Worked out Examples on Variation


In variation we will follow step-by-step some of the worked out examples on variation. Variations are classified into three types such as; direct, inverse and joint variation. Using variation, application to simple examples of time and work; time and distance; mensuration; physical laws and economics.


Step-by-step explanation on worked-out examples on variation: 

1. (i) If A varies inversely as B and A = 2 when B = 10, find A when B = 4. 

(ii) If x ∝ y² and x = 8 when y = 4, find y when x = 32. 

Solution: (i) Since A varies inversely as B 

Therefore A ∝ 1/B or, A = k ∙ 1/B ………………. (1), where k = constant of variation. 

Given A = 2 when B = 10. 

Putting these values in (1), we get, 

2 = k ∙ 1/10 

or, k = 20. 

Therefore, the law of variation is: A = 20 ∙ 1/B……………... (2) 

When B = 4, then from (2) we get, A = 20 ∙ ¼ = 5. 

Therefore, A = 5 when B = 4. 

(ii) Since, x ∝ y²

Therefore, x = m ∙ y² ……………… (1) 

where m = constant of variation. 

Given x = 8 when y = 4. 

Putting these values in (1), we get, 

8 = m ∙ 42 = 16m 

or, m = 8/16 

or, m = 1/2

Therefore the law of variation is: x = ½ ∙ y² ………….. (2) When x = 32, then from (2) we get, 

32 = 1/2 ∙ y² 

or, y² = 64 

or, y = ± 8. 

Hence, y = 8 or, - 8 when x = 32. 



2. x varies directly as the square of y and inversely as the cube root of z and x = 2, when y = 4, z = 8. What is the value of y when x = 3, and z = 27? 


Solution:

By the condition of the problem, we have,

x ∝ y² ∙ 1/∛z

Therefore x = k ∙ y² ∙ 1/∛z ……(1)

where k = constant, of variation.

Given x = 2 when y = 4, z = 8.

Putting these values in (1), we get,

2 = k ∙ 4² = 1/∛8 = k ∙ 16 ∙ 1/2 = 8k

or, k = 2/8 = 1/4

Therefore the law of variation is: x = 1/4 ∙ y² ∙ 1/3√z .... (2)

When x = 3, z = 27, then from (2) we get,

3 = 1/4 ∙ y² ∙ 1/∛27 = 1/4 ∙ y² ∙ 1/3

or, y² = 36

or, y = ± 6

Therefore, the required value of y is 6 or - 6.


3. Fill in the gaps: 


(i) If A ∝ B² then B ∝ …..

(ii) If P ∝ 1/√Q, then Q ∝ ……

(iii) If m ∝ ∛n, then n ∝ ……

Solution:

(i) Since A ∝ B²

Therefore, A = kB² [k = constant of variation]

or, B² = ( 1/k) A

or, B = ± (1/√K) √A

Therefore B ∝ √A since ± 1/√K = constant.

(ii) Since p ∝ 1/√Q

Therefore p = k ∙ 1/√Q [k = constant of variation]

Since, √Q = k/p

or, Q = k²/p²

Therefore, Q ∝ 1/p², as k² = constant.

(iii) Since, m ∝ ∛n

Therefore m = k ∙ ∛n [k = constant of variation]

or, m³ = k³ ∙ n

or, n = (1/k³) ∙ m³

Therefore n ∝ m³ as 1/k ³ = constant.



4. If a² ∝ bc, b² ∝ ca and c² ∝ ab, then find the relation between the three constants of variation.

Solution:

Since, a² ∝ bc
Therefore, a² = kbc …….(1) [k = constant of variation]

Again, b² ∝ ca

Therefore, b² = lca ……. (2) [l = constant of variation]

and c² ∝ ab

Therefore, c² = mab ……. (3) [m = constant of variation]

Multiplying both sides of (1), (2) and (3) we get,

a²b²c² = kbc ∙ lca ∙ mab = klm a²b²c²

or, klm = 1, which is the required relation between the three constants of variation.


Various types of worked-out examples on variation: 

5. If (A² + B²) ∝ (A² - B²), then show that A ∝ B.

Solution:

Since, A² + B² ∝ (A² - B²)

Therefore, A² + B² = k (A² - B²), where k = constant of variation.

or, A² - kA² = - kB² - B²

or, A² (1 - k) = - (k + 1)B²

or, A² = [(k + 1)/(k – 1)]B² = m²B² where m² = (k + 1)/(k – 1) = constant.

or, A = ± mB

Therefore A ∝ B, since ± m = constant. Proved.


6. If (x + y) ∝ (x – y), then show that,

(i) x² + y² ∝ xy

(ii) (ax + by) ∝ (px + qy), where a, b, p and q are constants.

Solution:

Since, (x + y) ∝ (x – y)

Therefore, x + y = k (x - y), where k = constant of variation.

or, x + y = kx - ky

or, y + ky = kx - x

or, y(1 + k) = (k – 1)x

or, y = [(k – 1)/( k + 1)] x = mx where m = (k - 1)/(k + 1) = constant.

