Chapter – 13 : Variation

Ganit Prakash Solution 10

Book Name	: Ganit Prakash
Subject	: Mathematics
Class	: 10 (Madhyamik)
Publisher	: Prof Nabanita Chatterjee
Chapter Name	: Variation (13th Chapter)

Question 4

If the square art paper having the sides of 60 cm length is covered with a coloured paper on its four sides, then let us write by calculating how much length of coloured paper is required. [Let me do it myself]

Solution:

Coloured paper is required = 4 × 60 cm of length = 240 cm in length. If the length of one side of square art paper is 50 cm, then the coloured paper is required to cover the art paper of 4 × 50 = 200 cm length.

Question 5

Similarly let us write by understanding that the number of pens (x) and total cost price of pen (y) are in direct variation. [Let me do it myself]

Solution:

No. of pens (x)

Total cost of pen (y)

Here No. of pen (x) & Total cost price (y) are in direct variation such that, $\frac{x}{y}$

= Constant, i.e., x ∝ y.

Question 6

Let us write four examples related to two variables, where the variables are in direct variations. [Let me do it myself]

A	90	30	12	15	156
B	60	20	8	10	104

We have got values of interrelated variables A and B :

Solution:

Four examples related to two variables, where the variables are in direct variation are:

(a) Distance (x) & Time (t) are in direct variation when speed in constant.

(b) Base (b) & Area (A) of triangle are in direct variation when height is constant.

(d) Temparature (T) in kelvin scale & volume of gas V are in direct variation.

Question 7

Let us find if there is any variation relation between P and Q. [Let me do it myself]

Solution:

Here the value of P increases/decreases, the value of Q increases/decreases:

Here, $\frac{P}{Q} = \frac{35}{15} = \frac{49}{21} = \frac{56}{24} = \frac{14}{6} = \frac{7}{3}$

$\therefore$ P ∝ Q & the constant of variation $= \frac{7}{3}.$

Application 4.

y varies directly with the square root of x and y = 9 when x = 4, let us write the value of varional constant; and let us express y as a function of x and if y = 8, let us write the value of x. [Let me do it myself]

Solution:

Here, y varies directly with square root of x

i.e., y ∝ $\sqrt{x}$ or, y = k $\sqrt{x}$

y = 9, when x = 4

$\therefore$ 9 = k $\sqrt{4} \quad \therefore$ k = $\frac{9}{2}$

$\therefore$ y $= \frac{9}{2} \sqrt{x}$

or, 8 $= \frac{9}{2} \sqrt{x} \quad$ or, $\sqrt{x} = \frac{8 × 2}{9} \quad \therefore \text{ x } = \frac{256}{81}$

$\therefore$ When y = 8, x = $\frac{256}{81} .$

Question 8

By counting we see that there are 24 coconut laddus in the tiffin box. We shall divide them equally without breaking any laddu, let us find how many laddus we shall each get.

Solution:

We shall each get 24 ÷ 2 = 12 laddus.

But Shibani too has joined with us in this activity.

So, we shall divide 24 laddus among 3 persons. Now each will get 8 laddus.

But it was expected that three friends would come in club room. If they will come we shall be 6 persons in total.

So, if 24 laddus are divided among 6 persons then each will get 4 laddus.

Application 5.

I arranged 36 buttons in a rectangular size and let us write by under. standing in each case, if there is any inverse variation between buttons lying along length and those of breadth. [Let me do it myself]

No. of buttons lying in length (x)	4	2	3	9
No. of buttons lying in breadth (y)	9	18	12	4

Here no. of buttons lying in length increases but no. of buttons lying in breadth decreases.

But x × y = (4 × 9); (2 × 18); (3 × 12); (9 × 4)

$\therefore \text{ x } ∝ \frac{1}{y}$ & variation constant = 36

Application 6.

Let us write two examples with two interrelated variables where the variables are in inverse variation. [Let me do it myself]

Solution:

(i) Pressure (P) & Volume (V) gas are in inverse proportion.

(ii) Resistance (R) & Cross Section (A) of wire are in inverse proportion.

Application 8.

The values of interrelated variables x and y are

x	3	2	6
y	18	27	9

Let us determine if there is any variation relation between x and y. [Let me do it myself]

Solution:

Here x.y = Constant (54) $\therefore \text{ x } ∝ \frac{1}{y}$

Application 10.

