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

Question 1. Following are the car parking charges near a railway station upto

Check if the parking charges are in direct proportion to the parking time.
Answer: To check for direct proportion, we compare the ratio of parking charges to parking time. For 4 hours, the charge is Rs. 60, making the ratio \( \frac{60}{4} = \frac{15}{1} \). For 8 hours, the charge is Rs. 100, so the ratio is \( \frac{100}{8} = \frac{25}{2} \). For 12 hours, the charge is Rs. 140, giving a ratio of \( \frac{140}{12} = \frac{35}{3} \). Finally, for 24 hours, the charge is Rs. 180, and the ratio becomes \( \frac{180}{24} = \frac{15}{2} \). Since these ratios are not equal, the parking charges do not vary in direct proportion to the parking time.
In simple words: We checked if the parking cost increases steadily with time. Since the ratio of cost to time was different for different durations, the charges don't follow a direct proportion.

Exam Tip: When checking for direct proportion, always calculate the ratio of the two quantities for each pair of values. If these ratios are not constant, then the quantities are not in direct proportion.

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?

Answer: Let's represent the red pigment parts as \( x_1, x_2, x_3, \dots \) and the base parts as \( y_1, y_2, y_3, \dots \). As the quantity of base gets larger, the necessary amount of red pigment will also increase.
\( \implies \) This shows that the quantities have a direct variation.
So, the ratio \( \frac{x_1}{y_1} = \frac{x_2}{y_2} \) will remain constant. Given \( x_1 = 1 \) and \( y_1 = 8 \), the constant ratio is \( \frac{1}{8} \). We can now find the missing values:
For \( x_2 = 4 \): \( \frac{4}{y_2} = \frac{1}{8} \). Solving for \( y_2 \) gives \( y_2 = 4 \times 8 = 32 \).
For \( x_3 = 7 \): \( \frac{7}{y_3} = \frac{1}{8} \). Solving for \( y_3 \) gives \( y_3 = 7 \times 8 = 56 \).
For \( x_4 = 12 \): \( \frac{12}{y_4} = \frac{1}{8} \). Solving for \( y_4 \) gives \( y_4 = 12 \times 8 = 96 \).
For \( x_5 = 20 \): \( \frac{20}{y_5} = \frac{1}{8} \). Solving for \( y_5 \) gives \( y_5 = 20 \times 8 = 160 \).
In simple words: When paint has more red color, it needs more base. The amount of red color and base always keep the same ratio. We used this rule to find the missing base amounts for different amounts of red pigment.

Exam Tip: For direct proportion problems, remember that the ratio \( \frac{x}{y} \) is constant. Set up equations using this constant ratio to find unknown values.

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: We know that the ratio of red pigment to base remains constant in this mixture. We have the formula \( \frac{x_1}{y_1} = \frac{x_2}{y_2} \). Here, \( x_1 = 1 \) part of red pigment needs \( y_1 = 75 \) mL of base. We want to find \( x_2 \) (red pigment) for \( y_2 = 1800 \) mL of base.
\( \implies \) Plugging in the values, we get \( \frac{1}{75} = \frac{x_2}{1800} \).
\( \implies \) To solve for \( x_2 \), we multiply both sides by 1800: \( x_2 = \frac{1800}{75} \).
\( \implies \) Performing the division, we find \( x_2 = 24 \).
Thus, we will require 24 parts of red pigment.
In simple words: We need to keep the red pigment and base mixed in the same way. If 1 part of red pigment needs 75 mL of base, then for 1800 mL of base, you will need 24 parts of red pigment to keep the mixture right.

Exam Tip: When solving proportion problems, ensure you set up the ratios correctly with corresponding quantities in the numerator and denominator. Convert units if necessary, though not in this case.

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: We observe that if the machine operates for more hours, it will fill a larger number of bottles. This means the number of bottles filled and the operating hours are in direct proportion.
\( \implies \) We can set up the proportion as \( \frac{\text{Number of bottles filled}}{\text{Number of hours}} = \text{constant} \).
\( \implies \) So, \( \frac{840}{6} = \frac{x}{5} \).
\( \implies \) Cross-multiplying, we get \( 6 \times x = 5 \times 840 \).
\( \implies \) To find \( x \), we perform the calculation: \( x = \frac{5 \times 840}{6} \).
\( \implies \) This simplifies to \( x = 5 \times 140 = 700 \).
Therefore, the machine will fill 700 bottles in five hours.
In simple words: More time means more bottles filled. We set up a ratio: 840 bottles in 6 hours is like 'x' bottles in 5 hours. Solving for 'x' gives us 700 bottles.

Exam Tip: Always identify whether the problem involves direct or inverse proportion first. Then, set up the ratios correctly before solving for the unknown value.

