blank should be filled with proportion.

To Find:

$40 \%$ of $250$ km.

Solution:

$40 \%$ of $250 = \frac{40}{100} \times 250$

Simplifying the expression:

$= \frac{4}{10} \times 250$

$= 4 \times 25$

$= 100$

So, $40 \%$ of $250$ km is $100$ km.

Solution:

Let the total distance of the journey be $x$ km.

According to the statement:

$25 \%$ of $x = 800$

$\frac{25}{100} \times x = 800$

$\frac{1}{4} \times x = 800$

$x = 800 \times 4$

$x = 3200$ km

Since the calculated total distance is $3200$ km and the statement claims it is $3000$ km, the statement is False.

Solution:

To convert a decimal into a percentage, we multiply it by $100$.

$0.05 \times 100 = \frac{5}{100} \times 100$

$= 5 \%$

Since $0.05$ is indeed equal to $5 \%$, the statement is True.

Given:

Selling Price ($S.P.$) of the sofa set = $\textsf{₹} 9600$

Profit percentage = $20 \%$

To Find:

Cost Price ($C.P.$) of the sofa set.

Solution:

We know the formula to find $C.P.$ when $S.P.$ and Profit $\%$ are given is:

$C.P. = \frac{100}{100 + \text{Profit} \%} \times S.P.$

Substituting the given values into the formula:

$C.P. = \frac{100}{100 + 20} \times 9600$

$C.P. = \frac{100}{120} \times 9600$

$C.P. = \frac{10}{12} \times 9600$

$C.P. = 10 \times \frac{\cancel{9600}^{800}}{\cancel{12}_{1}}$

$C.P. = 10 \times 800$

$C.P. = \textsf{₹} 8000$

Therefore, the cost price of the sofa set is $\textsf{₹} 8000$.

Given:

Principal ($P$) = $\textsf{₹} 75000$

Amount ($A$) = $\textsf{₹} 80000$

To Find:

Interest ($I$).

Solution:

We know that Amount is the sum of Principal and Interest. Therefore, Interest can be calculated as:

$\text{Interest} = \text{Amount} - \text{Principal}$

$\text{Interest} = 80000 - 75000$

$\text{Interest} = \textsf{₹} 5000$

Therefore, the interest paid by John is $\textsf{₹} 5000$.

Given:

Principal ($P$) = $\textsf{₹} 2500$

Simple Interest ($I$) = $\textsf{₹} 1500$

Time ($T$) = $4$ years

To Find:

Rate of interest per annum ($R$).

Solution:

We know the formula for Simple Interest is:

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

To find the Rate ($R$), we can rearrange the formula as:

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

Substituting the given values:

$R = \frac{1500 \times 100}{2500 \times 4}$

Using the cancellation method:

$R = \frac{1500 \times \cancel{100}^1}{\cancel{2500}_{25} \times 4}$

$R = \frac{1500}{25 \times 4}$

$R = \frac{1500}{100}$

$R = \frac{\cancel{1500}^{15}}{\cancel{100}_1}$

$R = 15 \%$

Therefore, the rate of interest per annum is $15 \%$.

Given:

From the provided graphic:

Top speed of Cheetah = $31.3$ m/s

Top speed of Tortoise = $0.08$ m/s

Time taken ($t$) = $1$ minute = $60$ seconds

To Find:

How much farther the cheetah travels compared to the tortoise (Difference in distance).

Solution:

We know that $\text{Distance} = \text{Speed} \times \text{Time}$.

1. Distance traveled by the Cheetah:

$D_1 = 31.3 \times 60$

$D_1 = 1878$ m

2. Distance traveled by the Tortoise:

$D_2 = 0.08 \times 60$

$D_2 = 4.8$ m

3. Difference in distance:

$\text{Difference} = D_1 - D_2$

$\text{Difference} = 1878 - 4.8$

$\text{Difference} = 1873.2$ m

Therefore, the cheetah travels $1873.2$ m farther than the tortoise.

Given:

Total length = $700$ m

Percentage = $20 \%$

To Find:

The value of $20 \%$ of $700$ m.

Solution:

We calculate the percentage as follows:

$20 \%$ of $700 = \frac{20}{100} \times 700$

$= 20 \times \frac{\cancel{700}^7}{\cancel{100}_1}$

$= 20 \times 7$

$= 140$ m

Correct Option: (d)

Given:

Total Income = $\textsf{₹} 1,60,000$

House rent percentage = $15 \%$

Child's education percentage = $10 \%$ of the remainder

To Find:

The money left with Gayatri.

Solution:

First, calculate the house rent:

House Rent = $15 \%$ of $1,60,000 = \frac{15}{100} \times 1,60,000$

$= 15 \times 1600 = \textsf{₹} 24,000$

Now, calculate the remainder income:

Remainder = $1,60,000 - 24,000 = \textsf{₹} 1,36,000$

Next, calculate the money spent on education:

Education Expense = $10 \%$ of $1,36,000 = \frac{10}{100} \times 1,36,000$

$= \textsf{₹} 13,600$

Finally, the money left is:

Money Left = Remainder - Education Expense

Money Left = $1,36,000 - 13,600$

Money Left = $\textsf{₹} 1,22,400$

Correct Option: (c)

Given:

Ratio of Fatima's income to her savings = $4 : 1$

To Find:

The percentage of money saved by her.

Solution:

The given ratio is $4 : 1$.

To find the percentage of savings, we first determine the total parts represented by the ratio:

Total parts = $4 + 1 = 5$

Now, the savings represent $1$ part out of the total $5$ parts.

Percentage of money saved = $\left( \frac{\text{Savings part}}{\text{Total parts}} \times 100 \right) \%$

Percentage of money saved = $\left( \frac{1}{5} \times 100 \right) \%$

Percentage of money saved = $\frac{1}{\cancel{5}_1} \times \cancel{100}^{20} \%$

Percentage of money saved = $20 \%$

Correct Option: (a)

Alternate Solution:

If we consider the ratio components as parts of a whole where the total income is divided into $5$ equal units ($4$ units for expenditure and $1$ unit for savings):

Income = $5$ units

Savings = $1$ unit

Percentage saved = $\frac{1}{5} \times 100 \% = 20 \%$

Solution:

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

$0.07 = 0.07 \times 100 \%$

$= \frac{7}{100} \times 100 \%$

$= 7 \%$

Correct Option: (b)

Given:

Scouts from Gujarat = $40 \%$ of total scouts.

Scouts from Ahmedabad = $20 \%$ of those from Gujarat.

To Find:

Percentage of total scouts from Ahmedabad.

Solution:

Percentage of scouts from Ahmedabad = $20 \%$ of $40 \%$

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

$= \frac{1}{5} \times 40 \%$

$= 8 \%$

Correct Option: (c)

Solution:

Let $x \%$ of $\textsf{₹} 4500$ be $\textsf{₹} 9000$.

$\frac{x}{100} \times 4500 = 9000$

$x \times 45 = 9000$

$x = \frac{9000}{45}$

$x = 200$

So, the percentage is $200 \%$.

Correct Option: (a)

Solution:

Converting decimal $5.2$ to percentage:

$5.2 \times 100 \%$

$= 520 \%$

Correct Option: (c)

Solution:

To convert the ratio $3 : 8$ into percentage:

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

$= \frac{300}{8} \%$

$= 37.5 \%$

Correct Option: (b)

Given:

Percentage = $225 \%$

To Find:

The equivalent ratio of the given percentage.

Solution:

To convert a percentage into a ratio, we first write it as a fraction by dividing by $100$ and then simplify it.

$225 \% = \frac{225}{100}$

Dividing both numerator and denominator by their highest common factor, which is $25$:

$\frac{\cancel{225}^{9}}{\cancel{100}_{4}} = \frac{9}{4}$

The fraction $\frac{9}{4}$ can be written in the form of a ratio as $9 : 4$.

Correct Option: (a)

Given:

Cost Price ($C.P.$) of the bicycle = $\textsf{₹} 1800$

Profit percentage = $12 \%$

To Find:

Selling Price ($S.P.$) of the bicycle.

Solution:

First, we calculate the profit amount:

$\text{Profit} = 12 \% \text{ of } 1800$

$\text{Profit} = \frac{12}{100} \times 1800$

$\text{Profit} = 12 \times 18 = \textsf{₹} 216$

Now, we find the Selling Price using the formula:

$S.P. = C.P. + \text{Profit}$

$S.P. = 1800 + 216$

$S.P. = \textsf{₹} 2016$

Correct Option: (b)

Given:

Cost Price ($C.P.$) = $\textsf{₹} 800$

Selling Price ($S.P.$) = $\textsf{₹} 1600$

To Find:

Profit percentage.

Solution:

First, calculate the profit amount:

$\text{Profit} = S.P. - C.P.$

$\text{Profit} = 1600 - 800 = \textsf{₹} 800$

Now, calculate the profit percentage:

$\text{Profit} \% = \left( \frac{\text{Profit}}{C.P.} \times 100 \right) \%$

$\text{Profit} \% = \left( \frac{800}{800} \times 100 \right) \%$

$\text{Profit} \% = 1 \times 100 \% = 100 \%$

Correct Option: (a)

Given:

$C.P.$ of buffalo = $\textsf{₹} 44000$, Loss on buffalo = $5 \%$

$C.P.$ of cow = $\textsf{₹} 18000$, Profit on cow = $10 \%$

To Find:

Net result (Total Profit or Total Loss).

Solution:

Step 1: Calculate loss on buffalo

$\text{Loss} = 5 \% \text{ of } 44000$

$\text{Loss} = \frac{5}{100} \times 44000 = 5 \times 440 = \textsf{₹} 2200$

Step 2: Calculate profit on cow

$\text{Profit} = 10 \% \text{ of } 18000$

$\text{Profit} = \frac{10}{100} \times 18000 = \textsf{₹} 1800$

Step 3: Calculate net result

Since the loss ($\textsf{₹} 2200$) is greater than the profit ($\textsf{₹} 1800$), there is a net loss.

$\text{Net Loss} = \text{Total Loss} - \text{Total Profit}$

$\text{Net Loss} = 2200 - 1800 = \textsf{₹} 400$

Correct Option: (c)

Solution:

Let Raman's income be $\textsf{₹} 100$.

Mohan's income is $25 \%$ more than Raman's income.

Mohan's income = $100 + (25 \% \text{ of } 100) = 100 + 25 = \textsf{₹} 125$.

Now, we need to find how much Raman's income is less than Mohan's income in terms of percentage.

Difference in income = $125 - 100 = \textsf{₹} 25$.

Percentage less = $\left( \frac{\text{Difference}}{\text{Mohan's income}} \times 100 \right) \%$

Percentage less = $\left( \frac{25}{125} \times 100 \right) \%$

$\frac{\cancel{25}^{1}}{\cancel{125}_{5}} \times 100 = \frac{100}{5} = 20 \%$

Correct Option: (c)

Given:

Principal ($P$) = $\textsf{₹} 30000$

Time ($T$) = $3$ years

Rate ($R$) = $15 \%$ p.a.

To Find:

Simple Interest ($I$).

Solution:

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

$I = \frac{30000 \times 15 \times 3}{100}$

$I = 300 \times 45$

$I = \textsf{₹} 13500$

Correct Option: (d)

Given:

Principal ($P$) = $\textsf{₹} 3000$

Time ($T$) = $2$ years

Rate ($R$) = $11 \%$ p.a.

To Find:

Total Amount ($A$).

Solution:

First, calculate the interest:

$I = \frac{3000 \times 11 \times 2}{100} = 30 \times 22 = \textsf{₹} 660$

Now, calculate the Amount:

$A = P + I$

$A = 3000 + 660 = \textsf{₹} 3660$

Correct Option: (b)

Given:

Principal ($P$) = $\textsf{₹} 12000$

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

Time ($T$) = $1$ month = $\frac{1}{12}$ year

To Find:

Interest ($I$).

Solution:

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

$I = \frac{12000 \times 10 \times \frac{1}{12}}{100}$

$I = \frac{12000 \times 10}{100 \times 12}$

$I = \frac{1000 \times 12 \times 10}{100 \times 12}$

$I = \frac{\cancel{12000}^{1000}}{\cancel{12}_{1}} \times \frac{10}{100}$

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

$I = \textsf{₹} 100$

Correct Option: (c)

Given:

For Rajni: Principal ($P_1$) = $\textsf{₹} 3000$, Rate ($R$) = $10 \%$ p.a., Time ($T_1$) = $3$ years.