Solution:

Let speed $= 4 \quad \& \quad$ Time = T

$\therefore$ V ∝ $\frac{1}{T}$ or, V = K. $\frac{1}{T}$ [Where K is a constant]

or, 60 $= \mathrm{K} \cdot \frac{1}{2} \quad \therefore \mathrm{K}$ = 120

When V = 80, $\quad$ V $= \frac{K}{T} \quad$ or, 80 $= \frac{120}{T} \quad \therefore$ T $= \frac{120}{80} = \frac{3}{2}$

$\therefore$ Samir babu takes $1 \frac{1}{2}$ hours to reach the station from his house.

Application 16.

If A² + B² ∝ A² – B², let us prove that A ∝ B [Let me do it myself]

Solution:

$\frac{A^2 + B^2}{A^2 - B^2} = \frac{K}{1}$ (Constant)

$\frac{A^2 + B^2 + A^2 - B^2}{A^2 + B^2 - A^2 + B^2} = \frac{K + 1}{K - 1}$

or, $\frac{2 A^2}{2 B^2} = \frac{K + 1}{K - 1}$

$\frac{A}{B} = \frac{\sqrt{K + 1}}{\sqrt{K - 1}}$ = P (Constant).

$\therefore$ A ∝ B

Question 14

Here, is V in joint variation with T and $\frac{1}{P}$ ?

V = R. $\frac{T}{P}$ [R = constant] in this relation we can say V is in joint variation with number of persons engaged in that work and number of days they worked. [Let me do it myself].

Solution:

Let total earning = (A); No. of men = B & No. of days = C.

$\therefore$ A ∝ B when C is constant

& A ∝ C when B is constant.

$\therefore$ A ∝ B C when B & C both vary.

We can say, A is in joint variation with B & C.

Application 19.

If 5 men can cultivate in 10 bighas of land in 9 days, let us calculate by theory of variation how long will be taken by 25 men for cultivating 10 bighas of land. [Let me do it myself]

Solution:

Let No. of men = A, No. of days = B, & Area of land = C

No. of men is in direct variation with area of land, when no. of days remains constant, i.e., A ∝ C when B is constant:

Again, No. of men is in inverse variation with no. of days when area of land remains constant, i.e., A ∝ $\frac{1}{B}$ when C is constant.

$\therefore$ According to the theorem on joint variation,

A ∝ $\frac{C}{B}$ when B & C both vary

$\therefore$ A = K. $\frac{C}{B}$ where K is a constant.

Given A = 5, B = 9, C = 10

5 = K $\frac{10}{9}$ or, 10 K = 9 × 5 $\quad \therefore$ K $= \frac{9 × 5}{10} = \frac{9}{2}$

$\therefore$ A = K $\frac{C}{B}$

25 = $\frac{9}{2} × \frac{10}{B}$ or, B $= \frac{9 × 10}{2 × 25} = \frac{9}{5} = 1 \frac{4}{5}$

Application 24.

If x ∝ y and y ∝ z, let us prove that x² + y² + z² ∝ xy + yz + zx [Let me do it myself]

Solution:

x ∝ y $\quad \therefore$ x = K₁y

& y ∝ z $\quad \therefore$ y = K₂z

$\therefore$ x = K₁y = K₁K₂z

Now, $\frac{x^2 + y^2 + z^2}{x y + y z + z x} = \frac{\left(K_1 K_2 z\right)^2 + \left(K_2 z\right)^2 + z^2}{K_1 K_2 z \cdot K_2 z + K_2 z \cdot z + z \cdot K_1 K_2 z}$

$= \frac{K_1^2 K_2^2 z^2 + K_2^2 z^2 + z^2}{K_1 K_2^2 z^2 + K_2 z^2 + K_1 K_2 z^2} = \frac{z^2\left(K_1^2 K_2^2 + K_2^2\right)}{z^2\left(K_1 K_2^2 + K_2 + K_1 K_2\right)}$ = P (Constant)

$\therefore x^2 + y^2 + z^2 ∝ x y + y z + z x$ Proved.

Application 27.

The volume of a sphere varies directly with the cube of length of its radius. The length of diameter of a solid sphere of lead is 14 cm. If the sphere is melted, let us find by applying the theory of variation how many spheres having the length of 3.5 cm radius can be made. (let the volume remain unchanged before or after melting.) [Let me do it myself]

Solution:

Let the volume of a sphere of radius r is v.

Given v ∝ r³ or v = K r³ where K is a constant.

Diameter of lead sphere = 14 cm

$\therefore$ Its radius (r) = 7 cm

$\therefore$ Volume = K(7)³ cu cm.

Volume of small sphere of radius 3.5 cm = K × (3.5)³ cu cm.

As the volume remains same before & after melting,

no. of small spheres $= \frac{K × 7^3}{\mathrm{~K} ×(3.5)^3} = \frac{\mathrm{K} × 7 × 7 × 7}{\mathrm{~K} × 3.5 × 3.5 × 3.5}$ = 8.

