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Exercise 13.1
Question 1:
Following are the car parking charges near a railway station up to:
Check if the parking charges are in direct proportion to the parking time.
Answer 1:
Here, the charges per hour are not same, i.e., C_{1} ≠ C_{2 }≠ C_{3} ≠C_{4}
Therefore, the parking charges are not in 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  —  —  —  — 
Answer 2:
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?
Answer 3:
Let the parts of red pigment mix with 1800 mL base be x
Parts of red pigment  1  X 
Parts of base  75  1800 
Hence, with base 1800 mL, 24 parts red pigment should be mixed.
Question 4:
A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Answer 4:
Let the number of bottles filled in five hours be x.
Hours  1  X 
Bottles  75  1800 
Hence, the number of bottles filled in five hours be 700.
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?
Answer 5:
Length  5  X 
Enlarged length  50,000  20,000 
Here length and enlarged length of bacteria are in direct proportion
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 if the ship is 28 m, how long is the model ship?
Answer 6:
Let the length of model ship be x
Length of actual ship (in m)  12  28 
Length of model ship (in cm)  9  X 
Here length of mast and actual length of ship are in direct proportion
Hence, the length of the model ship is 21 cm.
Question 7:
Suppose 2 kg of sugar contains 9 x 10^{6} crystals. How many sugar crystals are there in
(i) 5 kg of sugar?
(ii) 1.2 kg of sugar?
Answer 7:
Let sugar crystals be x.
Weight of sugar (in kg)  2  5 
No. of crystals  9x10^{6}  X 
Hence, the number of sugar crystals is 2.25 x 10^{7}.
(ii)
Let sugar crystals be x.
Weight of sugar (in kg)  2  1.2 
No. of crystals  9x10^{6}  X 
Hence, the number of sugar crystals is 5.4 x 10^{6}.
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?
Answer 8:
Let distance covered in the map be x.
Actual distance (in km)  18  72 
Distance covered in map (in cm)  1  X 
Here actual distance and distance covered in the map are in direct proportion.
Hence, distance covered in 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.
Answer 9:
Here height of the pole and length of the shadow are in direct proportion.
And 1 m = 100 cm
5 m 60 cm = 5 x 100 + 60 = 560 cm
3 m 20 cm = 3 x 100 + 20 = 320 cm
10 m 50 cm = 10 x 100 + 50 = 1050 cm
5 m = 5 x 100  500 cm
(i) Let the length of the shadow of another pole be x.
Height of pole (in cm)  560  1050 
Length of shadow (in cm)  320  X 
Hence, the length of the shadow of another pole be 6m.
Let the hight of the pole be x.
Height of pole (in cm)  560  x 
Length of shadow (in cm)  320  500 
Hence, the hight of the pole be 8m 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?
Answer 10:
Let distance covered in 5 hours be .r km.
1 hour = 60 minutes
5 hours = 5 x 60 = 300 minutes
Distance (in km]  14  X 
Time (in minutes)  25  300 
Here distance covered and time in direct proportion.
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