Chapter 10 Direct & Inverse Proportions (Class 8 - Maths NCERT Exemplar Solutions)

Given:

Solution:

Since $x$ and $y$ are directly proportional, their ratio must be constant:

Now, we check the ratio $\frac{x}{y}$ for each option:

(a) For $x = 1$ and $y = 3$, $\frac{x}{y} = \frac{1}{3}$ (Possible)

(b) For $x = 17$ and $y = 51$, $\frac{x}{y} = \frac{17}{51} = \frac{1}{3}$ (Possible)

(c) For $x = 30$ and $y = 10$, $\frac{x}{y} = \frac{30}{10} = 3$ (Not Possible, as $3 \neq \frac{1}{3}$)

(d) For $x = 6$ and $y = 18$, $\frac{x}{y} = \frac{6}{18} = \frac{1}{3}$ (Possible)

Final Answer: The correct option is (c) 30 and 10.

Given:

To Find:

The new average speed ($S_2$).

Solution:

Since the distance remains the same, speed and time are in inverse proportion. Thus, $S_1 T_1 = S_2 T_2$.

Final Answer: The correct option is (a) 80 km/h.

Solution:

(a) Incorrect: Volume depends on three sides ($V = l \times b \times h$). If only one side changes, it is direct, but usually volume is related to the cube of a side in a cube, not a cuboid generally.

(b) Correct: Since $\text{Distance} = \text{Speed} \times \text{Time}$, if time is fixed, $\text{Distance} \propto \text{Speed}$. As speed increases, distance travelled increases in the same proportion.

(c) Incorrect: Height and weight do not increase in a constant ratio for all individuals.

(d) Incorrect: More pipes will take less time to fill a tank. This is an inverse proportion.

Final Answer: The correct option is (b).

Given:

Solution:

Distance and time are in direct proportion when speed is constant:

Final Answer: The time taken is 9 hours.

Solution:

When two quantities $x$ and $y$ are in inverse proportion, their product remains constant.

Final Answer: The blank should be filled with $xy$.

Solution:

Let the side of the rhombus be $s$ and its perimeter be $P$.

Since the ratio of the perimeter to the side is constant, they are in direct proportion.

Final Answer: The blank should be filled with direct.

Solution:

For inverse proportion, the product $xy$ is constant. The ratio $\frac{x}{y}$ being constant is the condition for direct proportion.

Final Answer: The statement is False (F).

Given:

Solution:

The calculated cost matches the value given in the statement.

Final Answer: The statement is True (T).

Given:

Solution:

The number of persons and the time taken are in inverse proportion. As the number of persons decreases, the time taken increases.

Since $50 \text{ days} \neq 2 \text{ days}$, the statement is incorrect.

Final Answer: The statement is False (F).

Given:

To Find:

Number of days the provision will last ($y_2$).

Solution:

The number of cadets and the number of days the food lasts are in inverse proportion. This means if the number of cadets increases, the food will last for fewer days.

Simplifying the fraction:

Final Answer: The provision will last for $36$ days.

Given:

To Find:

Space required for 200 boxes ($y_2$).

Solution:

The number of boxes and the space they occupy are in direct proportion. As the number of boxes increases, the space occupied also increases.

Final Answer: The space required to keep 200 boxes is $50,000 \text{ cm}^3$.

Given:

To Find:

Final pressure ($P_2$).

Solution:

According to the problem, volume and pressure are in inverse proportion at constant temperature ($PV = k$).

Final Answer: The pressure of the gas will be $315 \text{ mm}$ of mercury.

Given:

Solution:

To compare the values, we find the CP and SP of a single lemon:

Since $\text{CP} > \text{SP}$, there is a loss.

Final Answer: The transaction resulted in a loss of $20\%$.

Solution:

Since $u$ and $v$ vary directly, their ratio must be constant:

$\frac{u}{v} = k$

$k = \frac{10}{15} = \frac{2}{3}$

Now, we check the ratio $\frac{u}{v}$ for all the options:

(a) $\frac{2}{3} = \frac{2}{3}$ (Possible)

(b) $\frac{8}{12} = \frac{2}{3}$ (Possible)

(c) $\frac{15}{20} = \frac{3}{4}$ (Not possible, since $\frac{3}{4} \neq \frac{2}{3}$)

(d) $\frac{25}{37.5} = \frac{250}{375} = \frac{2}{3}$ (Possible)

Final Answer: The correct option is (c) 15 and 20.

Solution:

Since $x$ and $y$ vary inversely, their product must be constant:

$x \times y = k$

$k = 10 \times 6 = 60$

Now, we check the product $xy$ for all the options:

(a) $12 \times 5 = 60$ (Possible)

(b) $15 \times 4 = 60$ (Possible)

(c) $25 \times 2.4 = 60$ (Possible)

(d) $45 \times 1.3 = 58.5$ (Not possible, since $58.5 \neq 60$)

Final Answer: The correct option is (d) 45 and 1.3.

Solution:

Under the assumption of uniform fertility, if the area of land increases, the yield (production) will also increase in the same ratio. Similarly, if the area decreases, the yield decreases.

Therefore, the area of land and the yield on it vary directly with each other.

Final Answer: The correct option is (a) directly with each other.

Solution:

The number of teeth in a person increases during childhood, remains constant for a long period during adulthood, and may decrease during old age. This change does not follow a constant ratio or a constant product over time.

Therefore, the number of teeth and the age of a person vary neither directly nor inversely.

Final Answer: The correct option is (c) neither directly nor inversely with each other.

Given:

Diesel consumed ($x_1$) = $54$ litres

Distance covered ($y_1$) = $297$ km

New Distance ($y_2$) = $550$ km

Solution:

The distance covered and the diesel required vary directly with each other.

$\frac{x_1}{y_1} = \frac{x_2}{y_2}$

$\frac{54}{297} = \frac{x_2}{550}$

$x_2 = \frac{54 \times 550}{297}$

$x_2 = \frac{2 \times 550}{11}$

$x_2 = 2 \times 50 = 100 \text{ litres}$

Final Answer: The correct option is (a) 100 litres.

Given:

Speed ($s_1$) = $48$ km/h

Time ($t_1$) = $10$ hours

New Time ($t_2$) = $8$ hours

Solution:

Speed and time are inversely proportional when the distance is constant.

$s_1 \times t_1 = s_2 \times t_2$

$48 \times 10 = s_2 \times 8$

$s_2 = \frac{48 \times 10}{8}$

$s_2 = 6 \times 10 = 60 \text{ km/h}$

Final Answer: The correct option is (a) 60 km/h.

Solution:

For two quantities to vary directly, their ratio must be constant ($\frac{x}{y} = k$). We will check the ratio for each case:

Case (a):

$\frac{x}{y} = \frac{0.5}{2} = 0.25$

$\frac{x}{y} = \frac{2}{8} = 0.25$

$\frac{x}{y} = \frac{8}{32} = 0.25$

$\frac{x}{y} = \frac{32}{128} = 0.25$

Since the ratio is constant ($0.25$) for all pairs, these quantities vary directly.

Case (b):

$\frac{p}{q} = \frac{1^2}{1^3} = 1$

$\frac{p}{q} = \frac{2^2}{2^3} = \frac{4}{8} = 0.5$

The ratios $1$ and $0.5$ are not equal, so they do not vary directly.

Case (c):

$\frac{r}{s} = \frac{2}{25} = 0.08$

$\frac{r}{s} = \frac{5}{10} = 0.5$

The ratios are not constant, so they do not vary directly.

