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NCERT Solutions for Class 9th Maths Chapter 2 Exercise Set 2.5 — Linear Relationships

Book page 27 Updated on2026-09-08

Q1.
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 ₹400. When she accessed 14 modules, her bill was ₹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.
Answer

Substitute the two observations into y = ax + b.

400 = 10a + b   …(1)
500 = 14a + b   …(2)

Subtracting (1) from (2) removes b:

100 = 4a
a = 25
From (1): b = 400 − 10(25) = 400 − 250 = 150
a = 25,   b = 150,   so y = 25x + 150

Check: x = 10 gives 250 + 150 = ₹400 ✓ and x = 14 gives 350 + 150 = ₹500. ✓

Why two observations are exactly enough: y = ax + b hides two unknowns, so we need two independent facts. Each bill gives one equation, and subtracting them eliminates b in one stroke — the difference in the bills, ₹100, is caused only by the extra 4 modules, so the rate must be 100 ÷ 4 = ₹25 per module. The fixed fee is then whatever is left over: ₹150. Geometrically, two points determine one straight line.
Tip: Subtracting the equations is quicker than substituting, and it is where the meaning is: a = (change in y) ÷ (change in x).
Q2.
A 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 ₹800. When she used it for 15 hours, her bill was ₹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.
Answer
800 = 10a + b   …(1)
1100 = 15a + b   …(2)

(2) − (1):

300 = 5a
a = 60
From (1): b = 800 − 10(60) = 800 − 600 = 200
a = 60,   b = 200,   so y = 60x + 200

Check: 60(15) + 200 = 900 + 200 = ₹1100. ✓

Reading the answer back into the situation: the court costs ₹60 per hour and the gym charges a fixed ₹200 a month whether the court is used or not. That reading is what makes the algebra worth doing: the extra ₹300 between the two bills bought exactly 5 extra hours, so an hour is ₹60; and ₹800 for 10 hours is ₹600 of court time plus ₹200 that was never about the court.
Try This: what would 20 hours cost? 60(20) + 200 = ₹1400. And for what bill does she get 25 hours? ₹1700.
Q3.
Consider 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.)
Answer

Here °F is the input and °C the output. The two fixed points give:

0 = 32a + b   …(1)   (ice melts)
100 = 212a + b   …(2)   (water boils)

(2) − (1):

100 = 180a
a = 100/180 = 5/9
From (1): b = −32a = −32 × 5/9 = −160/9
a = 5/9,   b = −160/9
°C = (5/9)°F − 160/9 = (5/9)(°F − 32)

Check: °F = 32 gives (5/9)(0) = 0 °C ✓; °F = 212 gives (5/9)(180) = 100 °C. ✓

Why the slope is 5/9: between melting and boiling, Celsius moves 100 degrees while Fahrenheit moves 212 − 32 = 180 degrees. The same physical interval is being cut into 100 parts on one scale and 180 on the other, so one Fahrenheit degree is 100/180 = 5/9 of a Celsius degree — and that ratio is the slope. The constant −160/9 is there only because the two scales put their zeros in different places; factoring it out as (5/9)(°F − 32) makes that shift visible.
Try This: a comfortable Delhi winter day at 68 °F is (5/9)(68 − 32) = (5/9)(36) = 20 °C. And the two scales agree at −40: check that −40 °F is −40 °C.
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