Ganit Prakash - Class-X - Variation
Let us work out 13 Algebra
📘 Let us work out 13 Solutions
Variation| A | 25 | 30 | 45 | 250 |
| B | 10 | 12 | 18 | 100 |
Analysis:
Let us check the ratio of the corresponding values of A and B \( \left(\frac{\text{A}}{\text{B}}\right) \).
\[\begin{array}{l} \frac{25}{10} = \frac{5}{2} \\ \frac{30}{12} = \frac{5}{2} \\ \frac{45}{18} = \frac{5}{2} \\ \frac{250}{100} = \frac{5}{2} \end{array}\]Since the ratio \( \frac{\text{A}}{\text{B}} \) is constant for all corresponding values, we can write:
\[\begin{array}{l} \frac{\text{A}}{\text{B}} = \frac{5}{2} \text{ (Constant)} \\ \Rightarrow \text{A} = \frac{5}{2}\text{B} \\ \Rightarrow \text{A} \propto \text{B} \end{array}\]Therefore, A and B are in direct variation.
The value of the variation constant is \( \frac{5}{2} \).
| x | 18 | 8 | 12 | 6 |
| y | 3 | \( \frac{27}{4} \) | \( \frac{9}{2} \) | 9 |
Analysis:
Let us check the product of the corresponding values of x and y \( (\text{xy}) \).
\[\begin{array}{l} 18 \times 3 = 54 \\ 8 \times \frac{27}{4} = 2 \times 27 = 54 \\ 12 \times \frac{9}{2} = 6 \times 9 = 54 \\ 6 \times 9 = 54 \end{array}\]Since the product \( \text{xy} \) is constant for all corresponding values, we can write:
\[\begin{array}{l} \text{xy} = 54 \text{ (Constant)} \\ \Rightarrow x = \frac{54}{y} \\ \Rightarrow x \propto \frac{1}{y} \end{array}\]Therefore, x and y are in inverse variation.
The relation of variation is \( x \propto \frac{1}{y} \).
Explanation:
Let the distance traveled be \( d \) and the time taken be \( t \).
Since the speed is constant, the distance traveled varies directly with the time taken.
\[\begin{array}{l} d \propto t \\ \Rightarrow d = kt \quad \text{[where } k \text{ is a non-zero variation constant]} \end{array}\]Given: \( d = 14 \) km, \( t = 25 \) minutes.
\[\begin{array}{l} 14 = k \times 25 \\ \Rightarrow k = \frac{14}{25} \end{array}\]Now, we need to find the distance \( d \) when time \( t = 5 \) hours.
First, convert hours to minutes: \( t = 5 \times 60 = 300 \) minutes.
\[\begin{array}{l} d = \frac{14}{25} \times 300 \\ d = 14 \times 12 \\ d = 168 \text{ km} \end{array}\]Therefore, the taxi will travel 168 km in 5 hours.
Explanation:
Let the number of children be \( C \) and the number of sweets each gets be \( S \).
Since the total number of sweets in the box is constant, if the number of children decreases, each child will get more sweets. Thus, they are in inverse variation.
\[\begin{array}{l} C \propto \frac{1}{S} \\ \Rightarrow CS = k \quad \text{[where } k \text{ is a non-zero variation constant representing total sweets]} \end{array}\]Given: \( C = 24 \), \( S = 5 \).
\[\begin{array}{l} k = 24 \times 5 = 120 \end{array}\]If the number of children is reduced by 4, the new number of children \( C' = 24 - 4 = 20 \).
Let the new number of sweets each gets be \( S' \).
\[\begin{array}{l} C' \times S' = k \\ \Rightarrow 20 \times S' = 120 \\ \Rightarrow S' = \frac{120}{20} \\ \Rightarrow S' = 6 \end{array}\]Therefore, each child would get 6 sweets.
Explanation:
Let the number of villagers (persons) be \( V \) and the number of days be \( D \).
Since the amount of work (digging the pond) is constant, the number of persons required varies inversely with the number of days.
\[\begin{array}{l} V \propto \frac{1}{D} \\ \Rightarrow VD = k \quad \text{[where } k \text{ is a non-zero variation constant representing total work]} \end{array}\]Given: \( V = 50 \), \( D = 18 \).
\[\begin{array}{l} k = 50 \times 18 = 900 \end{array}\]We need to find the number of persons \( V' \) required to complete the work in \( D' = 15 \) days.
\[\begin{array}{l} V' \times D' = k \\ \Rightarrow V' \times 15 = 900 \\ \Rightarrow V' = \frac{900}{15} \\ \Rightarrow V' = 60 \end{array}\]To finish the work in 15 days, 60 persons are needed in total.
Number of extra persons required \( = 60 - 50 = 10 \).
Therefore, 10 extra persons will be required.
Explanation:
Given that \( y \) varies directly with the square root of \( x \):
\[\begin{array}{l} y \propto \sqrt{x} \\ \Rightarrow y = k\sqrt{x} \quad \text{[where } k \text{ is a non-zero variation constant]} \end{array}\]When \( y = 9 \) and \( x = 9 \):
\[\begin{array}{l} 9 = k\sqrt{9} \\ \Rightarrow 9 = k \times 3 \\ \Rightarrow k = \frac{9}{3} = 3 \end{array}\]So, the relation is \( y = 3\sqrt{x} \).