For Mohini: Principal ($P_2$) = $\textsf{₹} 4000$, Rate ($R$) = $10 \%$ p.a., Time ($T_2$) = $2\frac{1}{2}$ years = $2.5$ years.

To Find:

The difference of the amounts received by them.

Solution:

1. Calculation for Rajni:

Simple Interest ($I_1$) = $\frac{3000 \times 10 \times 3}{100} = \textsf{₹} 900$

Amount received by Rajni ($A_1$) = $P_1 + I_1$

$A_1 = 3000 + 900 = \textsf{₹} 3900$

2. Calculation for Mohini:

Simple Interest ($I_2$) = $\frac{4000 \times 10 \times 2.5}{100} = 40 \times 25 = \textsf{₹} 1000$

Amount received by Mohini ($A_2$) = $P_2 + I_2$

$A_2 = 4000 + 1000 = \textsf{₹} 5000$

3. Difference of the amounts:

Difference = $A_2 - A_1$

Difference = $5000 - 3900 = \textsf{₹} 1100$

Correct Option: (d)

Given:

$90 \%$ of $x = 315$ km

To Find:

The value of $x$.

Solution:

According to the question:

$\frac{90}{100} \times x = 315$

$x = \frac{315 \times 100}{90}$

$x = \frac{315 \times 10}{9}$

Dividing $315$ by $9$:

$x = 35 \times 10$

$x = 350$ km

Correct Option: (b)

Given:

Selling Price ($S.P.$) = $\textsf{₹} 329$

Loss percentage = $6 \%$

To Find:

Cost Price ($C.P.$).

Solution:

We use the formula:

$C.P. = \frac{100}{100 - \text{Loss} \%} \times S.P.$

$C.P. = \frac{100}{100 - 6} \times 329$

$C.P. = \frac{100}{94} \times 329$

$C.P. = \frac{32900}{94}$

$C.P. = \textsf{₹} 350$

Correct Option: (d)

Solution:

The expression is: $\frac{25\% \times 50\% \times 100\%}{25 \times 50}$

Converting percentages into fractions:

$= \frac{\frac{25}{100} \times \frac{50}{100} \times \frac{100}{100}}{25 \times 50}$

$= \frac{\frac{1}{4} \times \frac{1}{2} \times 1}{1250}$

$= \frac{1/8}{1250}$

$= \frac{1}{8 \times 1250}$

$= \frac{1}{10000}$

Now, to convert this into percentage, we multiply by $100$:

$= \left( \frac{1}{10000} \times 100 \right) \%$

$= \frac{1}{100} \% = 0.01 \%$

Correct Option: (c)

Given:

Simple Interest ($I$) = $\textsf{₹} 126$

Time ($T$) = $2$ years

Rate ($R$) = $14 \%$ p.a.

To Find:

Principal (Sum, $P$).

Solution:

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

$P = \frac{126 \times 100}{14 \times 2}$

$P = \frac{12600}{28}$

Dividing by $14$:

$P = \frac{\cancel{126}^{9} \times 100}{\cancel{14}_{1} \times 2}$

$P = \frac{900}{2}$

$P = \textsf{₹} 450$

Correct Option: (c)

Solution:

The given figure is a $10 \times 10$ square grid, making a total of $100$ small squares.

Let's count the shaded squares ring by ring:

1. Outer shaded border: $10 + 10 + 8 + 8 = 36$ squares.

2. Inner shaded ring: $6 + 6 + 4 + 4 = 20$ squares.

3. Central shaded block: $2 \times 2 = 4$ squares.

Total shaded squares = $36 + 20 + 4 = 60$ squares.

Now, the number of unshaded squares = Total squares - Shaded squares

Unshaded squares = $100 - 60 = 40$ squares.

Percentage of unshaded region = $\left( \frac{40}{100} \times 100 \right) \% = 40 \%$

Correct Option: (c)

Solution:

The grid is a $10 \times 10$ square, containing $100$ small squares in total.

By observing the image, there are three distinct shaded blocks:

1. Top right shaded block: $4 \times 3 = 12$ squares.

2. Center shaded block: $4 \times 3 = 12$ squares.

3. Bottom left shaded block: $4 \times 3 = 12$ squares.

Total shaded squares = $12 + 12 + 12 = 36$ squares.

Percentage of shaded region = $\left( \frac{36}{100} \times 100 \right) \% = 36 \%$

Correct Option: (a)

Given:

Ratio = $2 : 3$

To Find:

Percentage equivalent of the given ratio.

Solution:

To convert a ratio into a percentage, we first express the ratio as a fraction and then multiply it by $100$.

Fraction form = $\frac{2}{3}$

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

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

The value can be written as $66\frac{2}{3} \%$ or approximately $66.67 \%$.

Thus, $2 : 3 = $ $66\frac{2}{3}$ %.

Given:

Percentage = $18\frac{3}{4} \%$

To Find:

The equivalent ratio.

Solution:

First, convert the mixed fraction into an improper fraction:

$18\frac{3}{4} = \frac{(18 \times 4) + 3}{4} = \frac{72 + 3}{4} = \frac{75}{4}$

Now, to remove the percentage sign, divide the fraction by $100$:

Fraction = $\frac{75}{4} \times \frac{1}{100}$

Fraction = $\frac{75}{400}$

Simplifying by dividing both numerator and denominator by $25$:

Ratio = $\frac{\cancel{75}^3}{\cancel{400}_{16}} = \frac{3}{16}$

In ratio form, it is $3 : 16$.

Thus, $18\frac{3}{4} \%$ = $3$ : $16$.

Given:

Total Amount = $\textsf{₹} 360$

Percentage = $30 \%$

To Find:

The value of $30 \%$ of the given amount.

Solution:

Value = $30 \%$ of $360$

Value = $\frac{30}{100} \times 360$

Value = $\frac{3 \times \cancel{10}}{10 \times \cancel{10}} \times 360$

Value = $3 \times 36$

Value = $\textsf{₹} 108$.

Given:

Distance = $50$ km

Percentage = $120 \%$

To Find:

The result of the percentage calculation.

Solution:

Result = $120 \%$ of $50$ km

Result = $\frac{120}{100} \times 50$

Result = $120 \times \frac{\cancel{50}^1}{\cancel{100}_2}$

Result = $\frac{120}{2}$

Result = $60$ km.

Thus, $120 \%$ of $50$ km = $60$ km.

Given:

Decimal value = $2.5$

To Find:

Equivalent percentage.

Solution:

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

Percentage = $2.5 \times 100 \%$

Percentage = $\frac{25}{10} \times 100 \%$

Percentage = $25 \times 10 \%$

Percentage = $250 \%$

Thus, $2.5$ = $250$ %.

Given:

Fraction = $\frac{8}{5}$

To Find:

Equivalent percentage.

Solution:

To convert a fraction into a percentage, we multiply it by $100$.

Percentage = $\left( \frac{8}{5} \times 100 \right) \%$

Percentage = $8 \times \frac{\cancel{100}^{20}}{\cancel{5}_1} \%$

Percentage = $8 \times 20 \%$

Percentage = $160 \%$

Thus, $\frac{8}{5}$ = $160$ %.

Solution:

By definition, a fraction with its denominator $100$ is called a per cent. The word "percent" is derived from the Latin "per centum," meaning "by the hundred."

Example: $\frac{5}{100} = 5 \%$

Therefore, the blank should be filled with fraction.

Given:

Part value = $15$ kg

Total value = $50$ kg

To Find:

Percentage of $15$ kg in $50$ kg.

Solution:

Percentage = $\left( \frac{\text{Part Value}}{\text{Total Value}} \times 100 \right) \%$

Substituting the values:

Percentage = $\left( \frac{15}{50} \times 100 \right) \%$

Percentage = $15 \times \frac{\cancel{100}^{2}}{\cancel{50}_{1}} \%$

Percentage = $15 \times 2 \%$

Percentage = $30 \%$

Thus, $15$ kg is $30$ % of $50$ kg.

Given:

Original weight = $60$ kg

New weight = $66$ kg

To Find:

Percentage increase in weight.

Solution:

First, find the increase in weight:

$\text{Increase} = \text{New weight} - \text{Original weight}$

$\text{Increase} = 66 - 60 = 6$ kg

Now, calculate the percentage increase:

$\text{Percentage Increase} = \left( \frac{\text{Increase}}{\text{Original weight}} \times 100 \right) \%$

$\text{Percentage Increase} = \left( \frac{6}{60} \times 100 \right) \%$

$\text{Percentage Increase} = \left( \frac{1}{10} \times 100 \right) \% = 10 \%$

Thus, the increase in weight is $10$ %.

Given:

Total students = $50$

Percentage of absent students = $8 \%$

To Find:

Number of students present.

Solution:

First, find the number of absent students:

Number of absent students = $8 \%$ of $50$

Number of absent students = $\frac{8}{100} \times 50$

Number of absent students = $\frac{8}{\cancel{2}} = 4$

Now, find the number of present students:

Number of present students = $\text{Total students} - \text{Absent students}$

Number of present students = $50 - 4 = 46$

Alternate Solution:

Percentage of present students = $100 \% - 8 \% = 92 \%$

Number of present students = $92 \%$ of $50$

Number of present students = $\frac{92}{100} \times 50 = \frac{92}{2} = 46$

Thus, the number of students present was $46$.

Given:

Marks obtained = $440$

Total marks = $500$

To Find:

Percentage of marks secured.

Solution:

Percentage = $\left( \frac{\text{Marks obtained}}{\text{Total marks}} \times 100 \right) \%$

Percentage = $\left( \frac{440}{500} \times 100 \right) \%$

Percentage = $\frac{440}{\cancel{500}_{5}} \times \cancel{100}^{1} \%$

Percentage = $\frac{440}{5} \%$

Percentage = $88 \%$

Thus, she secured $88$ % marks.

Given:

Total deposit = $\textsf{₹} 1500$

Percentage withdrawn = $40 \%$

To Find:

The balance remaining in the account.

Solution:

First, calculate the amount withdrawn:

Amount withdrawn = $40 \%$ of $1500$

Amount withdrawn = $\frac{40}{100} \times 1500$

Amount withdrawn = $40 \times 15 = \textsf{₹} 600$

Now, calculate the remaining balance:

$\text{Balance} = \text{Total deposit} - \text{Amount withdrawn}$

Balance = $\textsf{₹} 900$

Alternate Solution:

Percentage of balance remaining = $100 \% - 40 \% = 60 \%$

Balance = $60 \%$ of $1500$

Balance = $\frac{60}{100} \times 1500 = 60 \times 15 = \textsf{₹} 900$

Thus, the balance in her account is $\textsf{₹} 900$.

To Find:

A value which is $50 \%$ greater than $60$.

Solution:

First, find $50 \%$ of $60$:

$50 \% \text{ of } 60 = \frac{50}{100} \times 60 = \frac{1}{2} \times 60 = 30$

Now, add this value to the original number $60$:

Value = $60 + 30$

Value = $90$

Thus, $90$ is $50 \%$ more than $60$.

Given:

Selling Price ($S.P.$) = $\textsf{₹} 75$

Loss = $\textsf{₹} 8$

To Find:

Cost Price ($C.P.$) of the bat.

Solution:

We know that in case of a loss:

$C.P. = S.P. + \text{Loss}$

Substituting the values:

$C.P. = 75 + 8$

$C.P. = \textsf{₹} 83$

Thus, the cost price of the bat is $\textsf{₹} 83$.

Given:

Original price of $3$ kg sugar = $\textsf{₹} 120$

Percentage decrease = $20 \%$

To Find:

The new price of $3$ kg sugar.

Solution:

First, calculate the amount of decrease:

Decrease = $20 \%$ of $120$

Decrease = $\frac{20}{100} \times 120 = \frac{1}{5} \times 120 = \textsf{₹} 24$

Now, calculate the new price:

$\text{New Price} = \text{Original Price} - \text{Decrease}$

New Price = $120 - 24$

New Price = $\textsf{₹} 96$

Thus, the new price will be $\textsf{₹} 96$.

