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. ****If A varies directly as B and the value of A is 15 and B is 25, what is the equation that describes this direct variation of A and B?**

As A varies directly with B,

A = KB

or, 15 = K x 25

K = \(\frac{25}{15}\)

= \(\frac{5}{3}\)

So the equation that describes the direct variation of A and B is A = B.

**2. (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.**

**3. ****If a car runs at a constant speed and takes 3 hrs to run a distance of 150 km, what time it will take to run 100 km?**

**Solution:**

If T is the time taken to cover the distance and S is the distance and V is the speed of the car, the direct variation equation is S = VT where V is constant.

For the case given in the problem,

150 = V x 3

or, V = \(\frac{150}{3}\)

= 50

So speed of the car is 60 kmph and it is constant.

For 100 km distance

S = VT

or, 100 = 50 x T

T = \(\frac{100}{50}\)

= 2 hrs.

So it will take 2 hr.

**4. 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. **

**5. If a car runs at a speed of 60 kmph and takes 3 hrs to run a distance, what time it will take to run at a speed of 40 km?**

If T is the time taken to cover the distance and S is the distance and V is the speed of the car, the indirect variation equation is S= VT where S is constant and V and T are variables.

For the case given in the problem the distance that car covers is

S = VT = 60 x 3 = 180 km.

So at a speed of the car is 40 kmph and it will take

S = VT

or, 180 = 40 x T

or, T = \(\frac{180}{40}\)

= \(\frac{9}{2}\) hrs

= 4 hrs 30 mins.

**6. 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.

**7. ****The area of a triangle is jointly related to the height and the base of the triangle. If the base is increased 20% and the height is decreased by 10%, what will be the percentage change of the area?**

We know the area of triangle is half the product of base and height. So the joint variation equation for area of triangle is A = \(\frac{bh}{2}\) where A is the area, b is the base and h is the height.

Here \(\frac{1}{2}\) is the constant for the equation.

Base is increased by 20%, so it will be b x \(\frac{120}{100}\) = \(\frac{12b}{10}\).

Height is decreased by 10%, so it will be h x \(\frac{90}{100}\) = \(\frac{9h}{10}\).

So the new area after the changes of base and height is

\(\frac{\frac{12b}{10} \times \frac{9h}{10}}{2}\)

= (\(\frac{108}{100}\))\(\frac{bh}{2}\) = \(\frac{108}{100}\)A.

So the area of the triangle is decreased by 8%.

**8. 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:

**9. A rectangle’s length is doubled and width is halved, how much the area will increase or decrease?**

**Solution:**

Formula for the area is A = lw where A is area, l is length and w is width.

This is joint variation equation where 1 is constant.

If length is doubled, it will become 2l.

And width is halved, so it will become \(\frac{w}{2}\).

So the new area will be P = \(\frac{2l × w}{2}\) = lw.

So the area will be same if length is doubled and width is halved.

**
10. 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.*

**11.** 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:

**12.** 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.

**13.** 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.*

**14.** 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.*

**15.** 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**

**What is Variation?****Direct Variation****Inverse Variation****Joint Variation****Theorem of Joint Variation****Worked out Examples on Variation****Problems on Variation**

**11 and 12 Grade Math**** ****From Worked out Examples on Variation to HOME PAGE**

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