Case (d):

$\frac{u}{v} = \frac{2}{18} \approx 0.11$

$\frac{u}{v} = \frac{4}{9} \approx 0.44$

The ratios are not constant, so they do not vary directly.

Final Answer: The quantities in option (a) vary directly.

Solution:

For two quantities to vary inversely, their product must be constant ($x \times y = k$). We will check the product for each case from the previous question:

Case (a):

$x \times y = 0.5 \times 2 = 1$

$x \times y = 2 \times 8 = 16$

Products are not constant.

Case (b):

$p \times q = 1^2 \times 1^3 = 1$

$p \times q = 2^2 \times 2^3 = 4 \times 8 = 32$

Products are not constant.

Case (c):

$r \times s = 2 \times 25 = 50$

$r \times s = 5 \times 10 = 50$

$r \times s = 10 \times 5 = 50$

$r \times s = 25 \times 2 = 50$

$r \times s = 50 \times 0.5 = 25$

Since the last product ($25$) is different from $50$, this is not a perfect inverse variation.

Case (d):

$u \times v = 2 \times 18 = 36$

$u \times v = 4 \times 9 = 36$

$u \times v = 6 \times 6 = 36$

$u \times v = 9 \times 4 = 36$

$u \times v = 12 \times 3 = 36$

Since the product is constant ($36$) for all pairs, these quantities vary inversely.

Final Answer: The quantities in option (d) vary inversely.

Solution:

Two quantities vary inversely if an increase in one leads to a proportional decrease in the other, such that their product remains constant.

(a) Speed and distance covered: If speed increases, more distance is covered in a fixed time. This is a direct variation.

(b) Distance covered and taxi fare: Generally, more distance means more fare. This is a direct variation.

(c) Distance travelled and time taken: To cover more distance at a fixed speed, more time is required. This is a direct variation.

(d) Speed and time taken: To cover a fixed distance, if the speed increases, the time taken decreases. Their product (Speed $\times$ Time) equals the constant distance. This is an inverse variation.

Final Answer: The correct option is (d) speed and time taken.

Solution:

Given: $x$ and $y$ are in direct proportion.

$\frac{x}{y} = k$ (where $k$ is a constant)

We need to find the relationship between $\frac{1}{x}$ and $\frac{1}{y}$. Let us find their ratio:

$\frac{1/x}{1/y} = \frac{1}{x} \div \frac{1}{y}$

$\frac{1/x}{1/y} = \frac{1}{x} \times y = \frac{y}{x}$

Since $\frac{x}{y} = k$, then $\frac{y}{x} = \frac{1}{k}$. Since $\frac{1}{k}$ is also a constant, the ratio of $\frac{1}{x}$ to $\frac{1}{y}$ is constant.

Therefore, $\frac{1}{x}$ and $\frac{1}{y}$ are also in direct proportion.

Final Answer: $\frac{1}{x}$ and $\frac{1}{y}$ are in direct proportion.

Given:

Initial speed ($s_1$) = $12$ km/h

Initial time ($t_1$) = $20$ minutes

Target time ($t_2$) = $12$ minutes

Solution:

Since the distance to the school remains constant, speed and time vary inversely.

$s_1 \times t_1 = s_2 \times t_2$

$12 \times 20 = s_2 \times 12$

$s_2 = \frac{12 \times 20}{12}$

$s_2 = 20 \text{ km/h}$

Final Answer: The correct option is (c) 20 km/h.

Given:

Initial persons ($x_1$) = $100$

Initial days ($y_1$) = $24$

New number of persons ($x_2$) = $100 - 20 = 80$

Solution:

The number of persons and the number of days for which food lasts are in inverse proportion.

$x_1 \times y_1 = x_2 \times y_2$

$100 \times 24 = 80 \times y_2$

$y_2 = \frac{100 \times 24}{80}$

$y_2 = \frac{10 \times 24}{8}$

$y_2 = 10 \times 3 = 30 \text{ days}$

Final Answer: The correct option is (a) 30 days.

Solution:

By definition of direct proportion, two quantities $x$ and $y$ are said to vary directly if they increase or decrease together such that the ratio of their corresponding values remains constant.

$\frac{x}{y} = k$ (constant)

Final Answer: The correct option is (a) $\frac{x}{y}$ remains constant.

Solution:

By definition of inverse proportion, two quantities $p$ and $q$ are said to vary inversely if an increase in $p$ causes a proportional decrease in $q$ (and vice versa) such that their product remains constant.

$p \times q = k$ (constant)

Final Answer: The correct option is (c) p × q remains constant.

Given:

Distance in 1 hour = $10$ km

Time = $1$ hour = $60$ minutes

Solution:

First, we convert the distance into meters:

$10 \text{ km} = 10 \times 1000 = 10,000 \text{ meters}$

Since speed is constant, distance and time are in direct proportion.

$\text{Distance in 60 minutes} = 10,000 \text{ m}$

$\text{Distance in 1 minute} = \frac{10,000}{60} \text{ m}$

$\text{Distance in 1 minute} = \frac{1000}{6} = \frac{500}{3} \text{ m}$

Final Answer: The correct option is (d) $\frac{500}{3}$ m.

Solution:

Since $x$ and $y$ vary directly, their ratio must be constant:

$\frac{x}{y} = \frac{10}{14} = \frac{5}{7}$

We check the ratio $\frac{x}{y}$ for all options:

(a) $\frac{25}{35} = \frac{5}{7}$ (Possible)

(b) $\frac{35}{25} = \frac{7}{5}$ (Not possible, as $\frac{7}{5} \neq \frac{5}{7}$)

(c) $\frac{35}{49} = \frac{5}{7}$ (Possible)

(d) $\frac{15}{21} = \frac{5}{7}$ (Possible)

Final Answer: The correct option is (b) 35 and 25.

Solution:

The given equation is $x = 5y$.

$\frac{x}{y} = 5$

Since the ratio of $x$ and $y$ is a constant ($5$), they vary directly with each other.

Final Answer: The blank should be filled with directly.

Solution:

The given equation is $xy = 10$.

Since the product of $x$ and $y$ is a constant ($10$), they vary inversely with each other.

Final Answer: The blank should be filled with inversely.

Solution:

The symbol $\propto$ denotes proportionality. The notation $x \propto y$ represents that $x$ is proportional to $y$.

Final Answer: The blanks should be filled with direct and directly.

Solution:

The notation $x \propto \frac{1}{y}$ represents that $x$ is proportional to the reciprocal of $y$, which means they are in inverse proportion.

Final Answer: The blanks should be filled with inverse and inversely.

Solution:

When the product of two variables is equal to a constant $k$, the relationship is known as inverse variation.

Final Answer: The blank should be filled with inversely.

Solution:

In direct variation, the ratio of the two quantities remains constant.

$\frac{x}{y} = k$

Final Answer: The blank should be filled with $\frac{x}{y}$.

Solution:

When the ratio of two quantities is constant, they are said to be in direct proportion.

Final Answer: The blank should be filled with directly.

Solution:

When the product of two quantities is constant, an increase in one will cause a decrease in the other. This is called inverse variation.

Final Answer: The blank should be filled with inversely.

Given:

Number of pumps ($x_1$) = $12$

Time taken ($y_1$) = $20$ hours

New number of pumps ($x_2$) = $45$

Solution:

The number of pumps and the time taken to empty the reservoir vary inversely. More pumps will take less time.

$x_1 \times y_1 = x_2 \times y_2$

$12 \times 20 = 45 \times y_2$

$y_2 = \frac{12 \times 20}{45}$

$y_2 = \frac{240}{45}$

$y_2 = \frac{16}{3} = 5\frac{1}{3} \text{ hours}$

Final Answer: The blank should be filled with $5\frac{1}{3}$ (or approximately $5.33$).