LET US WORK OUT – 13

Question 1

Corresponding values of two variables A & B are :

A	25	20	45	250
B	10	12	18	100

If there is any relation of variation between A and B, let us determine it and write the value of variation constant.

Solution:

Here when values of A increases/decreases, the values of B increases/decreases.

$\frac{A}{B} = \frac{25}{10} = \frac{30}{12} = \frac{45}{18} = \frac{250}{100} = \frac{5}{2}$

$\therefore A ∝ B . \quad \& \text { Variable Constant } = \frac{5}{2}.$

Question 2

Corresponding values of two variable x & y are :

x	18	8	12	6
y	3	²⁷/₄	⁹/₂	9

Solution:

Here when the value of A increases/decreases the value of B decreases/increases.

x.y = 18 × 3 = 8 × $\frac{27}{4}$ = 12 × $\frac{9}{2}$ = 6 × 9 = 54

$\therefore$ x ∝ $\frac{1}{y}$ & Variation Constant = 54

Question 3

(i) A taxi of Bipin uncle travels 14 km path in 25 minutes. Let us calculate by applying the theory of variation how much path he will go in 5 hours by driving taxi with same speed.

Solution:

Let required time = T & required distance = S.

When time increases, distance increases when speed is constant.

$\therefore$ T & S are in direct variation.

$\therefore$ T ∝ S or, T = KS where K is constant.

Here T = 25 $\quad$ & = 14

$\therefore$ 25 = K × 14 or, $\frac{25}{14}$

$\therefore$ T $= \frac{25}{14}$ S

Now when T = 300 $\quad$ [ Show = 300 minutes]

$\therefore$ 300 = $\frac{25}{14}$ × S

$\therefore \text{ S } = \frac{300 × 14}{25}$ = 168 .

$\therefore$ He will go 168 km in 5 hours.

(ii) A box of sweets is divided among 24 children of class one our school, they will get 5 sweets each. Let us calculate by applying theory the of variation how many sweets would each get, if the number of the children is reduced by 4 .

Solution:

Let no. of children = C $\quad$ & no. of sweet each will get = S

As total no. of sweets are fixed.

When C increases/decreases, S decreases/increases.

$\therefore$ C & S are in increase variation

$\therefore$ C ∝ $\frac{1}{S}$ or, C = $\frac{K}{S}$ [Where K is a Constant]

Here C = 24 ; & S = 5

24 = $\frac{K}{5} \quad \therefore$ K = 24 × 5 = 120

2nd case, C = 24 – 4 = 20, K = 120, S = ?

20 = $\frac{120}{\mathrm{~S}}$ or, S = $\frac{120}{20}$ = 6

$\therefore$ Each student will get 6 sweets.

(iii) 50 villagers had taken 18 days to dig a pond. Let us calculate by using the theory of variation how many extra persons will be required to dig the pond in 15 days.

Solution:

No. of villagers = V & no. of days = D

When V increases/decreases, D decreases/increases.

$\therefore$ V & D are in inverse variation.

$\therefore$ V $\alpha \frac{1}{D} \quad$ or, V $= \frac{K}{D} \quad$ [Where K is a constant]

Here V = 50, D = 18 $\quad \therefore 50 \frac{K}{18} \quad \therefore$ K = 50 × 18 = 900

2nd case V = ? $\quad$ D = 15, K = 900

$\therefore$ V $= \frac{900}{15}$ = 60

$\therefore$ Extra villagers required = 60 – 50 = 10

Question 4

(i) y varies directly with the square root of x and y = 9 when x = 9. Let us find the value of x when y = 6.

Solution:

y varies directly with square root of x

i.e., y ∝ $\sqrt{x} \quad \therefore$ y = k $\sqrt{x}$

When y = 9, x = 9 $\quad \therefore$ 9 = K $\sqrt{9} \quad \therefore$ K $= \frac{9}{\sqrt{9}} = \frac{9}{3}$ = 3

$\therefore \text{ y } = 3 \sqrt{x}$

When y = 6, $\quad \therefore$ 6 = 3 $\sqrt{x}, \quad \therefore \sqrt{x} = \frac{6}{3}$ = 2

$\therefore$ x = 2² = 4 Ans.

(ii) x varies directly with y and inversely with z. When y = 4, z = 5, then x = 3. Again if y = 16, z = 30, let us write by calculating the value of x.

Solution:

Here x ∝ y when z is constant

& $x ∝ \frac{1}{z}$ when y is constant.

$\therefore x ∝ \frac{y}{z}$ when y & z are both variables.