Given:

Cost Price ($C.P.$) of the cow = $\textsf{₹} 9000$

Loss = $\textsf{₹} 900$

To Find:

Selling Price ($S.P.$) of the cow.

Solution:

We know that the formula for Selling Price in case of loss is:

$S.P. = C.P. - \text{Loss}$

Substituting the values:

$S.P. = 9000 - 900$

$S.P. = \textsf{₹} 8100$

Thus, the selling price of the cow is $\textsf{₹} 8100$.

Given:

Cost Price ($C.P.$) = $\textsf{₹} 700$

Selling Price ($S.P.$) = $\textsf{₹} 750$

To Find:

Profit percentage.

Solution:

First, we calculate the profit amount:

$\text{Profit} = S.P. - C.P.$

$\text{Profit} = 750 - 700 = \textsf{₹} 50$

Now, calculate the profit percentage:

$\text{Profit } \% = \left( \frac{\text{Profit}}{C.P.} \times 100 \right) \%$

$\text{Profit } \% = \left( \frac{50}{700} \times 100 \right) \%$

$\text{Profit } \% = \frac{50}{7} \%$

$\text{Profit } \% = 7\frac{1}{7} \%$ or approximately $7.14 \%$.

Thus, she earns a profit of $7\frac{1}{7}$ %.

Given:

Cost Price ($C.P.$) = $\textsf{₹} 400$

Selling Price ($S.P.$) = $\textsf{₹} 440$

To Find:

Type of transaction (Profit/Loss) and its percentage.

Solution:

Since $S.P. > C.P.$, there is a profit.

$\text{Profit} = S.P. - C.P.$

$\text{Profit} = 440 - 400 = \textsf{₹} 40$

Now, calculate profit percentage:

$\text{Profit } \% = \left( \frac{\text{Profit}}{C.P.} \times 100 \right) \%$

$\text{Profit } \% = \left( \frac{40}{400} \times 100 \right) \%$

$\text{Profit } \% = \frac{\cancel{40}^1}{\cancel{400}_{10}} \times 100 \%$

$\text{Profit } \% = 10 \%$

Thus, her profit % is 10.

Given:

Cost Price ($C.P.$) = $\textsf{₹} 60$

Selling Price ($S.P.$) = $\textsf{₹} 54$

To Find:

Type of transaction (Profit/Loss) and its percentage.

Solution:

Since $C.P. > S.P.$, there is a loss.

$\text{Loss} = C.P. - S.P.$

$\text{Loss} = 60 - 54 = \textsf{₹} 6$

Now, calculate loss percentage:

$\text{Loss } \% = \left( \frac{\text{Loss}}{C.P.} \times 100 \right) \%$

$\text{Loss } \% = \left( \frac{6}{60} \times 100 \right) \%$

$\text{Loss } \% = \frac{\cancel{6}^1}{\cancel{60}_{10}} \times 100 \%$

$\text{Loss } \% = 10 \%$

Thus, his loss % is 10.

Given:

Purchase price = $\textsf{₹} 50,59,700$

Repair costs = $\textsf{₹} 40,300$

Desired Profit = $5 \%$

To Find:

The Selling Price ($S.P.$).

Solution:

First, calculate the total Cost Price ($C.P.$) including repairs:

$\text{Total } C.P. = \text{Purchase Price} + \text{Repair costs}$

Total $C.P. = \textsf{₹} 51,00,000$

Now, calculate the profit amount:

$\text{Profit} = 5 \% \text{ of } 51,00,000$

$\text{Profit} = \frac{5}{100} \times 51,00,000 = 5 \times 51,000 = \textsf{₹} 2,55,000$

Now, find the Selling Price:

$S.P. = C.P. + \text{Profit}$

$S.P. = 51,00,000 + 2,55,000$

$S.P. = \textsf{₹} 53,55,000$

Thus, she should sell the house for $\textsf{₹} 53,55,000$.

Given:

$C.P.$ of $20$ lemons = $\textsf{₹} 10$

$S.P.$ of $5$ lemons = $\textsf{₹} 3$

To Find:

Nature of transaction and the percentage.

Solution:

First, find the Cost Price of $1$ lemon:

$C.P. \text{ of } 1 \text{ lemon} = \frac{10}{20} = \textsf{₹} 0.50$

Next, find the Selling Price of $1$ lemon:

$S.P. \text{ of } 1 \text{ lemon} = \frac{3}{5} = \textsf{₹} 0.60$

Since $S.P. > C.P.$, there is a profit.

$\text{Profit per lemon} = 0.60 - 0.50 = \textsf{₹} 0.10$

Now, calculate profit percentage:

$\text{Profit } \% = \left( \frac{\text{Profit}}{C.P.} \times 100 \right) \%$

$\text{Profit } \% = \left( \frac{0.10}{0.50} \times 100 \right) \%$

$\text{Profit } \% = \frac{1}{5} \times 100 \% = 20 \%$

Thus, profit in the transaction is 20 %.

Given:

Total oranges = $120$

Rate of purchase = $\textsf{₹} 4$ per orange

Sales: $60 \%$ at $\textsf{₹} 5$ each, remainder at $\textsf{₹} 3.50$ each.

To Find:

Net Profit or Loss percentage.

Solution:

1. Total Cost Price:

$C.P. = 120 \times 4 = \textsf{₹} 480$

2. Total Selling Price:

Oranges sold at ₹ 5 = $60 \%$ of $120$

$= \frac{60}{100} \times 120 = 72 \text{ oranges}$

$S.P. \text{ part 1} = 72 \times 5 = \textsf{₹} 360$

Remaining oranges = $120 - 72 = 48 \text{ oranges}$

$S.P. \text{ part 2} = 48 \times 3.50 = \textsf{₹} 168$

Total $S.P. = 360 + 168 = \textsf{₹} 528$

3. Profit and Percentage:

Since $S.P. > C.P.$, there is a profit.

$\text{Profit} = 528 - 480 = \textsf{₹} 48$

$\text{Profit } \% = \left( \frac{48}{480} \times 100 \right) \% = 10 \%$

Thus, his profit is 10 %.

Given:

Purchase quantity = $20$ kg at $\textsf{₹} 50$/kg

Rotten apples = $5 \%$

Selling rate = $\textsf{₹} 60$/kg

To Find:

Net Profit or Loss percentage.

Solution:

1. Total Cost Price:

$C.P. = 20 \times 50 = \textsf{₹} 1000$

2. Total Selling Price:

Rotten apples = $5 \%$ of $20$ kg

$= \frac{5}{100} \times 20 = 1 \text{ kg}$

Remaining apples = $20 - 1 = 19 \text{ kg}$

$S.P. = 19 \text{ kg} \times 60 = \textsf{₹} 1140$

3. Profit and Percentage:

Since $S.P. > C.P.$, there is a profit.

$\text{Profit} = 1140 - 1000 = \textsf{₹} 140$

$\text{Profit } \% = \left( \frac{140}{1000} \times 100 \right) \% = 14 \%$

Thus, his profit is 14 %.

Given:

Principal ($P$) = $\textsf{₹} 3000$

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

Time ($T$) = $3$ years

To Find:

Simple Interest ($I$).

Solution:

The formula for Simple Interest is:

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

Substituting the given values:

$I = \frac{3000 \times 10 \times 3}{100}$

$I = 30 \times 10 \times 3$

$I = \textsf{₹} 900$

Thus, the interest is $\textsf{₹} 900$.

Given:

Principal ($P$) = $\textsf{₹} 20,000$

Rate ($R$) = $8 \%$ p.a.

Time ($T$) = $6$ months = $\frac{6}{12}$ year = $0.5$ year

To Find:

Total Amount ($A$).

Solution:

First, calculate the Simple Interest ($I$):

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

$I = \frac{20000 \times 8 \times 0.5}{100}$

$I = 200 \times 4$

$I = \textsf{₹} 800$

Now, calculate the Amount ($A$):

$A = P + I$

$A = 20000 + 800$

$A = \textsf{₹} 20,800$

Thus, the amount obtained is $\textsf{₹} 20,800$.

Given:

Principal ($P$) = $\textsf{₹} 12500$

Rate ($R$) = $18 \%$ p.a.

Time ($T$) = $2$ years $4$ months = $2\frac{4}{12}$ years = $2\frac{1}{3}$ years = $\frac{7}{3}$ years

To Find:

Simple Interest ($I$).

Solution:

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

$I = \frac{12500 \times 18 \times \frac{7}{3}}{100}$

$I = \frac{125 \times 18 \times 7}{3}$

$I = 125 \times 6 \times 7$

$I = 125 \times 42$

$I = \textsf{₹} 5250$

Thus, the interest is $\textsf{₹} 5250$.

Given:

Part value = $25$ ml

Total value = $5$ litres

To Find:

Percentage of $25$ ml in $5$ litres.

Solution:

First, convert litres into millilitres ($1$ litre = $1000$ ml):

$5$ litres = $5000$ ml

Now, calculate the percentage:

Percentage = $\left( \frac{25}{5000} \times 100 \right) \%$

Percentage = $\frac{25}{50} \%$

Percentage = $0.5 \%$

Thus, $25$ ml is $0.5$ per cent of $5$ litres.

Solution:

Let the value of $A$ be $100$.

Given, $A$ is increased by $20 \%$ to get $B$:

$B = 100 + 20 = 120$

Now, $B$ is decreased by $50 \%$ to get $C$:

$C = 120 - 60 = 60$

We need to find what percent of $A$ is $C$:

$\text{Percentage} = \frac{C}{A} \times 100$

$\text{Percentage} = \frac{60}{100} \times 100 = 60 \%$

Thus, $60$ % of A is equal to C.

Solution:

In the formula for simple interest:

$T$ is Time period (usually in years)

$R\%$ is Rate of interest per annum and

$P$ is Principal.

Given:

Principal ($P$) = $\textsf{₹} 2100$

Rate ($R$) = $8 \%$ p.a.

To Find:

The difference in interest for $3$ years and $2$ years.

Solution:

The difference in interest for $3$ years and $2$ years is simply the interest for $1$ year ($3 - 2 = 1$).

Difference = Interest for $1$ year

Difference = $\frac{P \times R \times 1}{100}$

Difference = $\frac{2100 \times 8 \times 1}{100}$

Difference = $21 \times 8 = \textsf{₹} 168$

Thus, the difference of interest is $\textsf{₹} 168$.

Solution:

To convert any fraction into a percentage, we multiply the fraction by $100$ and attach the $\%$ symbol.

Example: To convert $\frac{1}{4}$ to percent: $\frac{1}{4} \times 100 = 25 \%$

Thus, the blank should be filled with multiply.

Solution:

To convert a decimal into a percentage, we multiply the decimal by $100$. Multiplying by $100$ is equivalent to shifting the decimal point two places to the right.

Example: $0.125 = 0.125 \times 100 \% = 12.5 \%$

Thus, the blank should be filled with right.

Given:

Principal ($P$) = $\textsf{₹} 2000$

Rate ($R$) = $6 \%$ p.a.

Time period 1 ($T_1$) = $1\frac{1}{2}$ years = $1.5$ years

Time period 2 ($T_2$) = $2$ years

Solution:

First, calculate interest for $1.5$ years ($I_1$):

$I_1 = \frac{2000 \times 6 \times 1.5}{100} = 20 \times 9 = \textsf{₹} 180$

Now, calculate interest for $2$ years ($I_2$):

$I_2 = \frac{2000 \times 6 \times 2}{100} = 20 \times 12 = \textsf{₹} 240$

Adding both interests:

Total Interest = $180 + 240 = \textsf{₹} 420$

Since the sum of the two interests equals $\textsf{₹} 420$, the missing word is sum or total.

Solution:

To convert the decimal $6.5$ into a percentage, we multiply it by $100$:

$6.5 \times 100 = 650 \%$

Comparing $650 \%$ with $100 \%$:

$650 \% > 100 \%$

Therefore, the value of $6.5$ is greater (or more) than $100 \%$.

Solution:

The number line is divided into $10$ equal parts between $0$ and $1$. Each part represents an increment of $\frac{1}{10}$ or $0.1$ or $10 \%$. We will convert the given values to find the missing blanks.