Solution:

Since $x$ varies inversely as $y$, the product $xy$ must be constant.

From the second column:

$k = x \times y = 60 \times 10 = 600$

Using this constant for the first column:

$x \times 2 = 600$

$x = \frac{600}{2} = 300$

Final Answer: The blank in the table should be filled with 300.

Solution:

Since $x$ varies directly as $y$, the ratio $\frac{x}{y}$ must be constant.

From the first column:

$\frac{x}{y} = \frac{12}{48} = \frac{1}{4}$

Using this constant for the second column:

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

$y = 6 \times 4 = 24$

Final Answer: The blank in the table should be filled with 24.

Solution:

We know the formula for distance is:

$\text{Distance} = \text{Speed} \times \text{Time}$

If speed is constant, then the ratio $\frac{\text{Distance}}{\text{Time}}$ is constant. Therefore, distance is directly proportional to time.

Final Answer: The blank should be filled with directly.

Solution:

By the definition of direct variation, the ratio of the two quantities must remain constant as they change together.

Final Answer: The blank should be filled with constant.

Solution:

In inverse variation, as one quantity increases, the other decreases in such a way that their product remains unchanged.

Final Answer: The blanks should be filled with their product and constant respectively.

Solution:

Direct variation implies that the quotient or the ratio of the corresponding values of the two quantities is always the same.

Final Answer: The blank should be filled with ratio (or quotient).

Solution:

Inverse variation is defined by the property that the product of the two quantities remains fixed.

Final Answer: The blank should be filled with product.

Solution:

The perimeter (circumference) $C$ of a circle is given by $C = \pi d$, where $d$ is the diameter.

$\frac{C}{d} = \pi$

Since $\pi$ is a constant, the ratio of perimeter to diameter is constant, which means they vary directly.

Final Answer: The blank should be filled with directly.

Given:

Distance in 60 minutes (1 hour) = $48$ km

Time = $12$ minutes

Solution:

Distance and time are in direct proportion.

$\frac{48}{60} = \frac{x}{12}$

$x = \frac{48 \times 12}{60}$

$x = \frac{48 \times 1}{5} = 9.6 \text{ km}$

Final Answer: The distance travelled is $9.6$ km.

Given:

Initial time = $3$ hours

Distance = $36$ km

Solution:

First, we find the initial speed:

$\text{Initial speed} = \frac{\text{Distance}}{\text{Time}} = \frac{36}{3} = 12 \text{ km/h}$

The speed is increased by $4$ km/h:

$\text{New speed} = 12 + 4 = 16 \text{ km/h}$

Now, calculate the new time for the same distance:

$\text{New time} = \frac{\text{Distance}}{\text{New speed}} = \frac{36}{16}$

$\text{New time} = \frac{9}{4} = 2.25 \text{ hours}$

Converting $0.25$ hours to minutes: $0.25 \times 60 = 15$ minutes.

Final Answer: The time taken is $2.25$ hours (or $2$ hours $15$ minutes).

Given:

Thickness of 12 sheets = $45$ mm

Number of sheets = $240$

Solution:

The number of sheets and thickness are in direct proportion.

$\frac{12}{45} = \frac{240}{x}$

$x = \frac{45 \times 240}{12}$

$x = 45 \times 20 = 900 \text{ mm}$

We need the answer in cm. Since $10 \text{ mm} = 1 \text{ cm}$:

$x = \frac{900}{10} = 90 \text{ cm}$

Final Answer: The thickness of the pile is $90$ cm.

Solution:

Since $x$ varies inversely as $y$, their product remains constant:

$x_1 y_1 = x_2 y_2$

$4 \times 6 = 3 \times y$

$24 = 3y$

$y = \frac{24}{3} = 8$

Final Answer: The blank should be filled with 8.

Solution:

In a direct proportion, the ratio between the corresponding values of two quantities remains equal and constant.

Final Answer: The blank should be filled with is equal to (or the symbol $=$).

Solution:

For inverse proportion, the product is constant ($a_1 b_1 = a_2 b_2$). This can be rearranged as a ratio:

$\frac{a_1}{a_2} = \frac{b_2}{b_1}$

Final Answer: The blanks should be filled such that the expression becomes $\frac{a_1}{a_2} = \frac{b_2}{b_1}$.

Given:

Number of stamps ($x_1$) = $15$

Area ($y_1$) = $60 \text{ cm}^2$

New number of stamps ($x_2$) = $120$

Solution:

The number of stamps and the area they occupy are in direct proportion.

$\frac{x_1}{y_1} = \frac{x_2}{y_2}$

$\frac{15}{60} = \frac{120}{y_2}$

$y_2 = \frac{120 \times 60}{15}$

$y_2 = 120 \times 4 = 480 \text{ cm}^2$

Final Answer: The area occupied will be $480 \text{ cm}^2$.

Given:

Number of persons ($x_1$) = $45$

Time ($y_1$) = $20$ days

New number of persons ($x_2$) = $75$

Solution:

The number of persons and time taken are in inverse proportion ($x_1 y_1 = x_2 y_2$).

$45 \times 20 = 75 \times y_2$

$y_2 = \frac{45 \times 20}{75}$

$y_2 = \frac{3 \times 20}{5} = 3 \times 4 = 12 \text{ days}$

Now, convert the time into hours ($1 \text{ day} = 24 \text{ hours}$):

$\text{Time in hours} = 12 \times 24 = 288 \text{ hours}$

Final Answer: The time taken will be 288 hours.

Given:

Distance ($d_1$) = $50$ m

Steps ($s_1$) = $75$

New steps ($s_2$) = $375$

Solution:

Steps and distance are in direct proportion.

$\frac{50}{75} = \frac{d_2}{375}$

$d_2 = \frac{50 \times 375}{75}$

$d_2 = 50 \times 5 = 250 \text{ m}$

Convert distance to kilometers ($1 \text{ km} = 1000 \text{ m}$):

$\text{Distance in km} = \frac{250}{1000} = 0.25 \text{ km}$

Final Answer: The distance travelled is $0.25$ km.

Solution:

The condition $xy = k$ defines inverse variation. For direct variation, the condition is $\frac{x}{y} = k$.

Final Answer: The statement is False (F).

Solution:

When speed is fixed, $\text{Distance} = \text{Speed} \times \text{Time}$. This means as time increases, the distance covered also increases. Therefore, they vary directly.

Final Answer: The statement is False (F).

Solution:

When distance is fixed, $\text{Speed} \times \text{Time} = \text{Constant}$. As speed increases, time taken decreases. Therefore, they vary inversely.

Final Answer: The statement is False (F).

Solution:

The area of a square ($A$) is given by $A = s^2$, where $s$ is the side. For direct variation, the ratio $\frac{A}{s}$ must be constant. Here, $\frac{A}{s} = \frac{s^2}{s} = s$. Since the ratio depends on the side and is not a constant value, they do not vary directly.

Final Answer: The statement is False (F).

Solution:

For an equilateral triangle with side $s$, the perimeter $P$ is given by:

$P = 3s$

This can be written as:

$\frac{P}{s} = 3$ (Constant)

Since the ratio of the perimeter to the side is a constant, they vary directly with each other, not inversely.

Final Answer: The statement is False (F).

Solution:

If $d$ varies directly as $t^2$, then the ratio of $d$ to $t^2$ must be constant:

$\frac{d}{t^2} = k$

The expression $dt^2 = k$ represents inverse variation between $d$ and $t^2$.

Final Answer: The statement is False (F).

