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Can you think of other way(s) to find a rational number between any two rational numbers?
An alternative way to find a rational number between any two given numbers is by using the mean or midpoint method. You simply add the two rational numbers together and then divide their sum by 2. The resulting value is guaranteed to lie exactly halfway between them on a number line. Because rationaRead more
An alternative way to find a rational number between any two given numbers is by using the mean or midpoint method. You simply add the two rational numbers together and then divide their sum by 2. The resulting value is guaranteed to lie exactly halfway between them on a number line. Because rational numbers are dense, you can repeat this averaging process indefinitely with any new endpoints to discover infinite intermediate rational numbers.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 3 The world of numbers (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-3/
See lessFind three rational numbers between 3.1415 and 3.1416.
The given boundary values are 3.1415 and 3.1416. To find rational numbers situated between them, we can look at the next decimal place value by imagining the numbers as 3.14150 and 3.14160. By choosing terminating decimal values that fall strictly between these two new limits, we easily create validRead more
The given boundary values are 3.1415 and 3.1416. To find rational numbers situated between them, we can look at the next decimal place value by imagining the numbers as 3.14150 and 3.14160. By choosing terminating decimal values that fall strictly between these two new limits, we easily create valid intermediate fractions. Three appropriate rational numbers that fit perfectly inside this specific numerical range are 3.14151, 3.14152, and 3.14153.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 3 The world of numbers (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-3/
See lessExplain, using a real-world example of debt, why subtracting a negative number is the same as adding a positive number (e.g., 10 – (–5) = 15).
Imagine you have 10 rupees in cash but also owe a friend 5 rupees, making your total financial position 10 minus 5. If your friend generously cancels or subtracts that 5 rupee debt, you no longer owe that money. Removing this negative financial burden increases your net worth back to 15 rupees. MathRead more
Imagine you have 10 rupees in cash but also owe a friend 5 rupees, making your total financial position 10 minus 5. If your friend generously cancels or subtracts that 5 rupee debt, you no longer owe that money. Removing this negative financial burden increases your net worth back to 15 rupees. Mathematically, removing a debt of 5 is written as 10 minus negative 5, which gives the exact same result as adding 5 rupees.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 3 The world of numbers (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-3/
See lessA learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observes that when she accessed 10 modules, her bill was rupees 400. When she accessed 14 modules, her bill was rupees 500. If the monthly bill y depends on the number of modules accessed, x, according to the relation y = ax + b, find the values of a and b.
To find the constants, form two linear equations using the given data points (10, 400) and (14, 500) according to the relation y = ax + b. This gives 400 = 10a + b and 500 = 14a + b. Subtracting the first equation from the second equation helps eliminate b, yielding 4a = 100, which means a = 25. SubRead more
To find the constants, form two linear equations using the given data points (10, 400) and (14, 500) according to the relation y = ax + b. This gives 400 = 10a + b and 500 = 14a + b. Subtracting the first equation from the second equation helps eliminate b, yielding 4a = 100, which means a = 25. Substituting a = 25 back into the first equation yields b = 150. Thus, a is 25 and b is 150
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2 Introduction to Linear Polynomials (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-2/
See lessA gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for 10 hours, her bill was rupees 800. When she used it for 15 hours, her bill was rupees 1100. If the monthly bill y depends on the hours of the use of the badminton court, x, according to the relation y = ax + b, find the values of a and b.
Using the linear relation y = ax + b, plug in the given variables where x represents the hours and y represents the total bill. This provides two distinct linear equations: 800 = 10a + b and 1100 = 15a + b. Subtracting the first equation from the second simplifies the system to 5a = 300, which solveRead more
Using the linear relation y = ax + b, plug in the given variables where x represents the hours and y represents the total bill. This provides two distinct linear equations: 800 = 10a + b and 1100 = 15a + b. Subtracting the first equation from the second simplifies the system to 5a = 300, which solves to a = 60. Substituting a = 60 back into the first equation gives b = 200. Therefore, a = 60 and b = 200.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2 Introduction to Linear Polynomials (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-2/
See less