GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1

   

Gujarat Board GSEB Textbook Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1

Question 1.
Following are the car parking charges near a railway station upto
GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 img 1
Check if the parking charges are in direct proportion to the parking time.
Solution:
We have:
GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 img 2
Since \(\frac{15}{1}\) ≠ \(\frac{15}{2}\) ≠ \(\frac{35}{3}\) ≠ \(\frac{15}{2}\)
∴ The parking charges are not in direct proportion with the parking time.

GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1

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?
GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 img 4
Solution:
Let the red pigment be represented by x1, x2, x3, … and the base represented by y1, y2, y3 …………..
As the base increases, the required number of red pigments will also increase.
∴ The quantities vary directly.
GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 img 5

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:
We have
\(\frac{x_{1}}{y_{1}}\) = \(\frac{1}{8}\) = \(\frac{x_{2}}{y_{2}}\)
Here, x1 = 1, y1 = 75 and y2 = 1800
∴ \(\frac{1}{75}\) = \(\frac{x_{2}}{1800}\)
⇒ x2 = \(\frac{1800}{75}\) = 24
Thus, the required red pigments = 24 parts.

GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1

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:
GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 img 6
For more number of hours, more number of bottles would be filled. Thus given quantities vary directly.
∴ \(\frac{840}{x}\) = \(\frac{6}{5}\)
or 6x = 5 × 840
⇒ x = \(\frac{5×840}{6}\)
= 5 × 140 = 700
Thus, the required number of bottles = 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?
GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 img 7
Solution:
Let the actual length of the bacteria be x cm
GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 img 8
The length increases with an increment in the number of times the paragraph enlarged.
∴ It is a case of direct proportion
∴ \(\frac{1}{50000}\) = \(\frac{x}{5}\) or x = \(\frac{5}{50000}\)
= \(\frac{1}{10000}\) cm = 10-4 cm
Again,
GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 img 9
We have \(\frac{5}{50000}\) = \(\frac{x}{20000}\)
or 50000 × x = 5 × 20000
or x = \(\frac{5×20000}{50000}\) = 2
Thus, the enlarged length = 2 cm

GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1

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 28m, how long is the model ship?
GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 img 10
Solution:
Let the required length of the model of the ship be x cm.
We have:
GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 img 11
Since, more the length of the ship, more would be the length of its mast.
∴ It is a case of direct variation.
∴ \(\frac{28}{x}\) = \(\frac{12}{9}\)
or x = \(\frac{28×9}{12}\) = 7 × 3 = 21
Thus, the required length of the model = 21 cm

Question 7.
Suppose 2 kg of sugar contains 9 × 106 crystals. How many sugar crystals are there in

  1. 5 kg of sugar? and
  2. 1.2 kg of sugar?

Solution:
Let the required number of sugar crystals be x in 5 kg of sugar.
We have:
1.
GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 img 12
Since, more the amount of sugar, more would be the number of sugar crystals.
∴ It is a case of direct variation
GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 img 13
Thus, the required number of sugar crystals = 2.25 × 107

2. Let the number of sugar crystals in 1.2 kg of sugar be y.
GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 img 14
∴ For a direct variation,
\(\frac{2}{1.2}\) = \(\frac{9 \times 10^{6}}{y}\)
or 2 × y = 1.2 × 9 × 106
or y = \(\frac{12}{10}\) × \(\frac{9}{2}\) × 106 = 5.4 × 106
Thus, the required number of sugar crystals = 5.4 × 106

GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1

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 the required distance covered in the map be x cm.
We have:
GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 img 15
Since, it is a case of direct variation,
∴ \(\frac{18}{72}\) = \(\frac{1}{x}\) or 18 × x = 1 × 72
or x = \(\frac{1×72}{18}\) = 1 × 4 = 4
Thus, the required distance 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.

  1. The length of the shadow cast by another pole 10 m 50 cm high and
  2. The height of a pole which casts a shadow 5 m long.

Solution:
1. Let the required length of shadow be x cm.
We have:
GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 img 16
Height of the pole Length of the shadow
As the height of the pole increases the length of its shadow also increases in the same ratio.
Therefore, it is a case of direct variation.
∴ \(\frac{560}{1050}\) = \(\frac{320}{x}\)
or 560 × x = 320 × 1050
or x = \(\frac{320×1050}{560}\)
= 4 × 150 = 600 cm
Thus, the required length of the shadow = 600 cm or 6 m

2. Let thc required height of the pole be y cm.
∴ We have:
GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 img 17
Since, it is a case of direct vanation.
∴ \(\frac{560}{y}\) = \(\frac{320}{500}\)
or 320 × y = 500 × 500
or y = \(\frac{500×560}{320}\)
= 125 × 7 = 875
Thus, the required height of the pole = 875 cm or 8 m 75 cm.

GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1

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:
Since thc speed is constant,
∴ For longer distance, more time will be required
So, it is a case of direct variation.
Let the required distance to be travelled in 5 hours (= 300 minutes) be x km.
We have:
GSEB Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 img 18
∴ \(\frac{14}{x}\) = \(\frac{25}{300}\)
or 25 × x = 14 × 300
or x = \(\frac{14×300}{25}\) = 14 × 2 = 168
∴ The required distance = 168 km.

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