WBBSE Solution For Class 10 Maths Algebra Chapter 4 Variation

Algebra Chapter 4 Variation What Is Variation

Variation is a relation between two variables, from which we can find the value of one of the variables with respect to the other.

That is, there is such a relation that if the changes of the value of one variable cause the change of the value of the other proportionally, then we say that one variable varies directly to the other, i.e., the simple variation relation exists between the variables.

For example, distance is in direct variation with time, i.e., in more time more distance, and in less time less distance should be traveled.

Similarly, the circumference and radius of a circle are in direct variation.

Variation sign 

To indicate the variation between two variables we generally use the sign “∝” Thus if x and y are in direct variation, we write x∝y.

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Algebra Chapter 4 Variation Different Types Of Variation

Variation is of three types, Namely

  1. Direct variation;
  2. Inverse variation; and
  3. Joint variation;

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1. Direct variation:

If two variables are interrelated in such a way that changes in value of any one of them directly cause a change in the value of the other proportionally, then this type of variation is called direct variation.

For example, the number of books and their cost prices varies directly, i.e., if the number of books be increased, then the total value of the books increases. Similarly, if the number of books be decreased, then the total value of the books also decreases.

[If the decrease or increase of one variable directly causes the decrease or increase of the other variable, then it can not be said certainly that the two variables are in direct variation unless this decrease or increase occurs at a fixed rate of proportion.]

2. Inverse Variation:

If the decrement of one of the variables causes the increment of the other variable and the increment of one of the variables causes the decrement of the other variable, then we say that the variables are in inverse variation.

For example, to reach a certain place, if the speed be increased, then the time to be required to reach the place should be decreased.

Further, for a fixed amount of food, if the number of man increases, then the number of days in which they can feed them, decreases and vice versa.

Thus if a and b are in inverse variation, then a ∝ \(\frac{1}{b}\)

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3. Joint variation:

Sometimes we see that the value of one variable depends on the values of more than one independent variables.

One variable is said to vary jointly as a number of other variables when it varies directly as their product.

Thus if x ∝ yz, i.e., x = kyz [k ≠0 being a constant], then x varies jointly as y and z.

For example, the area of a triangle varies jointly as its base and altitude.

If a ∝ bc, then we say that a varies jointly as b and c.

For example, the quantity of work jointly varies as the number of man engaged in the work and the number of days required to complete the work.

Variation constant:

If x ∝ y, then x = ky, here k≠ 0 is said to be variation constant.

That is, the certain rate, of proportion at which two variables vary is known as variation constant.

The value of the variation constant may or may not depend on the variables, but may depend on some other objects.

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Algebra Chapter 4 Variation Theorem Of Joint Variation

Theorem: If x varies as y when z remains unchanged and x varies as z when y remains unchanged, then x will vary as the product of y and z when both y and z vary.

Proof: By given condition, x∝ y, when z in constant ……. (1)

and x ∝ z, when y is constant…… (2)

∴ By (1) we get x = k1y where k1≠0 is the variation constant.

From (2) we get, k1y ∝ z  ⇒ k1∝ z  (as y is a constant)

⇒ k1 = k2z, k2 ≠ 0 is the variation constant.

∴ From (1) we get x = k2z. y [k1 = k2z]

⇒ x = k2yz ⇒ x ∝ yz [k2 ≠ 0 constant]

Hence, x ∝ y and x ∝ z ⇒ x ∝ yz (Proved)

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Algebra Chapter 4 Variation Properties Of Variation Examples

Example 1. If A ∝ B, then B ∝ A.

Solution:

Proof: Given that A ∝ B, ∴ A = k. B

or, B = \(\frac{1}{k}\) A = mA (when \(\frac{1}{k}\) = m) and m = constant

∴ B = mA [Where m is variation constant]

∴ B ∝ A. (Proved)

Example 2. If A ∝ B, then Am ∝ Bm

Solution:

Proof: Given that A ∝ B, ∴ A = kB (k = variation constant)

⇒ Am= (kB)m or, Am = km. Bm

⇒ Am = nBm when n = km

∴ Am ∝ Bn (n = km =constant) (Proved)

Example 3. If A ∝ B and B ∝ C, then A ∝ C.

Solution:

Proof: Given that A ∝ B ⇒ A = k1B (k1 = variation constant)

Again, B ∝ C ⇒ B = k2C (k2 = variation constant)

∴ From A = k1B, we get, A = k1k2C [B = k2C]

⇒ A = kC when k = k1k2

⇒ A ∝ c (k = variation constant)

Hence A ∝ C. (Proved)

Example 4. If A ∝ BC, then B ∝ \(\frac{A}{c}\) and C ∝ \(\frac{A}{b}\) 

Solution:

Proof: Given that A ∝ BC ⇒ A = k.BC (Where k = variation constant)

⇒ B = \(\frac{1}{k}.\frac{A}{C}\)

⇒ B = m. \(\frac{A}{C}\) [When m = \(\frac{1}{k}\)]

∴ B ∝ \(\frac{A}{C}\) [when m = \(\frac{1}{k}\)]

Again, A ∝ BC ⇒ A = k.BC (k = variation constant)

⇒ C = \(\frac{1}{k}.\frac{A}{B}\)

⇒ C = m. \(\frac{A}{B}\) [When m = \(\frac{1}{k}\)]

∴ C ∝ \(\frac{A}{B}\) [m =\(\frac{1}{k}\)= variation constant]

Hence B ∝ \(\frac{A}{C}\) and C ∝ \(\frac{A}{C}\)(Proved)

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Example 5. If A ∝ B, then AC ∝ BC (where C is either a constant or variable).

Solution:

Proof: Given that A ∝ B ⇒ A = k.B, (k = variation constant)

⇒ AC = k.BC

∴ AC ∝ BC (k = variation constant)

∴ AC ∝ BC. (Proved)

Example 6. If A ∝ c and B ∝ c, then (A ± B) ∝ c and AB ∝ c2.

Solution:

Proof: A ∝ C  ⇒ A = k1C {k1 = variation constant ≠ 0)

B ∝ C ⇒ B = k2C (k2 = variation constant ≠ 0)

Now, A ± B = k1C ± k2C = (k1± k2) C

∴ A ± B = kC [When k = k1± k2 = constant ]

∴ A ± B ∝ c (Proved) [k = variation constant ]

Again, A ∝ C  ⇒ A = k1C (k1 = variation constant)

and B ∝ c ⇒ B = k2C (k2 = variation constant)

∴ AB = k1C x k2C = k1k2C2 = kC2 [When k = k1k2 = constant]

∴ AB = kC2.

