WBBSE Solutions For Class 10 Maths Algebra Chapter 4 Variation

Algebra Chapter 4 Variation

If the variable x and y are related to each other in such a way that \(\frac{x}{y}\) = k (now zero constant), it is called that x and y are in direct variation and non-zero constant is said to be variation constant, x and y are in Direct variation and it can be written as x y and now zero constant is said to be variation constant.

⇒ If two variables x and y are related to each other in such a way that xy = k (non-zero constant), then it is said x and y are in inverse variation and written as x \(\frac{1}{y}\) and non zero constant is said to be variation constant.

⇒ If a variable in direct variation with the product of two or more variables, the first variable is said to be in joint variation with other variables.

Theorem on joint variation: If three variables x, y, z be such that x ∝ y when z is constant, x ∝ z when y is constant, then x ∝ yz when y and z both vary.

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WBBSE Solutions For Class 10 Maths Algebra Chapter 4 Variation

Algebra Chapter 4 Variation True Or False

Example 1. y ∝ \(\frac{1}{x}\) then \(\frac{y}{x}\) = non zero constant.

Solution: False

Example 2. If x ∝ z and y ∝ z then xy ∝ z.

Solution: True

Class 10 Maths Algebra Chapter 4 Solutions

Example 3. If x ∝ y2 and y = 2a when x = a then y2 = 4ax.

Solution: True

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Example 4. If \(\frac{x}{y}\) ∝ z, \(\frac{y}{z}\) ∝ x \(\frac{z}{x}\) ∝ y then xyz = 1.

Solution: False

Example 5. If x ∝ y, y ∝ z then x μ z.

Solution: True

Example 6. If x ∝ \(\frac{1}{y}\), y ∝ \(\frac{1}{z}\) then x ∝ \(\frac{1}{z}\)

Solution: False

Example 7. If A2 ∝ BC, B2 ∝ CA, C2 ∝ AB then the product of three variations constant = 1.

Solution: True

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Example 8. If x ∝ \(\frac{1}{z}\) when y is constant and x ∝ y when z is constant then x ∝ \(\frac{y}{z}\) then both y and z vary.

Solution: True

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Example 9. If x ∝ y then x + y ∝ \(\frac{1}{x-y}\)

Solution: False

Example 10. If x ∝ y then xn ∝ yn.

Solution: True

Example 11. If A2 + B2 ∝ A2 – B2 then A ∝ \(\frac{1}{B}\)

Solution: False

Example 12. If \(\frac{x}{y}\) ∝ x + y and \(\frac{y}{x}\) ∝ x-y then x2 – y = constant.

Solution: True

Class 10 Algebra Chapter 4 Solved Examples

Algebra Chapter 4 Variation Fill In The Blanks

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

Solution: z

Example 2. If x ∝ y, xn ∝ ______

Solution: yn

Example 3. If x ∝ y and x ∝ z then (y + z) ∝ _____

Solution: x

Example 4. If \(\frac{x}{y}\) constant, then x and y are in ______ variation.

Solution: Direct

Example 5. If xy = constant then x and y are in _______ variation.

Solution: Inverse

Example 6. If a variable is in direct variation with the product of two or more variables, the first variable is said to be in _______ variation with the other variables.

Solution: Joint

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Example 7. If V = R\(\frac{T}{P}\) (R = constant) in this relation we can say that V is in _______ variation with T and \(\frac{1}{P}\).

Solution: Joint

Example 8. If \(\frac{x}{y}\) ∝ x + y and \(\frac{y}{x}\) ∝ x -y then x2 – y2 = _______

Solution: Constant

Example 9. If x2 ∝ yz, y2 ∝ zx, z2 ∝ xy then the product of the three non-zero variation constants is _________

Solution: 1

Example 10. If x + y ∝ x- y then x _______

Solution: y

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Algebra Chapter 4 Variation Short Answer Type Questions

Example 1. If x ∝ y2 and y = 2a when x = a. Find the relation between x and y.

Solution: x = ky2 [x is a non-zero variation constant.]

∴ a = k (2a)2

or, k = \(\frac{1}{4a}\)

x= \(\frac{1}{4a}\)y2

∴ y2 = 4ax.

Example 2. If x ∝ y, y ∝ z and z ∝ x, find the product of three non-zero constants.

Solution: x = k1y, y = k2z, z = k3x [k1, k2, k3 are three non zero constants]

∴ xyz = k1k2k3 xyz

∴ k1k2k3 = 1

∴ Required product is 1.

