Exercise 13.1
Introduction - Direct and Inverse Proportions - Chapter 13 - NCERT Class 8th Maths
Q 1 - Ex 13.1 - Direct and Inverse Proportions - NCERT Maths Class 8th - Chapter 13
Following are the car parking charges near a railway station up to
4 hours Rs 60
8 hours Rs 100
12 hours Rs 140
24 hours Rs 180
Check if the parking charges are in direct proportion to the parking time.
Sol :
A table of the given information is formed as
Number of hours
|
4
|
8
|
12
|
24
|
Parking charges (in Rs)
|
60
|
100
|
140
|
180
|
As each ratio is not same, therefore, the parking charges are not in a direct proportion to the parking time.
Q 2 - Ex 13.1 - Direct and Inverse Proportions - NCERT Maths Class 8th - Chapter 13
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
|
…
|
…
|
…
|
…
|
The given mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. For more parts of red pigments, the parts of the base will also be more. Therefore, the parts of red pigments and the parts of base are in direct proportion. The given information in the form of a table is as follows.
Parts of red pigment
|
1
|
4
|
7
|
12
|
20
|
Parts of base
|
8
|
x1
|
x2
|
x3
|
x4
|
The table can be drawn as follows.
Parts of red pigment
|
1
|
4
|
7
|
12
|
20
|
Parts of base
|
8
|
32
|
56
|
96
|
160
|
Q 3 - Ex 13.1 - Direct and Inverse Proportions - NCERT Maths Class 8th - Chapter 13
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?
Sol :
Let the parts of red pigment required to mix with 1800 mL of base be x.
The given information in the form of a table is as follows.
Parts of red pigment
|
1
|
x
|
Parts of base (in mL)
|
75
|
1800
|
Therefore, we obtain
Thus, 24 parts of red pigments should be mixed with 1800 mL of base.
Q 4 - Ex 13.1 - Direct and Inverse Proportions - NCERT Maths Class 8th - Chapter 13
Question 4
A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Sol :
Let the number of bottles filled by the machine in five hours be x.
The given information in the form of a table is as follows.
Number of bottles
|
840
|
x
|
Time taken (in hours)
|
6
|
5
|
Thus, 700 bottles will be filled in 5 hours.
Q 5 - Ex 13.1 - Direct and Inverse Proportions - NCERT Maths Class 8th - Chapter 13
Question 5
A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?
Sol :
Let the actual length of bacteria be x cm and the enlarged length of bacteria be y cm, if the photograph is enlarged for 20,000 times.
The given information in the form of a table is as follows.
Length of bacteria (in cm)
|
5
|
x
|
y
|
Number of times photograph of Bacteria was enlarged
| 50000 |
1
|
20000
|
Therefore, we obtain
Hence, the actual length of bacteria is 10−4 cm.
Let the length of bacteria when the photograph of bacteria is enlarged 20, 000 times be y.
Hence, the enlarged length of bacteria is 2 cm.
Q 6 - Ex 13.1 - Direct and Inverse Proportions - NCERT Maths Class 8th - Chapter 13
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?
Sol :
Let the length of the mast of the model ship be x cm.
The given information in the form of a table is as follows:
-
|
Height of mast
|
Length of ship
|
Model ship
|
9 cm
|
x
|
Actual ship
|
12 m
|
28 m
|
Therefore, we obtain:
Thus, the length of the model ship is 21 cm.
Q 7 - Ex 13.1 - Direct and Inverse Proportions - NCERT Maths Class 8th - Chapter 13
Question 7
Suppose 2 kg of sugar contains 9 × 106 crystals.
How many sugar crystals are there in (i) 5 kg of sugar? (ii) 1.2 kg of sugar?
Sol :
(i) Let the number of sugar crystals in 5 kg of sugar be x.
The given information in the form of a table is as follows.
- Amount of sugar (in kg)25Number of crystals9 × 106x
Hence, the number of sugar crystals is 2.25 × 107.
(ii) Let the number of sugar crystals in 1.2 kg of sugar be y. The given information in the form of a table is as follows.
- Amount of sugar (in kg)21.2Number of crystals9 × 106y
Hence, the number of sugar crystals is.
Q 8 - Ex 13.1 - Direct and Inverse Proportions - NCERT Maths Class 8th - Chapter 13
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?
Sol :
Let the distance represented on the map be x cm.
The given information in the form of a table is as follows.
Distance covered on road in (in km)
|
18
|
72
|
Distance represented on map (in cm)
|
1
|
x
|
Hence, the distance represented on the map is 4 cm.
Q 9 - Ex 13.1 - Direct and Inverse Proportions - NCERT Maths Class 8th - Chapter 13
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.
Sol :
(i) Let the length of the shadow of the other pole be x m.
1 m = 100 cm
The given information in the form of a table is as follows.
- Height of pole (in m)5.6010.50Length of shadow (in m)3.20x
Thus, the height of an object and length of its shadow are directly proportional to each other. Therefore, we obtain
Hence, the length of the shadow will be 6 m.
(ii) Let the height of the pole be y m.
The given information in the form of a table is as follows.
- Height of pole (in m)5.60yLength of shadow (in m)3.205
Thus, the height of the pole is 8.75 m or 8 m 75 cm.
Q 10 - Ex 13.1 - Direct and Inverse Proportions - NCERT Maths Class 8th - Chapter 13
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?
Sol :
Let the distance travelled by the truck in 5 hours be x km.
We know, 1 hour = 60 minutes
∴ 5 hours = (5 × 60) minutes = 300 minutes
The given information in the form of a table is as follows.
Distance travelled (in km)
|
14
|
x
|
Time (in min)
|
25
|
300
|
Hence, the distance travelled by the truck is 168 km.
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