∴ AB ∝ C2 (Proved) [k = variation constant]

Example 7. If A ∝ B and C ∝ D, then AC ∝ BD and \(\frac{A}{C}\)∝ \(\frac{B}{D}\)

Solution:

Proof: A ∝ B ⇒ A = k1B (where k1= variation constant ≠ 0)

C ∝ D ⇒ C = k2D (where k2 = variation constant ≠ 0)

∴ AC = (k1B)(k2D) = k1k2 BD = kBD (when k1k2 = k = constant]

∴ AC = k.BD

∴ AC ∝ BD [k = variation constant ]

Again, A ∝ B ⇒ A = k1B (where k1≠ 0 = variation constant)

C ∝ D ⇒ C = k2D (where k2 ≠ 0 = variation constant)

\(\frac{\mathrm{A}}{\mathrm{C}}=\frac{k_1 \mathrm{~B}}{k_2 \mathrm{D}}=\frac{k_1}{k_2} \cdot \frac{\mathrm{B}}{\mathrm{D}}=k \cdot \frac{\mathrm{B}}{\mathrm{D}}\)

Maths Solutions Class 10 Wbbse

(when \(\frac{k_1}{k_2}\)=k = constant)

∴ \(\frac{\mathrm{A}}{\mathrm{C}} \doteq k \cdot \frac{\mathrm{B}}{\mathrm{D}}\) (when k≠0= variation constant)

∴ \(\frac{\mathrm{A}}{\mathrm{C}} \propto \frac{\mathrm{B}}{\mathrm{D}}\)

∴ AC ∝ BD and \(\frac{\mathrm{A}}{\mathrm{C}} \propto \frac{\mathrm{B}}{\mathrm{D}}\) (proved)

Example 8. If x ∝ y and y ∝ z, then (x2 + y2 + z2) ∝ (xy + yz + zx)

Solution:

Proof : x ∝ y ⇒ x = k1y (where k1≠0 = variation constant)

y ∝ z ⇒ y = k2z (where k2≠0= variation constant)

∴ x = k1.(k2z) = k1k2 z.

Now, x2+ y2+z2 = (k1k2z)2 + (k2z)2 +z2

= k12k22z2 + k22z2 + z2

= z2 (k12k22+ k22+1) …..(1)

and xy + yz + zx = (k1k2z)(k2z)+(k2z)z + z(k1k2z)

= k1k22z2 + k2z2 + k1k2z2

= k2z2(k1k2 +1 + k1)

= k2z2 (1 + k1+ k1k2) ……..(2)

∴ Dividing (1) by (2) we get,

\(\frac{x^2+y^2+z^2}{x y+y z+z x}\)

Maths Solutions Class 10 Wbbse

= \(\frac{z^2\left(k_1^2 k_2^2+k_2^2+1\right)}{k_2 z^2\left(1+k_{\mathrm{I}}+k_1 k_2\right)}\)

= \(\frac{1+k_2^2+k_1^2 k_2^2}{k_2\left(1+k_1+k_1 k_2\right)}\)

= k(let), [where k ≠0= \(\frac{1+k_2^2+k_1^2 k_2^2}{k_2\left(1+k_1+k_1 k_2\right)}\)]

∴ \(\frac{x^2+y^2+z^2}{x y+y z+z x}\) = k

⇒ x2 + y2+ z2 = k (xy + yz + zx) (where k ≠ 0 = variation constant)

∴ (x2 + y2 + z2) ∝ (xy + yz.+ zx). (Proved).

 

Algebra Chapter 4 Variation Multiple Choice Questions

Example 1. If (x + y) ∝ (x – y), then

  1. x = y
  2. x ∝ y
  3. x ∝ \(\frac{1}{y}\)
  4. y ∝ \(\frac{1}{x}\)

Solution: (x + y)∝(x- y)

∴ (x + y) = k (x – y) [where k = non-zero constant]

or, \(\frac{x+y}{x-1}=\frac{k}{1}\)

or, \(\frac{x+y+x-y}{x+y-x-y}=\frac{k+1}{k-1}\)

or, \(\frac{2 x}{2 y}=\frac{k+1}{k-1}\)

or, \(x=\left(\frac{k+1}{k-1}\right) x y\)

∴ x ∝ y [because \(\frac{k+1}{k-1}\)=constant]

∴ 2. x ∝ y

(x + y)∝(x- y) then x ∝ y

Example 2. If x ∝ y and y = 24 when x = 6; if y = 20, then the value of x is

  1. 3
  2. 4
  3. 5
  4. 6

Solution: x ∝ y ⇒ x = ky(where k= non-zero constant)…….. (1)

Given that y = 24 when x = 6

∴ 6 = k x 24 ⇒ k = \(\frac{6}{24}\) = \(\frac{1}{4}\)

∴ from (1) we get, x = \(\frac{1}{4}\) y …..(2)

Putting, y = 20 in (2) we get, x = \(\frac{1}{4}\) x 20 or, x = 5

∴ 3. 5

The value of x is 5

Example 3. If x ∝ y and y ∝ z then

  1. x ∝ z
  2. y ∝ xz
  3. z ∝\(\frac{x}{y}\)
  4. \(\frac{1}{x}\)∝\(\frac{1}{z}\)

Solution: y ∝ z

∴ y = kz [k = variation constant]

Again, x ∝ z (variation constant)

∴ x = my [ m = variation constant]

= m x kz, or, x = mk x z

∴ x ∝ z[mk= constant]

∴ 1. x ∝ z is correct.

x ∝ y and y ∝ z then x ∝ z

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Example 4. If a ∝ \(\frac{1}{b}\)

  1. ab = 1
  2. a = \(\frac{1}{b}\)
  3. b = \(\frac{1}{a}\)
  4. ab = non zero constant

Solution: a ∝ \(\frac{1}{b}\) ⇒ a = k. \(\frac{1}{b}\) [where k = non zero constant]

∴ ab = k ⇒ xy = non zero constant

∴ 4. ab = non zero constant

Example 5. 1. If x ∝ yz, then

  1. y = xz
  2. z ∝ xy
  3. z ∝ \(\frac{x}{y}\)
  4. z ∝ \(\frac{y}{x}\)

Solution: x ∝ yz

∴ x = kyz [k = variation constant]

or, z = \(\frac{1}{k}.\frac{x}{y}\)

or, z ∝ \(\frac{x}{y}\).\(\frac{1}{k}\)

∴ 3. z ∝ \(\frac{x}{y}\)

Example 6. If x2 -y2 ∝ xy, then (x+y) ∝ (x -y) = 

  1. x + y ∝ x – y
  2. \(\frac{x}{y}\) ∝ \(\frac{x}{y}\)
  3. x = \(\frac{1}{y}\)
  4. None of these

Solution: x2 + y2 = 2kxy [ 2k is a non-zero constant]

or, \(\frac{x^2+y^2}{2 x y}=k\)

or, \(\frac{x^2+y^2+2 x y}{x^2+y^2-2 x y}=\frac{k+1}{k-1}\) [by components dividends]