We need to find \( x \) when \( y = 6 \):
\[\begin{array}{l} 6 = 3\sqrt{x} \\ \Rightarrow \sqrt{x} = \frac{6}{3} \\ \Rightarrow \sqrt{x} = 2 \\ \Rightarrow x = 2^2 \\ \Rightarrow x = 4 \end{array}\]Therefore, the value of \( x \) is 4.
Explanation:
By the theorem of joint variation, since \( x \propto y \) and \( x \propto \frac{1}{z} \):
\[\begin{array}{l} x \propto \frac{y}{z} \\ \Rightarrow x = k \frac{y}{z} \quad \text{[where } k \text{ is a non-zero variation constant]} \end{array}\]When \( y = 4 \), \( z = 5 \), and \( x = 3 \):
\[\begin{array}{l} 3 = k \times \frac{4}{5} \\ \Rightarrow k = \frac{3 \times 5}{4} = \frac{15}{4} \end{array}\]So, the relation is \( x = \frac{15}{4} \cdot \frac{y}{z} \).
Now, if \( y = 16 \) and \( z = 30 \):
\[\begin{array}{l} x = \frac{15}{4} \times \frac{16}{30} \\ \Rightarrow x = \frac{15}{30} \times \frac{16}{4} \\ \Rightarrow x = \frac{1}{2} \times 4 \\ \Rightarrow x = 2 \end{array}\]Therefore, the value of \( x \) is 2.
Explanation:
By the theorem of joint variation, since \( x \propto y \) and \( x \propto \frac{1}{z} \):
\[\begin{array}{l} x \propto \frac{y}{z} \\ \Rightarrow x = k \frac{y}{z} \quad \text{[where } k \text{ is a non-zero variation constant]} \end{array}\]When \( y = 5 \), \( z = 9 \), and \( x = \frac{1}{6} \):
\[\begin{array}{l} \frac{1}{6} = k \times \frac{5}{9} \\ \Rightarrow k = \frac{1}{6} \times \frac{9}{5} \\ \Rightarrow k = \frac{9}{30} = \frac{3}{10} \end{array}\]So, the relation among the three variables is \( x = \frac{3y}{10z} \).
Now, we calculate \( x \) when \( y = 6 \) and \( z = \frac{1}{5} \):
\[\begin{array}{l} x = \frac{3}{10} \times \frac{6}{\frac{1}{5}} \\ \Rightarrow x = \frac{18}{10} \times 5 \\ \Rightarrow x = \frac{18}{2} \\ \Rightarrow x = 9 \end{array}\]Therefore, the value of \( x \) is 9.
📘 Question 5 Proofs
VariationProof:
Given \( x \propto y \), we can write:
\[\begin{array}{l} x = ky \quad \text{[where } k \text{ is a non-zero variation constant]} \end{array}\]We need to check the ratio of \( (x+y) \) to \( (x-y) \):
\[\begin{array}{l} \frac{x+y}{x-y} = \frac{ky+y}{ky-y} \\ = \frac{y(k+1)}{y(k-1)} \\ = \frac{k+1}{k-1} \end{array}\]Since \( k \) is a constant, \( \frac{k+1}{k-1} \) is also a non-zero constant (let's call it \( m \)).
\[\begin{array}{l} \frac{x+y}{x-y} = m \\ \Rightarrow x+y = m(x-y) \\ \Rightarrow x+y \propto x-y \quad \text{[Proved]} \end{array}\]Proof:
From the given variations, we introduce constants \( k_1 \) and \( k_2 \):
\[\begin{array}{l} A \propto \frac{1}{C} \Rightarrow A = \frac{k_1}{C} \quad \text{--- (1)} \\ C \propto \frac{1}{B} \Rightarrow C = \frac{k_2}{B} \quad \text{--- (2)} \end{array}\]Substitute the value of \( C \) from equation (2) into equation (1):
\[\begin{array}{l} A = \frac{k_1}{\left(\frac{k_2}{B}\right)} \\ \Rightarrow A = \frac{k_1 \cdot B}{k_2} \\ \Rightarrow A = \left(\frac{k_1}{k_2}\right)B \end{array}\]Since \( k_1 \) and \( k_2 \) are non-zero constants, their ratio \( \left(\frac{k_1}{k_2}\right) \) is also a non-zero constant.
\[\begin{array}{l} \therefore A \propto B \quad \text{[Proved]} \end{array}\]Explanation:
Let's introduce non-zero variation constants \( k_1, k_2, \) and \( k_3 \):
\[\begin{array}{l} a = k_1 b \quad \text{--- (1)} \\ b = \frac{k_2}{c} \quad \text{--- (2)} \\ c = k_3 d \quad \text{--- (3)} \end{array}\]Substitute \( c \) from (3) into (2):
\[\begin{array}{l} b = \frac{k_2}{k_3 d} \end{array}\]Now, substitute this expression for \( b \) into (1):
\[\begin{array}{l} a = k_1 \left( \frac{k_2}{k_3 d} \right) \\ \Rightarrow a = \left( \frac{k_1 k_2}{k_3} \right) \frac{1}{d} \end{array}\]Since \( \frac{k_1 k_2}{k_3} \) is a non-zero constant, we have:
\[\begin{array}{l} a \propto \frac{1}{d} \end{array}\]Answer: The relation is that \( a \) varies inversely with \( d \).