Example 3. If x ∝ \(\frac{1}{y}\), y ∝ \(\frac{1}{z}\) find if there be any relation direct or inverse variation between x and z.

Solution: \(x=\frac{k_1}{y}, y=\frac{k_2}{z}\)

⇒ [k1, k2 are non-zero constants]

∴ x = \(\frac{k_1}{\frac{k_2}{z}} \quad \text { or, } \quad x=\frac{k_1}{k_2} z\)

∴ x ∝ z [\(\frac{k_1}{k_2}\) constant]

⇒ There is a direct variation.

Example 4. If x ∝ yz and y ∝ zx, show that z is a nonzero constant.

Solution: x = k1yz, y = k2zx [k1k2 are non-zero constant.]

⇒ or, xy = k1k2 yz2x or, 1 = k1k2z2

∴ \(z= \pm \sqrt{\frac{1}{k_1 k_2}}\) = non zero constant.

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Example 5. If ba3 and an increases ratio of 2 : 3, find what ratio b will increase.

Solution: b = ka3 [k is a non zero variation constant]

\(\frac{b_1}{b_2}=\frac{k a_1^3}{k a_2^3}=\left(\frac{2}{3}\right)^3\)

 

⇒ b1: b2 = 8: 27.

Example 6. If \(\left(a x+\frac{b}{y}\right) \propto\left(c x+\frac{d}{y}\right)\) then show that xy = constant, (where a, b, c, d constant).

Solution: \(\left(a x+\frac{b}{y}\right)\) = k \(\left(c x+\frac{d}{y}\right)\)

⇒ or, x(a-kc) = (kd-b)\(\frac{1}{y}\)

⇒ or, \(x y=\frac{k d-b}{a-k c}=k_1=\text { constant }\)

⇒ when \(k_1=\frac{k d-b}{a-k c}\)

Example 7. If x ∝ y and y ∝ z then show that x + y ∝ z.

Solution: x ∝ y ⇒ x = k1y,

⇒ y ∝ z ⇒ y =k2z [k1 ,k2 are non zero constant]

⇒ Now, \(\frac{x+y}{z}=\frac{k_1 k_2 z+k_2 z}{z}=\frac{\left(k_1 k_2+k_2\right)}{z} \cdot z\)

= (k k k) = constant

∴ x + y ∝ z..

Example 8. If P2– Q2 ∝ PQ then show that (P + Q) ∝ (P- Q)

Solution: P2 + Q2 = 2KPQ [2k is a non-zero variation constant]

⇒  or, \(\frac{\mathrm{P}^2+\mathrm{Q}^2}{2 \mathrm{PQ}}=\mathrm{k}\)

⇒ or, \(\frac{\mathrm{P}^2+\mathrm{Q}^2+2 \mathrm{PQ}}{\mathrm{P}^2+\mathrm{Q}^2-2 \mathrm{PQ}}=\frac{k+1}{k-1}\) [by components dividends]

⇒ or, \(\frac{(\mathrm{P}+\mathrm{Q})^2}{(\mathrm{P}-\mathrm{Q})^2}=\frac{k+1}{k-1}\)

∴ \(\frac{\mathrm{P}+\mathrm{Q}}{\mathrm{P}-\mathrm{Q}}= \pm \sqrt{\frac{k+1}{k-1}}=\text { constant. }\)

⇒ p + Q ∝ P – Q

Example 9. x ∝ y when z constant and x ∝ \(\frac{1}{z}\) when y constant. If y = b when z = c, x = a then find the value of x when y = b2, z = c2.

Solution: x ∝ y (z constant.) x ∝ \(\frac{1}{z}\) (y constant.)

By compound variation, x ∝ \(\frac{y}{z}\) (y, z vary)

∴ x = k\(\frac{y}{z}\) [k is a non zero variation constant.]

⇒ Q = k\(\frac{b}{c}\)

⇒ k = k\(\frac{ac}{b}\)

\(x=\frac{a c}{b} \cdot \frac{y}{z}=\frac{a c}{b} \cdot \frac{b^2 b}{c^2 c}=\frac{a b}{c}\)

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Example 10. If \(x^3-\frac{1}{y^3} \propto x^3+\frac{1}{y^3}\) then show that x ∝ \(\frac{1}{y}\)

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

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

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

⇒ or, \(x^3 y^3=\frac{k+1}{1-k}\)

⇒ or, \(x y=\sqrt[3]{\frac{k+1}{1-k}}\) = constant

∴ \(\quad x \propto \frac{1}{y}\)

 

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