- NCERT Solutions for Class 8 Maths Chapter 11 Direct and Inverse Proportions Ex 11.1
- NCERT Solutions for Class 8 Maths Chapter 11 Direct and Inverse Proportions Ex 11.2
NCERT Solutions for Class 8 Maths Chapter 11 Direct and Inverse Proportions Ex 11.1
Ex 11.1 Class 8 Maths Question 1.
Following are the car parking charges near a railway station upto
4 hours ₹ 60
8 hours ₹ 100
12 hours ₹ 140
24 hours ₹ 180
Check if the parking charges are in direct proportion to the parking time.
Solution:
Hence, the parking charges are not in direct proportion to the parking time.
Ex 11.1 Class 8 Maths Question 2.
A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.
Parts of red pigment | 1 | 4 | 7 | 12 | 20 |
Parts of base | 8 | … | … | … | … |
Solution:
It is given that the parts of red pigment the parts of base are in direct proportion. Therefore, the ratio of the corresponding values remains constant.
Let the parts of red pigment be x and the parts of base be y, then x/y = 1/8
Or, y = 8x
Thus, the required entries for y are 32, 56, 96, 160.
Thus, table becomes
Parts of red pigment | 1 | 4 | 7 | 12 | 20 |
Parts of base | 8 | 32 | 56 | 96 | 160 |
Ex 11.1 Class 8 Maths Question 3.
In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?
Solution:
Let x parts of red pigment should be mixed with 1800 mL of base. Then, the above information can be put in the tabular form as follows:
Parts of red pigment | 1 | x |
Parts of base | 75 | 1800 |
Ex 11.1 Class 8 Maths Question 4.
A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Solution:
Let x bottles will be filled up in five hours. Then, the information can be put in the tabular form as follows:
Time (in hours) | 6 | 5 |
Bottles filled up | 840 | x |
We observe that the lesser the time consumed, the lesser the number of bottles filled up. So, it is a case of direct proportion.
Ex 11.1 Class 8 Maths Question 5.
A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?
Solution:
Let x be the enlarged length of the bacteria when its photograph is enlarged 20,000 times. Then, the information can be put in the tabular form as follows:
Enlarged length (in cm) | 5 | x |
Enlarged photograph | 50000 | 20000 |
Ex 11.1 Class 8 Maths Question 6.
In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship?
Solution:
Let the length of the model ship be x metres, if actual length is 28 m. Then, the information can be put in the tabular form as follows:
Length of model (in cm) | 9 | x |
Actual length (in cm) | 12 | 28 |
Ex 11.1 Class 8 Maths Question 7.
Suppose 2 kg of sugar contains 9 × 10^{6} crystals. How many sugar crystals are there in
(i) 5 kg of sugar?
(ii) 1.2 kg of sugar?
Solution:
Let x and y crystals are there in 5 kg of sugar and 1.2 kg of sugar respectively. Then, the given information can be put in the tabular form as follows:
Ex 11.1 Class 8 Maths Question 8.
Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?
Solution:
Let x cm be the distance covered in the map for the driving of 72 km on a road. Then, the given information can be put in the tabular form as follows.
Distance covered on the map (in cm) | 1 | x |
Distance covered on the road (in km) | 18 | 72 |
Ex 11.1 Class 8 Maths Question 9.
A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time (i) the length of the shadow cast by another pole 10 m 50 cm high (ii) the height of a pole which casts a shadow 5 m long.
Solution:
Let x m be the length of the shadow if the length of the pole is 10 m 50 cm.
Let y m be the length of the pole whose shadow is 5 m long.
Then, the given information can be put in the tabular form as follows:
Length of pole (in m) | 5.60 | 10.50 | y |
Length of its shadow (in m) | 3.20 | x | 5 |
Clearly, it is a case of direct proportion.
Ex 11.1 Class 8 Maths Question 10.
A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Solution:
Let the truck travels x km in 5 hours. Then, the given information can be put in the tabular form as follows:
Distance travelled (in km) | 14 | x |
Time (in hours) | 25/60 | 5 |
Clearly, it is a case of direct proportion.
Hence, the truck travels 168 km in 5 hours.
Related Links:
NCERT Solutions for Maths Class 9
NCERT Solutions for Maths Class 10
NCERT Solutions for Maths Class 11