## Maharashtra State Board Class 8 Maths Solutions Chapter 7 Variation Practice Set 7.2

Question 1.

The information about number of workers and number of days to complete a work is given in the following table. Complete the table.

Number of workers |
30 | 20 | __ | 10 | __ |

Days |
6 | 9 | 12 | 36 |

Solution:

Let, n represent the number of workers and d represent the number of days required to complete a work.

Since, number of workers and number of days to complete a work are in inverse poportion.

∴ \(\mathbf{n} \propto \frac{1}{\mathrm{d}}\)

∴ \(\mathrm{n}=\mathrm{k} \times \frac{1}{\mathrm{d}}\)

where k is the constant of variation.

∴ n × d = k …(i)

i. When n = 30, d = 6

∴ Substituting n = 30 and d = 6 in (i), we get

n × d = k

∴ 30 × 6 = k

∴ k = 180

Substituting k = 180 in (i), we get

∴ n × d = k

∴ n × d = 180 …(ii)

This is the equation of variation

ii. When d = 12, n = 7

∴ Substituting d = 12 in (ii), we get

n × d = 180

∴ n × 12 = 180

∴ n = \(\frac { 180 }{ 12 }\)

∴ n = 15

iii. When n = 10, d = ?

∴ Substituting n = 10 in (ii), we get

n × d = 180

10 × d = 180

∴ d = \(\frac { 180 }{ 10 }\)

∴ d = 18

iv. When d = 36, n = ?

∴ Substituting d = 36 in (ii), we get

n × d = 180

∴ n × 36 = 180

∴ n = \(\frac { 180 }{ 36 }\)

∴ n = 5

Number of workers |
30 | 20 | 15 | 10 | 5 |

Days |
6 | 9 | 12 | 18 | 36 |

Question 2.

Find constant of variation and write equation of variation for every example given below:

i. \(p \propto \frac{1}{q}\) ; if p = 15 then q = 4.

ii. \(z \propto \frac{1}{w}\) ; when z = 2.5 then w = 24.

iii. \(s \propto \frac{1}{t^{2}}\) ; if s = 4 then t = 5.

iv. \(x \propto \frac{1}{\sqrt{y}}\) ; if x = 15 then y = 9.

Solution:

i. \(p \propto \frac{1}{q}\) …[Given]

∴ p = k × \(\frac { 1 }{ q }\)

where, k is the constant of variation.

∴ p × q = k …(i)

When p = 15, q = 4

∴ Substituting p = 15 and q = 4 in (i), we get

p × q = k

∴ 15 × 4 = k

∴ k = 60

Substituting k = 60 in (i), we get

p × q = k

∴ p × q = 60

This is the equation of variation.

∴ The constant of variation is 60 and the equation of variation is pq = 60.

ii. \(z \propto \frac{1}{w}\) …[Given]

∴ z = k × \(\frac { 1 }{ w }\)

where, k is the constant of variation,

∴ z × w = k …(i)

When z = 2.5, w = 24

∴ Substituting z = 2.5 and w = 24 in (i), we get

z × w = k

∴ 2.5 × 24 = k

∴ k = 60

Substituting k = 60 in (i), we get

z × w = k

∴ z × w = 60

This is the equation of variation.

∴ The constant of variation is 60 and the equation of variation is zw = 60.

iii. \(s \propto \frac{1}{t^{2}}\) …[Given]

∴ \(s=k \times \frac{1}{t^{2}}\)

where, k is the constant of variation,

∴ s × t² = k …(i)

When s = 4, t = 5

∴ Substituting, s = 4 and t = 5 in (i), we get

s × t² = k

∴ 4 × (5)² = k

∴ k = 4 × 25

∴ k = 100

Substituting k = 100 in (i), we get

s × t² = k

∴ s × t² = 100

This is the equation of variation.