Explanation:
Let the three non-zero variation constants be \( k_1, k_2, \) and \( k_3 \).
\[\begin{array}{l} x = k_1 y \quad \text{--- (1)} \\ y = k_2 z \quad \text{--- (2)} \\ z = k_3 x \quad \text{--- (3)} \end{array}\]Multiplying the left-hand sides and the right-hand sides of these three equations together:
\[\begin{array}{l} (x) \cdot (y) \cdot (z) = (k_1 y) \cdot (k_2 z) \cdot (k_3 x) \\ \Rightarrow xyz = (k_1 k_2 k_3) \cdot xyz \end{array}\]Assuming \( x, y, \text{ and } z \) are non-zero, we can divide both sides by \( xyz \):
\[\begin{array}{l} 1 = k_1 k_2 k_3 \end{array}\]Answer: The product of the three constants of variation is 1.
Common setup for Question 6:
Given: \( x+y \propto x-y \)
\[\begin{array}{l} \Rightarrow x+y = k(x-y) \quad \text{[where } k \text{ is a non-zero variation constant]} \\ \Rightarrow \frac{x+y}{x-y} = k \end{array}\]Applying componendo and dividendo:
\[\begin{array}{l} \frac{(x+y) + (x-y)}{(x+y) - (x-y)} = \frac{k+1}{k-1} \\ \Rightarrow \frac{2x}{2y} = \frac{k+1}{k-1} \\ \Rightarrow \frac{x}{y} = \frac{k+1}{k-1} \end{array}\]Let \( \frac{k+1}{k-1} = m \) (where \( m \) is another non-zero constant). Then, \( x = my \).
We will use \( x = my \) to prove the parts of Question 6.
Proof:
We substitute \( x = my \) in the expression \( \frac{x^2+y^2}{xy} \):
\[\begin{array}{l} \frac{x^2+y^2}{xy} = \frac{(my)^2 + y^2}{(my)y} \\ = \frac{m^2y^2 + y^2}{my^2} \\ = \frac{y^2(m^2 + 1)}{my^2} \\ = \frac{m^2 + 1}{m} \end{array}\]Since \( m \) is a constant, \( \frac{m^2 + 1}{m} \) is also a constant. Let it be \( k_1 \).
\[\begin{array}{l} \frac{x^2+y^2}{xy} = k_1 \\ \Rightarrow x^2+y^2 = k_1 xy \\ \Rightarrow x^2+y^2 \propto xy \quad \text{[Proved]} \end{array}\]Proof:
We substitute \( x = my \) in the expression \( \frac{x^3+y^3}{x^3-y^3} \):
\[\begin{array}{l} \frac{x^3+y^3}{x^3-y^3} = \frac{(my)^3 + y^3}{(my)^3 - y^3} \\ = \frac{m^3y^3 + y^3}{m^3y^3 - y^3} \\ = \frac{y^3(m^3 + 1)}{y^3(m^3 - 1)} \\ = \frac{m^3 + 1}{m^3 - 1} \end{array}\]Since \( m \) is a constant, \( \frac{m^3 + 1}{m^3 - 1} \) is also a constant. Let it be \( k_2 \).
\[\begin{array}{l} \frac{x^3+y^3}{x^3-y^3} = k_2 \\ \Rightarrow x^3+y^3 = k_2(x^3-y^3) \\ \Rightarrow x^3+y^3 \propto x^3-y^3 \quad \text{[Proved]} \end{array}\]Proof:
We substitute \( x = my \) in the expression \( \frac{ax+by}{px+qy} \):
\[\begin{array}{l} \frac{ax+by}{px+qy} = \frac{a(my) + by}{p(my) + qy} \\ = \frac{y(am + b)}{y(pm + q)} \\ = \frac{am + b}{pm + q} \end{array}\]Since \( a, b, p, q, \text{ and } m \) are all constants, \( \frac{am + b}{pm + q} \) is a constant. Let it be \( k_3 \).
\[\begin{array}{l} \frac{ax+by}{px+qy} = k_3 \\ \Rightarrow ax+by = k_3(px+qy) \\ \Rightarrow ax+by \propto px+qy \quad \text{[Proved]} \end{array}\]Proof:
Given \( a^2+b^2 \propto ab \), we can write:
\[\begin{array}{l} a^2+b^2 = k \cdot ab \quad \text{[where } k \text{ is a non-zero variation constant]} \\ \Rightarrow \frac{a^2+b^2}{ab} = k \\ \Rightarrow \frac{a^2+b^2}{2ab} = \frac{k}{2} \end{array}\]Applying componendo and dividendo:
\[\begin{array}{l} \frac{a^2+b^2 + 2ab}{a^2+b^2 - 2ab} = \frac{k+2}{k-2} \\ \Rightarrow \frac{(a+b)^2}{(a-b)^2} = \frac{k+2}{k-2} \\ \Rightarrow \left(\frac{a+b}{a-b}\right)^2 = \frac{k+2}{k-2} \end{array}\]Taking the square root of both sides:
\[\begin{array}{l} \frac{a+b}{a-b} = \pm\sqrt{\frac{k+2}{k-2}} \end{array}\]Since \( k \) is a constant, \( \pm\sqrt{\frac{k+2}{k-2}} \) is also a constant. Let it be \( m \).