Given:

Height of tree ($x_1$) = $24$ m

Shadow of tree ($y_1$) = $15$ m

Shadow of pole ($y_2$) = $6$ m

Solution:

Under similar conditions, the height of an object and the length of its shadow are in direct proportion.

$\frac{x_1}{y_1} = \frac{x_2}{y_2}$

$\frac{24}{15} = \frac{x_2}{6}$

$x_2 = \frac{24 \times 6}{15}$

$x_2 = \frac{144}{15} = 9.6 \text{ m}$

The calculated height matches the statement.

Final Answer: The statement is True (T).

Solution:

If $x$ and $y$ are in direct proportion, then $\frac{x}{y} = k$. For $(x - 1)$ and $(y - 1)$ to be in direct proportion, the ratio $\frac{x - 1}{y - 1}$ must also equal the same constant $k$.

Let $x = 10$ and $y = 20$, so $k = 0.5$.

$\frac{10 - 1}{20 - 1} = \frac{9}{19} \approx 0.47 \neq 0.5$

Since the ratio changes, they are not in direct proportion.

Final Answer: The statement is False (F).

Solution:

If $x$ and $y$ are in inverse proportion, $xy = k$. For $(x + 1)$ and $(y + 1)$ to be in inverse proportion, their product $(x + 1)(y + 1)$ must remain constant.

Let $x = 2$ and $y = 6$, so $k = 12$.

$(2 + 1)(6 + 1) = 3 \times 7 = 21 \neq 12$

Since the product is not constant, they are not in inverse proportion.

Final Answer: The statement is False (F).

Solution:

Similar to the previous examples, adding or subtracting a constant from variables in an inverse proportion does not maintain the constant product property ($p \times q = k$).

$(p + 2)(q - 2) = pq - 2p + 2q - 4$

This expression will not result in the constant $k$ for all values of $p$ and $q$.

Final Answer: The statement is False (F).

Solution:

Let the angles be $A$, $B$, and $C$. If $\angle A$ is fixed:

$A + B + C = 180^\circ$

$B + C = 180^\circ - A = \text{Constant}$

Inverse variation requires the product of the quantities to be constant ($B \times C = k$). Here, the sum is constant, which does not define inverse proportion.

Final Answer: The statement is False (F).

Solution:

Direct variation requires more than just both quantities increasing; they must increase in a constant ratio. For example, in $y = x^2$, as $x$ increases, $y$ increases, but the ratio $\frac{y}{x}$ is not constant ($x$), so it is not a direct variation.

Final Answer: The statement is False (F).

Solution:

Inverse variation requires the product of the quantities to be constant. If one increases and the other decreases, they do not "always" vary inversely. For example, in $y = 10 - x$ (where $x < 10$), as $x$ increases, $y$ decreases, but their product $x(10 - x)$ is not constant.

Final Answer: The statement is False (F).

Given:

$x_1 = 6, y_1 = 8$

$x_2 = 8$

Solution:

In inverse variation:

$x_1 y_1 = x_2 y_2$

$6 \times 8 = 8 \times y_2$

$48 = 8 y_2$

$y_2 = \frac{48}{8} = 6$

The statement claims $y = 10$, which is incorrect.

Final Answer: The statement is False (F).

Solution:

The number of workers and the time taken to complete a job vary inversely. If the number of workers increases, the time required to finish the job decreases. Since they do not increase or decrease together in a constant ratio, it is not a case of direct proportion.

Final Answer: The statement is False (F).

Solution:

The formula for Simple Interest (SI) is:

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

If the Rate ($R$) and Time ($T$) are fixed, then $\frac{RT}{100}$ becomes a constant $k$. Thus, $\text{SI} = k \times P$. Since the ratio $\frac{\text{SI}}{P}$ remains constant, they are in direct proportion.

Final Answer: The statement is True (T).

Solution:

Assuming the fertility of the land is uniform, an increase in the area of cultivated land will result in a proportional increase in the amount of crop harvested. Therefore, it is a case of direct proportion.

Final Answer: The statement is True (T).

Solution:

(i) Vary Inversely: For a fixed distance, increasing the speed reduces the time taken.

(ii) Vary Directly: More distance covered requires more CNG.

(iii) Vary Inversely: More people working reduces the time taken to finish the work.

(iv) Vary Directly: As the income increases, the income tax amount also increases (as per Indian tax slabs).

(v) Vary Directly: For a constant speed, as the time increases, the distance travelled also increases.

Solution:

(i) Vary Directly: More students will consume more food.

(ii) Vary Directly: Larger wall area leads to a higher cost of white washing.

(iii) Vary Directly: More people working (for a fixed time) can complete a greater quantity of work.

(iv) Vary Directly: A higher rate of interest results in more simple interest for the same sum and time.

(v) Vary Directly: A larger sum invested will result in a higher compound interest earned.

Solution:

(i) Vary Directly: More quantity of rice will cost more.

(ii) Neither: A tree's growth is not proportional to time throughout its life; it eventually stops or slows down significantly.

(iii) Vary Inversely: If the price per shirt increases, the number of shirts you can buy with a fixed budget decreases.

(iv) Vary Directly: Larger area of land costs more.

(v) Vary Directly: A higher bill amount results in a higher sales tax (calculated as a percentage of the bill).

Given:

$x_1 = 20, y_1 = 600$

$x_2 = 400$

To Find:

The value of $y_2$.

Solution:

Since $x$ and $y$ vary inversely, their product remains constant:

$x_1 y_1 = x_2 y_2$

$20 \times 600 = 400 \times y_2$

$12,000 = 400 y_2$

$y_2 = \frac{12,000}{400}$

$y_2 = \frac{120}{4} = 30$

Final Answer: The value of $y$ is 30.

Given:

Initial value of $x$ ($x_1$) = $80$

Initial value of $y$ ($y_1$) = $160$

New value of $x$ ($x_2$) = $64$

Solution:

Since $x$ and $y$ vary directly, their ratio remains constant:

$\frac{x_1}{y_1} = \frac{x_2}{y_2}$

$\frac{80}{160} = \frac{64}{y_2}$

$\frac{1}{2} = \frac{64}{y_2}$

$y_2 = 64 \times 2$

$y_2 = 128$

Final Answer: The value of $y$ is 128.

Given:

$l_1 = 5, m_1 = \frac{2}{3}$

$m_2 = \frac{16}{3}$

Solution:

For direct variation, the ratio of the quantities is constant:

$\frac{l_1}{m_1} = \frac{l_2}{m_2}$

$l_1 \times m_2 = l_2 \times m_1$

$5 \times \frac{16}{3} = l_2 \times \frac{2}{3}$

Multiplying both sides by $3$ to simplify:

$5 \times 16 = l_2 \times 2$

$80 = 2 \times l_2$

$l_2 = \frac{80}{2} = 40$

Final Answer: The value of $l$ is 40.

Given:

$x_1 = 1.5, y_1 = 60$

$y_2 = 4.5$

Solution:

Since $x$ varies inversely as $y$, their product remains constant:

$x_1 \times y_1 = x_2 \times y_2$

$1.5 \times 60 = x_2 \times 4.5$

$90 = x_2 \times 4.5$

$x_2 = \frac{90}{4.5}$

$x_2 = \frac{900}{45} = 20$

Final Answer: The value of $x$ is 20.

Given:

Initial number of persons ($x_1$) = $300$

Initial number of days ($y_1$) = $42$

New number of persons ($x_2$) = $300 + 20 = 320$

Solution:

The number of persons and the number of days for which the flour lasts are in inverse proportion.