1st Mark: $10 \%$ is $\frac{1}{10}$ and $0.1$. (Given)

2nd Mark: Percent is $20 \%$. Fraction = $\frac{20}{100} = \frac{1}{5}$. Decimal = $0.2$.

3rd Mark: Percent is $30 \%$. Fraction = $\frac{3}{10}$. Decimal = $\frac{3}{10} = 0.3$.

4th Mark: Decimal is $0.4$. Percent = $0.4 \times 100 = 40 \%$. Fraction = $\frac{4}{10} = \frac{2}{5}$.

5th Mark: Percent is $50 \%$. Fraction = $\frac{50}{100} = \frac{1}{2}$. Decimal = $0.5$.

6th Mark: Fraction is $\frac{3}{5}$. Percent = $\frac{3}{5} \times 100 = 60 \%$. Decimal = $0.6$.

7th Mark: Decimal is $0.7$. Percent = $0.7 \times 100 = 70 \%$. Fraction = $\frac{7}{10}$.

8th Mark: Percent is $80 \%$. Fraction = $\frac{80}{100} = \frac{4}{5}$. Decimal = $0.8$.

9th Mark: Fraction is $\frac{9}{10}$. Percent = $\frac{9}{10} \times 100 = 90 \%$. Decimal = $0.9$.

10th Mark: Percent is $100 \%$. Fraction = $1$. Decimal = $1.0$.

The completed labels for the blanks (highlighted in bold) are as follows:

Solution:

The number line is divided into $8$ equal parts between $0$ and $1$. Each part represents an increment of $\frac{1}{8} = 0.125 = 12.5 \%$.

1st Mark: Fraction is $\frac{1}{8}$. Percent = $\frac{1}{8} \times 100 = 12.5 \%$. Decimal = $0.125$. (Given)

2nd Mark: Percent is $25 \%$. Fraction = $\frac{25}{100} = \frac{1}{4}$. Decimal = $0.25$.

3rd Mark: Percent is $37.5 \%$. Fraction = $\frac{37.5}{100} = \frac{375}{1000} = \frac{3}{8}$. Decimal = $0.375$.

4th Mark: Decimal is $0.5$. Percent = $0.5 \times 100 = 50 \%$. Fraction = $\frac{5}{10} = \frac{1}{2}$.

5th Mark: Fraction is $\frac{5}{8}$. Percent = $\frac{5}{8} \times 100 = 62.5 \%$. Decimal = $0.625$.

6th Mark: Decimal is $0.75$. Percent = $0.75 \times 100 = 75 \%$. Fraction = $\frac{75}{100} = \frac{3}{4}$.

7th Mark: Percent is $87.5 \%$. Fraction = $\frac{87.5}{100} = \frac{7}{8}$. Decimal = $0.875$.

8th Mark: End point $1$. Percent = $100 \%$. Fraction = $1$. Decimal = $1.0$.

The completed labels for the blanks (highlighted in bold) are as follows:

Solution:

To convert a fraction into a percentage, we multiply it by $100$.

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

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

Converting the improper fraction $\frac{200}{3}$ into a mixed fraction:

$\frac{200}{3} = 66\frac{2}{3} \%$

Since the calculated value matches the statement, the statement is True.

Solution:

An improper fraction is a fraction where the numerator is greater than or equal to the denominator (e.g., $\frac{5}{4}$, $\frac{3}{2}$, $\frac{1}{1}$).

The value of an improper fraction is always greater than or equal to $1$.

When multiplied by $100$ to find the percentage, the result will always be $100 \%$ or more.

Therefore, the percentage cannot be less than $100$.

Hence, the statement is False.

Solution:

First, we convert $4$ days into hours:

$1$ day = $24$ hours

$4$ days = $4 \times 24 = 96$ hours

Now, calculate $50 \%$ of $4$ days ($96$ hours):

$50 \%$ of $96 = \frac{50}{100} \times 96$

$= \frac{1}{2} \times 96 = 48$ hours

Since $8$ hours is not equal to $48$ hours, the statement is False.

Given:

Principal ($P$) = $\textsf{₹} 350$

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

Time ($T$) = $73$ days = $\frac{73}{365}$ year

Solution:

We know that $365$ is a multiple of $73$ ($73 \times 5 = 365$).

So, $T = \frac{73}{365} = \frac{1}{5}$ year.

Simple Interest ($I$) = $\frac{P \times R \times T}{100}$

$I = \frac{350 \times 5 \times \frac{1}{5}}{100}$

$I = \frac{350}{100} = \textsf{₹} 3.50$

Since the calculated interest is $\textsf{₹} 3.50$ and the statement claims it is $\textsf{₹} 35$, the statement is False.

Solution:

The standard formula for calculating simple interest is the product of Principal ($P$), Rate of interest ($R$), and Time period ($T$), divided by $100$.

$\text{Simple Interest} = \frac{P \times R \times T}{100}$

Since multiplication is commutative ($P \times R \times T = T \times P \times R$), the formula provided in the statement is correct.

Hence, the statement is True.

Solution:

To convert a percentage into a fraction, we divide by $100$:

$75 \% = \frac{75}{100}$

Simplifying the fraction by dividing both numerator and denominator by $25$:

$\frac{\cancel{75}^3}{\cancel{100}_4} = \frac{3}{4}$

Since $\frac{3}{4}$ is not equal to $\frac{4}{3}$ (it is actually the reciprocal), the statement is False.

Solution:

We calculate $12 \%$ of $120$:

$= \frac{12}{100} \times 120$

$= \frac{12 \times 12}{10}$

$= \frac{144}{10} = 14.4$

Since $14.4$ is not equal to $100$, the statement is False.

Given:

Marks obtained = $336$

Total marks = $600$

Solution:

Percentage = $\left( \frac{\text{Marks obtained}}{\text{Total marks}} \times 100 \right) \%$

Percentage = $\left( \frac{336}{600} \times 100 \right) \%$

Percentage = $\frac{336}{6} \%$

Dividing $336$ by $6$:

$336 \div 6 = 56 \%$

Since the calculated percentage is $56 \%$ and the statement claims it is $33.6 \%$, the statement is False.

Solution:

To convert a decimal into a percentage, we multiply the decimal by $100$ and append the percent sign ($\%$).

Calculation:

$0.018 \times 100 \%$

$= \frac{18}{1000} \times 100 \%$

$= \frac{18}{10} \%$

$= 1.8 \%$

Since $1.8 \%$ is not equal to $8 \%$, the statement is False.

Given:

Total amount = $\textsf{₹} 50$

Percentage = $50 \%$

Solution:

We calculate $50 \%$ of the given amount:

$50 \% \text{ of } \textsf{₹} 50 = \frac{50}{100} \times 50$

$= \frac{1}{2} \times 50$

$= \textsf{₹} 25$

Since the calculated value matches the statement, the statement is True.

Given:

Part value = $250$ cm

Total value = $1$ km

Solution:

First, we convert both quantities to the same unit (cm):

$1$ km = $1000$ m

$1$ m = $100$ cm

So, $1$ km = $1000 \times 100$ cm = $1,00,000$ cm

Now, we calculate $4 \%$ of $1,00,000$ cm:

$4 \% \text{ of } 1,00,000 = \frac{4}{100} \times 1,00,000$

$= 4 \times 1000 = 4000$ cm

Since $250$ cm is not equal to $4000$ cm, the statement is False.

Given:

Total students = $600$

Students who went for picnic = $126$

Solution:

Number of students who did not go = $\text{Total students} - \text{Students who went}$

Number of students who did not go = $600 - 126 = 474$

Now, calculate the percentage of students who did not go:

Percentage = $\left( \frac{474}{600} \times 100 \right) \%$

Percentage = $\frac{474}{6} \%$

The percentage is $79 \%$.

Since the calculated percentage is $79 \%$ and the statement claims it is $75$, the statement is False.

Given:

Selling Price ($S.P.$) = $\textsf{₹} 50$

Loss Percentage = $10 \%$

Solution:

Let's check if the Cost Price ($C.P.$) is $\textsf{₹} 60$.

If $C.P. = \textsf{₹} 60$ and Loss $= 10 \%$, then:

Loss amount = $10 \%$ of $60 = \textsf{₹} 6$

$S.P. = C.P. - \text{Loss}$

$S.P. = 60 - 6 = \textsf{₹} 54$

However, the given $S.P.$ is $\textsf{₹} 50$. Since $54 \neq 50$, the statement is False.

Given:

Cost Price ($C.P.$) = $\textsf{₹} 2000$

Gain Percentage = $10 \%$

Solution:

First, calculate the gain amount:

$\text{Gain} = 10 \% \text{ of } 2000 = \frac{10}{100} \times 2000 = \textsf{₹} 200$

Now, calculate the Selling Price ($S.P.$):

$S.P. = C.P. + \text{Gain}$

$S.P. = 2000 + 200 = \textsf{₹} 2200$

Since the calculated $S.P.$ is $\textsf{₹} 2200$ and the statement claims it is $\textsf{₹} 2010$, the statement is False.

Given:

Cost Price ($C.P.$) = $\textsf{₹} 650$

Selling Price ($S.P.$) = $\textsf{₹} 585$

Solution:

Since $C.P. > S.P.$, the transaction results in a loss, not a profit.

$\text{Loss} = C.P. - S.P.$

$\text{Loss} = 650 - 585 = \textsf{₹} 65$

$\text{Loss } \% = \left( \frac{\text{Loss}}{C.P.} \times 100 \right) \%$

$\text{Loss } \% = \left( \frac{65}{650} \times 100 \right) \% = 10 \%$

Although the magnitude ($10$) is correct, the statement calls it a profit when it is actually a loss. Therefore, the statement is False.

Given:

Selling Price ($S.P.$) = $\textsf{₹} 3320$

Gain amount = $\textsf{₹} 320$

Solution:

First, we find the Cost Price ($C.P.$):

$C.P. = S.P. - \text{Gain}$

$C.P. = 3320 - 320 = \textsf{₹} 3000$

Now, calculate the Selling Price required for a $10 \%$ gain:

New Gain = $10 \% \text{ of } 3000 = \textsf{₹} 300$

New $S.P. = C.P. + \text{New Gain}$

New $S.P. = 3000 + 300 = \textsf{₹} 3300$

Since the calculated $S.P.$ matches the statement, the statement is True.

Given:

Principal ($P$) = $\textsf{₹} 1200$

Time ($T$) = $1\frac{1}{2}$ years = $1.5$ years

Rate ($R$) = $15 \%$ p.a.

Solution:

We calculate the Simple Interest ($I$):

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

$I = \frac{1200 \times 15 \times 1.5}{100}$

$I = 12 \times 15 \times 1.5$

$I = 180 \times 1.5$

$I = \textsf{₹} 270$

Since the calculated interest is $\textsf{₹} 270$ and the statement claims it is $\textsf{₹} 180$, the statement is False.

Given:

Principal ($P$) = $\textsf{₹} 800$

Time ($T$) = $3$ years

Rate ($R$) = $12 \%$ p.a.

Solution:

First, we calculate the Interest ($I$):

$I = \frac{800 \times 12 \times 3}{100} = 8 \times 36 = \textsf{₹} 288$

Now, calculate the Amount ($A$):

$A = P + I$

$A = 800 + 288 = \textsf{₹} 1088$

Since the calculated amount is $\textsf{₹} 1088$ and the statement claims it is $\textsf{₹} 896$, the statement is False.

Given:

Principal ($P$) = $\textsf{₹} 6400$

Rate ($R$) = $15 \%$ p.a.

Time for Feroz ($T_1$) = $3.5$ years

Time for Rashmi ($T_2$) = $5$ years

Solution:

Difference in time ($\Delta T$) = $5 - 3.5 = 1.5$ years

Difference in interest = $\frac{P \times R \times \Delta T}{100}$

Difference = $\frac{6400 \times 15 \times 1.5}{100}$

Difference = $64 \times 15 \times 1.5$

Difference = $960 \times 1.5 = \textsf{₹} 1440$

Since the calculated difference is $\textsf{₹} 1440$ and the statement claims it is $\textsf{₹} 150$, the statement is False.

