In variation we will follow stepbystep 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.
Stepbystep explanation on workedout 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 workedout 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 workedout 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
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