Answer \( {\color{orange} (i)} \ \color{red} \text{in direct proportion} \)

Answer \( {\color{orange} (iv)} \ \color{red} (xy) \text{ remains constant} \)

Inverse proportion, \( x \times y = (xy) \text{ constant} \).

Solution

If \( p \) and \( q \) vary directly, then \( \displaystyle \frac{p}{q} = K \text{ (constant)} \)

\[ \begin{align*} \frac{p}{q} \implies \frac{10}{15} & = \color{green} \frac{2}{3} \\ \\ (i) \quad \frac{15}{20} & = \color{red} \frac{3}{4} \text{ ❌} \\ \\ (ii) \quad \frac{20}{30} & = \color{green} \frac{2}{3} \text{ ✅} \\ \\ (iii) \quad \frac{2}{3} & = \color{green} \frac{2}{3} \text{ ✅} \\ \\ (iv) \quad \frac{5}{7.5} = \frac{50}{75} &\implies \color{green} \frac{2}{3} \text{ ✅} \\ \\ \end{align*} \]

Answer \( {\color{orange} (i)} \ \color{red} 15 \text{ and } 20 \)

Solution

\[ \begin{align*} Given: \\ x & = 24 \\ k &= 3 \\ \\ \text{In direct } & variation, \\ \\ \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{24}{y} &= 3 \\ \\ \frac{24}{3} &= y \\ \\ y &= 8 \\ \\ \end{align*} \]

Answer \({\color{orange} (iii)} \ \color{red} 8 \)

Answer \( {\color{orange} (iv)} \ \color{red} \text{Number of workers and time taken to finish a job} \)

If number of workers increases, time taken to finish a job decreases ⇒ inverse proportion.

Solution

\[ \begin{array}{|c|c|c|} \hline l & 5 & x \\ \hline m & \displaystyle \frac{2}{3} & \displaystyle \frac{16}{3} \\ \hline \end{array} \] \[ \begin{align*} \text{Direct} & \text{ variation} \\ \frac{l}{m} &= k \text{ (constant)} \\ \\ \frac{5}{\frac{2}{3}} &= \frac{x}{\frac{16}{3}} \\ \\ \frac{5 \times 3}{2} &= x \times \frac{3}{16} \\ \\ \frac{15}{2} &= x \times \frac{3}{16} \\ \\ \frac{\cancel{15}^5}{\cancel2_1} \times \frac{\cancel{16}^8}{\cancel3_1} &= x \\ \\ 5 \times 8 &= x \\ 40 &= x \\ \color{green} x &= \color{green} 40 \end{align*} \]

Answer \( \color{red} l = 40 \)

Solution

\[ \begin{align*} \text{Speed} &= 36 \, km/hr \\ \\ \text{Convert speed from } & {\color{green} km/hr} \text{ to } \color{green} m/s \\ \\ &= \cancel{36}^2 \times\frac{5}{\cancel{18}_1} \\ & = 2 \times 5 \\ \color{green} Speed &= \color{green} 10 \text{ m/s} \\ Distance &= 20 \, m \\ \\ \color{red} Time &= \color{red} \frac{Distance}{Speed} \\ \\ & = \frac{\cancel{20}^2}{\cancel{10}_1} \\ \\ & = 8 \times 3 \\ \color{green} Time &= \color{green} 2 \,seconds \end{align*} \]

Answer \( \color{red} 2 \, \text{seconds} \)

Solution

Let \( x \) be the number of sweets distributed among 40 children. \[ \begin{array}{|c|c|} \hline \text{No. of Children} & 50 & 40 \\ \hline \text{No. of Sweets} & 4 & x \\ \hline \end{array} \]

As the number of children decreases, the sweeets given to children will increase. It is a case of inverse variation.

\[ \begin{align*} \implies a \times b &= k \text{ (constant)} \\ \\ 50 \times 4 &= 40 \times x \\ \\ x &= \frac{\cancel{50}^5 \times \cancel4^1}{\cancel{40}_{\cancel{10}_1}} \\ \\ \color{green} x &= \color{green} 5 \end{align*} \]

Answer Each child will get \( \color{red} 5 \, sweets. \)

Solution

\[ \begin{array}{|c|c|c|} \hline \displaystyle \frac{1}{x} & \displaystyle \frac{1}{2} & \displaystyle \frac{1}{a} \\ \hline y & 20 & 1.25 \\ \hline \end{array} \] \[ \begin{align*} \text{Direct} & \text{ variation} \\ \frac{ \frac{1}{x}}{y} &= k \text{ (constant)} \\ \\ \frac{ \frac{1}{2}}{20} &= \frac{ \frac{1}{a}}{1.25} \\ \\ \frac{1}{2 \times 20} &= \frac{1}{a \times 1.25} \\ \\ \frac{1}{40} &= \frac{1}{a \times 1.25} \\ \\ a &= \frac{40 \times \color{green} 100}{1.25 \times \color{green} 100} \\ \\ a &= \frac{\cancel{40}^{8} \times \cancel{100}^4}{\cancel{125}_{\cancel{25}_1}} \\ \\ \color{green} a &= \color{green} 32 \end{align*} \]

