Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 7.3 8th Std Maths Answers Solutions Chapter 7 Variation.
Practice Set 7.3 8th Std Maths Answers Chapter 7 Variation
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.