∴ The constant of variation is 100 and the equation of variation is st² = 100.

iv. \(x \propto \frac{1}{\sqrt{y}}\) …[Given]

∴ \(x=\mathrm{k} \times \frac{1}{\sqrt{y}}\)

where, k is the constant of variation,

∴ x × √y = k …(i)

When x = 15, y = 9

∴ Substituting x = 15 and y = 9 in (i), we get

x × √y = k

∴ 15 × √9 = k

∴ k = 15 × 3

∴ k = 45

Substituting k = 45 in (i), we get

x × √y = k

∴ x × √y = 45 .

This is the equation of variation.

∴ The constant of variation is k = 45 and the equation of variation is x√y = 45.

Question 3.

The boxes are to be filled with apples in a heap. If 24 apples are put in a box then 27 boxes are needed. If 36 apples are filled in a box how many boxes will be needed?

Solution:

Let x represent the number of apples in each box and y represent the total number of boxes required.

The number of apples in each box are varying inversely with the total number of boxes.

∴ \(x \infty \frac{1}{y}\)

∴ \(x=k \times \frac{1}{y}\)

where, k is the constant of variation,

∴ x × y = k …(i)

If 24 apples are put in a box then 27 boxes are needed.

i.e., when x = 24, y = 27

∴ Substituting x = 24 and y = 27 in (i), we get

x × y = k

∴ 24 × 27 = k

∴ k = 648

Substituting k = 648 in (i), we get

x × y = k

∴ x × y = 648 …(ii)

This is the equation of variation.

Now, we have to find number of boxes needed

when, 36 apples are filled in each box.

i.e., when x = 36,y = ?

∴ Substituting x = 36 in (ii), we get

x × y = 648

∴ 36 × y = 648

∴ y = \(\frac { 648 }{ 36 }\)

∴ y = 18

∴ If 36 apples are filled in a box then 18 boxes are required.

Question 4.

Write the following statements using symbol of variation.

- The wavelength of sound (l) and its frequency (f) are in inverse variation.
- The intensity (I) of light varies inversely with the square of the distance (d) of a screen from the lamp.

Solution:

- \(l \propto \frac{1}{\mathrm{f}}\)
- \(\mathrm{I} \propto \frac{1}{\mathrm{d}^{2}}\)

Question 5.

\(x \propto \frac{1}{\sqrt{y}}\) and when x = 40 then y = 16. If x = 10, find y.

Solution:

\(x \propto \frac{1}{\sqrt{y}}\)

∴ \(x=\mathrm{k} \times \frac{1}{\sqrt{y}}\)

where, k is the constant of variation.

∴ x × √y = k …(i)

When x = 40, y = 16

∴ Substituting x = 40 andy = 16 in (i), we get

x × √y = k

∴ 40 × √16 = k

∴ k = 40 × 4

∴ k = 160

Substituting k = 160 in (i), we get

x × √y = k

∴ x × √y = 160 …(ii)

This is the equation of variation.

When x = 10,y = ?

∴ Substituting, x = 10 in (ii), we get

x × √y = 160

∴ 10 × √y = 160

∴ √y = \(\frac { 160 }{ 10 }\)

∴ √y = 16

∴ y = 256 … [Squaring both sides]

Question 6.

x varies inversely as y, when x = 15 then y = 10, if x = 20, then y = ?

Solution:

Given that,

\(x \propto \frac{1}{\sqrt{y}}\)

∴ \(x=\mathrm{k} \times \frac{1}{\sqrt{y}}\)

where, k is the constant of variation.

∴ x × y = k …(i)

When x = 15, y = 10

∴ Substituting, x = 15 and y = 10 in (i), we get

x × y = k

∴ 15 × 10 = k

∴ k = 150

Substituting, k = 150 in (i), we get

x × y = k

∴ x × y = 150 …(ii)

This is the equation of variation.

When x = 20, y = ?

∴ substituting x = 20 in (ii), we get

x × y = 150

∴ 20 × y = 150

∴ y = \(\frac { 150 }{ 20 }\)

∴ y = 7.5