$x_1 \times y_1 = x_2 \times y_2$

$300 \times 42 = 320 \times y_2$

$y_2 = \frac{300 \times 42}{320}$

$y_2 = \frac{30 \times 42}{32}$

$y_2 = \frac{15 \times 42}{16} = \frac{15 \times 21}{8}$

$y_2 = \frac{315}{8} = 39.375 \text{ days}$

Final Answer: The flour will last for 39.375 days.

Given:

Initial time ($t_1$) = $9$ months

Initial persons ($p_1$) = $560$

New time ($t_2$) = $5$ months

Solution:

The time taken and the number of persons are in inverse proportion.

$t_1 \times p_1 = t_2 \times p_2$

$9 \times 560 = 5 \times p_2$

$p_2 = \frac{9 \times 560}{5}$

$p_2 = 9 \times 112 = 1008 \text{ persons}$

To find the extra persons required:

$\text{Extra persons} = 1008 - 560 = 448$

Final Answer: He should employ 448 extra persons.

Given:

Words typed ($w_1$) = $108$

Time taken ($t_1$) = $6$ minutes

New time ($t_2$) = $\frac{1}{2}$ hour = $30$ minutes

Solution:

The number of words typed and time taken vary directly.

$\frac{w_1}{t_1} = \frac{w_2}{t_2}$

$\frac{108}{6} = \frac{w_2}{30}$

$18 = \frac{w_2}{30}$

$w_2 = 18 \times 30 = 540$

Final Answer: She would type 540 words in half an hour.

Given:

Initial time ($t_1$) = $40$ minutes

Initial average speed ($s_1$) = $60$ km/h

New time ($t_2$) = $25$ minutes

Solution:

Speed and time are in inverse proportion for a fixed distance.

$s_1 \times t_1 = s_2 \times t_2$

$60 \times 40 = s_2 \times 25$

$2400 = 25 \times s_2$

$s_2 = \frac{2400}{25} = 96 \text{ km/h}$

Final Answer: The average speed should be 96 km/h.

Solution:

Since $l$ varies directly as $m$, their ratio is constant.

(i) Equation relating $l$ and $m$:

$l \propto m$

$l = km$

(ii) Finding the constant of proportion ($k$):

Given: $l = 6, m = 18$

$k = \frac{l}{m}$

$k = \frac{\cancel{6}^1}{\cancel{18}_3} = \frac{1}{3}$

(iii) Finding $l$ when $m = 33$:

$l = km$

$l = \frac{1}{3} \times 33 = 11$

(iv) Finding $m$ when $l = 8$:

$l = km$

$8 = \frac{1}{3} \times m$

$m = 8 \times 3 = 24$

Final Answers: (i) $l = km$, (ii) $k = \frac{1}{3}$, (iii) $l = 11$, (iv) $m = 24$.

Given:

Initial Principal ($P_1$) = $\textsf{₹} 2,000$

Interest earned ($I_1$) = $\textsf{₹} 500$

New Principal ($P_2$) = $\textsf{₹} 36,000$

To Find:

Interest earned on the new deposit ($I_2$).

Solution:

Since the time period (3 years) and the rate of interest are the same, the simple interest earned varies directly with the principal deposited.

$\frac{I_1}{P_1} = \frac{I_2}{P_2}$

$\frac{500}{2000} = \frac{I_2}{36000}$

$\frac{1}{4} = \frac{I_2}{36000}$

$I_2 = \frac{36000}{4} = 9,000$

Final Answer: A deposit of $\textsf{₹} 36,000$ would earn $\textsf{₹} 9,000$ interest.

Given:

Initial length ($l_1$) = $16$ cm

Initial mass ($m_1$) = $192$ g

New mass ($m_2$) = $105$ g

To Find:

New length ($l_2$).

Solution:

The mass and length are in direct proportion.

$\frac{l_1}{m_1} = \frac{l_2}{m_2}$

$\frac{16}{192} = \frac{l_2}{105}$

$\frac{1}{12} = \frac{l_2}{105}$

$l_2 = \frac{105}{12}$

$l_2 = 8.75 \text{ cm}$

Final Answer: The length of the rod is $8.75$ cm.

Solution:

Since $a$ and $b$ are in inverse proportion, their product remains constant ($ab = k$).

From the given data pairs:

$k = 12 \times 30 = 360$

Finding $x$:

$x \times 5 = 360$

$x = \frac{360}{5} = 72$

Finding $y$:

$8 \times y = 360$

$y = \frac{360}{8} = 45$

Final Answer: The values are $x = 72$ and $y = 45$.

Given:

Initial steps ($s_1$) = $250$

Initial distance ($d_1$) = $200$ m

New steps ($s_2$) = $350$

To Find:

Distance covered in 350 steps ($d_2$).

Solution:

Steps and distance are in direct proportion.

$\frac{s_1}{d_1} = \frac{s_2}{d_2}$

$\frac{250}{200} = \frac{350}{d_2}$

$\frac{5}{4} = \frac{350}{d_2}$

$d_2 = \frac{350 \times 4}{5}$

$d_2 = 70 \times 4 = 280 \text{ m}$

Final Answer: The distance travelled in 350 steps is $280$ metres.

Given:

Distance covered ($d_1$) = $225$ km

Petrol consumed ($p_1$) = $25$ litres

New distance to cover ($d_2$) = $540$ km

Solution:

The distance covered and the petrol required are in direct proportion.

$\frac{d_1}{p_1} = \frac{d_2}{p_2}$

$\frac{225}{25} = \frac{540}{p_2}$

$9 = \frac{540}{p_2}$

$p_2 = \frac{540}{9} = 60 \text{ litres}$

Final Answer: The car will require $60$ litres of petrol to cover $540$ km.

For $x$ and $y$ to be in direct proportion, the ratio $\frac{y}{x}$ (or $\frac{x}{y}$) must be constant for all pairs.

(i) Calculation:

$\frac{y}{x} = \frac{12}{3} = 4$

$\frac{y}{x} = \frac{24}{6} = 4$

$\frac{y}{x} = \frac{45}{15} = 3$

Since the ratio is not constant ($4 \neq 3$), $x$ and $y$ are not in direct proportion.

(ii) Calculation:

$\frac{y}{x} = \frac{24}{4} = 6$

$\frac{y}{x} = \frac{42}{7} = 6$

$\frac{y}{x} = \frac{60}{10} = 6$

$\frac{y}{x} = \frac{96}{16} = 6$

Since the ratio is constant ($6$) for all pairs, $x$ and $y$ are in direct proportion.

(iii) Calculation:

$\frac{y}{x} = \frac{1.5}{1} = 1.5$

$\frac{y}{x} = \frac{6}{4} = 1.5$

$\frac{y}{x} = \frac{13.5}{9} = 1.5$

$\frac{y}{x} = \frac{30}{20} = 1.5$

Since the ratio is constant ($1.5$) for all pairs, $x$ and $y$ are in direct proportion.

Since $a$ and $b$ vary inversely, their product must be constant ($ab = k$).

(i) Calculation:

$k = 6 \times 18 = 108$

$p = \frac{108}{8} = 13.5$

$q = \frac{108}{39} = \frac{36}{13} \approx 2.77$

$r = \frac{108}{25} = 4.32$

(ii) Calculation:

$k = 6 \times 15 = 90$

$x = \frac{90}{2} = 45$

$y = \frac{90}{12.5} = 7.2$

$z = \frac{90}{10} = 9$

(iii) Calculation:

$k = 6 \times 10 = 60$

$l = \frac{60}{5} = 12$

$m = \frac{60}{9} = \frac{20}{3} \approx 6.67$

$n = \frac{60}{25} = 2.4$

The length of the cloth and its cost are in direct proportion.