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: Let the bacteria's actual length be \( x \) cm. When a photograph is enlarged, its length increases proportionally to the enlargement factor. This represents a case of direct proportion.
To find the actual length:

The ratio of actual length to enlargement factor should be constant. So, \( \frac{1 \text{ (actual size)}}{x \text{ (actual length)}} = \frac{50,000 \text{ (enlarged times)}}{5 \text{ cm (enlarged length)}} \).
\( \implies \) This gives us \( \frac{1}{50000} = \frac{x}{5} \).
\( \implies \) Solving for \( x \): \( x = \frac{5}{50000} = \frac{1}{10000} \text{ cm} = 10^{-4} \text{ cm} \).
Thus, the actual length of the bacteria is \( 10^{-4} \text{ cm} \).
To find the new enlarged length:

We use the same principle of direct proportion. If 50,000 times enlargement gives 5 cm, then 20,000 times enlargement will give \( x \) cm.
\( \implies \) We set up the proportion: \( \frac{5 \text{ cm}}{50000 \text{ times}} = \frac{x \text{ cm}}{20000 \text{ times}} \).
\( \implies \) This means \( \frac{5}{50000} = \frac{x}{20000} \).
\( \implies \) Cross-multiplying, we get \( 50000 \times x = 5 \times 20000 \).
\( \implies \) Solving for \( x \): \( x = \frac{5 \times 20000}{50000} = \frac{100000}{50000} = 2 \).
Therefore, the enlarged length would be 2 cm if enlarged 20,000 times.
In simple words: First, we found the actual tiny size of the bacteria by working backwards from its enlarged picture. Then, we calculated how big the picture would be if we enlarged it a different number of times, keeping the same scale.

Exam Tip: When dealing with scale or enlargement problems, always maintain consistency in your units and ensure the ratio is set up correctly for direct proportionality.

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?
Answer: Let \( x \) cm be the length of the model ship that we need to find. We are comparing the lengths and mast heights of the actual ship and its model. This situation represents a direct variation, as a longer ship will have a taller mast proportionally.

We can set up the proportion:
\( \frac{\text{Length of actual ship}}{\text{Length of model ship}} = \frac{\text{Height of actual mast}}{\text{Height of model mast}} \)
\( \implies \) Given the length of the actual ship is 28 m and its mast height is 12 m. The model's mast height is 9 cm. We need to find the model's length in cm.
\( \implies \) So, \( \frac{28}{x} = \frac{12}{9} \).
\( \implies \) To solve for \( x \), we can cross-multiply: \( 12 \times x = 28 \times 9 \).
\( \implies \) Then, \( x = \frac{28 \times 9}{12} \).
\( \implies \) Simplifying the calculation, \( x = \frac{252}{12} = 21 \).
Therefore, the required length of the model ship is 21 cm.
In simple words: The model ship and the real ship are scaled versions of each other. We used the known heights of their masts to figure out the right length for the model ship based on the real ship's length.

Exam Tip: When working with models and actual objects, make sure to set up the ratios accurately. If different units are used for corresponding parts (like meters for the actual ship and centimeters for the model), ensure that the proportional relationship is applied consistently across those units.

Question 7. Suppose 2 kg of sugar contains \( 9 \times 10^6 \) crystals. How many sugar crystals are there in
1. 5 kg of sugar? and
2. 1.2 kg of sugar?
Answer: The number of sugar crystals is directly proportional to the weight of sugar. This means as the amount of sugar increases, the number of crystals also grows at a steady rate.
**1. For 5 kg of sugar:**
Let \( x \) be the number of sugar crystals in 5 kg of sugar.
We can set up the proportion based on direct variation: \( \frac{\text{Initial Weight}}{\text{New Weight}} = \frac{\text{Initial Crystals}}{\text{New Crystals}} \).

\( \implies \frac{2}{5} = \frac{9 \times 10^6}{x} \)
\( \implies \) Cross-multiplying, we get \( 2 \times x = 5 \times 9 \times 10^6 \).
\( \implies \) Solving for \( x \): \( x = \frac{5 \times 9 \times 10^6}{2} = \frac{45 \times 10^6}{2} \).
\( \implies \) This gives \( x = 22.5 \times 10^6 \), which in standard form is \( 2.25 \times 10^7 \).
Therefore, 5 kg of sugar contains \( 2.25 \times 10^7 \) crystals.
**2. For 1.2 kg of sugar:**
Let \( y \) be the number of sugar crystals in 1.2 kg of sugar.
Using the same direct variation principle:

\( \implies \frac{2}{1.2} = \frac{9 \times 10^6}{y} \)
\( \implies \) Cross-multiplying, we get \( 2 \times y = 1.2 \times 9 \times 10^6 \).
\( \implies \) Solving for \( y \): \( y = \frac{1.2 \times 9 \times 10^6}{2} = \frac{10.8 \times 10^6}{2} \).
\( \implies \) This results in \( y = 5.4 \times 10^6 \).
Thus, 1.2 kg of sugar contains \( 5.4 \times 10^6 \) crystals.
In simple words: Since sugar crystals increase with more sugar, we used a simple comparison method. We found how many crystals are in 5 kg of sugar and then how many are in 1.2 kg, based on the initial information.