\[\begin{array}{l} \frac{a+b}{a-b} = m \\ \Rightarrow a+b = m(a-b) \\ \Rightarrow a+b \propto a-b \quad \text{[Proved]} \end{array}\]Proof:
Given \( x^3+y^3 \propto x^3-y^3 \), we can write:
\[\begin{array}{l} \frac{x^3+y^3}{x^3-y^3} = k \quad \text{[where } k \text{ is a non-zero constant]} \end{array}\]Applying componendo and dividendo:
\[\begin{array}{l} \frac{(x^3+y^3) + (x^3-y^3)}{(x^3+y^3) - (x^3-y^3)} = \frac{k+1}{k-1} \\ \Rightarrow \frac{2x^3}{2y^3} = \frac{k+1}{k-1} \\ \Rightarrow \frac{x^3}{y^3} = \frac{k+1}{k-1} \end{array}\]Taking the cube root of both sides:
\[\begin{array}{l} \frac{x}{y} = \sqrt[3]{\frac{k+1}{k-1}} \end{array}\]Since \( k \) is a constant, \( \sqrt[3]{\frac{k+1}{k-1}} \) is also a constant. Let it be \( m \).
So, \( \frac{x}{y} = m \Rightarrow x = my \).
Now, substitute \( x = my \) in the expression \( \frac{x+y}{x-y} \):
\[\begin{array}{l} \frac{x+y}{x-y} = \frac{my+y}{my-y} \\ = \frac{y(m+1)}{y(m-1)} \\ = \frac{m+1}{m-1} \end{array}\]Since \( m \) is a constant, \( \frac{m+1}{m-1} \) is also a constant. Let it be \( k_1 \).
\[\begin{array}{l} \frac{x+y}{x-y} = k_1 \\ \Rightarrow x+y = k_1(x-y) \\ \Rightarrow x+y \propto x-y \quad \text{[Proved]} \end{array}\]Explanation:
Let \( D \) be the number of days, \( F \) be the number of farmers, and \( L \) be the area of land in bighas.
The number of days varies directly with the area of land (when the number of farmers is constant): \( D \propto L \)
The number of days varies inversely with the number of farmers (when the area of land is constant): \( D \propto \frac{1}{F} \)
By the theorem of joint variation:
\[\begin{array}{l} D \propto \frac{L}{F} \\ \Rightarrow D = k \frac{L}{F} \quad \text{[where } k \text{ is a non-zero variation constant]} \end{array}\]Given: \( F = 15 \), \( L = 18 \), and \( D = 5 \).
\[\begin{array}{l} 5 = k \times \frac{18}{15} \\ \Rightarrow 5 = k \times \frac{6}{5} \\ \Rightarrow k = \frac{5 \times 5}{6} = \frac{25}{6} \end{array}\]So, the relation is \( D = \frac{25}{6} \cdot \frac{L}{F} \).
Now, we need to find \( D \) when \( F = 10 \) and \( L = 12 \):
\[\begin{array}{l} D = \frac{25}{6} \times \frac{12}{10} \\ \Rightarrow D = \frac{25}{6} \times \frac{6}{5} \\ \Rightarrow D = \frac{25}{5} \\ \Rightarrow D = 5 \end{array}\]Therefore, 5 days will be required.
Explanation:
Let \( V \) be the volume and \( r \) be the radius of the sphere.
\[\begin{array}{l} V \propto r^3 \\ \Rightarrow V = kr^3 \quad \text{[where } k \text{ is a non-zero variation constant]} \end{array}\]The diameters of the three spheres are \( 1\frac{1}{2} \) m \( = \frac{3}{2} \) m, \( 2 \) m, and \( 2\frac{1}{2} \) m \( = \frac{5}{2} \) m.
So, their radii are:
\[\begin{array}{l} r_1 = \frac{3/2}{2} = \frac{3}{4} \text{ m} \\ r_2 = \frac{2}{2} = 1 \text{ m} \\ r_3 = \frac{5/2}{2} = \frac{5}{4} \text{ m} \end{array}\]Let the radius of the new sphere be \( R \).
Total volume of 3 spheres = Volume of new sphere
\[\begin{array}{l} V_1 + V_2 + V_3 = V_{new} \\ \Rightarrow k(r_1)^3 + k(r_2)^3 + k(r_3)^3 = kR^3 \\ \Rightarrow k \left[ \left(\frac{3}{4}\right)^3 + 1^3 + \left(\frac{5}{4}\right)^3 \right] = kR^3 \\ \end{array}\]Dividing both sides by \( k \):
\[\begin{array}{l} \frac{27}{64} + 1 + \frac{125}{64} = R^3 \\ \Rightarrow \frac{27 + 64 + 125}{64} = R^3 \\ \Rightarrow \frac{216}{64} = R^3 \\ \Rightarrow R^3 = \left(\frac{6}{4}\right)^3 \\ \Rightarrow R^3 = \left(\frac{3}{2}\right)^3 \\ \Rightarrow R = \frac{3}{2} = 1.5 \text{ m} \end{array}\]The diameter of the new sphere = \( 2R = 2 \times 1.5 = 3 \) m.
Therefore, the length of the diameter of the new sphere is 3 metres.
Explanation:
Let \( y = A + B \), where \( A \) varies directly with \( x \) and \( B \) varies inversely with \( x \).
\[\begin{array}{l} A \propto x \Rightarrow A = k_1 x \\ B \propto \frac{1}{x} \Rightarrow B = \frac{k_2}{x} \end{array}\]Where \( k_1 \) and \( k_2 \) are non-zero variation constants.