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

∴ \(\frac{x+y}{x-y}= \pm \sqrt{\frac{k+1}{k-1}}=\mathrm{c}\)= constant

∴ x + y ∝ x – y

∴ 1. x + y ∝ x – y

Example 7. If x ∝ y2 and y = 8 when x = 16; If x = 64 then the positive values of y is

  1. 8
  2. 16
  3. 32
  4. 64

Solution: x ∝ y2

∴ x = ky2 (where k = non-zero constant)…..(1)

As per question, y = 8 when x = 16

∴ from (1) we get, 16 = k.82

or, 16 = 64k  or, k = \(\frac{16}{24}\) = \(\frac{1}{4}\)

∴ x = \(\frac{1}{4}\) y2 [from(1)]….(2)

Now, putting x = 64 in (2) we get, 64 = \(\frac{1}{4}\) y2

or, y2 = 256 or, y = √256 = ±16

∴ y = 16 (taking positive value)

∴ 2. 16

 

Algebra Chapter 4 Variation Write True Or False

Example 1. If x ∝ \(\frac{1}{y}\) is a non xero variation constant

Solution: x ∝ \(\frac{1}{y}\) ⇒ xy = k (k is anon zero variation constant)

∴ The statement is true

Example 2. If x ∝ \(\frac{1}{y}\) and y ∝ \(\frac{1}{z}\), then xy ∝ z

Solution: \(x=\frac{k_1}{y}, y=\frac{k_2}{z}\) (k1, k2 non zero variation constant)

\(x=\frac{k_1}{y}=\frac{k_1}{k_2}=\frac{k_1}{k_2} \cdot z\)

∴ x ∝ z

∴ The statement is false

Example 3. If \(y-z \propto \frac{1}{z}, z-x \propto \frac{1}{y}, x-y \propto \frac{1}{z}\), sum of three variations constant is-1.

Solution: \(y-z=\frac{\kappa_1}{x}, \quad z-x=\frac{\kappa_2}{y}, \quad x-y=\frac{k_3}{z}\)

(k1,k2,k3 are non-zero variation constants)

Now, k1 + k2 + k3 = xy – xz + yz – xy + xz – yz = 0

∴ The statement is false.

Example 4. If x ∝ y then \(\frac{x}{y}\) = non zero variation constant

Solution: x ∝ y  ∴ x = ky (k = non zero variation)

or, \(\frac{x}{y}\) = k

∴ \(\frac{x}{y}\) = non-zero variation constant

∴ The statement is true.

 

Algebra Chapter 4 Variation Fill In The Blanks

Example 1. If x ∝ z and y ∝ z, then xy ∝√z = ______

Solution: x ∝ z  ∴ x = k1z [k1= variation constant] …..(1)

and y ∝ z  ∴ y = k2z [k2= varaition constant]….(2)

∴ x x y = k1z x k2z [By multiplication 1 and 2]

or, xy = k1k2z or, xy ∝ z2   [k1k2= constant]     

∴ xy ∝ z2

xy ∝√z = xy ∝ z2

Example 2. If x ∝ y, then xn∝ ______

Solution: x ∝ y  ⇒ x= ky (k = non zero variation constant)

⇒ \((x)^n=(k y)^n\) (takimg n th power)

⇒ \(x^n=k^n\)

⇒ \(x^n=m y^n\) (when m=\(k^n \neq 0\) = variation constant)

⇒ xn ∝ yn (m = non zero variation constant)

∴ xn ∝ yn

xn∝ = xn ∝ yn

Example 3. If x ∝ y, then xn  ∝ ______

Solution: x ∝ y ∴ x = ky [k = variation constant]

or, xn = kn.yn ∴ xn ∝ yn [kn = constant]

∴ xn ∝ yn

xn  ∝ = xn ∝ yn

Example 4. If x ∝ y, then y ∝ _____

Solution: x ∝ y ∴ x = kxy (k = variation constant)

or, y = \(\frac{1}{k}\) ∴ y ∝ x \(\frac{1}{k}\)

∴ y ∝ x

y ∝ =  y ∝ x

 

Algebra Chapter 4 Variation Short Answer Type Questions

Example 1. If x ∝ y, y ∝ z, and z ∝ x  then find the product of three non-zero varaition constant.

Solution: x ∝ y ⇒ x = k1y (k1 = non zero variation constant)

⇒ k1 = \(\frac{x}{y}\)….(1)

y ∝ z ⇒ y = k2z (k2= non zero variation constant)

⇒ k2 = \(\frac{y}{z}\)….(2)

z ∝ x ⇒ z = k3x (k3 = non zero variation constant)

⇒ k3 = \(\frac{z}{x}\)….(3)

Now, multiplying (1), (2) and 93) we get, k1k2k3 = \(\frac{x}{y}\) x \(\frac{y}{z}\) x \(\frac{z}{x}\).

or, k1k2k3 = 1

Hence the product of non zero variation constant is 1

Example 2. If x ∝ \(\frac{1}{y}\) and y ∝ \(\frac{1}{z}\) then find if there be any relation of direct or inverse variation between x and z.

Solution: x ∝ \(\frac{1}{y}\) ⇒ x = k1 \(\frac{1}{y}\) [where k1 ≠ 0 = varation constant) ….(1)

Again, y ∝ \(\frac{1}{z}\) ⇒ y = k2 \(\frac{1}{z}\) [where k2 ≠ 0 = varation constant) ….(2)

Now, pitting, y = k2 \(\frac{1}{z}\) in (1) we get,

\(x=\frac{k_1}{k_2 \cdot \frac{1}{z}}=\frac{k_1}{k_2} \cdot z=\dot{k} z\) [when \( k=\frac{k_1}{k_2} \neq 0\)]

 

∴ x = k. z (where k ≠0 = variation constant)

∴ x ∝ z(k ≠0 = variation constant)

∴ x and z are in direct variation.

Example 3. If x ∝ yz and y ∝ zx, then prove that z is anon zero varaition constant.

Solution: x ∝ yz ⇒ x = k1 .yz (k1 ≠0 = variation constant)

⇒ y = \(\frac{x}{k_1 z}\) …(1)

Again, y∝ zx ⇒ y = k2.zx (k2 ≠0 = variation constant)

⇒ \(\frac{x}{k_1 z}=k_2 \cdot z x\) …..[from (1)]

⇒ z2=k1k2

⇒ \(z=\sqrt{k_1 k_2}\) ….(2)

Since k1, k2 are non-zero variation constants,

∴ √k1k2 is a non-zero variation constant.

⇒ z is a non-zero variation constant.

Hence z is a non-zero variation constant (proved).

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Example 4. If x ∝ y2  and y ∝ zx, then prove that z is a non zero varaition constant.