Answer \(\color{red} x = 32 \)

Solution

\[ \begin{array}{|c|c|} \hline l & 6 & 4 \\ \hline \sqrt m & \sqrt 4 = 2 & \sqrt m \\ \hline \end{array} \] \[ \begin{align*} \text{Inverse} & \text{ variation} \\ \implies l \times \sqrt m &= k \text{ (constant)} \\ \\ 6 \times 2 &= 4 \times \sqrt m \\ \\ \sqrt m &= \frac{\cancel{12}^3}{\cancel4_1} \\ \\ \sqrt m &= 3 \\ m &= 3^2 \\ \color{green} m &= \color{green} 9 \end{align*} \]

Answer \(\color{red} m = 9 \)

Solution

\[ \begin{align*} \color{green} \text{Direct variation} \implies & \color{green}\frac{x}{y} = k \text{ (constant)} \\ \\ \frac{\cancel{12}^4}{\cancel9_3} & = \frac{4}{3} \\ \\ \frac{\cancel 3^1}{\cancel{36}_{12}} & = \frac{1}{12} \\ \\ \frac{27}{4} & = \frac{27}{4} \\ \\ \frac{\cancel6^1}{\cancel{18}_3} & = \frac{1}{3} \\ \\ \frac{\cancel2^1}{\cancel{54}_{27}} & = \frac{1}{27} \\ \\ \text{They are not } & \text{directly proportional.} \\ \\ \color{green} \text{Inverse variation} \implies & \color{green} x \times y = k \text{ (constant)} \\ \\ 12 \times 9 & = \color{red} 108 \\ 3 \times 36 & = \color{red} 108 \\ 27 \times 4 & = \color{red} 108 \\ 6 \times 18 & = \color{red} 108 \\ 2 \times 54 & = \color{red} 108 \\ \\ \text{They are inversely} & \text{ proportional.} \end{align*} \]

Answer They vary inversely.

Solution

\[ \begin{align*} \color{green} \text{Direct variation} \implies & \color{green}\frac{x}{y} = k \text{ (constant)} \\ \\ \frac{4}{2} &= 2 \\ \\ \frac{\cancel{20}^5}{\cancel8_2} &= \frac{5}{2} \\ \\ \frac{30}{10} &= 3 \\ \\ \frac{\cancel{40}^8}{\cancel{15}_3} &= \frac{8}{3} \\ \\ \frac{60}{20} &= 3 \\ \\ \frac{\cancel{80}^{16}}{\cancel{25}_5} &= \frac{16}{5} \\ \\ \text{They are not directly } & \text{proportional.} \\ \\ \color{green} \text{Inverse variation} \implies & \color{green} x \times y = k \text{ (constant)} \\ \\ 4 \times 2 &= 8 \\ 20 \times 8 &= 160 \\ 30 \times 10 &= 300 \\ 40 \times 15 &= 600 \\ 60 \times 20 &= 1200 \\ 80 \times 25 &= 2000 \\ \\ \text{They are not inversely} & \text{ proportional.} \end{align*} \]

Answer They vary in neither way.

Solution

\[ \begin{align*} \color{green} \text{Direct variation} \implies & \color{green}\frac{x}{y} = k \\ \\ \frac{\cancel2^1}{\cancel{10}_5} &= \color{red} \frac{1}{5} \\ \\ \frac{\cancel9^1}{\cancel{45}_5} &= \color{red} \frac{1}{5} \\ \\ \frac{\cancel7^1}{\cancel{35}_5} &= \color{red} \frac{1}{5} \\ \\ \frac{\cancel4^1}{\cancel{20}_5} &= \color{red} \frac{1}{5} \\ \\ \frac{\cancel3^1}{\cancel{15}_5} &= \color{red} \frac{1}{5} \\ \\ \text{They vary } & \color{red} directly \\ \\ \color{green} \text{Inverse variation} \implies & \color{green} x \times y = k \\ \\ 2 \times 10 &= 20 \\ 9 \times 45 &= 405 \\ 7 \times 35 &= 245 \\ 4 \times 20 &= 80 \\ 3 \times 15 &= 45 \\ \text{They are not inversely} & \text{ proportional.} \end{align*} \]

Answer They vary directly.