$\therefore x = K \cdot \frac{y}{z}$ (Where K is a Constant)

Given x = 3, y = 4 & z = 5

$\therefore 3 = \mathrm{K} \frac{4}{5} \quad \text{ or, } \mathrm{K} = \frac{3 × 5}{4} = \frac{15}{4}$

When y = 16, z = 30, x = ?

x = K. $\frac{y}{z}$

x = $\frac{15}{4} × \frac{16}{30}$ = 2

$\therefore$ x = 2 Ans.

(iii) x varies directly with y and inversely with z. When y = 5, z = 9 then x = $\frac{1}{6}$ . Let us find the relation among three variables x, y and z and if y = 6 and z = $\frac{1}{5}$ , let us write by calculating the value of x.

Solution:

x ∝ y when z is constant

x ∝ $\frac{1}{z}$ when y is constant.

$\therefore x ∝ \frac{y}{z},$ when both y & z are variables:

or, x = K, $\quad$ y × $\frac{1}{z} = \text{ K }\frac{y}{z}$

Given x $= \frac{1}{6}$ , y = 5 & z = 9

$\therefore \frac{1}{6}$ = K. $\frac{5}{9}$ $\quad \therefore$ K = $\frac{9}{5 × 6} = \frac{3}{10}$

Now, x $= \frac{3}{10} \cdot \frac{y}{z}$

x $= \frac{3 y}{10 z}$

Question 5

(i) If x ∝ y, let us show that x + y ∝ x – y.

Solution:

x ∝ y $\quad \therefore$ x = K y

Now, $\frac{x + y}{x y} = \frac{K y + y}{K y - y} = \frac{y(K + 1)}{y(K - 1)} = \frac{K + 1}{K - 1} = P \text{ (let) [Constant] } \therefore$ x + y ∝ x – y Proved.

(ii) $A ∝ \frac{1}{C}, C ∝ \frac{1}{B}$ let us show that A ∝ B.

Solution:

A ∝ $\frac{1}{C} \quad \therefore$ A = K₁ $\therefore = \frac{K_1}{C}$

Again, $C ∝ \frac{1}{B} \quad \therefore$ C = K₂ × $\frac{1}{B} = \frac{K_2}{B}$

$\therefore A = \frac{K_1}{C} = \frac{K_1}{K_2 / B} = \frac{K_1}{K_2} B = PB [\text{ Where Constant P } = \frac{K_1}{K_2} ]$

$\therefore$ A ∝ B Proved.

(iii) If a ∝ b, b ∝ $\frac{1}{c}$ and c ∝ d, let us write the relation of variation between a and b.

Solution:

a ∝ b or, a = K₁ b

& b ∝ $\frac{1}{C}$ or, b = K₂ × $\frac{1}{C} = \frac{K_2}{C}$

& C ∝ d or, C = K₃d

$\therefore$ a = K₁b

$= \mathrm{K}_1 \frac{\mathrm{K}_2}{\mathrm{C}}$

$= \frac{K_1 K_2}{K_3 d}$

a $= \frac{p}{d} \quad$ where P $= \frac{K_1 K_2}{K_3}$ = Constant

$\therefore \text{ a } ∝ \frac{1}{d}$

$\therefore$ a & d are in inverse variation. Ans.

(iv) If x ∝ y, y ∝ z and z ∝ x, let us find the relation among three constants variation.

Solution:

x ∝ y $\quad$ or, x = K_1 y

y ∝ z or, y = K_2 z

z ∝ x $\quad$ or, z = K₃x $\quad$ [Where $K_1, K_2 \& K_3$ are constants of variation]

Now, x = K₁y

or, x = K₁K₂Z

or, x = K₁K₂K₃x

$\therefore$ K₁K₂K₃= 1. Ans.

Question 6

If x + y ∝ x – y, let us show that

(i) x² + y² ∝ xy

Solution:

x + y ∝ x – y.

or, x + y = K(x – y) $\quad$ [Where K is a constant]

$\therefore \frac{x + y}{x - y} = \frac{K}{1} or, \frac{(x + y)^2}{(x - y)^2} = \frac{K^2}{1}$

or, $\frac{(x + y)^2 + (x - y)^2}{(x + y)^2 - (x - y)^2} = \frac{K^2 + 1}{K^2 - 1}$

or, $\frac{2 x}{2 y} = \frac{K + 1}{K - 1}$

or, $\frac{2 x^2 + 2 y^2}{4 x y} = \frac{K^2 + 1}{K^2 - 1}$

or, $\frac{2\left(x^2 + y^2\right)}{24 x y} = \frac{k^2 + 1}{K^2 - 1}$

$\therefore x^2 + y^2 = \frac{2\left(k^2 + 1\right)}{k^2 - 1}$

$\therefore x^2 + y^2 ∝ x y \quad [Where \frac{2\left(\mathrm{~K}^2 + 1\right)}{\mathrm{K}^2 - 1}$ is Constant.]