Exam Tip: When solving direct proportion problems with multiple parts, remember to apply the same constant ratio for each scenario. Be careful with calculations involving powers of 10 and scientific notation.

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: Let \( x \) cm be the distance Rashmi covers on the map. The map's scale shows a direct relationship between actual distance and map distance: 1 cm on the map equals 18 km in reality.

\( \implies \) We can set up a direct proportion: \( \frac{\text{Actual distance 1}}{\text{Actual distance 2}} = \frac{\text{Map distance 1}}{\text{Map distance 2}} \).
\( \implies \) So, \( \frac{18 \text{ km}}{72 \text{ km}} = \frac{1 \text{ cm}}{x \text{ cm}} \).
\( \implies \) This simplifies to \( \frac{18}{72} = \frac{1}{x} \).
\( \implies \) Cross-multiplying, we get \( 18 \times x = 1 \times 72 \).
\( \implies \) Solving for \( x \): \( x = \frac{72}{18} = 4 \).
Therefore, the distance covered on the map will be 4 cm.
In simple words: The map's scale tells us how much real distance 1 cm represents. We used this scale to find out how many centimeters on the map would show a 72 km journey.

Exam Tip: For map scale problems, always ensure you correctly identify the corresponding real-world and map distances to set up the direct proportion accurately.

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.
Answer: Since the sun's angle is constant at the same time, the height of a pole and the length of its shadow are in direct proportion.
**Original Pole:** Height \( 5 \text{ m } 60 \text{ cm} = 560 \text{ cm} \). Shadow length \( 3 \text{ m } 20 \text{ cm} = 320 \text{ cm} \).
**1. Find shadow length for a 10 m 50 cm pole:**
Let \( x \) cm be the shadow length. The new pole height is \( 10 \text{ m } 50 \text{ cm} = 1050 \text{ cm} \).

\( \implies \) Set up the proportion: \( \frac{\text{Initial Height}}{\text{New Height}} = \frac{\text{Initial Shadow}}{\text{New Shadow}} \).
\( \implies \frac{560}{1050} = \frac{320}{x} \)
\( \implies \) Cross-multiplying, we get \( 560 \times x = 320 \times 1050 \).
\( \implies \) Solving for \( x \): \( x = \frac{320 \times 1050}{560} \).
\( \implies \) This calculation gives \( x = 600 \text{ cm} \).
Therefore, the required length of the shadow is 600 cm or 6 m.
**2. Find pole height for a 5 m shadow:**
Let \( y \) cm be the height of the pole. The new shadow length is \( 5 \text{ m} = 500 \text{ cm} \).

\( \implies \) Set up the proportion: \( \frac{\text{Initial Height}}{\text{New Height}} = \frac{\text{Initial Shadow}}{\text{New Shadow}} \).
\( \implies \frac{560}{y} = \frac{320}{500} \)
\( \implies \) Cross-multiplying, we get \( 320 \times y = 560 \times 500 \).
\( \implies \) Solving for \( y \): \( y = \frac{560 \times 500}{320} \).
\( \implies \) This calculation gives \( y = 875 \text{ cm} \).
Thus, the required height of the pole is 875 cm or 8 m 75 cm.
In simple words: Since the sun's light hits everything at the same angle, taller objects make longer shadows. We used this rule to find the shadow length for a taller pole and the height of a pole that makes a specific shadow length.

Exam Tip: Always convert all measurements to a consistent unit (like centimeters) before setting up and solving direct proportion problems involving mixed units.

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: Since the truck's speed remains constant, the distance it travels is directly proportional to the time taken. This means that if the truck travels for more time, it will cover a greater distance.
First, we convert 5 hours into minutes: \( 5 \text{ hours} = 5 \times 60 = 300 \text{ minutes} \).
Let \( x \) km be the distance the truck can travel in 300 minutes.

\( \implies \) We set up the direct proportion: \( \frac{\text{Initial Distance}}{\text{New Distance}} = \frac{\text{Initial Time}}{\text{New Time}} \).
\( \implies \frac{14 \text{ km}}{x \text{ km}} = \frac{25 \text{ minutes}}{300 \text{ minutes}} \).
\( \implies \) This simplifies to \( \frac{14}{x} = \frac{25}{300} \).
\( \implies \) Cross-multiplying, we get \( 25 \times x = 14 \times 300 \).
\( \implies \) Solving for \( x \): \( x = \frac{14 \times 300}{25} \).
\( \implies \) This gives \( x = 14 \times 12 = 168 \).
Therefore, the required distance the truck can travel is 168 km.
In simple words: Because the truck drives at a steady speed, it will cover more ground if it drives longer. We first changed 5 hours to minutes, then used a direct comparison to find out how far it would go in that longer time.

Exam Tip: Always make sure all time units are consistent (e.g., convert hours to minutes) before performing calculations in direct proportion problems.