## Maharashtra State Board Class 8 Maths Solutions Chapter 7 Variation Practice Set 7.3

Question 1.

Which of the following statements are of inverse variation?

i. Number of workers on a job and time taken by them to complete the job.

ii. Number of pipes of same size to fill a tank and the time taken by them to fill the tank.

iii. Petrol filled in the tank of a vehicle and its cost.

iv. Area of circle and its radius.

Solution:

i. Let, x represent number of workers on a job, and y represent time taken by workers to complete the job.

As the number of workers increases, the time required to complete the job decreases.

∴ \(x \propto \frac{1}{y}\)

ii. Let, n represent number of pipes of same size to fill a tank and t represent time taken by the pipes to fill the tank.

As the number of pipes increases, the time required to fill the tank decreases.

∴ \(\mathrm{n} \propto \frac{1}{\mathrm{t}}\)

iii. Let, p represent the quantity of petrol filled in a tank and c represent the cost of the petrol.

As the quantity of petrol in the tank increases, its cost increases.

∴ p ∝ c

iv. Let, A represent the area of the circle and r represent its radius.

As the area of circle increases, its radius increases.

∴ A ∝ r

∴ Statements (i) and (ii) are examples of inverse variation.

Question 2.

If 15 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours?

Solution:

Let, n represent the number of workers building the wall and t represent the time required.

Since, the number of workers varies inversely with the time required to build the wall.

∴ \(\mathrm{n} \propto \frac{1}{\mathrm{t}}\)

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

where k is the constant of variation

∴ n × t = k …(i)

15 workers can build a wall in 48 hours,

i.e., when n = 15, t = 48

∴ Substituting n = 15 and t = 48 in (i), we get

n × t = k

∴ 15 × 48 = k

∴ k = 720

Substituting k = 720 in (i), we get

n × t = k

∴ n × t = 720 …(ii)

This is the equation of variation.

Now, we have to find number of workers required to do the same work in 30 hours.

i.e., when t = 30, n = ?

∴ Substituting t = 30 in (ii), we get

n × t = 720

∴ n × 30 = 720

∴ n = \(\frac { 720 }{ 30 }\)

∴ n = 24

∴ 24 workers will be required to build the wall in 30 hours.

Question 3.

120 bags of half litre milk can be filled by a machine within 3 minutes find the time to fill such 1800 bags?

Solution:

Let b represent the number of bags of half litre milk and t represent the time required to fill the bags.

Since, the number of bags and time required to fill the bags varies directly.

∴ b ∝ t

∴ b = kt …(i)

where k is the constant of variation.

Since, 120 bags can be filled in 3 minutes

i.e., when b = 120, t = 3

∴ Substituting b = 120 and t = 3 in (i), we get

b = kt

∴ 120 = k × 3

∴ k = \(\frac { 120 }{ 3 }\)

∴ k = 40

Substituting k = 40 in (i), we get

b = kt

∴ b = 40 t …(ii)

This is the equation of variation.

Now, we have to find time required to fill 1800 bags

∴ Substituting b = 1800 in (ii), we get

b = 40 t

∴ 1800 = 40 t

∴ t = \(\frac { 1800 }{ 40 }\)

∴ t = 45

∴ 1800 bags of half litre milk can be filled by the machine in 45 minutes.

Question 4.

A car with speed 60 km/hr takes 8 hours to travel some distance. What should be the increase in the speed if the same distance is

to be covered in \(7\frac { 1 }{ 2 }\) hours?

Solution:

Let v represent the speed of car in km/hr and t represent the time required.

Since, speed of a car varies inversely as the time required to cover a distance.

∴ \(v \propto \frac{1}{t}\)

∴ \(\mathbf{v}=\mathbf{k} \times \frac{1}{\mathbf{t}}\)

where, k is the constant of variation.

∴ v × t = k …(i)

Since, a car with speed 60 km/hr takes 8 hours to travel some distance.

i.e., when v = 60, t = 8

∴ Substituting v = 60 and t = 8 in (i), we get

v × t = k

∴ 60 × 8 = t

∴ k = 480

Substituting k = 480 in (i), we get

v × t = k

∴ v × t = 480 …(ii)

This is the equation of variation.

Now, we have to find speed of car if the same distance is to be covered in \(7\frac { 1 }{ 2 }\) hours.

i.e., when t = \(7\frac { 1 }{ 2 }\) = 7.5 , v = ?

∴ Substituting, t = 7.5 in (ii), we get

v × t = 480

∴ v × 7.5 = 480

\(v=\frac{480}{7.5}=\frac{4800}{75}\)

∴ v = 64

The speed of vehicle should be 64 km/hr to cover the same distance in 7.5 hours.

∴ The increase in speed = 64 – 60

= 4km/hr

∴ The increase in speed of the car is 4 km/hr.