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A 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 lessConsider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by °C = a °F + b. Find a and b, given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit and water boils at 100 degrees Celsius and 212 degrees Fahrenheit. (Hint: When °C = 0, °F = 32 and when °C = 100, °F = 212. Use this information to find a and b and thus, the linear relationship between °C and °F.)
Substitute the temperature coordinates into the linear relationship C = aF + b. For the melting point, 0 = 32a + b, which means b = -32a. For the boiling point, 100 = 212a + b. Substituting b into the second equation yields 100 = 212a - 32a, which simplifies to 100 = 180a. This gives a = 100/180 = 5Read more
Substitute the temperature coordinates into the linear relationship C = aF + b. For the melting point, 0 = 32a + b, which means b = -32a. For the boiling point, 100 = 212a + b. Substituting b into the second equation yields 100 = 212a – 32a, which simplifies to 100 = 180a. This gives a = 100/180 = 5/9. Using a to find b gives b = -160/9. The values are a = 5/9 and b = -160/9.
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 lessWrite a polynomial of degree 3 in the variable x, in which the coefficient of the x2 term is –7.
A polynomial of degree 3 must have a maximum exponent of 3 on its variable. The general expression can be represented as ax3 + bx2 + cx + d where a cannot be zero. Following the specific instruction that the coefficient of the x2 term must be exactly -7, we can choose arbitrary values for the otherRead more
A polynomial of degree 3 must have a maximum exponent of 3 on its variable. The general expression can be represented as ax3 + bx2 + cx + d where a cannot be zero. Following the specific instruction that the coefficient of the x2 term must be exactly -7, we can choose arbitrary values for the other coefficients. An appropriate and simple polynomial satisfying these conditions is x3 – 7×2 + x + 1.
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 lessIf we multiply a number by 5/2 and add 2/3 to the product, we get –7/12. Find the number.
Let the required number be represented by the variable x. Translating the word problem into a mathematical equation results in 5/2 x + 2/3 = -7/12. To solve for x, subtract 2/3 from both sides of the expression, making 5/2 x = -7/12 - 8/12, which combines to -15/12. Multiplying both sides by 2/5 isoRead more
Let the required number be represented by the variable x. Translating the word problem into a mathematical equation results in 5/2 x + 2/3 = -7/12. To solve for x, subtract 2/3 from both sides of the expression, making 5/2 x = -7/12 – 8/12, which combines to -15/12. Multiplying both sides by 2/5 isolates the variable, resulting in x = (-15/12) multiplied by (2/5), which simplifies perfectly to -1/2.
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