(ii) x³ + y³ ∝ x³ – y³

Solution:

x + y $\quad$ x – y $\quad \therefore$ x + y = K(x – y) (Where K is constant)

$\frac{x + y}{x - y} = \frac{K}{1} \quad or, \frac{x + y + x - y}{x + y - x + y} = \frac{K + 1}{K - 1} \quad \text{ or, } \frac{2 x}{2 y} = \frac{K + 1}{K - 1}$ = P (let)

$\therefore \frac{x}{y}$ = P $\quad \therefore x = P y \quad$ (Where P is Constant)

Now, $\frac{x^3 + y^3}{x^3 - y^3} = \frac{(P y)^3 + y^3}{(P y)^3 - y^3} = \frac{y^3 + \left(P^3 + 1\right)}{y^3 - \left(P^3 - 1\right)}$ = Constant.

(iii) ax + by ∝ px + ay [where a, b, p, q are non-zero constants]

Solution:

x + y ∝ x – y or, x + y = K(x – y)

or, $\frac{x + y}{x - y} = \frac{K}{1} or, \frac{x + y + x - y}{x + y - x + y} = \frac{K + 1}{K - 1} \text{ or, } \frac{2 x}{2 y} = \frac{K + 1}{K - 1} = R (let) (Where \mathrm{K} \ \mathrm{ R are Constants }) \quad \therefore \mathrm{x} = \mathrm{Ry}$

$\frac{a x + b y}{P x + q y} = \frac{a R y + b y}{P R y + q y} = \frac{y(a R + b)}{y(P R + q)} = \frac{a R + b}{P R + q}$ = Constant

$\therefore$ ax + by ∝ px + qy.

Question 7

(i) If a² + b² ∝ ab, let us prove that a + b ∝ a – b.

Solution:

a² + b² ∝ ab

or, a² + b² = Kab

a² + b² = $\frac{K}{2} \quad$ 2ab

or, $\frac{a^2 + b^2}{2 a b} = \frac{k}{2}$

or, $\frac{a^2 + b^2 + 2 a b}{a^2 + b^2 - 2 a b} = \frac{K + 2}{K - 2}$

or, $\frac{(a + b)^2}{(a - b)^2} = \frac{K + 2}{K - 2}$

or, $\frac{a + b}{a - b} = \frac{\sqrt{K + 2}}{\sqrt{K - 2}}$ = Constant

$\therefore$ a + b ∝ a – b Proved.

(ii) If x³ + y³ ∝ x³ – y³, let us prove that x + y ∝ x – y.

Solution:

x³ + y³ ∝ x³ – y³

$\therefore$ x³ + y³ = K(x³ – y³) (where K is a Constant)

or, $\frac{x^3 + y^3}{x^3 - y^3} = \frac{k}{1}$

or, $\frac{x^3 + y^3 + x^3 - y^3}{x^3 + y^3 - x^3 + y^3} = \frac{k + 1}{k - 1}$

or, $\frac{2 x^3}{2 y^3} = \frac{K + 1}{K - 1}$

or, $\frac{x^3}{y^3} = \frac{k + 1}{k - 1}$

$\frac{x}{y} = \frac{\sqrt[3]{K + 1}}{\sqrt[3]{K - 1}}$ = R (let) (Where R is Constant)

$\therefore$ (x + y) ∝ (x – y) Proved.

Question 8

If 15 farmers can cultivate 18 bighas of land in 5 days, let us determine by using the theory of variation the number of days required by 10 farmers to cultivate 12 bighas of land.

Solution:

No. of farmers = A, no. of days = B & area of land = C

No. of days is in inverse variation with no. of farmers, when area of land remains constant

i.e., B ∝ $\frac{1}{A}$ when C is constant

Again, No. of days is indirect variation with area of land; when No. of farmers remains constant

$\therefore$ B ∝ C when A is Constant.

According to the theorem on joint variation,

B ∝ $\frac{C}{A}$ when C & A both vary

$\therefore$ B = K $\frac{C}{A}$ where K is a constant of variation.

Given A = 15, B = 5, & C = 18.

5 = K $\frac{18}{15}$ or, K $= \frac{15 \times 5}{18} = \frac{25}{6}$

B = K $\frac{C}{A}$

$= \frac{25}{6} \times \frac{12}{10}$ = 5

$\therefore$ No. of days = 5. Ans.

