Class 8 maths exercise 13.1 solutions

  1. Exercise 13.1 Class 8 Maths? – Tiwari Academy Discussion
  2. NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions
  3. NCERT Solutions for Class 8 Maths Chapter 13 Exercise 13.1
  4. NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.2
  5. NCERT Solutions for Class 8 Maths Chapter 13
  6. NCERT Solutions for Class 8 Maths chapter


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Exercise 13.1 Class 8 Maths? – Tiwari Academy Discussion

NCERT Exercise Solution Class 8 Maths Exercise 13.1 Class 8 Chapter 13 Exercise 13.1 Direct and Inverse Proportions Class 8 Exercise 13.1 Solution in Hindi Class 8 Exercise 13.1 Solution in English 8th Maths Solutions Exercise 13.1 for terminal Exam Exercise 13.1 Solution for first term exams 2022-2023

NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions

NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions • • • NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Exercise 13.1 Ex 13.1 Class 8 Maths Question 1. Following are the car parking charges near a railway station up to. 4 hours – ₹ 60 8 hours – ₹ 100 12 hours – ₹ 140 24 hours – ₹ 180 Check if the parking charges are in direct proportions to the parking time. Solution: We have the ratio of time period and the parking charge. Hence the given quantities are not directly proportional. Ex 13.1 Class 8 Maths 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. Solution: Let the number to be filled in the blanks be a, b, c and d respectively. Ex 13.1 Class 8 Maths 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: Let the required red pigment be x part. Hence, the required amount of red pigment = 24 parts. Ex 13.1 Class 8 Maths 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: Let the required number of bottles be x. Hence the required number of bottles = 700. Ex 13.1 Class 8 Maths 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? I...

NCERT Solutions for Class 8 Maths Chapter 13 Exercise 13.1

On a graph paper, let X’OX and YOY’ be the coordinate axes. Let p be a point on the graph paper such that p is at distance of a units from the y-axis and b units from the x-axis. Then, we say that the coordinates of p are p (a, b). Here a is called the x-coordinate or abscissa of p, while b is called the y-coordinate or ordinate of p. Plot each of the following points on a graph paper: (i) A (5, 2) (ii) B (-2, 4) (iii) C (-4, – 6) (iv) D (4, – 3) Let X’ OX and YOY’ be the coordinate axes, (i) On the x-axis, take 5 units to the right of the y-axis and then on the y-axis. take 2 units above the x-axis. Thus, we get the point A (5, 2). (ii) On the x-axis, take 2 units to the left of the y-axis and then on the y-axis, take 4 units above the x-axis. Thus, we get the point B (-2, 4). (iii) On the x-axis, take 4 units to the right of the y-axis and then on the y-axis, take 6 units below the x-axis. Thus, we get the point C (-4, – 6). (iv) On the x-axis take 4 units to the right of the y-axis and then on the y-axis, take 3 units below the x-axis. Thus, we get the point D (4, – 3). For a square we have, perimeter = (4 x side of the square). (a) Consider the relation between the perimeter and the side of a square, given by P = 4a. Draw a graph of the above function. side of the square (b) From the graph, find the value of p, when (i) a = 5 (ii) a = 6 The given function is p = 4a. For different values of a, the corresponding values of p On a graph paper, plot the points O (0, 0), A (...

NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.2

NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.2 are part of • Board CBSE Textbook NCERT Class Class 8 Subject Maths Chapter Chapter 13 Chapter Name Direct and Inverse Proportions Exercise Ex 13.2 Number of Questions Solved 11 Category NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.2 Ex 13.2 Class 8 Maths Question 1. Which of the following are in inverse proportion? (i) The number of workers on a joh and the time to complete the job. (ii) The time taken for a journey and the distance travelled in a uniform speed. (iii) Area of cultivated land and the crop harvested. (iv) The time taken for a fixed journey and the speed of the vehicle. (v) The population of a country and the area of land per person. Solution: (i) We know that more is the number of workers to do a job, less is the time taken to finish the job. So, it is a case of inverse proportion. (ii) Clearly, more is the time taken, more is the distance travelled in a uniform speed. So, it is a case of direct proportion. (iii) Clearly, more is the area cultivated land, more is the crop harvested. So, it is a case of direct proportion. (iv) We know that more is the speed of the vehicle, less is the time to cover a fixed distance. So, it is a case of inverse proportion. (v) Clearly, more is the population, less is the area of land per person in a country. So, it is a case of inverse proportion. Ex 13.2 Class 8 Maths Question 2. In a Television game show,...

NCERT Solutions for Class 8 Maths Chapter 13

NCERT Solutions for Class 8 Maths Chapter 13 - Exercise 13.1 - Direct and Inverse Proportions A direct proportion refers to the direct relationship between two quantities. While an inverse proportion points to the indirect relationship between two quantities. Exercise 13.1 1. Following are the car parking charges near a railway station upto 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. Solution We will calculate car parking charges to check if they are in direct proportion with time or not. Charges to park car for 1 hour in the first case = 60/4 = Rs. 15 Charges to park car for 1 hour in the second case = 100/8 = Rs. 12.5 Charges to park car for 1 hour in the third case = 140/12 = Rs. 11.67 Charges to park car for 1 hour in the fourth case = 180/24 = Rs. 7.50 Clearly, parking charges are different in all the cases. Therefore, it is not in direct proportion to the parking time. 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 8 … … … … Solution Since, the mixture is prepared in direct proportion, the ratio will remain same in all the cases, say ‘k’. So, by first column, k = a 1/b 1 = 1/8 By second column, k = a 2/b 2 = 1/8 4/b 2 = 1/8 b 2 = 4×8 b 2 = 32 By third column, k = a 3/b 3 = 1/8 7/b 3 = 1/8 b 3 = 7×8 b 3 = 56 By four...

NCERT Solutions for Class 8 Maths chapter

NCERT Solution for class 8 maths chapter-13 Direct and Inverse Proportions NCERT solutions for class 8 maths chapter 13 Direct and Inverse Proportions is prepared by academic team of Physics Wallah. We have prepared NCERT solutions for all exercise of chapter 13. Given below is step by step solutions of all questions given in NCERT textbook for chapter 13. Read chapter 13 theory make sure you have gone through the theory part of chapter 13 from NCERT textbook and you have learned the formula of the given chapter. Physics Wallah prepared a detail notes and additional questions for class 8 maths with short notes of all maths formula of class 8 maths. do read these contents before moving to solve the exercise of NCERT chapter 13 NCERT Solutions for Class 8 Maths Exercise 13.1 Question 1. Following are he 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. Solution : Charges per hour: 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. Solution : 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 : Let the parts of red pigment mix with 1800 mL base be x. ...

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