Solution: x ∝ y2 ⇒ x = k. y2…..(1)  (when k ≠ 0 = varaition constant)

If x = a, y = 2a, ∴ from (1) we get, a = k (2a)2

⇒ a = k.4a2 ⇒ k = \(\frac{a}{4 a^2}=\frac{1}{4 a}\)

∴ from (1) we get. x =\(\frac{1}{4a}\) y2 ⇒ y2 = 4ax.

Hence the required relation between x and y is y2 = 4ax.

Example 5. If b ∝ a3 and a increases in the ratio of 2: 3 then find in what ratio b will be increases.

Solution: b ∝ a3 ⇒ b = k.a3 (when k ≠ 0 = varaition constant) ….(1)

Now, a increases in the ratio 2: 3

Now putting a x \(\frac{3}{2}\) = \(\frac{3a}{2}\) in stead of a in (1) we get

\(b=k \cdot\left(\frac{3 a}{2}\right)^3 \Rightarrow b=k \cdot \frac{27 a^3}{8}\)….(2)

 

Now,  from (1) and (2) we get, the required ratio

= \(\frac{k a^3}{k \cdot \frac{27 a^3}{8}}=\frac{8}{27}=8: 27\)

Hence b will be increased in the ratio 8: 27

Example 6. If (y-z) ∝ \(\frac{1}{x}\), (z-x) ∝ \(\frac{1}{y}\) and (x-y) ∝ \(\frac{1}{z}\) then find the sum of three variation constants

Solution: (y-z) ∝ \(\frac{1}{x}\) ⇒ y – z = k1 \(\frac{1}{x}\) (where k1≠ 0 = variation constant)

⇒ k1 = x(y-z) ….(1)

(z-x) ∝ \(\frac{1}{y}\)

⇒ z – x = k2 \(\frac{1}{y}\) (where k2≠ 0 = variation constant)

⇒ k2 = y(z-x) ….(2)

Also,

(x-y) ∝ \(\frac{1}{z}\)

⇒ x – y = k3 \(\frac{1}{z}\) (where k3≠ 0 = variation constant)

⇒ k3 = z(x-y) ….(3)

Now, adding (1) + (2) + (3) we get

k1 + k2 + k3 = x (y – z) + y (z – x) + z (x – y)

= xy – xz + yz – xy + zx -yz = 0

∴ k1 + k2 + k3 = 0.

Hence the sum of three variation constants = 0

Example 7. If (x + y) ∝ (x- y), then prove that (x2 + y2) ∝ xy.

Solution: (x + y)∝ (x- y) ⇒ x + y = k (x- y) (where k ≠0 = variation constant)

⇒ (x + y)2 = k2 (x- y)2 (squaring both the sides)

⇒  x2 + 2xy+ y2 = k2 (x2-2xy + y2)

⇒ x2 + y2 = k2 (x2+ y2)—2k2xy- 2xy.

⇒ (x2 + y2) – k2 (x2 + y2) =- 2k2xy – 2xy

⇒ x2+ y2– k2 (x2 + y2) = – xy (2k2 + 2)

⇒ (x2 +y2)(1- k2) =- (2k2 + 2) xy

⇒ (x2+ y2)(k2– 1) = (2k2 + 2) xy

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

⇒ \(x^2+y^2=\frac{2 k^2+2}{k^2-1} \cdot x y\)

⇒ \(x^2+y^2=m x y\)

(where m= \(\frac{2 k^2+2}{k^2-1} \neq 0\))

⇒ \(x^2+y^2 \propto x y\) [because m ≠0 = variation constant]

∴ \(\left(x^2+y^2\right) \propto x y\)(Proved)

Wbbse Class 10 Maths Solutions

Example 8. If (x + y)∝(x – y), then show that (x3 + y3)∝(x3 – y3).

Solution: (x + y)∝(x – y) ⇒ (x+y) = k(x-y) (where k ≠0= varaition constant)

⇒ \(\frac{x+y}{x-y}=k\)

⇒ \(\frac{x+y+x-y}{x+y-x+y}=\frac{k+1}{k-1}\) [by componendo and dividendo]

⇒ \(\frac{2 x}{2 y}=\frac{k+1}{k-1}\)

⇒ \(\frac{x}{y}=\frac{k+1}{k-1}=m (let)\)

[when \(\frac{k+1}{k-1}=m\)]

⇒ \(\frac{x}{y}^3\) =m3(Cubing both sides)

⇒ \(\frac{x^3}{y^3}=m^3\)

⇒ \(\frac{x^3+y^3}{x^3-y^3}=\frac{m^3+1}{m^3-1}\) [by componendo and dividendo]

⇒ \(\frac{x^3+y^3}{x^3-y^3}=n\) [where \(n=\frac{m^3+1}{m^3-1} \neq 0\)

⇒ \(x^3+y^3=n\left(x^3-y^3\right) and n \neq 0=\) variation constant

⇒ \(\left(x^3+y^3\right) \propto\left(x^3-y^3\right)\) (Proved)

 

Algebra Chapter 4 Variation Long Answer Type Questions

Example 1. A taxi of Bipinbabu travels 14 km path in 25 minutes. Applying theory of variation find how much path he will go in 5 hours by driving taxi with same speed.

Solution:

Given:

A taxi of Bipinbabu travels 14 km path in 25 minutes. Applying theory of variation

Let the distance = S and time = t.

We know that the distance travelled and the time required are always, in direct variation.

∴ S ∝ t ⇒ S = kt ……(1) (where k≠ 0 = variation constant)

As per question, S = 14 km and t = 25 minutes.

∴ from (1) we get, 14 = k.25 ⇒ k = \(\frac{14}{25}\)

∴ from (1), S = \(\frac{14}{25}\) t …..(2)

Now, 5 hours = 5 x 60 minutes = 300 minutes.

∴ from (2) we get, S = \(\frac{14}{25}\) x 300 or, S = 168.

∴ The required path = 168 km.

Example 2. A box pf sweets is divided among 24 children of class one of your school, they will get 5 sweets each. Calculate by applying theory of variation how many sweets would each get, if the number of the children is reduced by 4. 

Solution:

Given:

A box pf sweets is divided among 24 children of class one of your school, they will get 5 sweets each.

Let the number of children = B and the number of sweets obtained by each child = A.

∴ B varies inversely as A, since if the number of children increases, the number of sweets obtained by each will be decreased.

∴ B ∝ \(\frac{1}{A}\) ⇒ B = k . \(\frac{1}{A}\) …..(1)

(where k ≠ 0 = v.c) [v.c. = variation constant]

As per the question, if B = 24, then A = 5.

∴ We get from (1), 24 = k.\(\frac{1}{5}\) ⇒ k = 120.

∴ by (1) we get, B = 120. \(\frac{1}{A}\)

⇒ B = \(\frac{120}{A}\) ….(2)

Then if B = 24 – 4 = 20, we get from (2), 20 = \(\frac{120}{A}\) or, A = \(\frac{120}{20}\) = 6.

