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NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8

Document Description: NCERT Solutions: Direct & Inverse Proportions - 1 for Class 8 2022 is part of Mathematics (Maths) Class 8 preparation. The notes and questions for NCERT Solutions: Direct & Inverse Proportions - 1 have been prepared according to the Class 8 exam syllabus. Information about NCERT Solutions: Direct & Inverse Proportions - 1 covers topics like and NCERT Solutions: Direct & Inverse Proportions - 1 Example, for Class 8 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for NCERT Solutions: Direct & Inverse Proportions - 1.

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Question 1. Observe the following tables and find if x and y are directly proportional.
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
i.e. Each ratio is the same.
∴ x and y are directly proportional

NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
i.e. All the ratios are not the same.
∴ x and y are not directly proportional.

NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
i.e. All the ratios are not the same.
∴ x and y are not directly proportional.

Question 2. Principal = Rs 1000, Rate = 8% per annum. Fill in the following table and find which type of interest (simple or compound) change in direct proportion with time period.
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
Solution. Case of Simple Interest [P = Rs 1000, r = 8% p.a.]

NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8

∵ In each case the ratio NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8 is the same.
∴ The simple interest changes in direct proportion with time period.
Case of Compound Interest [P = Rs 1000, r = 8% p.a.]

NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8


NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8  is not the same in each case.
∴ The compound interest does not change in direct proportion with time period.

THINK, DISCUSS AND WRITE
Question: If we fix time period and the rate of interest, simple interest changes proportionally with principal. Would there be a similar relationship for compound interest? Why?
Solution. For simple interest,
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
Since rate and time period are constant, then SI changes directly according to P.
Thus, the simple interest changes in direct proportion with the principal.
For compound interest,
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
Since, rate and time period are constant, then
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
i.e. CI changes with P.
Thus, the compound interest also changes directly in proportion with principal.

EXERCISE 13.1
Question 1. Following are the car parking charges near a railway station upto
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
Check if the parking charges are in direct proportion to the parking time.

Solution: We have:
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8

Since     NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8

∴ The parking charges are not in direct proportion with 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.
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
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.
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8

Now

NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8

NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
Thus, the required parts of base are: 32, 56, 96 and 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?
Solution:
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
Here, x1 = 1, y1 = 75 and y2 = 1800
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
Thus, the required red pigments = 24 parts.

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: 

Number of bottles filledNumber of hours
8406
x5


For more number of hours, more number of bottles would be filled. Thus given quantities vary directly.
                                                       NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
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?
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
Solution: Let the actual length of the bacteria = x cm

Number of times photograph enlargedLength (in cm)
1x
50,0005


The length increases with an increment in the number of times the paragraph enlarged.
∴ It is a case of direct proportion
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
Again,
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8

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?
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
Solution: Let the required length of the model of the ship = x cm
We have:

Length of the shipHeight of the mast
2812
x9


Since, more the length of the ship, more would be the length of its mast.
∴ It is a case of direct variation.
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
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
(i) 5 kg of sugar? and (ii) 1.2 kg of sugar?
Solution: Let the required number of sugar crystals be x in 5 kg of sugar.
We have:
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
Since, more the amount of sugar, more would be the number of sugar crystals.
∴  It is a case of direct variation.
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
Thus, the required number of sugar crystals = 2.25 * 107
(ii) Let the number sugar crystals in 1.2 kg of sugar be y.
∴ We have:
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
∴ For a direct variation,
                                       NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8

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:
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
Since, it is a case of direct variation,

NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
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(i) the length of the shadow cast by another pole 10 m 50 cm high and (ii) the height of a pole which casts a shadow 5 m long.
Solution: (i) Let the required length of shadow be x cm.
We have:
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
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.
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
Thus, the required length of the shadow = 600 cm or 6 m

(ii) Let the required height of the pole be y cm.
∴ We have:
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
Since, it is a case of direct variation,
NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
Thus, the required height of the pole = 875 cm 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?
Solution: Since the 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:

Distance (in km)Time (in minutes)

14

x

25

300


NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8
∴ The required distance = 168 km

The document NCERT Solutions: Direct & Inverse Proportions - 1 Notes | Study Mathematics (Maths) Class 8 - Class 8 is a part of the Class 8 Course Mathematics (Maths) Class 8.
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