Given:

Total lemons = $720$

Cost Price ($C.P.$) rate = $\textsf{₹} 120$ per $100$ lemons

Solution:

1. Total Cost Price:

$C.P. = \frac{120}{100} \times 720 = 1.2 \times 720 = \textsf{₹} 864$

2. Total Selling Price:

Rotten lemons = $10 \%$ of $720 = 72$

$S.P.$ of rotten lemons = $\frac{50}{100} \times 72 = 0.5 \times 72 = \textsf{₹} 36$

Remaining lemons = $720 - 72 = 648$

$S.P.$ of remaining lemons = $\frac{125}{100} \times 648 = 1.25 \times 648 = \textsf{₹} 810$

Total $S.P. = 36 + 810 = \textsf{₹} 846$

3. Profit or Loss:

Since $C.P. > S.P.$ ($864 > 846$), the vendor suffered a loss.

Loss = $864 - 846 = \textsf{₹} 18$

Since there is a loss and not a profit, the statement is False.

(i) 8% of ₹ x is ₹ 100

$\frac{8}{100} \times x = 100$

$x = \frac{100 \times 100}{8} = \frac{10000}{8}$

$x = 1250$

So, the value of $x$ is $\textsf{₹} 1250$.

(ii) 32% of x kg is 400 kg

$\frac{32}{100} \times x = 400$

$x = \frac{400 \times 100}{32} = \frac{40000}{32}$

$x = 1250$

So, the value of $x$ is $1250$ kg.

(iii) 35% of ₹ x is ₹ 280

$\frac{35}{100} \times x = 280$

$x = \frac{280 \times 100}{35}$

$x = 8 \times 100 = 800$

So, the value of $x$ is $\textsf{₹} 800$.

(iv) 45% of marks x is 405

$\frac{45}{100} \times x = 405$

$x = \frac{405 \times 100}{45}$

$x = 9 \times 100 = 900$

So, the value of $x$ is $900$ marks.

Given:

Total squares in a $10 \times 10$ grid = $100$

Total value of the grid = $300$

Value of each small square = $3$

(a) Shade 25% of the grid. What is 25% of 300? Compare the two answers.

Number of squares to be shaded = $25 \%$ of $100 = 25$ squares.

Value of shaded region = $25 \times 3 = 75$.

Calculation: $25 \%$ of $300 = \frac{25}{100} \times 300 = 25 \times 3 = 75$.

Comparison: Both answers are the same ($75$). This shows that $25 \%$ of the total value is equal to the total value of $25 \%$ of the squares.

(b) What is the value of 25 squares?

Value of $1$ square = $3$

Value of $25$ squares = $25 \times 3 = 75$

(c) Shade 17% of the grid. What is 17% of 300? Compare the two answers.

Number of squares to be shaded = $17 \%$ of $100 = 17$ squares.

Value of shaded region = $17 \times 3 = 51$.

Calculation: $17 \%$ of $300 = \frac{17}{100} \times 300 = 17 \times 3 = 51$.

Comparison: Both answers are the same ($51$).

(d) What is the value of $\frac{1}{10}$ of the grid?

Number of squares in $\frac{1}{10}$ of the grid = $\frac{1}{10} \times 100 = 10$ squares.

Value of $10$ squares = $10 \times 3 = 30$.

Solution:

To convert a fraction into a per cent, we multiply it by $100$.

Percentage = $\left( \frac{1}{6} \times 100 \right) \%$

Percentage = $\frac{100}{6} \%$

Simplifying the fraction:

Percentage = $\frac{\cancel{100}^{50}}{\cancel{6}_{3}} \%$

Percentage = $16\frac{2}{3} \%$ or approximately $16.67 \%$

Solution:

Percentage = $\left( \frac{9}{40} \times 100 \right) \%$

Percentage = $\frac{9 \times \cancel{100}^{5}}{\cancel{40}_{2}} \%$

Percentage = $\frac{45}{2} \%$

Percentage = $22.5 \%$

Solution:

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

Percentage = $1 \%$

Solution:

To convert a per cent into a fraction, we divide it by $100$.

$80 \% = \frac{80}{100}$

Simplifying to lowest terms:

$= \frac{\cancel{80}^{4}}{\cancel{100}_{5}}$

$= \frac{4}{5}$

Solution:

First, convert the mixed fraction to an improper fraction:

$33\frac{1}{3} \% = \frac{100}{3} \%$

Now, convert the per cent into a fraction:

$= \frac{100}{3} \times \frac{1}{100}$

$= \frac{\cancel{100}^{1}}{3 \times \cancel{100}_{1}}$

$= \frac{1}{3}$

In ratio form, it is $1 : 3$.

Solution:

First, convert to an improper fraction:

$16\frac{2}{3} \% = \frac{50}{3} \%$

Convert the per cent into a fraction:

$= \frac{50}{3} \times \frac{1}{100}$

$= \frac{\cancel{50}^{1}}{3 \times \cancel{100}_{2}}$

$= \frac{1}{6}$

In ratio form, it is $1 : 6$.

Solution:

$150 \% = \frac{150}{100}$

Simplifying to lowest terms:

$= \frac{\cancel{150}^{3}}{\cancel{100}_{2}}$

$= \frac{3}{2}$

In ratio form, it is $3 : 2$.

To Find:

$52 \%$ of $700$ using two different methods.

Solution:

Way 1: Sachin's Method (Unitary Approach)

First, we find $1 \%$ of $700$ by multiplying it by $0.01$ (or dividing by $100$):

$1 \%$ of $700 = 0.01 \times 700 = 7$

Now, multiply the result by $52$:

$52 \%$ of $700 = 7 \times 52$

$52 \%$ of $700 = 364$

Way 2: Sanjana's Method (Direct Decimal Approach)

In this method, we combine the steps by converting the percentage directly into a decimal ($0.52$) and multiplying it by the number:

$52 \%$ of $700 = 0.52 \times 700$

$= \frac{52}{100} \times 700$

$= 52 \times 7$

$= 364$

Conclusion:

Both methods yield the same result, i.e., $364$. Generally, Way 2 (Sanjana's method) is found to be easier and faster as it involves a single multiplication step.

Solution:

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

$0.089 = (0.089 \times 100) \%$

$= \left( \frac{89}{1000} \times 100 \right) \%$

$= \frac{89}{10} \%$

$= 8.9 \%$

Thus, $0.089$ written as a per cent is $8.9 \%$.

Solution:

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

$1.56 = (1.56 \times 100) \%$

$= 156 \%$

Thus, $1.56$ written as a per cent is $156 \%$.

Solution:

$15 \%$ of $20 = \frac{15}{100} \times 20$

Using the cancellation method:

$= \frac{15}{\cancel{100}_5} \times \cancel{20}^1$

$= \frac{15}{5}$

$= 3$

Thus, $15 \%$ of $20$ is $3$.

Solution:

$800 \%$ of $800 = \frac{800}{100} \times 800$

$= 8 \times 800$

$= 6400$

Thus, $800 \%$ of $800$ is $6400$.

Solution:

$100 \%$ of $500 = \frac{100}{100} \times 500$

$= 1 \times 500$

$= 500$

Thus, $100 \%$ of $500$ is $500$.

Given:

Total time = $1$ hour

Part time = $30$ minutes

To Find:

Percentage of $30$ minutes in $1$ hour.

Solution:

First, we convert both quantities to the same unit. We know that:

Now, calculate the percentage:

Percentage = $\left( \frac{\text{Part time}}{\text{Total time}} \times 100 \right) \%$

Percentage = $\left( \frac{30}{60} \times 100 \right) \%$

Percentage = $\frac{\cancel{30}^1}{\cancel{60}_2} \times 100 \%$

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

Percentage = $50 \%$

Therefore, $30$ minutes is $50 \%$ of $1$ hour.

Given:

Total time = $1$ day

Part time = $1$ minute

To Find:

Percentage of $1$ minute in $1$ day.

Solution:

First, we convert $1$ day into minutes:

$1 \text{ day} = 24 \text{ hours}$

$1 \text{ hour} = 60 \text{ minutes}$

So, $1 \text{ day} = 24 \times 60 \text{ minutes}$

Total time = $1440 \text{ minutes}$

Now, calculate the percentage:

Percentage = $\left( \frac{1}{1440} \times 100 \right) \%$

Percentage = $\frac{100}{1440} \%$

Simplifying the fraction:

Percentage = $\frac{\cancel{100}^5}{\cancel{1440}_{72}} \%$

Percentage = $\frac{5}{72} \%$

Therefore, $1$ minute is $\frac{5}{72} \%$ of $1$ day.

Given:

Total distance = $1$ km

Part distance = $1000$ metres

Solution:

First, we convert km to metres:

$1$ km = $1000$ metres

Now, calculate the percentage:

Percentage = $\left( \frac{1000 \text{ m}}{1000 \text{ m}} \times 100 \right) \%$

Percentage = $1 \times 100 \%$

Percentage = $100 \%$

Therefore, $1000$ metres is $100 \%$ of $1$ km.

Solution:

We need to calculate $8 \%$ of $25$ kg.

Value = $\frac{8}{100} \times 25$

Value = $8 \times \frac{\cancel{25}^1}{\cancel{100}_4}$

Value = $\frac{8}{4}$

Value = $2$ kg

Therefore, $8 \%$ of $25$ kg is $2$ kg.

Given:

Total Value (Whole) = $\textsf{₹} 80$

Part Value = $\textsf{₹} 100$

Solution:

Percentage = $\left( \frac{\text{Part Value}}{\text{Total Value}} \times 100 \right) \%$

Percentage = $\left( \frac{100}{80} \times 100 \right) \%$

Percentage = $\frac{\cancel{100}^5}{\cancel{80}_4} \times 100 \%$

Percentage = $5 \times \frac{\cancel{100}^{25}}{\cancel{4}_1} \%$

Percentage = $5 \times 25 \%$

Percentage = $125 \%$

Therefore, $\textsf{₹} 100$ is $125 \%$ of $\textsf{₹} 80$.

Given:

Percentage of men = $45 \%$

Percentage of women = $40 \%$

To Find:

Percentage of children in the town.

Solution:

The total population of the town is considered as $100 \%$.

Total percentage of men and women = Percentage of men + Percentage of women

Total percentage of men and women = $45 \% + 40 \% = 85 \%$

Now, calculate the percentage of children:

Percentage of children = $100 \% - 85 \%$

Percentage of children = $15 \%$

Therefore, the percentage of children in the town is $15 \%$.

Given:

Total strength of the school = $2000$

Percentage of girls = $40 \%$

To Find:

Number of boys in the school.

Solution:

Percentage of boys = $\text{Total percentage} - \text{Percentage of girls}$

Percentage of boys = $100 \% - 40 \% = 60 \%$

Now, we find the number of boys:

Number of boys = $60 \%$ of $2000$

Number of boys = $\frac{60}{100} \times 2000$

Number of boys = $60 \times 20 = 1200$

Therefore, there are $1200$ boys in the school.

Alternate Solution:

Number of girls = $40 \%$ of $2000 = \frac{40}{100} \times 2000 = 800$

Number of boys = $\text{Total students} - \text{Number of girls}$

Number of boys = $2000 - 800 = 1200$

Given:

Percentage of Calcium = $10 \%$

Percentage of Carbon = $3 \%$

Total weight of chalk = $2\frac{1}{2}$ kg = $2.5$ kg

To Find:

Amount of Carbon and Calcium in grams.

Solution:

First, convert the total weight from kg to grams:

Total weight = $2.5 \times 1000$ g = $2500$ g

1. Amount of Calcium:

Amount of Calcium = $10 \%$ of $2500$ g

Amount of Calcium = $\frac{10}{100} \times 2500 = 250$ g

2. Amount of Carbon:

Amount of Carbon = $3 \%$ of $2500$ g

Amount of Carbon = $\frac{3}{100} \times 2500 = 3 \times 25 = 75$ g

Therefore, the amount of calcium is $250$ g and the amount of carbon is $75$ g.

Given:

Total mass of mortar = $800$ kg

Percentage of sand = $55 \%$

Percentage of cement = $33 \%$

To Find:

The mass of lime in the mortar.