Hence each child will get 6 sweets.

Example 3. 50 villagers had taken: 18 days to dig a pond. Calculate by using theory of variation how many extra persons will be required to dig the pond in 15 days. 

Solution: Let the number of persons = M and the number of days = D.

∴ If the number of persons increases, the number of days will decrease.

∴ M varies inversely as D.

∴ M ∝ \(\frac{1}{D}\)

⇒  M = k. \(\frac{1}{D}\) [where k ≠ 0=v.c.]

⇒ M = \(\frac{k}{D}\) ….(1)

As per question, if M = 50, then D = 18.

∴ from (1) we get, 50 = \(\frac{K}{18}\) ⇒ k = 900.

∴ M = \(\frac{900}{D}\) …..(2) [from (1)]

Now, putting D = 15 we get, m = \(\frac{900}{15}\) = 60.

∴ The extra persons = 60 – 50 = 10.

Hence the required number of extra persons = 10.

Wbbse Class 10 Maths Solutions

Example 4. y varies directly with square root of xand y = 9 when x = Find the value of x when y = 6.

Solution: As per question, y ∝ √x ⇒ y = k√x …..(1)

(where k = non-zero variation constant)

When x = 9, y = 9

∴ 9 = k√9 [by (1)] .

or, 9 = k.3 or, k = \(\frac{9}{3}\) = 3.

∴ from (1) we get, y = 3√x …..(2)

Now, when y = 6, from (2) we get, 6 = 3√x

or, √x =\(\frac{6}{3}\) = 2

or, x = (2)2 = 4 [Squaring both the sides]

∴ The required value of x = 4.

Example 5. x varies directly with y and inversely with z. When y = 5, z = 9, then x = \(\frac{1}{6}\). Find the relation among three variables x, y and z and if y = 6 and z = \(\frac{1}{5}\), then calculate the value of x.

Solution: x varies directly with yand inversely with z.

∴ x ∝ \(\frac{y}{z}\)

⇒ x = k. \(\frac{y}{z}\) ….(1) (where k ≠ 0= varaition constant)

As per question, if y = 5 and z = 9 then x  = \(\frac{1}{6}\)

∴ from (1) we get, \(\frac{1}{6}\) = k. \(\frac{5}{9}\)

or, k = \(\frac{9}{5 \times 6}=\frac{3}{10}\)

∴ (1) becomes, x = \(\frac{3}{10}\). \(\frac{y}{z}\) ….(2)

Hence the required relation between x, y and z is x = \(\frac{3y}{10z}\)

Now, if y = 6 and z = \(\frac{1}{5}\), then from (2) we get,

\(x=\frac{3}{10} \times \frac{6}{\frac{1}{5}} \text { or, } x=\frac{3 \times 6 \times 5}{10 \times 1}=9\)

 

Hence the required path of x = 9.

Example 6. x varies directly with y and inversely with z. If y = 4, z = 5, then x = 3. Again if y = 16, z = 3, then find tha value of x.

Solution: x varies directly with y  and inversely with z.

∴ x ∝ \(\frac{y}{z}\)

⇒ x = k. \(\frac{y}{z}\)…..(1) (where k≠0 = varaition constant)

As per question, if y = 4, z=5, then x =3

∴ from(1) we get, 3 = k. \(\frac{4}{5}\) or, k = \(\frac{15}{4}\)

Then (1) becomes, x = \(\frac{15}{4}\). \(\frac{y}{z}\)….(2)

Now, putting y = 16  and z = 30 in (2) we get, \(x=\frac{15}{4} \times \frac{16}{30}\) or, x = 2

Hence the required value of x = 2

Wbbse Class 10 Maths Solutions

Example 7. If(x+y) ∝ (x-y) then prove that (ax + by) ∝ (px + qy) where a, b, p, q are non-zero variation constant. 

Solution: (x+ y)∝ (x- y) ⇒  x + y = k (x – y), (where k ≠ 0 = variation constant)

\(\begin{aligned}
\Rightarrow \frac{x+y}{x-y}=k & \Rightarrow \frac{x+y+x-y}{x+y-x+y}=\frac{k+1}{k-1} \text { (by componendo and dividendo) } \\
& \Rightarrow \frac{2 x}{2 y}=\frac{k+1}{k-1} \\
& \Rightarrow \frac{x}{y}=\frac{k+1}{k-1} \\
& \Rightarrow \frac{a x}{b y}=\frac{a(k+1)}{b(k-1)} \\
& \Rightarrow \frac{a x+b y}{b y}=\frac{a(k+1)+b(k-1)}{b(k-1)} \cdots \ldots \ldots \ldots(1)
\end{aligned}\)

 

Again, \(\begin{aligned}
\frac{x}{y}=\frac{k+1}{k-1} & \Rightarrow \frac{p x}{q y}=\frac{p(k+1)}{q(k-1)} \\
& \Rightarrow \frac{p x+q y}{q y}=\frac{p(k+1)+q(k-1)}{q(k-1)}
\end{aligned}\)…..(2)

Now, dividing (1) by (2) we get,

\(\frac{\frac{a x+b y}{b y}}{\frac{p x+q y_a}{q y}}=\frac{\frac{a(k+1)+b(k-1)}{b(k-1)}}{\frac{p(k+1)+q(k-1)}{q(k-1)}} \text {. }\)

 

or, \(\quad \frac{a x+b y}{p x+q y} \times \frac{q}{b}=\frac{a(k+1)+b(k-1)}{p(k+1)+q(k-1)} \times \frac{q}{b}\)

or, \(\frac{a x+b y}{p x+q y}=\frac{a(k+1)+b(k-1)}{p(k+1)+q(k-1)}\)

or, \(\frac{a x+b y}{p x+q y}=\frac{a(k+1)+b(k-1)}{p(k+1)+q(k-1)}=m(let)\)

∴ a, b, p, q, k are non-zero variation constant.

∴ m is also a non-zero constant.

∴ \(\frac{a x+b y}{p x+q y}=m \neq 0\)

⇒ ax + by = m (px + qy)

⇒ (ar + 6y) ∝ (px + qy) (m = nonzero variation constant.)

Hence (ax + by) ∝ (px + qy). (proved)

[ Aliter: (x + y) ∝ (x – y) ⇒ x+y = k(x-y)

\(\begin{aligned}
& \Rightarrow \frac{x+y}{x-y}=k \\
& \Rightarrow \frac{x+y+x-y}{x+y-x+y}=\frac{k+1}{k-1} \text { [by componendo and dividendo] } \\
& \Rightarrow \frac{2 x}{2 y}=m(\text { let })\left[\text { when } m=\frac{k+1}{k-1} \neq 0\right] \\
& \Rightarrow \frac{x}{y}=m \Rightarrow x=m y
\end{aligned}\)

 

Now, \(\frac{a x+b y}{p x+q y}=\frac{a \times m y+b y}{p \times m y+q y}=\frac{y(a m+b)}{y(p m+q)}\)

= \(\frac{a m+b}{p m+q}=k(l e t)\left[\text { when } k=\frac{a m+b}{p m+q} \neq 0\right]\)

∴ \(\frac{a x+b y}{p x+q y}=k\)

⇒ a x+by=k (px+q y)

∴ (ax+by) ∝ (p x+q y) (proved) [k ≠0 = 0variation constant]

Class 10 Maths Wbbse Solutions

Example 8. If a2 + b2 ∝ ab, then prove that (a + b) ∝ (a – b).