(i) Now, (x² + y²)/xy = {x² + (mx)²}/(x ∙ mx) = {x² ( 1 + m²)/(x² ∙ m)} = (1 + m²)/m

or, (x² + y²) /xy = n where n = (1 + m²)/m = constant, since m = constant.

Therefore, x² + y² ∝ xy. Proved.

(ii) We have, (ax + by)/(px + qy) = (ax + b ∙ mx)/(px + q ∙ mx) = {x (a + bm)}/{x (p + qm)}

or, (ax + by)/(px + qy) = (a + bm)/(p + qm) = constant, since a, b, p, q and m are constants.

Therefore, (ax + by) ∝ (px + qy). Proved.


More worked-out examples on variation:

7. b is equal to the sum of two quantities, one of which varies directly as a and the other inversely as the square of a². If b= 49 when a = 3 or 5, find the relation between a and b.

Solution:

By the condition of the problem, we assume,

b = x + y ……... (1)

where, x ∝ a and y ∝ 1/a²

Therefore x = ka and y = m ∙ 1/a²

where k and m are constants of variation.

Putting the values of x and y in (1), we get,

B = ka + m/a² ………. (2)

Given, b = 49 when a = 3.

Hence, from (2) we get,

49 = 3k + m/9

or, 27k + m = 49 × 9 …….... (3)

Again, b = 49 when a 5.

Hence, from (2) we get,

49 = 5k + m/25

or, 125k + m = 49 × 25 …….... (4)

Subtracting (3) from (4) we get,

98k = 49 × 25 - 49 × 9 = 49 × 16

or, k = (49 × 16)/98 = 8

Putting the value of k in (3) we get,

27 × 8 + m = 49 × 9

or, m = 49 × 9 - 27 × 8 = 9 × 25 = 225.

Now, substituting the values of k and m in (2) we get,

b = 8a + 225/a²

which is the required relation between a and b.


8. If(a - b) ∝ c when b is constant and (a - c) ∝ b when c is constant, show that, (a - b - c) ∝ bc when both b and c vary.

Solution:

Since (a - b) ∝ c when b is constant

Therefore, a - b = kc [where, k = constant of variation] when b is constant

or, a - b - c = kc - c = (k - 1) c when b is constant.

Therefore a - b - c ∝ c when b is constant [since (k - 1) = constant] ….... (1)

Again, (a - c ) ∝ b when c is constant.

Therefore a - c = mb [where, m = constant of variation] when c is constant.

or, a - b - c = mb - b = (m - 1) b when c is constant.

Therefore a - b - c ∝ b when c is constant [since, (m - 1) = constant]..... (2)

From (1) and (2), using the theorem of joint variation, we get, a - b - c ∝ bc when both b and c vary. Proved.



9. If x, y, z be variable quantities such that y + z - x is constant and (x + y - z)(z + x - y) ∝ yz, prove that, x + y + z ∝ yz.

Solution:

By question, y + z - x = constant c (say)

Again, (x + y - z) (z + x - y) ∝ yz

Therefore (x + y - z) (z + x - y) = kyz, where k = constant of variation

or, {x + (y - z)} {x - (y- z)} = kyz

or, x² - (y - z) ² = kyz

or, x² - {(y + z)² - 4yz} = kyz

or, x² - (y + z)² + 4yz = kyz

or, (y + z)² - x² = (4 - k)yz

or, (y + z + x) (y + z - x) = (4 - k)yz

or, (x + y + z) ∙ c = (4 - k)yz [since, y + z - x = c]

or, x + y + z = {(4 - k)/c} yz = myz

where m = (4 - k)/c = constant, since k and c are both constants.

Therefore, x + y + z ∝ yz. Proved.


10. If (x + y + z) (y + z - x) (z + x - y) (x + y - z) ∝ y²z² then show that either y² + z² = x² or, y² + z² - x ² ∝ yz.

Solution:

Since (x + y + z) (y + z - x) (z + x - y) (x + y - z) ∝ y²z²

Therefore (y + z + x) (y + z - x) {x - (y - z)} {x + (y - z)} = ky²z²

where k = constant of variation

or, [(y + z) ² - x²] [x² - (y - z) ²] = ky²z²

or, [2yz + (y² + z² - x² )] [2yz - (y² + z² - x²)] = ky²z²

or, 4y²z² - (y² + z² - x²)² = ky²z²

or, (y² + z² - x²)² = (4 - k)y²z² = m²y²z²

where m² = 4 - k constant

or, y² + z² - x² = ± myz.

Clearly, y² + z² - x² = 0 when m = 0 i.e., when k = 4.

and, y² + z² - x² ∝ yz when m ≠ 0 i.e., when k < 4.

Therefore either, y² + z² = x²

or, y² + z² - x² ∝ yz. Proved.


 Variation





11 and 12 Grade Math 

From Worked out Examples on Variation to HOME PAGE


New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.