$\text{Cost per metre} = \frac{337.50}{25} = \textsf{₹} 13.50$

(i) Cost of 40 metres:

$\text{Cost} = 40 \times 13.50 = \textsf{₹} 540$

(ii) Length for $\textsf{₹} 810$:

$\text{Length} = \frac{810}{13.50} = 60 \text{ metres}$

Initial time ($t_1$) = $4$ hours

Initial pumps ($p_1$) = $8$

New time ($t_2$) = $2\frac{2}{3}$ hours = $\frac{8}{3}$ hours

Solution:

The time taken and the number of pumps vary inversely.

$p_1 \times t_1 = p_2 \times t_2$

$8 \times 4 = p_2 \times \frac{8}{3}$

$32 = \frac{8p_2}{3}$

$p_2 = \frac{32 \times 3}{8}$

$p_2 = 4 \times 3 = 12 \text{ pumps}$

Final Answer: 12 pumps are required to fill the pool in $2\frac{2}{3}$ hours.

Given:

Weight of iron ($x_1$) = $27$ kg

Cost of iron ($y_1$) = $\textsf{₹} 1,080$

New weight of iron ($x_2$) = $120$ kg

To Find:

The cost of $120$ kg of iron ($y_2$).

Solution:

The weight of iron and its cost vary directly with each other. As the weight increases, the cost also increases.

$\frac{x_1}{y_1} = \frac{x_2}{y_2}$

$\frac{27}{1080} = \frac{120}{y_2}$

$y_2 = \frac{120 \times 1080}{27}$

$y_2 = 120 \times 40$

$y_2 = \textsf{₹} 4,800$

Final Answer: The cost of $120$ kg of iron will be $\textsf{₹} 4,800$.

Given:

Height of Qutub Minar ($h_1$) = $72$ m

Length of shadow of Qutub Minar ($s_1$) = $80$ m

Length of shadow of electric pole ($s_2$) = $1000$ cm = $10$ m

To Find:

The height of the electric pole ($h_2$).

Solution:

At a particular time, the height of an object and the length of its shadow vary directly.

$\frac{h_1}{s_1} = \frac{h_2}{s_2}$

$\frac{72}{80} = \frac{h_2}{10}$

$h_2 = \frac{72 \times 10}{80}$

$h_2 = \frac{72}{8}$

$h_2 = 9$ m

Final Answer: The height of the electric pole is $9$ m.

Given:

Initial number of girls ($x_1$) = $50$

Number of days food lasts ($y_1$) = $40$

New number of girls ($x_2$) = $50 + 30 = 80$

To Find:

The number of days the provision will last ($y_2$).

Solution:

The number of girls and the time the food lasts vary inversely. More girls will consume the food faster.

$x_1 \times y_1 = x_2 \times y_2$

$50 \times 40 = 80 \times y_2$

$2000 = 80 \times y_2$

$y_2 = \frac{2000}{80}$

$y_2 = 25$ days

Final Answer: The food provisions will last for $25$ days.

Solution:

A shade is considered "lighter" if the ratio of blue paint to white paint is smaller. Let us count the containers from the image and calculate the ratios:

Comparing the Mixtures:

(i) Between A and B: Mixture A ($0.75$) has a lower ratio than B ($1.00$). So, Mixture A is a lighter shade.

(ii) Between C and D: Mixture D ($0.40$) has a lower ratio than C ($1.00$). So, Mixture D is a lighter shade.

(iii) Between E and F: Mixture F ($2.00$) has a lower ratio than E ($6.00$). So, Mixture F is a lighter shade.

(iv) Between G and H: Mixture G ($1.00$) has a lower ratio than H ($1.33$). So, Mixture G is a lighter shade.

Lightest Blue Shade:

The lightest blue shade is the one with the minimum ratio of blue paint to white paint.

Comparing all ratios: $0.75, 1.00, 1.00, 0.40, 6.00, 2.00, 1.00, 1.33$.

The minimum value is $0.40$, which belongs to Mixture D.

Darkest Blue Shade Calculation:

The darkest blue shade is the one with the maximum ratio. The maximum ratio is $6.00$ for Mixture E.

$\text{Ratio of Blue to White} = 6 : 1$

$\text{Total parts} = 6 + 1 = 7$

The building requires $105$ litres of paint. Since each container is $1$ litre:

$\text{Quantity of one part} = \frac{105}{7} = 15 \text{ litres}$

$\text{Number of Blue containers} = 6 \times 15 = 90$

$\text{Number of White containers} = 1 \times 15 = 15$

Final Answer: The lightest shade is D. To paint the building with the darkest shade (E), 90 blue containers and 15 white containers are required.

Solution:

Based on the observation of the quilt pattern, we have the following count of squares for each colour in one quilt:

- Number of Purple squares (dark pattern): $12$

- Number of White squares (flower pattern): $16$

- Number of Blue squares (light pattern): $20$

Total squares in one quilt = $12 + 16 + 20 = 48$ squares.

Five Ratio Statements:

1. The ratio of Purple squares to White squares is $12 : 16$, which simplifies to $3 : 4$.

2. The ratio of White squares to Blue squares is $16 : 20$, which simplifies to $4 : 5$.

3. The ratio of Purple squares to Blue squares is $12 : 20$, which simplifies to $3 : 5$.

4. The ratio of Purple squares to the Total squares is $12 : 48$, which simplifies to $1 : 4$.

5. The ratio of Blue squares to the Total squares is $20 : 48$, which simplifies to $5 : 12$.

Calculation for 12 Quilts:

Since the number of squares of each colour is in direct proportion to the number of quilts, we multiply the count of each colour in one quilt by 12.

For Purple squares:

$12 \times 12 = 144$

For White squares:

$16 \times 12 = 192$

For Blue squares:

$20 \times 12 = 240$

Final Answer: In 12 such quilts, there will be 144 Purple squares, 192 White squares, and 240 Blue squares.

Given:

Initial number of children ($x_1$) = $10$

Sweets per child ($y_1$) = $4$

New number of children ($x_2$) = $8$

Solution:

The number of children and the number of sweets each child gets are in inverse proportion, as the total number of sweets remains constant.

$x_1 y_1 = x_2 y_2$

$10 \times 4 = 8 \times y_2$

$40 = 8y_2$

$y_2 = \frac{40}{8} = 5$

Final Answer: Each child will get 5 sweets.

Given:

Initial number of cows ($x_1$) = $44$

Number of days ($y_1$) = $9$

New number of days ($y_2$) = $12$

Solution:

The number of cows and the number of days required to graze the field are in inverse proportion. More cows will take fewer days to graze the same field.

$x_1 y_1 = x_2 y_2$

$44 \times 9 = x_2 \times 12$

$x_2 = \frac{44 \times 9}{12}$

$x_2 = 11 \times 3 = 33 \text{ cows}$

To find how many less cows are needed:

$\text{Decrease in cows} = 44 - 33 = 11$

Final Answer: 11 less cows will graze the field in 12 days.

Given:

Initial number of persons ($x_1$) = $30$

Number of days ($y_1$) = $17$

Target number of days ($y_2$) = $10$

Solution:

The number of persons and the number of days to reap the field are in inverse proportion.

$x_1 y_1 = x_2 y_2$

$30 \times 17 = x_2 \times 10$

$x_2 = \frac{30 \times 17}{10} = 3 \times 17 = 51 \text{ persons}$

To find how many more persons are required:

$\text{More persons} = 51 - 30 = 21$

Final Answer: 21 more persons should be engaged.

