Exercise 13.1
Question 1
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 |
The ratio of parking charges to the respective number of hours (Rs/ hour) can be calculated as
As each ratio is not same, therefore, the parking charges are not in a direct proportion to the parking time.
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 | … | … | … | … |
Sol :
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 | x_{1} | x_{2} | x_{3} | x_{4} |
According to direct proportion,
The table can be drawn as follows.
Parts of red pigment | 1 | 4 | 7 | 12 | 20 |
Parts of base | 8 | 32 | 56 | 96 | 160 |
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 |
The parts of red pigment and the parts of base are in direct proportion.
Therefore, we obtain
Thus, 24 parts of red pigments should be mixed with 1800 mL of base.
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 |
The number of bottles and the time taken to fill these bottles are in direct proportion. Therefore, we obtain
Thus, 700 bottles will be filled in 5 hours.
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 |
The number of times the photograph of bacteria was enlarged and the length of bacteria are in direct proportion.
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.
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 |
We know that the dimensions of the actual ship and the model ship are directly proportional to each other.
Therefore, we obtain:
Thus, the length of the model ship is 21 cm.
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?
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)
2
5
Number of crystals
9 × 10^{6}
x
The amount of sugar and the number of crystals it contains are directly proportional to each other. Therefore, we obtain
Hence, the number of sugar crystals is 2.25 × 10^{7}.
(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)
2
1.2
Number of crystals
9 × 10^{6}
y
Hence, the number of sugar crystals is.
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 |
The distances covered on road and represented on map are directly proportional to each other. Therefore, we obtain
Hence, the distance represented on the map is 4 cm.
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)
5.60
10.50
Length of shadow (in m)
3.20
x
More the height of an object, more will be the length of its shadow.
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)
5.60
y
Length of shadow (in m)
3.20
5
The height of the pole and the length of the shadow are directly proportional to each other. Therefore,
Thus, the height of the pole is 8.75 m or 8 m 75 cm.
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 |
The distance travelled by the truck and the time taken by the truck are directly proportional to each other. Therefore,
Hence, the distance travelled by the truck is 168 km.