So, the relation can be written as:
\[\begin{array}{l} y = k_1 x + \frac{k_2}{x} \quad \text{--- (i)} \end{array}\]Case 1: When \( x = -1 \), \( y = 1 \)
\[\begin{array}{l} 1 = k_1(-1) + \frac{k_2}{-1} \\ \Rightarrow 1 = -k_1 - k_2 \\ \Rightarrow k_1 + k_2 = -1 \quad \text{--- (ii)} \end{array}\]Case 2: When \( x = 3 \), \( y = 5 \)
\[\begin{array}{l} 5 = k_1(3) + \frac{k_2}{3} \\ \Rightarrow 15 = 9k_1 + k_2 \quad \text{--- (iii)} \end{array}\]Now, subtract equation (ii) from equation (iii):
\[\begin{array}{l} (9k_1 + k_2) - (k_1 + k_2) = 15 - (-1) \\ \Rightarrow 8k_1 = 16 \\ \Rightarrow k_1 = 2 \end{array}\]Putting \( k_1 = 2 \) in equation (ii):
\[\begin{array}{l} 2 + k_2 = -1 \\ \Rightarrow k_2 = -3 \end{array}\]Substituting the values of \( k_1 \) and \( k_2 \) in equation (i):
\[\begin{array}{l} y = 2x - \frac{3}{x} \end{array}\]Therefore, the relation between \( x \) and \( y \) is \( y = 2x - \frac{3}{x} \).
Proof:
Given \( a \propto b \) and \( b \propto c \), we can introduce non-zero variation constants \( k_1 \) and \( k_2 \):
\[\begin{array}{l} a = k_1 b \\ b = k_2 c \end{array}\]Substituting \( b \) into the first equation:
\[\begin{array}{l} a = k_1 (k_2 c) = k_1 k_2 c \end{array}\]Let \( k_1 k_2 = k_3 \) (another non-zero constant). So, \( a = k_3 c \).
Now, let's find the ratio of \( a^3b^3 + b^3c^3 + c^3a^3 \) to \( abc(a^3+b^3+c^3) \):
\[\begin{array}{l} \frac{a^3b^3 + b^3c^3 + c^3a^3}{abc(a^3+b^3+c^3)} \\ = \frac{(k_3 c)^3(k_2 c)^3 + (k_2 c)^3 c^3 + c^3(k_3 c)^3}{(k_3 c)(k_2 c)(c) [ (k_3 c)^3 + (k_2 c)^3 + c^3 ]} \\ = \frac{k_3^3 c^3 \cdot k_2^3 c^3 + k_2^3 c^3 \cdot c^3 + c^3 \cdot k_3^3 c^3}{k_3 k_2 c^3 [ k_3^3 c^3 + k_2^3 c^3 + c^3 ]} \\ = \frac{k_3^3 k_2^3 c^6 + k_2^3 c^6 + k_3^3 c^6}{k_3 k_2 c^6 (k_3^3 + k_2^3 + 1)} \\ = \frac{c^6 (k_3^3 k_2^3 + k_2^3 + k_3^3)}{c^6 k_3 k_2 (k_3^3 + k_2^3 + 1)} \\ = \frac{k_3^3 k_2^3 + k_2^3 + k_3^3}{k_3 k_2 (k_3^3 + k_2^3 + 1)} \end{array}\]Since \( k_2 \) and \( k_3 \) are constants, the entire right-hand side is a constant, let's call it \( m \).
\[\begin{array}{l} \frac{a^3b^3 + b^3c^3 + c^3a^3}{abc(a^3+b^3+c^3)} = m \\ \Rightarrow a^3b^3 + b^3c^3 + c^3a^3 = m \cdot abc(a^3+b^3+c^3) \\ \Rightarrow a^3b^3 + b^3c^3 + c^3a^3 \propto abc(a^3+b^3+c^3) \quad \text{[Proved]} \end{array}\]Explanation:
Let the total expense be \( E \). According to the problem, \( E \) is the sum of two parts, say \( A \) and \( B \).
\[\begin{array}{l} E = A + B \end{array}\]Given that \( A \propto x \) and \( B \propto x^2 \), we can write:
\[\begin{array}{l} A = k_1 x \\ B = k_2 x^2 \quad \text{[where } k_1 \text{ and } k_2 \text{ are non-zero variation constants]} \end{array}\]So, the relation becomes:
\[\begin{array}{l} E = k_1 x + k_2 x^2 \quad \text{--- (i)} \end{array}\]Case 1: When \( x = 100 \) dcm, \( E = 5000 \)
\[\begin{array}{l} 5000 = k_1(100) + k_2(100)^2 \\ \Rightarrow 5000 = 100k_1 + 10000k_2 \\ \Rightarrow 50 = k_1 + 100k_2 \quad \text{[Dividing by 100] --- (ii)} \end{array}\]Case 2: When \( x = 200 \) dcm, \( E = 12000 \)
\[\begin{array}{l} 12000 = k_1(200) + k_2(200)^2 \\ \Rightarrow 12000 = 200k_1 + 40000k_2 \\ \Rightarrow 60 = k_1 + 200k_2 \quad \text{[Dividing by 200] --- (iii)} \end{array}\]Subtracting equation (ii) from equation (iii):
\[\begin{array}{l} (k_1 + 200k_2) - (k_1 + 100k_2) = 60 - 50 \\ \Rightarrow 100k_2 = 10 \\ \Rightarrow k_2 = \frac{10}{100} = \frac{1}{10} \end{array}\]Substitute \( k_2 = \frac{1}{10} \) in equation (ii):
\[\begin{array}{l} k_1 + 100\left(\frac{1}{10}\right) = 50 \\ \Rightarrow k_1 + 10 = 50 \\ \Rightarrow k_1 = 40 \end{array}\]So, the general relation is \( E = 40x + \frac{1}{10}x^2 \).