Solution: (a2+ b2) = k. ab [when

= (a2 +b2^ = k.ab [when k ≠ 0 = variation constant]

∴ a2 + b2 = k.ab …….(1)

⇒ (a + b)2 – 2ab = k.ab

⇒ (a + b)2 = kab + 2ab

⇒ (a + b)2 = k+2)ab ….(2)

Again, a2 + b2 = k.ab

⇒ a2 + b2-2ab = k. ab- 2ab

⇒ (a – b)2 = (k- 2) ab…..(3)

Now, dividing (2) by (3) we get, \(\frac{(a+b)^2}{(a-b)^2}=\frac{(k+2) a b}{(k-2) a b}\)

or, \(\frac{(a+b)^2}{(a-b)^2}=\frac{(k+2)}{(k-2)}\) = m(let) ≠ 0 [\(\frac{(k+2)}{(k-2)}\) ≠ 0]

or, (a + b)2 = m (a – b)2

or, a + b = √m(a-b)

or, (a +b) ∝ (a – b) [√m≠ 0 = variation constant]

Hence (a + b)∝ (a – b) . (Proved)

Example 9. If (x3+ y3) ∝ (x3– y3), then prove that (x+ y) ∝ (x – y).

Solution: (x3+ y3) ∝ (x3-y3)

⇒ x3+y3 = k(x3-y3) [when k ≠ 0 = variation constant]

⇒ \(\frac{x^3+y^3}{x^3-y^3}=k\)

⇒ \(\frac{x^3+y^3+x^3-y^3}{x^3+y^3-x^3+y^3}=\frac{k+1}{k-1}\) [by componendo and dividendo]

⇒ \(\frac{2 x^3}{2 y^3}\)=\(\frac{k+1}{k-1}\)

⇒ \(\frac{x^3}{y^3}=\frac{k+1}{k-1}\)

⇒ \(\left(\frac{x}{y}\right)^3=\frac{k+1}{k-1}\)

⇒ \(\frac{x}{y}=\sqrt[3]{\frac{k+1}{k-1}}=m(\text { let }) \neq 0\)

⇒ \(\frac{x+y}{x-y}=\frac{m+1}{m-1} \text { [by componendo and dividendo] } \\\)

⇒ \(\frac{x+y}{x-y}=n\left[\text { when } n=\frac{m+1}{m-1} \neq 0\right]\)

⇒ x+y = n(x-y)

∴ (x + y) ∝ (x-y) (Proved) [n ≠ 0 = variation constant]

Example 10. If 15 framers can cultivate 8 bighas of land in 5 days, determine by using theory of varaition the number of days required by 10 framers to cultivate 12 bighas of land.

Solution: Let the number of farmers = F, the amount of land = G and the number of days = D.

if the number of farmers be increased, the amount of cultivated land also increases, but the number of days will decrease.

∴ F varies directly with G, but inversely with D.

\(\mathrm{F} \propto \frac{\mathrm{G}}{\mathrm{D}} \Rightarrow \mathrm{F}=k \cdot \frac{\mathrm{G}}{\mathrm{D}}\)

[when k ≠ 0 = variation constant]

∴ \(\mathrm{F}=k \cdot \frac{\mathrm{G}}{\mathrm{D}}\) …..(1)

As per question, if F = 15, G = 18 then, D = 5

∴ from (1) we get, 15 = \(k \cdot \frac{18}{5} \Rightarrow k=\frac{75}{18}=\frac{25}{6}\)

∴ (1) becomes, \(F=\frac{25}{6} \cdot \frac{\mathrm{G}}{\mathrm{D}}\) ….(2)

Now, putting F = 10, G = 12 in (2) we get, 10 = \(\frac{25}{6}\). \(\frac{12}{D}\)or, 10D = 50 or, D = \(\frac{50}{10}\) = 5.

Hence the required number of days = 5.

Class 10 Maths Wbbse Solutions

Example 11. Volume of a sphere varies directly with the cube of its radius. Three solid spheres having length of 1\(\frac{1}{2}\), 2 and 2\(\frac{1}{2}\) metres diameter are melted and a new sphere is formed. Find the length of diameter of the new sphere. [Consider that the volume of sphere remains same before and after melting.]

Solution :

Given:

Volume of a sphere varies directly with the cube of its radius. Three solid spheres having length of 1\(\frac{1}{2}\), 2 and 2\(\frac{1}{2}\) metres diameter are melted and a new sphere is formed.

Let the volume of the sphere = V and radius = R.

As per question, V ∝ R3 or, V = k. R3 (1)[ When k = non-zero variation constant ]

Now, putting \(\mathrm{R}=\frac{1 \frac{1}{2}}{2}=\frac{3}{4}, \mathrm{R}=\frac{2}{2}=1 \text { and } \mathrm{R}=\frac{2 \frac{1}{2}}{2}=\frac{5}{4}\) respectively in (1)

we get, V1 = k. \(\left(\frac{3}{4}\right)^3\) or, 64 V1 = 27k ….(2)

V2 = k. (1)3 or, V2 = k …..(3)

and V3 = k. \(\left(\frac{5}{4}\right)^3\) or, 64 V3 = 125k ….(4)

when volumes of the first, second and third spheres are V1, V2 and V3 respectively.

∴ Total volume, V = V1+V2 + V3.

= \(\frac{27 k}{64}+k+\frac{125 k}{64}=\frac{27 k+64 k+125 k}{64}\)

= \(\frac{216 k}{64}=\frac{54 k}{16}=\frac{27 k}{8}\)

Now putting V = \(\frac{27 k}{8}\) in (1) we get, \(\frac{27 k}{8}\) = k. R3

or, \(\mathrm{R}^3=\frac{27}{8}[because k \neq 0] \text { or, } \mathrm{R}^3=\left(\frac{3}{2}\right)^3 \text { or, } \mathrm{R}=\frac{3}{2}\)

∴ The radius of the new sphere = \(\frac{3}{2}\) metres

∴ Diameter of new sphere = 2 x \(\frac{3}{2}\) metres = 3 metres

Hence the required diameter of the new sphere = 3 metres.

Example 12. 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. Find the relation between x and y.