Solution:

First, find the percentage of lime in the mortar:

Percentage of lime = $100 \% - (\text{Percentage of sand} $$ + \text{Percentage of cement})$

Percentage of lime = $100 \% - (55 \% + 33 \%) = 100 \% - 88 \% = 12 \%$

Now, calculate the mass of lime:

Mass of lime = $12 \%$ of $800$ kg

Mass of lime = $\frac{12}{100} \times 800 = 12 \times 8 = 96$ kg

Therefore, the mass of lime in the mortar is $96$ kg.

Given:

Number of tables bought = $24$

Cost Price ($C.P.$) per table = $\textsf{₹} 450$

To Find:

Gain or Loss percentage.

Solution:

1. Total Cost Price:

Total $C.P. = 24 \times 450$

Total $C.P. = \textsf{₹} 10,800$

2. Total Selling Price:

$S.P.$ of $16$ tables = $16 \times 600 = \textsf{₹} 9,600$

Remaining tables = $24 - 16 = 8$

$S.P.$ of $8$ tables = $8 \times 400 = \textsf{₹} 3,200$

Total $S.P. = 9,600 + 3,200 = \textsf{₹} 12,800$

3. Gain or Loss:

Since $S.P. > C.P.$, there is a gain.

Gain = $12,800 - 10,800 = \textsf{₹} 2,000$

Gain $\%$ = $\left( \frac{\text{Gain}}{C.P.} \times 100 \right) \%$

Gain $\%$ = $\frac{2000}{10800} \times 100 = \frac{2000}{108} = \frac{500}{27}$

Gain $\%$ $\approx$ $18.52 \%$ (or $18\frac{14}{27} \%$).

Given:

Percentage deposited = $20 \%$

Percentage of remainder spent = $20 \%$

Final amount left = $\textsf{₹} 4800$

To Find:

Original amount Medha had.

Solution:

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

Amount deposited in bank = $20 \%$ of $x = 0.20x$

Remainder amount = $x - 0.20x = 0.80x$

Amount spent from remainder = $20 \%$ of $0.80x = 0.20 \times 0.80x = 0.16x$

Final amount left = $0.80x - 0.16x = 0.64x$

According to the question:

$x = \frac{4800}{0.64} = \frac{4800 \times 100}{64}$

$x = \frac{480000}{64}$

Dividing by $8$:

$x = \frac{60000}{8} = 7500$

Therefore, Medha originally had $\textsf{₹} 7500$.

Given:

Percentage increase = $12 \%$

Current cost = $\textsf{₹} 896$

To Find:

The original cost of the flower vase.

Solution:

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

According to the question, the original cost plus $12 \%$ of the original cost equals the current cost.

$x + (12 \% \text{ of } x) = 896$

$x + 0.12x = 896$

$1.12x = 896$

$x = \frac{896}{1.12}$

$x = \frac{89600}{112}$

Performing division:

$x = 800$

Therefore, the original cost was $\textsf{₹} 800$.

Given:

Total amount borrowed = $\textsf{₹} 12000$

Part 1: Principal ($P_1$) = $\textsf{₹} 4000$, Rate ($R_1$) = $18 \%$ p.a., Time ($T$) = $3$ years

Part 2: Principal ($P_2$) = $\textsf{₹} 12000 - \textsf{₹} 4000 = \textsf{₹} 8000$, Rate ($R_2$) = $15 \%$ p.a., Time ($T$) = $3$ years

To Find:

Total interest after $3$ years.

Solution:

First, we calculate the simple interest for Part 1 ($I_1$):

$I_1 = \frac{P_1 \times R_1 \times T}{100}$

$I_1 = \frac{4000 \times 18 \times 3}{100} = 40 \times 54 = \textsf{₹} 2160$

Next, we calculate the simple interest for Part 2 ($I_2$):

$I_2 = \frac{P_2 \times R_2 \times T}{100}$

$I_2 = \frac{8000 \times 15 \times 3}{100} = 80 \times 45 = \textsf{₹} 3600$

Now, we find the total interest:

$\text{Total Interest} = I_1 + I_2$

$\text{Total Interest} = 2160 + 3600$

Therefore, the total interest after 3 years is $\textsf{₹} 5760$.

Given:

Distance travelled by car = $60$ km

Distance travelled by train = $240$ km

To Find:

Percentage of total journey travelled by car and by train.

Solution:

First, calculate the total journey distance:

Total distance = $60$ km + $240$ km = $300$ km

Now, calculate the percentage of journey by car:

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

$\text{Percentage by car} = \left( \frac{60}{300} \times 100 \right) \% = \frac{60}{3} \% = 20 \%$

Next, calculate the percentage of journey by train:

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

$\text{Percentage by train} = \left( \frac{240}{300} \times 100 \right) \% = \frac{240}{3} \% = 80 \%$

Therefore, the man travelled $20 \%$ of the journey by car and $80 \%$ by train.

Given:

Selling Price ($S.P.$) = $\textsf{₹} 1440$

Loss percentage = $10 \%$

To Find:

Cost Price ($C.P.$) of the chair.

Solution:

Let the Cost Price be $C.P$. We know that:

$C.P. = \frac{100}{100 - \text{Loss } \%} \times S.P.$

$C.P. = \frac{100}{100 - 10} \times 1440$

$C.P. = \frac{100}{90} \times 1440$

$C.P. = 100 \times \frac{\cancel{1440}^{16}}{\cancel{90}_{1}}$

$C.P. = 100 \times 16 = 1600$

Therefore, the shopkeeper bought the chair for $\textsf{₹} 1600$.

Given:

Rate of interest ($R$) = $12 \%$ p.a.

Interest ($I$) = $\textsf{₹} 1620$

Time period ($T$) = From May 2006 to April 2008 = $2$ years

To Find:

Principal ($P$), i.e., money invested.

Solution:

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

Rearranging the formula to find $P$:

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

$P = \frac{1620 \times 100}{12 \times 2}$

$P = \frac{162000}{24}$

Dividing by $12$ first:

$P = \frac{\cancel{1620}^{135} \times 100}{\cancel{12}_{1} \times 2} = \frac{13500}{2}$

$P = 6750$

Therefore, the money invested by Dhruvika was $\textsf{₹} 6750$.

Given:

Initial loss percentage = $25 \%$

Actual loss percentage = $20 \%$

Difference in Selling Prices ($S.P.s$) = $\textsf{₹} 4000$

To Find:

Cost Price ($C.P.$) of the scooter.

Solution:

Let the Cost Price ($C.P.$) of the scooter be $x$.

$S.P. \text{ at } 25 \% \text{ loss} = x - 25 \% \text{ of } x = 0.75x$

$S.P. \text{ at } 20 \% \text{ loss} = x - 20 \% \text{ of } x = 0.80x$

The difference between these two $S.P.s$ is $\textsf{₹} 4000$.

$0.05x = 4000$

$x = \frac{4000}{0.05}$

$x = \frac{4000 \times 100}{5}$

$x = 800 \times 100 = 80000$

Therefore, the cost price of the scooter is $\textsf{₹} 80,000$.

Given:

Total population = $8000$

Literacy percentage = $80 \%$

Percentage of women among literates = $40 \%$

To Find:

The ratio of literate women to the total population.

Solution:

First, find the total number of literate people:

$\text{Number of literates} = 80 \% \text{ of } 8000 = \frac{80}{100} \times 8000 = 6400$

Next, find the number of literate women:

$\text{Number of literate women} = 40 \% \text{ of } 6400 = \frac{40}{100} \times 6400 = 2560$

Now, find the ratio of literate women to the total population:

$\text{Ratio} = \frac{\text{Literate women}}{\text{Total population}}$

$\text{Ratio} = \frac{2560}{8000}$

Simplifying the fraction:

$\text{Ratio} = \frac{256}{800}$

$\text{Ratio} = \frac{\cancel{256}^{8}}{\cancel{800}_{25}} = \frac{8}{25}$

Therefore, the ratio is $8 : 25$.

Given:

Number of tickets of $\textsf{₹} 400$ = $250$

Tax on $\textsf{₹} 400$ ticket = $40 \%$

Number of tickets of $\textsf{₹} 100$ = $500$

Tax on $\textsf{₹} 100$ ticket = $20 \%$

To Find:

Total entertainment tax collected.

Solution:

First, we calculate the tax collected from the $\textsf{₹} 400$ tickets:

Tax per ticket = $40 \%$ of $\textsf{₹} 400$ = $\frac{40}{100} \times 400 = \textsf{₹} 160$

Total tax from 250 tickets = $250 \times 160$

Tax from first category = $\textsf{₹} 40,000$

Now, we calculate the tax collected from the $\textsf{₹} 100$ tickets:

Tax per ticket = $20 \%$ of $\textsf{₹} 100$ = $\textsf{₹} 20$

Total tax from 500 tickets = $500 \times 20 = \textsf{₹} 10,000$

Total entertainment tax collected = $\textsf{₹} 40,000 + \textsf{₹} 10,000$

Total tax = $\textsf{₹} 50,000$

Given:

Original Income = $\textsf{₹} 50,000$

Original Expenditure = $80 \%$ of income

Income Increase = $20 \%$

Expenditure Increase = $20 \%$

To Find:

New Savings.

Solution:

Step 1: Calculate original expenditure

Original Expenditure = $\frac{80}{100} \times 50,000 = \textsf{₹} 40,000$

Step 2: Calculate new income

Increase in Income = $20 \%$ of $50,000 = \frac{20}{100} \times 50,000 = \textsf{₹} 10,000$

New Income = $50,000 + 10,000 = \textsf{₹} 60,000$

Step 3: Calculate new expenditure

Increase in Expenditure = $20 \%$ of $40,000 = \frac{20}{100} \times 40,000 = \textsf{₹} 8,000$

New Expenditure = $40,000 + 8,000 = \textsf{₹} 48,000$

Step 4: Calculate new savings

New Savings = New Income $-$ New Expenditure

New Savings = $60,000 - 48,000$

New Savings = $\textsf{₹} 12,000$

Given:

Number of papers = $3$

Maximum marks per paper = $100$

Marks in Paper 1 = $53$

Marks in Paper 2 = $75$

Target overall percentage = $70 \%$

To Find:

Marks required in the third paper.

Solution:

Total maximum marks = $3 \times 100 = 300$

Required total marks to get $70 \%$ = $70 \%$ of $300$

Required total = $\frac{70}{100} \times 300 = 210$ marks

Sum of marks obtained in first two papers = $53 + 75$

Marks required in the third paper = Required total $-$ Marks obtained

Marks required = $210 - 128$

Marks required = $82$

(a) Calculate an approximate ratio of systolic to diastolic blood pressures in the normal range.

In the normal range, the upper limits are approximately $120$ for systolic and $80$ for diastolic.

Ratio = $\frac{\text{Systolic}}{\text{Diastolic}}$

Ratio = $\frac{120}{80}$

Simplifying the fraction:

Ratio = $\frac{\cancel{120}^{3}}{\cancel{80}_{2}} = \frac{3}{2}$ or $3 : 2$ (which is $1.5$).

(b) If Manohar’s systolic blood pressure is 102 mm Hg, use the ratio from part (a) to predict his diastolic blood pressure.

Using the ratio $\frac{\text{Systolic}}{\text{Diastolic}} = \frac{3}{2}$:

$\frac{102}{\text{Diastolic}} = \frac{3}{2}$

$3 \times \text{Diastolic} = 102 \times 2$

$3 \times \text{Diastolic} = 204$

$\text{Diastolic} = \frac{204}{3}$

$\text{Diastolic} = 68$ mm Hg

Predicted Diastolic blood pressure = $68$ mm Hg.

(c) Calculate ratio of average systolic to average diastolic blood pressure in the prehypertension category.

Systolic Range (Prehypertension): $120$ to $139$

Average Systolic = $\frac{120 + 139}{2} = \frac{259}{2} = 129.5$

Diastolic Range (Prehypertension): $80$ to $89$

Average Diastolic = $\frac{80 + 89}{2} = \frac{169}{2} = 84.5$

Ratio = $\frac{129.5}{84.5}$

Multiplying by 10 to remove decimals: $\frac{1295}{845}$

Dividing both by 5: $\frac{\cancel{1295}^{259}}{\cancel{845}_{169}}$

The ratio is approximately $1.53 : 1$ (or $259 : 169$).

(a) Solution:

Given:

Length of the King Cobra = $558$ cm

Percentage of python's length represented by Cobra = $60 \%$

To Find:

The length of the largest reticulated python.