Given:

Initial time ($t_1$) = $20$ minutes

Initial speed ($s_1$) = $6$ km/h

New time ($t_2$) = $24$ minutes

Solution:

Speed and time are in inverse proportion when the distance is constant.

$s_1 t_1 = s_2 t_2$

$6 \times 20 = s_2 \times 24$

$s_2 = \frac{120}{24} = 5 \text{ km/h}$

Final Answer: Her speed should be 5 km/h.

Given:

Speed 1 ($s_1$) = $10$ km/h

Speed 2 ($s_2$) = $16$ km/h

Time difference = $8$ min late to $10$ min early = $18$ minutes

Solution:

Let the distance to the school be $D$ km.

Time taken at speed $s_1$ ($t_1$) = $\frac{D}{10}$ hours

Time taken at speed $s_2$ ($t_2$) = $\frac{D}{16}$ hours

The difference between these times is $18$ minutes, which is $\frac{18}{60} = \frac{3}{10}$ hours.

$\frac{D}{10} - \frac{D}{16} = \frac{3}{10}$

Taking LCM of $10$ and $16$, which is $80$:

$\frac{8D - 5D}{80} = \frac{3}{10}$

$\frac{3D}{80} = \frac{3}{10}$

$D = \frac{3}{10} \times \frac{80}{3} = 8 \text{ km}$

Now, calculate time taken at $10$ km/h:

$t_1 = \frac{8}{10} \text{ hours} = 48 \text{ minutes}$

Since he is $8$ minutes late at this speed, the actual time allowed to reach school is:

$\text{Allowed Time} = 48 - 8 = 40 \text{ minutes}$

The school starts 40 minutes after he leaves at 8:20 a.m.:

$\text{School Start Time} = 8:20 \text{ a.m.} + 40 \text{ minutes} = 9:00 \text{ a.m.}$

Final Answer: The school starts at 9:00 a.m.

Solution:

According to the definitions and mathematical properties of direct and inverse proportions, the correct matching is provided below:

1. x and y vary inversely to each other — This is a conceptual statement regarding the relationship between the two variables. It matches with H. x and y vary inversely.

2. Mathematical representation of inverse variation of quantities p and q — Inverse variation is represented as one quantity being proportional to the reciprocal of the other. It matches with D. $p \propto \frac{1}{q}$.

3. Mathematical representation of direct variation of quantities m and n — Direct variation is represented as one quantity being proportional to the other. It matches with G. $m \propto n$.

4. When x = 5, y = 2.5 and when y = 5, x = 10 — Here, the ratio $\frac{x}{y} = \frac{5}{2.5} = 2$ and $\frac{10}{5} = 2$. Since the ratio is constant, it matches with F. x and y are directly proportional.

5. When x = 10, y = 5 and when x = 20, y = 2.5 — Here, the product $x \times y = 10 \times 5 = 50$ and $20 \times 2.5 = 50$. Since the product is constant, it matches with C. $xy = \text{Constant}$.

6. x and y vary directly with each other — By definition, for direct variation, the ratio of corresponding values is always the same. It matches with A. $\frac{x}{y} = \text{Constant}$.

7. If x and y vary inversely then on decreasing x — In an inverse relationship, as one value decreases, the other must increase to maintain a constant product. It matches with B. y will increase in proportion.

8. If x and y vary directly then on decreasing x — In a direct relationship, both values move in the same direction to maintain a constant ratio. It matches with E. y will decrease in proportion.

Final Answer (Mapping):

1 — H; 2 — D; 3 — G; 4 — F; 5 — C; 6 — A; 7 — B; 8 — E

Given:

Weight of fish ($w_1$) = $75$ g

Amount of protein ($p_1$) = $20$ g

New weight of fish ($w_2$) = $225$ g

To Find:

Amount of protein in 225 g of fish ($p_2$).

Solution:

The amount of protein in the fish varies directly with the total weight of the fish.

$\frac{p_1}{w_1} = \frac{p_2}{w_2}$

$\frac{20}{75} = \frac{p_2}{225}$

$p_2 = \frac{20 \times 225}{75}$

$p_2 = 20 \times 3$

$p_2 = 60 \text{ g}$

Final Answer: There are $60$ grams of protein in $225$ gm of fish.

Given:

Distance on map = $3.5$ cm

Map scale: $1$ cm = $10$ km

To Find:

Actual distance between the villages.

Solution:

The actual distance is directly proportional to the distance measured on the map.

$\text{Actual distance} = \text{Map distance} \times \text{Scale factor}$

$\text{Actual distance} = 3.5 \times 10$

$\text{Actual distance} = 35 \text{ km}$

Final Answer: The actual distance between the villages is $35$ km.

Given:

Length of the tank's shadow ($s_1$) = $21 \text{ m}$

Height of the tree ($h_2$) = $9.5 \text{ m}$

Length of the tree's shadow ($s_2$) = $8 \text{ m}$

To Find:

Height of the water tank ($x$ or $h_1$).

Solution:

As given, the heights of the objects and the lengths of their shadows are in direct proportion.

$\frac{\text{Height of tank}}{\text{Shadow of tank}} = \frac{\text{Height of tree}}{\text{Shadow of tree}}$

$\frac{x}{21} = \frac{9.5}{8}$

$x = \frac{9.5 \times 21}{8}$

$x = \frac{199.5}{8}$

$x = 24.9375 \text{ m}$

Final Answer: The height of the water tank is $24.9375 \text{ m}$.

Solution:

First, we calculate the speed of each elevator using the formula $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$.

Elevator A:

$\text{Speed}_A = \frac{435}{29} = 15 \text{ m/s}$

Elevator B:

$\text{Speed}_B = \frac{448}{28} = 16 \text{ m/s}$

Elevator C:

$\text{Speed}_C = \frac{130}{10} = 13 \text{ m/s}$

Elevator D:

$\text{Speed}_D = \frac{85}{5} = 17 \text{ m/s}$

Comparing the speeds: $17 > 16 > 15 > 13$. Thus, Elevator D is the fastest and Elevator C is the slowest.

Distance travelled in 140 seconds:

For Elevator B:

$\text{Distance} = \text{Speed} \times \text{Time}$

$\text{Distance}_B = 16 \times 140 = 2240 \text{ m}$

For Elevator C:

$\text{Distance}_C = 13 \times 140 = 1820 \text{ m}$

Comparison:

$\text{Difference} = 2240 - 1820 = 420 \text{ m}$

Final Answer: Elevator D is the fastest and C is the slowest. In 140 sec, Elevator B travelled more than Elevator C by $420 \text{ m}$.

Given:

Dimensions of the Volleyball Court: $\text{Width}_1 = 9 \text{ m}$, $\text{Length}_1 = 18 \text{ m}$

Dimensions of the Swimming Pool: $\text{Length}_2 = 75 \text{ m}$

To Find:

Width of the swimming pool ($w$).

Solution:

Since the dimensions are directly proportional, the ratio of width to length remains constant.

$\frac{\text{Width}_1}{\text{Length}_1} = \frac{\text{Width}_2}{\text{Length}_2}$

$\frac{9}{18} = \frac{w}{75}$

$\frac{1}{2} = \frac{w}{75}$

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

$w = 37.5 \text{ m}$

Final Answer: The width of the swimming pool is $37.5 \text{ m}$.

Given:

Milk required for the recipe ($m_1$) = $1$ cup

Chocolates required for the recipe ($c_1$) = $1.5$ cups

Available chocolates ($c_2$) = $7.5$ cups

To Find:

Number of cups of milk needed ($m_2$).

Solution:

The amount of milk and chocolates required for muffins are in direct proportion.