Question 9

Volume of a sphere varies directly with the cube of its radius. Three solid sphere having length of $1 \frac{1}{2}$ , 2 and $2 \frac{1}{2}$ metre diameter are melted and a new solid sphere is formed. Let us find the length of diameter of the new sphere. [let us consider that the volume of sphere remains same before and after melting]

Solution:

Let the volume of sphere of radius r be v.

Given v ∝ r³ or v = K r³ where K is the constant of variation.

The radii of 3 spheres are $\frac{3}{4}, 1 \ \frac{5}{4}$ m.

$\therefore$ Volume of 1st sphere $= \mathrm{K} \times\left(\frac{3}{4}\right) ^3$ cu.m.

Volume of 2nd sphere $= \mathrm{K} \times(1) 3$ cu.m.

Volume of 3rd sphere $= K \times\left(\frac{5}{4}\right) ^3$ cu.m.

$\therefore$ If the radius of the new sphere = R

$\therefore$ Its volume = KR³

According to the problem,

KR³ $= \text{ K }(\frac{3}{4})^ 3 + K(1)^3 + K \times(\frac{5}{4}) ^3$

KR³ $= \text{ K }[\frac{27}{64} + 1 + \frac{125}{64}]$

$\therefore$ R³ = $\frac{216}{64}$

$\therefore \text{ R } = \frac{6}{4} = \frac{3}{2}$

$\therefore$ Diameter of the new sphere = 2R = 2 × $\frac{3}{2}$ = 3 m.

Question 10

y is a sum of two variables, one of which varies directly with x and another varies inversely with x. When x = – 1, then y = 1 and when x = 3, then y = 5. Let us find the relation between x and y.

Solution:

Let y = a + b where = a ∝ x & b ∝ $\frac{1}{x}$

$\therefore$ a = K₁ x & b = b $= \frac{K_2}{x}$

$\therefore$ y = K₁ x + $\frac{K_2}{x}$ ————————- (i)

When x = – 1, y = 1

$\therefore$ 1 = – K₁ – K₂ or, K₁ + K₂ = – 1 —————— (ii)

Again, x = 3, y = 5

$\therefore 5 = 3 \mathrm{~K}_1 + \frac{\mathrm{K} _2}{3} \text { or, } 9 \mathrm{~K}_1 + \mathrm{K}_2$ = 15 —————— (iii)

$\therefore$ (iii) – (ii)

$9K_1 + K_2 = 15\\ K_1 + K_2 = - 1\\ \underline{\text{(-) \text{ \quad(-)} \text{\quad(+)}}\quad}\\ 8K_1 = 16\\$

$\therefore$ K₁ = $\frac{16}{8}$ = 2

$\therefore$ K₂ = – 3

$\therefore$ y = 2x + $\left(\frac{ - 3}{x}\right)$

or, y = 2x – $\frac{3}{x}$

Question 11

If a ∝ b, b ∝ c, let us show that a³b³ + b³c³ + c³a³ ∝ abc (a³ + b³ + c³)

Solution:

a ∝ b $\quad \therefore$ a = K₁b

& b ∝ c $\quad \therefore$ b = K₂C $\quad \therefore$ a = K₁ $\quad$ K₂C = K₁ K₂ C (Where K₁, K₂ are constants of variation)

Now, $\frac{a^{3} b^{3} + b^{3} c^{3} + c^{3} a^{3}}{a b c(a^{3} + b^{3} + c^{3})}$

$= \frac{(\mathrm{K}_{1} \mathrm{~K}_{2} \mathrm{C})^{3}.(\mathrm{K}_{2} \mathrm{C})^{3} + (\mathrm{K}_{2} \mathrm{C})^{3} \mathrm{C}^{3} + \mathrm{C}^{3}(\mathrm{~K}_{1} \mathrm{~K}_{2} \mathrm{C})^{3}}{\mathrm{~K}_{1} \mathrm{~K}_{2} \mathrm{C}. \mathrm{K}_{2} \mathrm{C}. \mathrm{C}[(\mathrm{K}_{1} \mathrm{~K}_{2} \mathrm{C})^{3} + \mathrm{K}_{2} \mathrm{C}^{3} + \mathrm{C}^{3}]}$

$= \frac{\mathrm{K}_{1}{ }^{3} \mathrm{~K}_{2}{ }^{3} \mathrm{C}^{3} \mathrm{~K}_{2}{ }^{3} \mathrm{C}^{3} + \mathrm{K}_{2}{ }² \mathrm{C}^{3}. \mathrm{C}^{3} + \mathrm{K}_{1}{ }^{3} \mathrm{~K}_{2}{ }^{3} \mathrm{C}^{3}}{\mathrm{~K}_{1} \mathrm{~K}_{2}{ }² \mathrm{C}^{3}(\mathrm{~K}_{1}{ }^{3} \mathrm{~K}_{2}{ }^{3} \mathrm{C}^{3} + \mathrm{K}_{2}{ }^{3} \mathrm{C}^{3} + \mathrm{C}^{3})}$