Solution:

Let the length of the largest reticulated python be $x$ cm.

According to the question:

$60 \%$ of $x = 558$

$\frac{60}{100} \times x = 558$

$x = \frac{558 \times 100}{60}$

$x = \frac{558 \times 10}{6}$

$x = 93 \times 10$

$x = 930$ cm

Therefore, the length of the largest reticulated python is $930$ cm.

(b) Solution:

Given:

From the image of the children on the seesaw:

Mass on the left ($\text{mass 1}$) = $24$ kg

Distance of child 1 from fulcrum ($\text{length 1}$) = $3$ m

Distance of child 2 from fulcrum ($\text{length 2}$) = $2$ m

To Find:

Mass of the child on the right ($\text{mass 2}$).

Solution:

The balancing condition is given as:

$\frac{\text{mass 1}}{\text{length 2}} = \frac{\text{mass 2}}{\text{length 1}}$

Substituting the known values:

$\frac{24}{2} = \frac{\text{mass 2}}{3}$

$12 = \frac{\text{mass 2}}{3}$

$\text{mass 2} = 12 \times 3$

$\text{mass 2} = 36$ kg

Therefore, the mass of the child on the right is $36$ kg.

(c) Solution:

Given:

Scale of the model = $2$ cm : $0.0000001$ mm

Length of the DNA model = $17$ cm

To Find:

The actual length of the DNA chain.

Solution:

Let the actual length be $y$ mm.

Using the unitary method:

If $2$ cm model length corresponds to $0.0000001$ mm actual length,

Then $1$ cm model length corresponds to $\frac{0.0000001}{2}$ mm.

So, $17$ cm model length corresponds to $\frac{0.0000001}{2} \times 17$ mm.

Actual length = $8.5 \times 0.0000001$ mm

Actual length = $0.00000085$ mm

Therefore, the length of the actual DNA chain is $0.00000085$ mm.

Solution:

First, we identify the vowels (A, E, I, O, U) and consonants in each word provided in the cloud image. Note: 'Y' is treated as a consonant in this context.

1. Hypotenuse (H-Y-P-O-T-E-N-U-S-E)

Total letters = $10$

Vowels = O, E, U, E (Total = $4$)

Consonants = H, Y, P, T, N, S (Total = $6$)

(a) Ratio (Consonants : Vowels) = $6 : 4 = \mathbf{3 : 2}$

(b) Percentage of Consonants = $\left( \frac{6}{10} \times 100 \right) \% = \mathbf{60 \%}$

2. Congruence (C-O-N-G-R-U-E-N-C-E)

Total letters = $10$

Vowels = O, U, E, E (Total = $4$)

Consonants = C, N, G, R, N, C (Total = $6$)

(a) Ratio (Consonants : Vowels) = $6 : 4 = \mathbf{3 : 2}$

(b) Percentage of Consonants = $\left( \frac{6}{10} \times 100 \right) \% = \mathbf{60 \%}$

3. Perpendicular (P-E-R-P-E-N-D-I-C-U-L-A-R)

Total letters = $13$

Vowels = E, E, I, U, A (Total = $5$)

Consonants = P, R, P, N, D, C, L, R (Total = $8$)

(a) Ratio (Consonants : Vowels) = $\mathbf{8 : 5}$

(b) Percentage of Consonants = $\left( \frac{8}{13} \times 100 \right) \% \approx \mathbf{61.54 \%}$

4. Transversal (T-R-A-N-S-V-E-R-S-A-L)

Total letters = $11$

Vowels = A, E, A (Total = $3$)

Consonants = T, R, N, S, V, R, S, L (Total = $8$)

(a) Ratio (Consonants : Vowels) = $\mathbf{8 : 3}$

(b) Percentage of Consonants = $\left( \frac{8}{11} \times 100 \right) \% \approx \mathbf{72.73 \%}$

5. Correspondence (C-O-R-R-E-S-P-O-N-D-E-N-C-E)

Total letters = $14$

Vowels = O, E, O, E, E (Total = $5$)

Consonants = C, R, R, S, P, N, D, N, C (Total = $9$)

(a) Ratio (Consonants : Vowels) = $\mathbf{9 : 5}$

(b) Percentage of Consonants = $\left( \frac{9}{14} \times 100 \right) \% \approx \mathbf{64.29 \%}$

Given:

Percentage of defective T-shirts = $0.06 \%$

Student's claim = $6$ out of every $100$

Solution:

To find how many T-shirts are defective out of $100$, we convert the percentage into a number:

$0.06 \%$ of $100 = \frac{0.06}{100} \times 100 = 0.06$

So, $0.06 \%$ means $0.06$ T-shirts out of $100$, which is equivalent to $6$ out of $10,000$.

The Student's Error:

The student confused $0.06 \%$ with $6 \%$. A claim of "$6$ out of every $100$" represents $6 \%$, whereas the actual defective rate is much lower at $0.06 \%$.

Solution:

Two ratios are said to be in proportion if they are equal.

Let's check if $\frac{3}{4} = \frac{9}{16}$.

By cross-multiplication:

$3 \times 16 = 48$

$4 \times 9 = 36$

Since $48 \neq 36$, the ratios are not proportional.

The Student's Error:

The student likely thought that since $3^2 = 9$ and $4^2 = 16$, the ratios must be equivalent. However, to maintain proportionality, the numerator and denominator must be multiplied by the same number, not squared. To make $\frac{3}{4}$ have a numerator of $9$, it should be multiplied by $3$ ($ \frac{3 \times 3}{4 \times 3} = \frac{9}{12} $).

Given:

Price of $4$ T-shirts = $\textsf{₹} 1024$

Student's claim for unit price = $\textsf{₹} 25.6$

Solution:

To find the correct unit price, we divide the total cost by the number of T-shirts:

$\text{Correct Unit Price} = \frac{1024}{4}$

The correct unit price is $\textsf{₹} 256$.

The Student's Error:

The student made a decimal point error. They likely divided $102.4$ by $4$ or simply placed the decimal incorrectly after dividing. The student's answer is $10$ times smaller than the correct value.

Given:

Ratio of two varieties = $5 : 4$

Cost of variety 1 = $\textsf{₹} 200/\text{kg}$

Cost of variety 2 = $\textsf{₹} 300/\text{kg}$

Selling Price of blend = $\textsf{₹} 275/\text{kg}$

To Find:

Profit or Loss percentage.

Solution:

Let the merchant blend $5$ kg of variety 1 and $4$ kg of variety 2.

Total weight of the blend = $5 + 4 = 9$ kg

1. Calculation of Total Cost Price ($C.P.$):

$C.P. \text{ of variety 1} = 5 \times 200 = \textsf{₹} 1000$

$C.P. \text{ of variety 2} = 4 \times 300 = \textsf{₹} 1200$

Total $C.P. = 1000 + 1200 = \textsf{₹} 2200$

2. Calculation of Total Selling Price ($S.P.$):

Total $S.P. = \text{Total weight} \times \text{S.P. rate}$

Total $S.P. = 9 \times 275 = \textsf{₹} 2475$

3. Calculation of Profit/Loss:

Since $S.P. > C.P.$, there is a profit.

$\text{Profit} = S.P. - C.P. = 2475 - 2200 = \textsf{₹} 275$

$\text{Profit } \% = \left( \frac{\text{Profit}}{C.P.} \times 100 \right) \%$

$\text{Profit } \% = \left( \frac{275}{2200} \times 100 \right) \%$

$\text{Profit } \% = \frac{\cancel{275}^{25}}{\cancel{2200}_{22 \times 100}} \times 100$

$\text{Profit } \% = \frac{275}{22} = 12.5 \%$

Therefore, the merchant makes a $12.5 \%$ profit.

Given:

Original length of cloth = $5$ m

Shrinkage percentage = $10 \%$

To Find:

New length of cloth after washing.

Solution:

First, we calculate the length that the cloth shrinks:

$\text{Shrinkage} = 10 \% \text{ of } 5 \text{ m}$

$\text{Shrinkage} = \frac{10}{100} \times 5 = 0.1 \times 5 = 0.5 \text{ m}$

Now, we find the new length:

$\text{New Length} = \text{Original Length} - \text{Shrinkage}$

$\text{New Length} = 5 - 0.5 = 4.5 \text{ m}$

Therefore, the cloth will be $4.5$ m long after washing.

Given:

Marks obtained by Nancy = $426$ out of $600$

Marks obtained by Rohit = $560$ out of $800$

To Find:

Whose performance is better by comparing their percentages.

Solution:

To compare the performance, we need to calculate the percentage of marks obtained by both Nancy and Rohit.

1. Percentage obtained by Nancy:

$\text{Percentage} = \left( \frac{\text{Marks Obtained}}{\text{Total Marks}} \times 100 \right) \%$

$\text{Percentage} = \left( \frac{426}{600} \times 100 \right) \%$

$\text{Percentage} = \frac{426}{6} \%$

$\text{Percentage} = 71 \%$

2. Percentage obtained by Rohit:

$\text{Percentage} = \left( \frac{560}{800} \times 100 \right) \%$

$\text{Percentage} = \frac{560}{8} \%$

$\text{Percentage} = 70 \%$

Comparing the two percentages, we see that $71 \% > 70 \%$.

Therefore, Nancy's performance is better.

Given:

Principal Donation ($P$) = $\textsf{₹} 5,00,000$

Rate of Interest ($R$) = $12 \%$ per annum

Time ($T$) = $1$ year

Value of 2nd Scholarship = $\textsf{₹} 20,000$

Value of 3rd Scholarship = $\textsf{₹} 15,000$

To Find:

The value of the 1st scholarship.

Solution:

First, we calculate the total interest earned on the donation in one year:

$\text{Interest} = \frac{P \times R \times T}{100}$

$\text{Interest} = \frac{5,00,000 \times 12 \times 1}{100}$

$\text{Interest} = 5,000 \times 12 = \textsf{₹} 60,000$

The total interest is used to award the three scholarships. Therefore:

$\text{Total Interest} = \text{1st Scholarship} + \text{2nd Scholarship} + \text{3rd Scholarship}$

$60,000 = \text{1st Scholarship} + 20,000 + 15,000$

$60,000 = \text{1st Scholarship} + 35,000$

$\text{1st Scholarship} = 60,000 - 35,000$

$\text{1st Scholarship} = \textsf{₹} 25,000$

Therefore, the value of the first scholarship is $\textsf{₹} 25,000$.

Given:

Number of subjects = $5$

Marks in each subject = $100$

Marks in Mathematics = $99 \%$ of $100 = 99$

Marks in Hindi = $76 \%$ of $100 = 76$

Marks in English = $61 \%$ of $100 = 61$

Marks in Science = $84 \%$ of $100 = 84$

Marks in Social Science = $95 \%$ of $100 = 95$

To Find:

The aggregate percentage of marks obtained by Ambika.

Solution:

First, we find the total marks obtained by Ambika:

$\text{Total marks obtained} = 99 + 76 + 61 + 84 + 95$

$\text{Total marks obtained} = 415$

Now, we find the total maximum marks:

$\text{Total maximum marks} = 5 \times 100 = 500$

Finally, we calculate the aggregate percentage:

$\text{Aggregate Percentage} = \left( \frac{\text{Total Marks Obtained}}{\text{Total Maximum Marks}} \times 100 \right) \%$

$\text{Aggregate Percentage} = \left( \frac{415}{500} \times 100 \right) \%$

$\text{Aggregate Percentage} = \frac{415}{5} \%$

$\text{Aggregate Percentage} = 83 \%$

Therefore, the aggregate percentage of marks obtained by Ambika is $83 \%$.

Given:

Simple Interest ($I$) = $\textsf{₹} 9600$

Rate ($R$) = $16 \%$ per annum

Time ($T$) = $2$ years

To Find:

The Principal sum ($P$).

Solution:

The formula for Simple Interest is:

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

Rearranging the formula to find the Principal ($P$):

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

Substituting the given values:

$P = \frac{9600 \times 100}{16 \times 2}$

$P = \frac{9600 \times 100}{32}$

Dividing $9600$ by $32$:

$P = 300 \times 100$

$P = 30,000$

Therefore, the sum of money lent out is $\textsf{₹} 30,000$.

