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NCERT Class 9 Mathematics Exercise Set 2.5 Solutions: Introduction to Linear Polynomials
AI-written, checked on 23 Oct 2026
Solutions written by Shishya, checked — not the board's official answers
Step-by-step working for every question of Exercise Set 2.5 in NCERT's Class 9 Mathematics book, chapter “Introduction to Linear Polynomials”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.
How these were checked: the solutions were drafted with AI and then read and corrected on 23 Oct 2026. Every numerical final answer is recomputed from the question's own numbers by a calculator check that runs before each release; proofs and explanations were read, not machine-checked. The NCERT textbook is the authority — if something here disagrees with it, follow the book.
Tap “why?” under a step to see the textbook rule it uses — from Shishya's fixed list of NCERT Class 9-10 rules, not written by AI for this page. A step without one has no rule from the list yet.
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Ex 2.5 Q1
A learning platform bill y = ax + b for x modules: 10 modules cost ₹400, 14 modules ₹500. Find a and b.
- The cost of x modules is given by the linear equation y = ax + b.
- For 10 modules costing ₹400, substitute x = 10 and y = 400 to get 10a + b = 400.
why?
Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
- For 14 modules costing ₹500, substitute x = 14 and y = 500 to get 14a + b = 500.
why?
Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
- Subtract the first equation from the second to eliminate b: (14a + b) − (10a + b) = 500 − 400.
why?
Once one variable has the same coefficient in both equations, adding or subtracting the equations removes it, leaving one equation in one unknown.
- Simplify the result to find 4a = 100.
why?
Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
- Divide by 4 to find a = 25.
- Substitute a = 25 into the first equation: 10(25) + b = 400.
why?
Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
- Solve for b: 250 + b = 400, so b = 150.
Answer: a = 25, b = 150
Ex 2.5 Q2
A badminton-court bill y = ax + b for x hours: 10 hours cost ₹800, 15 hours ₹1100. Find a and b.
- The cost equation is given as y = ax + b, where x is the number of hours and y is the cost in rupees.
- For 10 hours, the cost is ₹800, so substitute x = 10 and y = 800 into the equation to get 800 = 10a + b.
why?
Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
- For 15 hours, the cost is ₹1100, so substitute x = 15 and y = 1100 into the equation to get 1100 = 15a + b.
why?
Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
- Subtract the first equation from the second: (1100 − 800) = (15a + b) − (10a + b).
- Simplify the subtraction to get 300 = 5a.
why?
Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
- Divide both sides by 5 to find a = 60.
why?
Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.
- Substitute a = 60 into the first equation: 800 = 10(60) + b.
why?
Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
- Calculate 10 × 60 = 600, so the equation becomes 800 = 600 + b.
- Subtract 600 from both sides to find b = 200.
why?
Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.
Answer: a = 60, b = 200
Ex 2.5 Q3
°C = a °F + b, with 0 °C = 32 °F and 100 °C = 212 °F. Find a and b.
- The relationship between Celsius (C) and Fahrenheit (F) is given by the equation C = aF + b.
- Using the first condition, 0 °C corresponds to 32 °F, so substitute C = 0 and F = 32 into the equation to get 0 = 32a + b.
why?
Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
- Using the second condition, 100 °C corresponds to 212 °F, so substitute C = 100 and F = 212 into the equation to get 100 = 212a + b.
why?
Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
- Subtract the first equation from the second to eliminate b: 100 − 0 = (212a + b) − (32a + b).
why?
Once one variable has the same coefficient in both equations, adding or subtracting the equations removes it, leaving one equation in one unknown.
- Simplify the subtraction to find 100 = 180a.
why?
Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
- Solve for a by dividing both sides by 180, which gives a = 100/180 = 5/9.
why?
Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.
- Substitute a = 5/9 back into the first equation 0 = 32a + b to find b.
why?
Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
- Calculate 32 × (5/9) = 160/9, so the equation becomes 0 = 160/9 + b.
- Solve for b by subtracting 160/9 from both sides, giving b = −160/9.
why?
Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.
Answer: a = 5/9, b = −160/9
Other exercises in this chapter
- Exercise Set 2.1 — 5 questions
- Exercise Set 2.2 — 7 questions
- Exercise Set 2.3 — 5 questions
- Exercise Set 2.4 — 4 questions
- Exercise Set 2.6 — 1 questions
- End-of-chapter exercises — 14 questions
← Introduction to Linear Polynomials: notes, practice and all exercises
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