$\frac{m_1}{c_1} = \frac{m_2}{c_2}$

$\frac{1}{1.5} = \frac{m_2}{7.5}$

$m_2 = \frac{1 \times 7.5}{1.5}$

$m_2 = \frac{75}{15} = 5 \text{ cups}$

Final Answer: Riya will need 5 cups of milk.

Solution:

By observing the given patterns in the image:

For Pattern A:

Number of blue dots = $4$

Number of red dots = $2$

For Pattern B:

Pattern B consists of four tiles of Pattern A. Therefore:

Number of blue dots = $4 \times 4 = 16$

Number of red dots = $2 \times 4 = 8$

Checking for Proportion:

To determine if they are in direct proportion, we compare the ratio of red dots to blue dots for both patterns:

$\text{Ratio in Pattern A} = \frac{2}{4} = \frac{1}{2}$

$\text{Ratio in Pattern B} = \frac{8}{16} = \frac{1}{2}$

Since the ratio of the corresponding values is constant ($\frac{1}{2} = \frac{1}{2}$), the number of red dots and blue dots are in direct proportion.

Constant of Proportion:

The constant of proportion ($k$) is the constant ratio between the two quantities:

$k = \frac{2}{4} = \frac{1}{2}$ (or $0.5$)

Final Answer: Yes, they are in direct proportion. The constant of proportion is $k = \frac{1}{2}$.

Given:

Speed of the ball = $120$ km/h

Distance = $20$ m

Solution:

First, we convert the speed from km/h to m/s:

$120 \text{ km/h} = 120 \times \frac{5}{18} \text{ m/s}$

$\text{Speed} = \frac{20 \times 5}{3} = \frac{100}{3} \text{ m/s}$

Now, we find the time using the formula:

$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$

$\text{Time} = \frac{20}{100/3}$

$\text{Time} = 20 \times \frac{3}{100}$

$\text{Time} = \frac{60}{100} = 0.6 \text{ seconds}$

Final Answer: The ball takes 0.6 seconds to reach the batsman.

Solution:

Given that $x$ and $y$ are inversely proportional to each other. This means their product remains constant:

$x \times y = \text{Constant}$

In an inverse variation, if one quantity increases by a certain percentage, the other quantity must decrease to ensure that the product of the two variables does not change.

According to the properties of inverse proportion for such variations, an increase in the value of $x$ by $p\%$ is balanced by a corresponding decrease in the value of $y$ to maintain the equilibrium of the product.

Therefore, the percentage decrease in $y$ will be equal to the percentage increase in $x$.

$\text{Percentage decrease in } y = p\%$

Final Answer: If $x$ increases by $p\%$, then $y$ will decrease by $p\%$.

Solution:

By observing the given pattern of the harmonium keyboard provided in the image:

Number of white keys = $10$

Number of black keys = $7$

Total number of keys = $10 + 7 = 17$

(a) Ratio of white keys to black keys:

$\text{Ratio} = \frac{\text{Number of white keys}}{\text{Number of black keys}}$

$\text{Ratio} = \frac{10}{7}$

$\text{Ratio} = 10 : 7$

(b) Ratio of black keys to all keys:

$\text{Ratio} = \frac{\text{Number of black keys}}{\text{Total number of keys}}$

$\text{Ratio} = \frac{7}{17}$

$\text{Ratio} = 7 : 17$

(c) Black keys in 14 such patterns:

Since the number of black keys is directly proportional to the number of patterns:

$\text{Black keys in 1 pattern} = 7$

$\text{Black keys in 14 patterns} = 7 \times 14$

$\text{Total black keys} = 98$

Final Answer: (a) The ratio of white keys to black keys is $10 : 7$. (b) The ratio of black keys to all keys is $7 : 17$. (c) There would be $98$ black keys in 14 such patterns.

Solution:

To determine the type of property, let us check the ratio of distance ($y$) to litres of gas ($x$) for each pair:

$\frac{15}{1} = 15$

$\frac{7.5}{0.5} = 15$

$\frac{30}{2} = 15$

$\frac{37.5}{2.5} = 15$

$\frac{45}{3} = 15$

$\frac{75}{5} = 15$

Since the ratio $\frac{y}{x}$ is constant ($k = 15$), the table indicates the property of direct proportion.

Distance covered in 8 litres:

Since $\text{Distance} = 15 \times \text{Litres of gas}$:

$\text{Distance} = 15 \times 8 = 120 \text{ km}$

Final Answer: The table indicates direct proportion and the car will cover $120$ km in 8 litres of gas.

Solution:

Kritika has $\frac{1}{6}$ cup of sugar syrup, while the original recipe requires $\frac{1}{3}$ cup. We first find the scaling factor:

$\text{Scaling Factor} = \frac{\text{Amount Available}}{\text{Recipe Amount}} = \frac{1/6}{1/3} = \frac{1}{6} \times 3 = \frac{1}{2}$

Since she is using half the amount of syrup, she must use half of every other ingredient to maintain the proportion.

New Quantities:

- Quick cooking oats: $1 \times \frac{1}{2} = \frac{1}{2}$ cup

- Bread flour: $2 \times \frac{1}{2} = 1$ cup

- Sugar syrup: $\frac{1}{6}$ cup (Given)

- Cooking oil: $1 \times \frac{1}{2} = \frac{1}{2}$ tablespoon

- Water: $1\frac{1}{3} = \frac{4}{3}$ cups; $\frac{4}{3} \times \frac{1}{2} = \frac{2}{3}$ cup

- Yeast: $3 \times \frac{1}{2} = 1\frac{1}{2}$ tablespoons

- Salt: $1 \times \frac{1}{2} = \frac{1}{2}$ teaspoon

Final Answer: Kritika should use the scaled quantities listed above.

Given:

Student-Teacher ratio = $35 : 1$

Increase in students = $280$

Solution:

The number of teachers required is directly proportional to the number of students. Let the number of new teachers to be appointed be $x$.

$\frac{\text{Students}}{\text{Teachers}} = \frac{35}{1}$

$\frac{280}{x} = \frac{35}{1}$

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

$x = \frac{\cancel{280}^{8}}{\cancel{35}_{1}} = 8$

Final Answer: The school will have to appoint 8 new teachers.

Given:

$50 \text{ miles} = 80 \text{ kilometres}$

Target distance = $200$ km

Solution:

The conversion between miles and kilometres is a direct proportion.

$\frac{\text{Miles}}{\text{Kilometres}} = \frac{50}{80}$

$\text{Miles for 200 km} = \frac{50}{80} \times 200$

$\text{Miles} = \frac{5}{8} \times 200$

$\text{Miles} = 5 \times 25 = 125 \text{ miles}$

Final Answer: Kusum would have to go 125 miles.

Given:

Money raised = $\textsf{₹} 250$ for $60$ posters.

(a) Solution:

The money raised is in direct proportion to the number of posters sold.

$\text{Ratio} = \frac{\textsf{₹} 250}{60 \text{ posters}}$

$\text{Money for 102 posters} = \frac{250}{60} \times 102$

$\text{Money} = \frac{25}{6} \times 102 = 25 \times 17$

$\text{Money} = \textsf{₹} 425$

(b) Solution:

To check if they can raise exactly $\textsf{₹} 2,000$, we find the number of posters ($x$):

$\frac{250}{60} = \frac{2000}{x}$

$x = \frac{2000 \times 60}{250}$

$x = \frac{200 \times 60}{25} = 8 \times 60$

$x = 480 \text{ posters}$

Since $480$ is a whole number, it is possible to raise exactly $\textsf{₹} 2,000$.

Final Answer: (a) They would make $\textsf{₹} 425$. (b) Yes, they would need to sell 480 posters.