Given:

Cost Price for Harish = $\textsf{₹} 900$

Profit percentage for Harish = $5 \%$

Loss percentage for Archana = $20 \%$

To Find:

The price Babita paid for the gas-chullah.

Solution:

1. Transaction between Harish and Archana:

Profit amount for Harish = $5 \%$ of $900$

$\text{Profit} = \frac{5}{100} \times 900 = \textsf{₹} 45$

Selling Price for Harish (which is the Cost Price for Archana) = $900 + 45 = \textsf{₹} 945$

2. Transaction between Archana and Babita:

Archana's Cost Price = $\textsf{₹} 945$

Loss amount for Archana = $20 \%$ of $945$

$\text{Loss} = \frac{20}{100} \times 945 = \frac{1}{5} \times 945$

$\text{Loss} = \textsf{₹} 189$

Selling Price for Archana (which is the price Babita pays) = $945 - 189$

Price paid by Babita = $\textsf{₹} 756$

Therefore, Babita got the gas-chullah for $\textsf{₹} 756$.

Solution:

Let's solve each entry in Column I to find its match in Column II:

(i) $3:5 = \frac{3}{5} \times 100 \% = 60 \%$        $\rightarrow$ (E)

(ii) $2.5 = 2.5 \times 100 \% = 250 \%$        $\rightarrow$ (H)

(iii) $100 \% = \frac{100}{100} = 1$, which can be written as $3:3$        $\rightarrow$ (O)

(iv) $\frac{2}{3} = \frac{2}{3} \times 100 \% = \frac{200}{3} \% = 66\frac{2}{3} \%$        $\rightarrow$ (J)

(v) $6\frac{1}{4} \% = \frac{25}{4} \% = \frac{25}{4 \times 100} = \frac{1}{4 \times 4} = \frac{1}{16}$        $\rightarrow$ (G)

(vi) $12.5 \% = \frac{12.5}{100} = 0.125$        $\rightarrow$ (L)

(vii) $S.P.$ when $C.P. = \textsf{₹} 50$ and loss $= 6 \%$:

(viii) $S.P.$ when $C.P. = \textsf{₹} 50$ and profit $= \textsf{₹} 4$:

(ix) Profit% when $C.P. = \textsf{₹} 40$ and $S.P. = \textsf{₹} 50$:

(x) Profit% when $C.P. = \textsf{₹} 50$ and $S.P. = \textsf{₹} 60$:

(xi) Interest when $P = \textsf{₹} 800, R = 10 \%, T = 2$ years:

(xii) Amount when $P = \textsf{₹} 150, R = 6 \%, T = 1$ year:

Matching Summary Table:

Given:

Percentage for Accent and Presentation = $20 \%$

Percentage for Subject Matter = $60 \%$ of the remaining marks

Marks for Rebuttal = $8$

To Find:

Total marks.

Solution:

Let the total marks be $x$.

Marks for Accent and Presentation = $20 \%$ of $x = 0.20x$

Remaining marks = $x - 0.20x = 0.80x$

Marks for Subject Matter = $60 \%$ of $0.80x = 0.48x$

Remaining marks for Rebuttal = $0.80x - 0.48x = 0.32x$

According to the question, the marks for rebuttal is $8$:

Therefore, the total marks are $25$.

Given:

Total Amount = $\textsf{₹} 10000$

Condition: Simple Interest on Part 1 ($I_1$) = Simple Interest on Part 2 ($I_2$)

Part 1: Time ($T_1$) = $4$ years, Rate ($R_1$) = $12 \%$

Part 2: Time ($T_2$) = $4.5$ years, Rate ($R_2$) = $16 \%$

To Find:

The two parts of the money.

Solution:

Let the first part be $\textsf{₹} x$.

Then, the second part will be $\textsf{₹} (10000 - x)$.

Simple Interest on first part ($I_1$):

Simple Interest on second part ($I_2$):

Since $I_1 = I_2$:

So, the first part is $\textsf{₹} 6000$.

The second part = $10000 - 6000 = $ $\textsf{₹} 4000$.

Given:

Principal ($P$) = $\textsf{₹} 9000$

Amount ($A$) = $\textsf{₹} 18000$

Time ($T$) = $8$ years

To Find:

Rate of interest per annum ($R$).

Solution:

First, we find the Simple Interest ($I$):

Now, using the Simple Interest formula:

Therefore, the rate of interest is $12.5 \%$ per annum.

Given:

To Find:

Time period ($T$) in years.

Solution:

We know the formula for Simple Interest:

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

Substituting the given values into the formula:

$4.05P = \frac{P \times 13.5 \times T}{100}$

Cancelling $P$ from both sides:

$4.05 = \frac{13.5 \times T}{100}$

$4.05 \times 100 = 13.5 \times T$

$405 = 13.5 \times T$

$T = \frac{405}{13.5}$

$T = \frac{4050}{135}$

Dividing $4050$ by $135$:

$T = 30$

Therefore, the simple interest will be $4.05$ times the principal in $30$ years.

Given:

Let the sum (Principal) be $\textsf{₹} P$.

Case 1: $T_1 = 8$ years, $R_1 = 12\%$ p.a.

Case 2: $T_2 = 5$ years, $R_2 = 14\%$ p.a.

Difference in Simple Interest ($SI_1 - SI_2$) = $\textsf{₹} 3120$

To Find:

The sum ($\textsf{₹} P$).

Solution:

Calculating interest for Case 1:

$SI_1 = \frac{P \times 12 \times 8}{100} = \frac{96P}{100}$

Calculating interest for Case 2:

$SI_2 = \frac{P \times 14 \times 5}{100} = \frac{70P}{100}$

According to the question:

$SI_1 - SI_2 = 3120$

$\frac{96P}{100} - \frac{70P}{100} = 3120$

$26P = 3120 \times 100$

$P = \frac{312000}{26}$

$P = 12000$

Therefore, the required sum is $\textsf{₹} 12,000$.

Given:

Let the sum (Principal) be $\textsf{₹} P$.

Case 1: $T_1 = 2.5$ years, $R_1 = 12\%$ p.a.

Case 2: $T_2 = 4.5$ years, $R_2 = 8\%$ p.a.

Condition: $SI_1 = SI_2 - 300$

To Find:

The sum ($\textsf{₹} P$).

Solution:

Calculating interest for Case 1:

$SI_1 = \frac{P \times 12 \times 2.5}{100} = \frac{30P}{100} = 0.30P$

Calculating interest for Case 2:

$SI_2 = \frac{P \times 8 \times 4.5}{100} = \frac{36P}{100} = 0.36P$

According to the given condition:

$SI_2 - SI_1 = 300$

$0.36P - 0.30P = 300$

$P = \frac{300}{0.06}$

$P = \frac{30000}{6}$

$P = 5000$

Therefore, the required sum is $\textsf{₹} 5,000$.

Solution:

To answer these questions, let us consider a specific case. Suppose a male student has a height of 5' (152 cm). Referring to the Ideal Height and Weight Proportion table:

1. Your ideal desired weight is __________ kg.

From the table, for a height of 152 cm (Men), the ideal weight is $48$ kg.

2. The quantity of calories you need to eat is _______ .

Calories needed = Ideal weight $\times$ 55.11

Calories = $48 \times 55.11 = 2645.28$

Total calories needed $\approx$ $2645.28$ calories.

3. The quantity of protein needed is ________ g.

Calories from protein = $15\%$ of total calories

Calories from protein = $0.15 \times 2645.28 = 396.792$ cal

Quantity of protein (in grams) = $\frac{\text{Calories from protein}}{4}$

Quantity of protein = $\frac{396.792}{4} = 99.198$ g

Protein required $\approx$ $99.20$ g.

4. The quantity of fat required is ___________ g.

Calories from fat = $30\%$ of total calories

Calories from fat = $0.30 \times 2645.28 = 793.584$ cal

Quantity of fat (in grams) = $\frac{\text{Calories from fat}}{9}$

Quantity of fat = $\frac{793.584}{9} = 88.176$ g

Fat required $\approx$ $88.18$ g.

5. The quantity of carbohydrates required is __________ g.

Calories from carbohydrates = $55\%$ of total calories

Calories from carbohydrates = $0.55 \times 2645.28 = 1454.904$ cal

Quantity of carbohydrates (in grams) = $\frac{\text{Calories from carbohydrates}}{4}$

Quantity of carbohydrates = $\frac{1454.904}{4} = 363.726$ g

Carbohydrates required $\approx$ $363.73$ g.

Given:

Total number of students studying English, Maths or both = $150$

Percentage of students studying English = $62 \%$

Percentage of students studying Maths = $68 \%$

To Find:

Number of students studying both subjects.

Solution:

Let the total percentage of students be $100 \%$.

According to the principle of set theory, the percentage of students studying both subjects can be found by adding the individual percentages and subtracting the total percentage.

Percentage of students studying both = (Percentage in English + Percentage in Maths) - Total Percentage

Percentage of students studying both = $(62 \% + 68 \%) - 100 \%$

Percentage of students studying both = $130 \% - 100 \% = 30 \%$

Now, we find the actual number of students:

Number of students studying both = $30 \%$ of $150$

Number of students studying both = $\frac{30}{100} \times 150$

Number of students studying both = $3 \times 15 = 45$

Therefore, $45$ students are studying both subjects.

(a) Mean, Median, and Mode:

Mean: Sum of all areas divided by $10$.

$\text{Sum} = 8,800,000 + 1,300,000 + 1,250,000 + 850,000 + 580,000 $$ + 370,000 + 320,000 + 310,000 + 310,000 + 260,000 = 14,350,000$

$\text{Mean} = \frac{14,350,000}{10} = 1,435,000$ km$^{2}$.

Median: Areas in descending order are already provided. The median is the average of the 5th and 6th terms.

$\text{Median} = \frac{580,000 + 370,000}{2} = \frac{950,000}{2} = 475,000$ km$^{2}$.

Mode: The value that occurs most frequently.

Since $310,000$ occurs twice, the Mode = $310,000$ km$^{2}$.

(b) How many times the size of the Gobi Desert is the Namib Desert?

Ratio = $\frac{\text{Area of Namib}}{\text{Area of Gobi}} = \frac{310,000}{1,300,000}$

Ratio = $\frac{\cancel{310,000}^{31}}{\cancel{1,300,000}_{130}} \approx 0.238$

The Namib Desert is approximately $0.24$ times the size of the Gobi Desert.

(c) What percentage of the deserts listed are in Asia?

Deserts in Asia: Gobi, Arabian, Takla Makan, Kara Kum, Thar (Total = $5$).

Percentage = $\left( \frac{5}{10} \times 100 \right) \% = 50 \%$

(d) What percentage of the total area of the deserts listed is in Asia?

Area in Asia = $1,300,000 + 850,000 + 320,000 + 310,000 $$ + 260,000 = 3,040,000$ km$^{2}$.

Percentage = $\left( \frac{3,040,000}{14,350,000} \times 100 \right) \%$

Percentage = $\frac{3040}{143.5} \% \approx 21.18 \%$

Given:

Total land area = $148,428,950$ km$^{2}$

Percentage of Asia = $30 \%$

Solution:

Land area of Asia = $30 \%$ of $148,428,950$

Land area of Asia = $\frac{30}{100} \times 148,428,950$

Land area of Asia = $3 \times 14,842,895$

Land area of Asia = $44,528,685$ km$^{2}$.

Solution:

By analyzing the fractions provided in the Tangram shape:

Total Red (R) parts = $\frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}$

Total Blue (B) parts = $\frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$

Total Green (G) parts = $\frac{1}{16} + \frac{1}{16} = \frac{2}{16} = \frac{1}{8}$

(i) Percentage of Red (R):

$\frac{3}{8} \times 100 \% = \frac{300}{8} \% = 37.5 \%$

(ii) Percentage of Blue (B):

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

(iii) Percentage of Green (G):

$\frac{1}{8} \times 100 \% = 12.5 \%$

Verification of Sum:

Sum = $37.5 \% + 50 \% + 12.5 \%$

Sum = $100 \%$ (Verified)

Rearrangement:

If we rearrange the same pieces, the percentage of colours will NOT change because the total area and the area of individual pieces remain constant regardless of their position.