Solution

\[ \begin{array}{|c|c|} \hline x & 25 & 15 \\ \hline y & 3 & y \\ \hline \end{array} \] \[ \begin{align*} \text{Inverse} & \text{ variation} \\ \implies x \times y &= k \text{ (constant)} \\ \\ 25 \times 3 &= 15 \times y \\ \\ y &= \frac{\cancel{25}^5 \times \cancel3^1}{\cancel{15}_{\cancel5_1}} \\ \\ \color{green} y &= \color{green} 5 \end{align*} \]

Answer \(\color{red} y = 5\)

Solution

\[ \begin{array}{|c|c|} \hline x & x \\ \hline y & 45 \\ \hline \end{array} \] \[ \begin{align*} \text{Inverse} & \text{ variation} \\ \implies x \times y &= k \\ x \times 45 &= 9 \\ x &= \frac{\cancel9^1}{\cancel{45}_5} \\ \\ \color{green} x &= \color{green} \frac{1}{5} \end{align*} \]

Answer \(\color{red} x = \displaystyle \frac{1}{5}\)

Solution

Let \( x \) be the cost of 8 oranges. \[ \begin{align*} 1 \text{ dozen} & = 12 \ oranges \\ 3 \text{ dozen} & = 36 \ oranges \end{align*} \] \[ \begin{array}{|c|c|c|} \hline \text{No. of oranges} & 36 & 8 \\ \hline \text{Cost (₹)} & 54 & x \\ \hline \end{array} \]

As the number of oranges decreases, the cost also decreases. It is a case of direct variation.

\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{36}{54} &= \frac{8}{x} \\ \\ 36 \times x &= 54 \times 8 \\ \\ x &= \frac{\cancel{54}^6 \times \cancel{8}^2}{\cancel{36}_{\cancel9_1}} \\ \\ \color{green} x &= \color{green} 12 \end{align*} \]

Answer Cost of 8 oranges \( = \color{red} \text{₹} 12 \)

Solution

Let \( x \) be the time taken to cover 100 km. \[ \begin{align*} 1 \text{ hr } 30 \text{ min} &= 1.5 \text{ hours } \end{align*} \] \[ \begin{array}{|c|c|c|} \hline \text{Distance (km)} & 60 & 100 \\ \hline \text{Time (hrs)} & 1.5 & x \\ \hline \end{array} \]

It is a case of direct variation. As the distance increases, time also increases.

\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{60}{1.5} &= \frac{100}{x} \\ \\ 60x &= 1.5 \times 100 \\ \\ x &= \frac{\cancel{150}^5}{\cancel{60}_2} \\ \\ x &= \frac{5}{2} \\ \\ x &= 2.5 \text{ hours } \\ \\ \color{green} 2.5 \text{ hours } &= \color{green} 2 \text{ hr } 30 \text{ min} \end{align*} \]

Answer Time taken to cover \( 100 \ km = \color{red} 2 \text{ hr } 30 \text{ min} \)

Solution

Let \( x \) be the weight that produces an extension of \( 9.8 \ cm \) \[ \begin{array}{|c|c|c|} \hline \text{Extension (cm)} & 1.4 & 9.8 \\ \hline \text{Weight (kg)} & 75 & x \\ \hline \end{array} \]

As the extension increases, the weight also increases. It is a case of direct variation.

\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{1.4}{75} &= \frac{9.8}{x} \\ \\ 1.4 \times x &= 75 \times 9.8 \\ \\ x &= \frac{75 \times 9.8 \times \color{green} 10 }{1.4 \times \color{green} 10} \\ \\ x &= \frac{75 \times \cancel{98}^7}{\cancel{14}_1} \\ \\ x &= 75 \times 7 \\ \color{green} x &= \color{green} 525 \end{align*} \]

Answer Weight \( = \color{red} 525 \text{ kg} \)

Solution

Let \( x \) be the amount of dust picked up in 15 days. \[ \begin{array}{|c|c|c|} \hline \text{No. of days} & 25 & 15 \\ \hline \text{Dust (pounds)} & 6 \times 10^8 & x \\ \hline \end{array} \]

It is a case of direct variation. As time increases, the dust collected also increases.

\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{25}{6 \times 10^8} &= \frac{15}{x} \\ \\ 25 \times x &= 15 \times 6 \times 10^8 \\ \\ x &= \frac{\cancel{15}^3 \times \cancel6^{1.2} \times 10^8}{\cancel{25}_{\cancel5_1}} \\ \\ x &= 3 \times 1.2 \times 10^8 \\ \color{green} x &= \color{green} 3.6 \times 10^8 \end{align*} \]

Answer Dust picked up in 15 days \( = \color{red} 3.6 \times 10^8 \ pounds \)