Solution: Let the variables are pand q,where p ∝ x and q ∝ \(\frac{1}{x}\)

⇒ p = k1x and q =\(\frac{k_2}{x}\), [k1, k2 ≠0]

As per question y  = p +q

⇒ \(y=k_1 x+\frac{k_2}{x}\) …..(1)

Now, if x = 1, then y = -1 \(k_1 1+\frac{k_2}{1}\)

or, k1 + k2 = -1 ….(2)

Also, if x = 3, then y = 5,

∴ from (1) we get, 5 = k1.3 + \(\frac{k_2}{3}\) or, 15 = 9k1 + k2 or, 9k1 +k2 = 15 ……(3)

Then subtracting (2) from (3) we get 8k1 = 16 ⇒ k1= 2

From (2) we get, k1 = – 1 -2 = – 3.

∴ (1) becomes y = 2x- \(\frac{3}{x}\)

Hence the required relation between x and y is y = 2x- \(\frac{3}{x}\)

Class 10 Maths Wbbse Solutions

Example 13. If a ∝ b and b ∝ c, then show that (a3b3 +b3c3 + c3a3) ∝ abc (a3 + b3 + c3).

Solution: a ∝ b ⇒ a – k1b (k1= non – zero variation constant)

b ∝ c ⇒ b = k2c (k2 = non – zero variation constant)

∴ a k1k2c.

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

= \(\frac{\left(k_1 k_2 c\right)^3\left(k_2 c\right)^3+\left(k_2 c\right)^3\left(c^3\right)+c^3 \cdot\left(k_1 k_2 c\right)^3}{k_1 k_2 c \cdot k_2 c \cdot c\left\{\left(k_1 k_2 c\right)^3+\left(k_2 c\right)^3+c^3\right\}}\)

= \(\frac{k_1^3 k_2{ }^6 c^6+k_2^3 c^6+k_1^3 k_2^3 c^6}{k_1 k_2{ }^2 c^3\left(k_1^3 k_2^3 c^3+k_2^3 c^3+c^3\right)}\)

= \(\frac{k_2^3 c^6\left(k_1^3 k_2^3+1+k_1^3\right)}{k_1 k_2^2 c^6\left(k_1^3 k_2^3+k_2^3+1\right)}\)

= \(\frac{k_2\left(k_1^3 k_2^3+k_1^3+1\right)}{k_1\left(k_1^3 k_2^3+k_2^3+1\right)}\)

= Non zero constant [k1 and k2 ≠ 0]

= k(let)

∴ a3b3+ b3c3 + c3a3 = k{abc (a3 + b3 + c3)}

∴ a3b3 + b3c3 + c3a3 ∝ abc (a3 + b3 + c3) [k ≠ 0 = variation constant.] (Proved)

Wbbse Class 10 Maths Solutions

Example 14. 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 200 dcm depths are ₹ 5000 and ₹ 12000 respectively, calculate the expenses of digging a well of 250 dcm depth.

Solution:

Given:

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 200 dcm depths are ₹ 5000 and ₹ 12000 respectively

Let total expenses = ₹ E, two parts of which are p and q, where p ∝ x and q ∝ x2

∴ p = k1x, k1 = non -zero variation and constant and q = k2x2, k2 =  non-zero variation constant

∴ E = p + q = k1x + k2x2

(where k1, k2 ≠ 0 = constant) ….(1)

As per question, expenses of 100 dcm is ₹ 5000

∴ from(1) we get, 5000 = k1 x 100 + k2 x (100)2

or, 50 = k1 + 100 k2

∴ k1 + 100 k2 = 50 …….(2)

Again, expenses of 200 dcm is ₹ 12000

∴ from (1) we get, 12000 = k1 x 200 + k2 (200)2

or, 12000 =  200 k1 + 40000 k2 or, 60 = k1 + 200 k2

∴ k1+ k2 200 = 60 …….(3)

Now, subtracting (2) from (3) we get, 100 k2 = 10 or, k2 = \(\frac{10}{100}\) = \(\frac{1}{10}\)

Then from (2) we get, \(k_1+100 \times \frac{1}{10}=50 \text { or, } k_1+10=50\)

or, k1 = 40

∴ becomes, \(\mathrm{E}=40 x+\frac{1}{10} x^2\) ….(4)

Now, putting x = 250 in (4) we get,

\(\begin{aligned}
\mathrm{E} & =40 \times 250+\frac{1}{10} \times(250)^2 . \\
& =10000+\frac{1}{4} \times 62500 .
\end{aligned}\)

= 1000 + 6250 = 16250

Hence to dig a well of depth 250 dcm the expenses is ₹16250

Class 10 Maths Wbbse Solutions

Example 15. Volume of a cylinder is in joint variation with square of the length of radius of base and its height. Ratio of radii of bases of two cylinders is 2 : 3 and ratio of their heights is 5 : 4 then find the ratio of their volumes.

Solution:

Given:

Volume of a cylinder is in joint variation with square of the length of radius of base and its height. Ratio of radii of bases of two cylinders is 2 : 3 and ratio of their heights is 5 : 4

Let the volume of the cylinder = V,

Radius of the base = R and height = H.

As per question, V ∝ R2H ⇒ V = k.R2H (where k ≠ 0 = variation constant).

Now, the volumes of the cylinders be V1 and V2 and their radii are 2r and 3r respectively and heights are 5h and 4h respectively. [ratio of radii = 2:3 and ratio of heights = 5:4]

∴ from (1) we get, V1 = k. (2r)2. 5h = 20 kr2h [R = 2r and H = 5h] and V2 = k. (3r)2. 4h = 36 kr2h [R = 3r and H = 4h]

∴ \(\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{20 k r^2 h}{36 k r^2 h}=\frac{5}{9}\)

∴ V1 : V2 = 5: 9

∴ The ratio of the volumes of the cylinders is 5 : 9.

Example 16. An agricultural co-operative society of village Narayanpur 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. Calculate by using the theory of variation, the number of ploughs work equally with one tractor.

Solution:

Given:

An agricultural co-operative society of village Narayanpur 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 the quantity of land = G bighas,

Number of Ploughs = P and time of cultivation = D days.

We know that if number of ploughs is constant, then the quantity of land varies directly with time of cultivation, i.e., G ∝ D when P is constant.

Again, if time of cultivation is constant, the quantity of land varies directly with number of ploughs, i.e.., G ∝ P, when D is constant.

Then by theorem of joint variation, G ∝ PD (where P and D are both variables) ⇒ G = k. PD

(where k ≠ 0 = variation constant)

∴ G = k. PD …….. (1)

As per question, if G = 2400 and P = 25, then D = 36

∴ from (1) we get, 2400 = k. 25 x 36.

or, k = \(\frac{2400}{25 \times 36}=\frac{96}{36}=\frac{8}{3}\)

∴ (1) becomes G = \(\frac{8}{3}\) PD ………(2)

Now putting G = \(\frac{2400}{2}\) =1200 and D = 30 in (2) we get,

\(1200=\frac{8}{3} \times P \times 30 \text { or, } \mathrm{P}=\frac{1200 \times 3}{8 \times 30}=15\)

Class 10 Maths Wbbse Solutions

Hence the required number of ploughs work equally with one tractor is 15.