$= \frac{C^{6}(K_{1}^{3} K_{2}^{6} + K_{2}^{3} + K_{1}^{3} K_{2}^{3})}{K_{1} K_{2}² C^{3} \times C^{3}(K_{1}^{3} K_{2}^{3} + K_{2}^{3} + 1)}$

$= \frac{\mathrm{K}_{1}^{3} \mathrm{~K}_{2}^{6} + \mathrm{K}_{2}^{3} + \mathrm{K}_{1}^{3} \mathrm{~K}_{2}^{3}}{\mathrm{~K}_{1} \mathrm{~K}_{2}² \mathrm{~K}_{1}^{3} \mathrm{~K}_{2}^{3} \mathrm{~K} + \mathrm{K}_{2}^{3} + 1}$ = Constant

$\therefore a^{3} b^{3} + b^{3} c^{3} + c^{3} a^{3} ∝ a b c(a^{3} + b^{3} + c^{3})$ Proved.

Question 12

To dig a well of x dcm deep one part of the total expenses varies directly with x and other part varies directly with x². If the expenses of digging wells of 100 dcm and 200dcm depths are Rs. 5,000 and Rs. 12,000 respectively, let us write by calculating the expenses of digging a well of 250 dcm depth.

Solution:

Let total expenses to dig a well = Rs. y.

Of which part A is one part & other part B.

$\therefore$ y = A + B

Now, A ∝ x or, A = K₁x

& B ∝ x² $\quad$ or, B = K₂x²

$\therefore$ y = K₁x + K₂ x ————(i)

When x = 100 dcm, y = 5,000

& when x = 200 dcm, y = 12,000

$\therefore$ 5,000 = K₁ 100 + K₂ 10,000 or, K₁ + 100 K₂ = 50 —————- (ii)

& 12,000 = K. 200 + K₂. 40,000 or, K₁ + 200 K₂ = 60 —————- (iii)

(iii) – (ii), 100 K₂ = 10 $\quad \therefore$ K₂ $= \frac{1}{10}$

From (ii) K₁ + 10 = 50 $\quad \therefore$ K₁ = 40

From (i) y = 40 x + $\frac{x²}{10}$ —————- (iv)

Now, when x = 250 dcm

y = 40 × 250 + $\frac{(250)²}{10}$ = 10,000 + 6250 = 16250

Total expenses = Rs. 16,250.

Question 13

Volume of a cylinder is in joint variation with the square of the length of radius heights is 5 : 4, let us find the ratio of their volumes. cylinders is 2 : 3 and ratio of their height is 5 : 4, let us find the ratio of their volume

Solution:

Let volume of cylinder = V, their volumes.

$\therefore$ V ∝ R^² HV = KR²H (Where K is a constant of variation) 1st case let radius = 2r & Height = 5

2nd case radius = 3r & Height = 4 h

1st case volume = V₁ = K.(2r)². 5h = K. 4r². 5h

2nd case volume = V₂ = K.(3r)². 4h = K. 9r². 4b

$\therefore \frac{V_{1}}{V_{2}} = \frac{K. 4 r². 5 h}{K. 9 r². 4 h} = \frac{4 \times 5}{9 \times 4} = \frac{5}{9}$

$\therefore$ Ratio of their volumes = 5 : 9.

Question 14

An agricultural Co-operative Society of the village of Pachla has purchased a tractor. Previously 2400 bighas of land were cultivated by 25 ploughs in 36 days. Now half of the land can be cultivated only by that tractor in 30 days. Let us calculate by using the theory of variation, the number of ploughs work equally with one tractor.

Solution:

Let no. of ploughs = N, area of Land = A, no. of days = D.

$\therefore$ N ∝ A & N ∝ $\frac{1}{D}$

According to the theorem of joint variation, N ∝ $\frac{1}{D}$

or, N = K $\frac{1}{\mathrm{D}}$ (Where K is a constant of variation)

Now, A = 2400, N = 25 ; D = 36

$\therefore$ 25 = K : $\frac{2400}{36} \quad \therefore$ K $= \frac{25 \times 36}{2400} = \frac{3}{8}$

Now, N = K. $\frac{A}{D}$

$= \frac{3}{8} \times \frac{1200}{30}$ as, A $= \frac{2400}{2} = 1200 \quad$ & D = 300

N = 15

$\therefore$ No. of ploughs = 15

$\therefore$ 15 plough work equally with one tractor.

