To locate 6 rational numbers between the whole numbers 3 and 4, we rewrite them as equivalent fractions with a common denominator greater than 6, such as 7. This transforms 3 into 21/7 and 4 into 28/7. Now, we can choose any six consecutive fractional values that lie between these two new boundary pRead more
To locate 6 rational numbers between the whole numbers 3 and 4, we rewrite them as equivalent fractions with a common denominator greater than 6, such as 7. This transforms 3 into 21/7 and 4 into 28/7. Now, we can choose any six consecutive fractional values that lie between these two new boundary points. The resulting six rational numbers are 22/7, 23/7, 24/7, 25/7, 26/7, and 27/7.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 3 The world of numbers (2026-27):
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):
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):
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):
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):
Find 6 rational numbers between 3 and 4.
To locate 6 rational numbers between the whole numbers 3 and 4, we rewrite them as equivalent fractions with a common denominator greater than 6, such as 7. This transforms 3 into 21/7 and 4 into 28/7. Now, we can choose any six consecutive fractional values that lie between these two new boundary pRead more
To locate 6 rational numbers between the whole numbers 3 and 4, we rewrite them as equivalent fractions with a common denominator greater than 6, such as 7. This transforms 3 into 21/7 and 4 into 28/7. Now, we can choose any six consecutive fractional values that lie between these two new boundary points. The resulting six rational numbers are 22/7, 23/7, 24/7, 25/7, 26/7, and 27/7.
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 lessCan 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 less