Example 17. 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. Prove that the square of volume of sphere varies directly with cube of its surface area. 

Solution:

Given:

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 volume of sphere = V, Area of surface = A and Radius = R.

As per question, V ∝ R3

⇒  V = k1R3 (when k1 ≠ 0 = variation constant.)

∴ V = k1R3…………(1)

Also, A ∝ R2 ⇒ A = k2.R2 ……..(2) (when k2 ≠ 0 = variation constant).

Now, from (1) we get, R3 \(=\frac{\mathrm{V}}{k_1} \text { or, } \mathrm{R}=\sqrt[3]{\frac{\mathrm{V}}{k_1}}\)

∴ from(2) we get, \(\mathrm{A}^3=k_2^3 \cdot \frac{\mathrm{V}^2}{k_1^2} \text { or, } \quad \mathrm{V}^2=\frac{k_1^2}{k_2^3} \cdot \mathrm{A}^3\)

or, V2 = k.A3 [when \(k=\frac{k_1^2}{k_2^3}\) ≠ 0, k1,k2 ≠0]

⇒ V2 ∝ A3 .[k = non-zero variation constant ] .

Hence the square of volume of sphere varies directly with cube of its surface area. (Proved)

Example 18. The value of diamond varies as the square of its weight. A diamond broke into three pieces whose weights are in the ratio 3:4:5. For this, there is a loss of ₹9400. Find the value of the original diamond.

Solution:

Given :

The value of diamond varies as the square of its weight. A diamond broke into three pieces whose weights are in the ratio 3:4:5. For this, there is a loss of ₹9400.

Let the weight of the diamond = m gm and its value = ₹ A.

As per question, A ∝ m2 ∴ a = km2 [k ≠ 0 =variation constant]

Let the weights of three pieces of diamond be 3n gm, 4n gm and 5n gm.

∴ The weight of the original diamond = (3 n + 4n + 5n) gm = 12n gm.

∴ The values of the three pieces are respectively

A1 = ₹ k (3n)2 = ₹ 9kn2

A2 =₹ k (4n)2 = ₹ 16kn2

A3 = ₹ k (5n)2 = ₹ 25kn2 (where k ≠ 0 = variation constant)

Also, the value of the original diamond = ₹ k.(12n)2 = ₹ 144 kn2. .

∴ Loss for breaking = ₹ {144 kn2 – (9kn2 + 16kn2 + 25kn2)} = ₹ 94 kn2

As per question, 94 kn2 = 9400 ∴kn2 = 100.

∴ The value of the original diamond = ₹ 144 kn2

= ₹ 144 x 100 = ₹ 14400.

Hence the yalue of the original diamond = ₹ 14400.

Class 10 Maths Wbbse Solutions

Example 19. A heavy particle while falling freely from rest under gravitational altraction the distance through which it falls varies directly as the square of the time taken to fall up to the distance. The particle falls through 125 metres in 5 seconds. Through what distance will it fall in 10 seconds ? What distance will it traverse in the 10th second? 

Solution:

Given

A heavy particle while falling freely from rest under gravitational altraction the distance through which it falls varies directly as the square of the time taken to fall up to the distance. The particle falls through 125 metres in 5 seconds.

Let distance = hm and the time required to traverse that distance = t second.

As per question, h ∝ t2 or, h = kt2 (where k ≠ 0 = variation constant)

∴ h = kt2 ……(1)

Now, if t = 5, then h = 125, ∴ 125 = k.(5)2 ⇒ k = \(\frac{125}{5}\) = 5.

∴ from (1) we get, h = 5t2 …….(2)

∴ putting t = 10 in (2) we get, h = 5 x (10)2

= 5 x 100 = 500.

Again, putting t = 9 we get, h = 5 x (9)2 = 5 x 81 = 405.

Now distance traverse in 10th second

= (Distance travelled in 10 seconds) – (Distance travelled in 9 seconds)

= (500 – 405) m = 95 m.

∴ The particle falls 500 m in 10 seconds and 95 m in 10th second.

Example 20. A locomotive engine without wagons can go 24 miles an hour and its speed is diminished as the square root of the number of wagons attached with 4 wagons its speed is 20 miles an hour. Find the maximum number of wagons with which the engine can move. 

Solution:

Given:

A locomotive engine without wagons can go 24 miles an hour and its speed is diminished as the square root of the number of wagons attached with 4 wagons its speed is 20 miles an hour.

Let the speed diminished by m mile / hour and the number of wagons = n.

As per question, m ∝ √n or, m = k. √n…….(1) (when k ≠ 0 = variation constant)

Now, if 4 wagons be attached with the engine, its speed diminishes by

(24 – 20) mile / hour = 4 mile / hour.

∴ from (1) we get, 4 = k. √4 or, 2k = 4 or, k = 2.

∴ (1) becomes, m = 2√n ……(2)

Putting m = 24 in (2) we get, 24 = 2√n

⇒ √n = 12 ⇒ n = (12)2 = 144.

∴ the engine will be speedless if 144 wagons are attached with it.

Hence the required maximum number of wagons = 144 – 1 = 143.

Example 21. The weekly expenses of a hostel are partly constant and partly vary directly as the number of inmates. If the expenses are ₹ 2000 when the number of inmates is 120 and ₹ 1700 when the number is Find the number of inmates when the expenses are ₹1880.

Solution:

Given

The weekly expenses of a hostel are partly constant and partly vary directly as the number of inmates. If the expenses are ₹ 2000 when the number of inmates is 120 and ₹ 1700

Let the expenses be ₹ yand the number of inmates = x.

As per question, y = k1+ k2x ……… (1)

[where k1 = constant and k2 ≠ 0 = variation constant ]

As per question, if x = 120, then y = 2000.

∴ from (1) we get, 2000 = k1+ k2 x 120

or, k1 + 120 k2= 2000 …..(2)

Again, if x = 100, then y = 1700.

∴ from (1) we get, 1700 = k1+ k2x 100

or, k1 + 100 k2 = 1700 …..(3)

Now, subtracting (3) from (2) we get,

20k2 = 300 or, k2 = 15.

∴ from (2) we get, k1 + 120 x 15 = 2000

⇒ k1 = 2000 – 1800

⇒ k1 = 200

∴ from (1) we get, y = 200 + 15x……….(4)

so, putting y = 1880 in (4) we get, 1880 = 200 + 15x or, 15a = 1880 – 200

or, 15x =1680 or, x = \(\) = 112.

Hence the required number of inmates =112.

 

 

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