Question 15

Volume of a sphere varies directly with cube of length of its radius and surface area of sphere varies directly with the square of the length of radius. Let us prove that the square of volume of sphere varies directly with cube of its surface area.

Solution:

Let volume of sphere = V, radius = R & surface area of sphere = S

$\therefore \mathrm{V} ∝ \mathrm{R}^{3} \quad \& \mathrm{~S} = \mathrm{R}²$

or, V = m R³ or; S = nR² (Where m & n are constants of variation)

or, R³ $= \frac{V}{m}$ or, R² $= \frac{S}{n}$

R $= (\frac{V}{m})^{1 / 3}$

$\therefore \frac{S}{n} = (\frac{V}{m})^{2 / 3} \quad$ or, S = n. $(\frac{V}{m})^{2 / 3} \quad$ or, S³ $= \frac{n^{3}}{m²}$ . V²

$\therefore$ V² $= \frac{m²}{n^{3}}$ S³

$\therefore$ V² ∝ S³( as $\frac{\mathrm{m}²}{\mathrm{n}^{3}}$ = Constant)

Very short answer type questions (V.S.A.) :

(A) M.C.Q.

Question 1

x ∝ $\frac{1}{y}$ , then

x $= \frac{1}{y}$
y $= \frac{1}{x}$
xy = 1
xy = non-zero constant.

Solution:

(d) xy = non-zero constant.

Question 2

If x ∝ y, then

$x^{2} ∝ y^{3}$
$x^{3} ∝ y^{2}$
$x ∝ y^{3}$
$x^{2} ∝ y^{2}$

Solution:

(d) $x^{2} ∝ y^{2}$

Question 3

If x ∝ y and y = 8 when x = 2; if y = 16, then the value of x is

Solution:

(b) 4

Question 4

If x ∝ y² and y = 4 when x = 8; if x = 32, then the value of y is

Solution:

(b) 8

Question 5

If y – z ∝ $\frac{1}{x}$ , x – x ∝ $\frac{1}{y}$ and x – y ∝ $\frac{1}{z}$ , sum of three variation constants is

0
1
– 1
2

Solution:

(a) 0

(B) Let us write whether the following statements are true or false: –

Question 1

If y ∝ $\frac{1}{x}$ , $\frac{y}{x}$ = non – zero constant.

Solution:

FALSE

Question 2

If x ∝ z and y ∝ z then xy ∝ z.

Solution:

TRUE

(C) Let us fill in the blanks :

Question 1

If x ∝ $\frac{1}{y}$ and y ∝ $\frac{1}{z}$ , then x ∝ ____________________.

Solution:

Question 2

If x ∝ y, xⁿ ∝ ________________.

Solution:

Y^n.

Question 3

If x ∝ y and x ∝ z, then (y + z) ∝ _________________.

Solution:

Short answer type questions (S.A.) :

Question 1

If x ∝ y^{^2} and y = 2 a when x = a; x and y let us find the relation between x and y.

Solution:

x = Ky²[K = Constant of variation ]

or, a = K(2a)² $\quad \therefore$ K $= \frac{a}{4 a^{2}} = \frac{1}{4 a}$

$\therefore \text{ x }= \frac{1}{4 a} y^{2}$ , i.e., y² = 4ax

Question 2

If x ∝ y, y ∝ z and z ∝ x, let us find the product of three non-zero constants.

Solution:

x = K₁y ; y = K₂z ; z = K₃x

$\therefore \mathrm{x} = \mathrm{K}_{1} \mathrm{y} = \mathrm{K}_{1} \mathrm{~K}_{2} \mathrm{z} = \mathrm{K}_{1} \mathrm{~K}_{2} \mathrm{~K}_{3} \mathrm{x}$

$\therefore \mathrm{K}_{1} \mathrm{~K}_{2} \mathrm{~K}_{3}$ = 1.

Question 3

If x ∝ $\frac{1}{y}$ and y ∝ $\frac{1}{z}$ , let us find if there be any relation of direct or inverse variation between x and z.

Solution:

x $= \frac{K_{1}}{y}$ —————— (i)

y $= \frac{K_{2}}{Z}$ ————– (ii) where K₁ & K₂ are constants of variation.

x $= \frac{K_{1}}{K_{2}} / z = \frac{K_{2}}{K_{1}} z \quad \therefore$ x ∝ z Proved.

Question 4

If x ∝ yz and y ∝ zx, let us show that z is a non-zero constant.

Solution:

x = K₁ yz, y = K₂ zx

$\therefore \mathrm{xy} = \mathrm{K}_{1} \mathrm{~K}_{2} \mathrm{xyz}^{2}$