Case 3: To find expenses for digging a well of \( x = 250 \) dcm depth:
\[\begin{array}{l} E = 40(250) + \frac{1}{10}(250)^2 \\ E = 10000 + \frac{1}{10}(62500) \\ E = 10000 + 6250 \\ E = 16250 \end{array}\]Therefore, the expenses of digging a 250 dcm deep well is ₹ 16250.
Explanation:
Let \( V \) be the volume, \( r \) be the radius of the base, and \( h \) be the height of the cylinder.
According to the given condition of joint variation:
\[\begin{array}{l} V \propto r^2 h \\ \Rightarrow V = k r^2 h \quad \text{[where } k \text{ is a non-zero variation constant]} \end{array}\]Let the radii of the two cylinders be \( r_1 \) and \( r_2 \), and their heights be \( h_1 \) and \( h_2 \).
Given ratio of radii: \( \frac{r_1}{r_2} = \frac{2}{3} \)
Given ratio of heights: \( \frac{h_1}{h_2} = \frac{5}{4} \)
Let their respective volumes be \( V_1 \) and \( V_2 \). Then:
\[\begin{array}{l} V_1 = k r_1^2 h_1 \\ V_2 = k r_2^2 h_2 \end{array}\]The ratio of their volumes is:
\[\begin{array}{l} \frac{V_1}{V_2} = \frac{k r_1^2 h_1}{k r_2^2 h_2} \\ = \left(\frac{r_1}{r_2}\right)^2 \times \left(\frac{h_1}{h_2}\right) \\ = \left(\frac{2}{3}\right)^2 \times \left(\frac{5}{4}\right) \\ = \frac{4}{9} \times \frac{5}{4} \\ = \frac{5}{9} \end{array}\]Therefore, the ratio of their volumes is 5 : 9.
Explanation:
Let \( L \) be the land cultivated (in bighas), \( P \) be the number of ploughs, and \( D \) be the number of days.
The number of ploughs varies directly with the land to be cultivated (when days are constant): \( P \propto L \)
The number of ploughs varies inversely with the number of days (when the land is constant): \( P \propto \frac{1}{D} \)
By the theorem of joint variation:
\[\begin{array}{l} P \propto \frac{L}{D} \\ \Rightarrow P = k \cdot \frac{L}{D} \quad \text{[where } k \text{ is a non-zero variation constant]} \end{array}\]Given initial condition: \( L = 2400 \), \( P = 25 \), and \( D = 36 \).
\[\begin{array}{l} 25 = k \times \frac{2400}{36} \\ \Rightarrow 25 = k \times \frac{200}{3} \\ \Rightarrow k = \frac{25 \times 3}{200} = \frac{75}{200} = \frac{3}{8} \end{array}\]So, the relation is \( P = \frac{3}{8} \cdot \frac{L}{D} \).
For the tractor:
It cultivates "half of the land" \( \Rightarrow L' = \frac{2400}{2} = 1200 \) bighas.
It takes "30 days" \( \Rightarrow D' = 30 \) days.
Let's calculate the equivalent number of ploughs (\( P' \)) required to do this exact work:
\[\begin{array}{l} P' = \frac{3}{8} \times \frac{1200}{30} \\ \Rightarrow P' = \frac{3}{8} \times 40 \\ \Rightarrow P' = \frac{120}{8} \\ \Rightarrow P' = 15 \end{array}\]Since one tractor can do the work of 15 ploughs in the same amount of time, we can conclude:
Answer: One tractor works equally with 15 ploughs.
Proof:
Let \( V \) be the volume, \( S \) be the surface area, and \( r \) be the radius of the sphere.
From the given conditions, we can write the following variation relations with non-zero constants \( k_1 \) and \( k_2 \):
\[\begin{array}{l} V \propto r^3 \Rightarrow V = k_1 r^3 \quad \text{--- (i)} \\ S \propto r^2 \Rightarrow S = k_2 r^2 \quad \text{--- (ii)} \end{array}\]We need to find the relationship between the square of the volume (\( V^2 \)) and the cube of the surface area (\( S^3 \)).
Let's calculate \( V^2 \) and \( S^3 \):
\[\begin{array}{l} V^2 = (k_1 r^3)^2 = k_1^2 r^6 \\ S^3 = (k_2 r^2)^3 = k_2^3 r^6 \end{array}\]Now, let's find the ratio of \( V^2 \) to \( S^3 \):
\[\begin{array}{l} \frac{V^2}{S^3} = \frac{k_1^2 r^6}{k_2^3 r^6} \\ \Rightarrow \frac{V^2}{S^3} = \frac{k_1^2}{k_2^3} \end{array}\]Since \( k_1 \) and \( k_2 \) are non-zero constants, their combination \( \frac{k_1^2}{k_2^3} \) is also a non-zero constant (let's call it \( m \)).
\[\begin{array}{l} \frac{V^2}{S^3} = m \\ \Rightarrow V^2 = m \cdot S^3 \\ \Rightarrow V^2 \propto S^3 \quad \text{[Proved]} \end{array}\]Therefore, the square of the volume of the sphere varies directly with the cube of its surface area.
📘 16. Very short answer type question (V.S.A.)
(A) M.C.Q.Explanation:
If \( x \) varies inversely with \( y \), we can introduce a non-zero variation constant \( k \):
\[\begin{array}{l} x \propto \frac{1}{y} \\ \Rightarrow x = k \times \frac{1}{y} \quad \text{[where } k \text{ is a non-zero constant]} \\ \Rightarrow xy = k \end{array}\]Correct Option: (d) \( xy = \text{non-zero constant} \)
Explanation:
If \( x \) varies directly with \( y \), we can write:
\[\begin{array}{l} x = ky \quad \text{[where } k \text{ is a non-zero constant]} \end{array}\]Squaring both sides:
\[\begin{array}{l} x^2 = (ky)^2 \\ \Rightarrow x^2 = k^2 y^2 \end{array}\]Since \( k \) is a constant, \( k^2 \) is also a constant. Therefore:
\[\begin{array}{l} x^2 \propto y^2 \end{array}\]Correct Option: (d) \( x^2 \propto y^2 \)
Explanation:
\[\begin{array}{l} x \propto y \Rightarrow x = ky \quad \text{--- (1)} \end{array}\]Given \( y = 8 \) when \( x = 2 \):
\[\begin{array}{l} 2 = k(8) \\ \Rightarrow k = \frac{2}{8} = \frac{1}{4} \end{array}\]Substitute \( k \) in equation (1):
\[\begin{array}{l} x = \frac{1}{4}y \end{array}\]Now, if \( y = 16 \):
\[\begin{array}{l} x = \frac{1}{4} \times 16 \\ \Rightarrow x = 4 \end{array}\]Correct Option: (b) 4
Explanation:
\[\begin{array}{l} x \propto y^2 \Rightarrow x = ky^2 \quad \text{--- (1)} \end{array}\]Given \( y = 4 \) when \( x = 8 \):
\[\begin{array}{l} 8 = k(4)^2 \\ \Rightarrow 8 = 16k \\ \Rightarrow k = \frac{8}{16} = \frac{1}{2} \end{array}\]Substitute \( k \) in equation (1):
\[\begin{array}{l} x = \frac{1}{2}y^2 \end{array}\]Now, if \( x = 32 \):
\[\begin{array}{l} 32 = \frac{1}{2}y^2 \\ \Rightarrow y^2 = 64 \\ \Rightarrow y = \pm 8 \end{array}\]Looking at the options, 8 is provided.
Correct Option: (b) 8
Explanation:
Let the three variation constants be \( k_1 \), \( k_2 \), and \( k_3 \).
\[\begin{array}{l} y-z = \frac{k_1}{x} \Rightarrow k_1 = x(y-z) = xy - xz \\ z-x = \frac{k_2}{y} \Rightarrow k_2 = y(z-x) = yz - xy \\ x-y = \frac{k_3}{z} \Rightarrow k_3 = z(x-y) = xz - yz \end{array}\]Now, finding the sum of the three variation constants:
\[\begin{array}{l} k_1 + k_2 + k_3 = (xy - xz) + (yz - xy) + (xz - yz) \\ = xy - xy - xz + xz + yz - yz \\ = 0 \end{array}\]Correct Option: (a) 0
📘 16. Very short answer type question (V.S.A.)
(B) True or FalseIf \( y \propto \frac{1}{x} \), \( \frac{y}{x} = \text{non-zero constant} \).
Explanation:
If \( y \propto \frac{1}{x} \), then \( y = \frac{k}{x} \), which means \( xy = k \) (where \( k \) is a non-zero constant). This represents an inverse variation.
The statement \( \frac{y}{x} = \text{constant} \) represents a direct variation (\( y \propto x \)).
Answer: False
If \( x \propto z \) and \( y \propto z \) then \( xy \propto z^2 \)
Explanation:
From the given variations:
\[\begin{array}{l} x = k_1 z \\ y = k_2 z \end{array}\]Multiplying both equations:
\[\begin{array}{l} x \cdot y = (k_1 z)(k_2 z) \\ \Rightarrow xy = (k_1 k_2)z^2 \end{array}\]Since \( k_1 k_2 \) is a constant, \( xy \propto z^2 \).
Answer: True
📘 16. Very short answer type question (V.S.A.)
(C) Fill in the blanksExplanation:
\[\begin{array}{l} x \propto \frac{1}{y} \Rightarrow x = \frac{k_1}{y} \\ y \propto \frac{1}{z} \Rightarrow y = \frac{k_2}{z} \end{array}\]Substituting the value of \( y \) into the first equation:
\[\begin{array}{l} x = \frac{k_1}{\left(\frac{k_2}{z}\right)} \\ \Rightarrow x = \left(\frac{k_1}{k_2}\right)z \end{array}\]Since \( \frac{k_1}{k_2} \) is a constant, \( x \propto z \).
Answer: \( z \)
Explanation:
\[\begin{array}{l} x \propto y \Rightarrow x = ky \end{array}\]Raising both sides to the power of \( n \):
\[\begin{array}{l} x^n = (ky)^n \\ \Rightarrow x^n = k^n y^n \end{array}\]Since \( k^n \) is a constant, \( x^n \propto y^n \).
Answer: \( y^n \)
Explanation:
If \( x \propto y \), then \( y \propto x \), which means \( y = k_1 x \).
If \( x \propto z \), then \( z \propto x \), which means \( z = k_2 x \).
Adding the two equations:
\[\begin{array}{l} y + z = k_1 x + k_2 x \\ \Rightarrow y + z = (k_1 + k_2)x \end{array}\]Since \( k_1 + k_2 \) is a constant, \( (y+z) \propto x \).
Answer: \( x \)
📘 17. Short answer type question (S.A.)
VariationExplanation:
Given \( x \propto y^2 \), we can write the relation with a non-zero variation constant \( k \):
\[\begin{array}{l} x = ky^2 \quad \text{--- (1)} \end{array}\]Substitute \( x = a \) and \( y = 2a \) into equation (1):
\[\begin{array}{l} a = k(2a)^2 \\ \Rightarrow a = k(4a^2) \\ \Rightarrow k = \frac{a}{4a^2} \\ \Rightarrow k = \frac{1}{4a} \quad [\text{assuming } a \neq 0] \end{array}\]Now, substitute the value of \( k \) back into equation (1):
\[\begin{array}{l} x = \left(\frac{1}{4a}\right)y^2 \\ \Rightarrow y^2 = 4ax \end{array}\]Therefore, the relation between \( x \) and \( y \) is \( y^2 = 4ax \).
Explanation:
Let the three non-zero variation constants be \( k_1, k_2, \) and \( k_3 \).
\[\begin{array}{l} x = k_1 y \quad \text{--- (1)} \\ y = k_2 z \quad \text{--- (2)} \\ z = k_3 x \quad \text{--- (3)} \end{array}\]Multiplying the left-hand sides and the right-hand sides of these three equations together:
\[\begin{array}{l} (x) \cdot (y) \cdot (z) = (k_1 y) \cdot (k_2 z) \cdot (k_3 x) \\ \Rightarrow xyz = (k_1 k_2 k_3) \cdot xyz \end{array}\]Assuming \( x, y, \text{ and } z \) are non-zero, we divide both sides by \( xyz \):
\[\begin{array}{l} 1 = k_1 k_2 k_3 \end{array}\]Therefore, the product of the three non-zero constants is 1.
Explanation:
Using non-zero variation constants \( k_1 \) and \( k_2 \):
\[\begin{array}{l} x = \frac{k_1}{y} \quad \text{--- (1)} \\ y = \frac{k_2}{z} \quad \text{--- (2)} \end{array}\]Substitute the value of \( y \) from equation (2) into equation (1):
\[\begin{array}{l} x = \frac{k_1}{\left(\frac{k_2}{z}\right)} \\ \Rightarrow x = \left(\frac{k_1}{k_2}\right)z \end{array}\]Since \( k_1 \) and \( k_2 \) are non-zero constants, their ratio \( \frac{k_1}{k_2} \) is also a non-zero constant (say, \( m \)).
\[\begin{array}{l} x = mz \\ \Rightarrow x \propto z \end{array}\]Therefore, there is a direct variation between \( x \) and \( z \).
Proof:
Using non-zero variation constants \( k_1 \) and \( k_2 \):
\[\begin{array}{l} x = k_1 yz \quad \text{--- (1)} \\ y = k_2 zx \quad \text{--- (2)} \end{array}\]Substitute the expression for \( y \) from equation (2) into equation (1):
\[\begin{array}{l} x = k_1 (k_2 zx) z \\ \Rightarrow x = k_1 k_2 z^2 x \end{array}\]Assuming \( x \neq 0 \), we can divide both sides by \( x \):
\[\begin{array}{l} 1 = k_1 k_2 z^2 \\ \Rightarrow z^2 = \frac{1}{k_1 k_2} \\ \Rightarrow z = \pm \sqrt{\frac{1}{k_1 k_2}} \end{array}\]Since \( k_1 \) and \( k_2 \) are constants, \( \pm \sqrt{\frac{1}{k_1 k_2}} \) is also a constant. Moreover, it is non-zero because \( k_1 \) and \( k_2 \) are non-zero constants.
Therefore, \( z \) is a non-zero constant. [Proved]
Explanation:
Given \( b \propto a^3 \), we can write:
\[\begin{array}{l} b = ka^3 \quad \text{[where } k \text{ is a non-zero constant]} \end{array}\]Let the initial value of \( a \) be \( a_1 \) and the final value be \( a_2 \).
It is given that \( a \) increases in the ratio 2 : 3. So, we can write:
\[\begin{array}{l} \frac{a_1}{a_2} = \frac{2}{3} \Rightarrow a_1 = 2x \text{ and } a_2 = 3x \quad \text{[where } x \text{ is a common multiplier]} \end{array}\]Let the corresponding initial and final values of \( b \) be \( b_1 \) and \( b_2 \).
\[\begin{array}{l} b_1 = k(a_1)^3 = k(2x)^3 = 8kx^3 \\ b_2 = k(a_2)^3 = k(3x)^3 = 27kx^3 \end{array}\]Now, we find the ratio of \( b_1 \) to \( b_2 \):
\[\begin{array}{l} \frac{b_1}{b_2} = \frac{8kx^3}{27kx^3} = \frac{8}{27} \end{array}\]Therefore, \( b \) will be increased in the ratio 8 : 27.
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